Yours are the best electronics instructional videos on the internet. If I magically had access to these videos back when I was a kid I'd be a retired physicist by now with a 60 year hobby in amateur radio. But better late than never. You're filling in many of the gaps in my Swiss cheese of electronics knowledge. Thank you.
When I left college as an electronic engineer I built thyristors and breakover diodes in a semiconductor plant. Your videos and explanations are reminiscent of the Forrest Mims books back in the day. Excellent and well put together videos!
@@w2aew I still have the books! Bought from Radio Shack in the 70s and diligently studied. I remember my interview at the semiconductor plant in 1982. My boss to be gave me a sketch pad and asked me random questions like "draw me a single transistor class A amplifier circuit with example biasing resistor values." That "back of the envelope" method helped to get me the job and I hope your videos help get new kids coming through a grip on electronics. Breaking things down into smaller parts helps to get a handle on the common techniques.
I've been a student of EE for 25 years and your style is the most impressive I've seen so far! I love the simply laid out notes and common themes. I'm on the binge and I want a vid specifically detailing the inside of an opamp from you. All the others I watch aren't cutting it!
I recently retired and now have time to learn about electronics. You are a good teacher, and I appreciate your videos. Thanks for all your hard work. 73!
This channel is fantastic! The explanations of important fundamentals are clear and the practical applications are always mentioned. I follow along and experiment with each topic covered. Thank you so much for your effort and I wish you all the best!
Your VOM is in parallel with the current source and this causes a reading error. I guess it is a 20kohm/V meter, so in the 10V range it has an impedance to 200k, that matches the impedance you calculated.
Yikes!! You are absolutely right!! Although I grew up using analog meters like the 260, how quickly we can forget about the meter's impedance after getting spoiled by the 10Mohm input impedance of modern DMMs. Ugh, what a bonehead error! I'll have to do a follow up video (with egg on my face). Thank you for keeping me honest!
As a subscriber of w2aew's videos I'd like to say kudos for the sharp eye and a professional and polite comment on what was done wrong. You are the antithesis of stereotypical youtube comments. Thank you.
***** This is good stuff! You, one of the gods of youtube electronics, made a simple to overlook mistake and from just a comment made a whole video explaining the error and give credit to the one who found it. I'm impressed at the level of sportsmanship/professionalism here. Thanks for this. This type of discourse is REAL education.
the slope of the IV characteristics is 1/R @ 6 min. So for a ideal constant current source, output impedance is infinite (1/infinite is 0 which means flat line).
@ 12:00 mark, I am interested to know why using the current sink with an impedance of 190k is an advantage over simply using a 190k resistor in the amplifier application you mention.
Because, in order to use a 190k resistor, the collector current would have to be very low in order to not saturate the transistor. The low collector current will lower the transconductance (gm), and therefore lower the gain...
Thanks for the video. (My Fluke 87vMax multimeter is back for a safety recall, if I ever see my meter again, if they repair it. Check yours?) This setup gave me some issues because I was down to my intrepid Cen-Tech P37772 meter with the patent infringment yellow bumper, and worn selector switch, which also doesn't do microamps. So I needed to scale the current up the setup. I also used two power supplies to keep my bias divider in one place, but that might be cheating. What I found: I was displeased with the linearity (low input-impedence) Why I found it: likely because I was against the compliance voltage for my altered setup. What made recovered the fun: I replaced R2, with a diode, which brought up the bias voltage, and linearized the setup better. Also, placing 1 though 5 LEDs in the current source with very little change in apparent brightness made it fun, once again, all the way past 12 volts. They didn't smoke at 15. What opportunity I found: By watching the LED brightness when changing the voltage, I could see the limits of the voltage compliance. Oddly, the 2n222 handled the current quite well. Apparently, some have a max continous current of 600mA, others 800mA (likely with some kind of clip-on heatsink.) What opportunity I missed: while I practiced calculating the input impedance a few the circuit, I should have for this setup, but didn't. Thanks, again for the videos.
Great video. Earlier today I built most of the circuits you showed, my favorite was the voltage divider biased BJT with two diodes in place of R2 in the divider. This circuit was remarkably stable, even when I changed the supply voltage or dropped in transistors with different betas. One configuration I tested, a 500 uA sink, showed less than 10 uA change with a 100% swing on the supply voltage. Remarkable.
Yes, this is exactly what you'd expect - nice job. Note that I made a mistake with my use of the old Simpson 260. It's input impedance was too low to be used to measure the voltage on the current source output. See my next video for details...
Just been building your second "sink" circuit - 0.6V across a 120 ohm emitter Resistor with a view to create a 5mA constant sink. That particular circuit was in an exam paper , GCE A level - June 1981 AEB Electronic Systems Paper 1. Testing the sink current in an actual circuit I'm getting about 10mA. The paper also asks for graphs of collector current and output voltage when you charge a 1000uF capacitor.
@@w2aew I thought it could be my "random" silicon npn transistor - I used a C4881 (TO-220) which measured 171 beta on my Peak tester. It appears to have survived being soldered on to a strip-board. The bias config is 2 diodes in series with a 300-ohm 1-watt resistor - dropping 15V - 1.2V.
@@armandine2 Always a good idea to backup your measurements with a second method. For example, if you're measuring 10mA collector current, double check that by measuring the voltage across the emitter resistor, as well as the resistor value - all should lead to the same conclusion - if not, then something is wrong.
11:33 .... re: Using the current sink as a load for an amplifier. Dumb question: Is this a good strategy for a LNA block that must amplify very low-V (uV) raw sensor output (assuming the opamp ckt is well designed for noise)?
Great lesson on current sources for tabletop circuits. My area of application is industrial and automotive where temperatures change from -20C to 70C at best. My designs have to cope with -40C to 80C. These basic current controllers won’t be useable without temperature compensation. Just consider the 0.6v Vbe reference voltage these circuits use, over a small 50C temperature range the voltage will change by 2mV per degree, that’s 100mV change for a 600mV reference! Even the self heating can render the current out of range. Throw in todays trend to run everything from a single battery or a USB voltage and you suddenly have a much more interesting design. I consider any circuit design to be incomplete if it doesn’t have some temperature compensation or a good reason not to use it. I’m sure you have a few design options for temperature correction, perhaps an update video one day.
Really well done videos! I do have a question....At about 16:06 in the video you say that you take the first current source output and tie it to the PNP current mirror input. This happens to be the emitter on the PNP....my question is : What did you tie the collectors of the PNP mirror to?
Take a careful look at the schematic at 15:47 - the input of the PNP current mirror is the base-collection junction labeled 'IN". The emitters are tied to the positive power supply.
@@w2aew ...my apologies......In prior video you always had a number (ie +10V, etc) labeled and I didn't recognize the 'circle' without the label....my bad....thanks
Can you explain why it was necessary to put a 470 ohm resistor across the outputs of the power supply to keep it sourcing current? Since a bench power supply typically has readouts for volts and current, I suppose one could "adjust" those readings to compensate for the resistor, correct?
In this setup, the power-supply would have to "sink" current (current flowing into the + terminal), and linear power supplies like this can't sink current. So, by placing a small resistor across the supply, the output will always "source" current and force the set voltage across the resistor. There is no adjustment to any readings necessary since we're not measuring the output current of the supply.
@@w2aew Thank you for your quick reply! I have a switching power supply. Can I ask a follow-up question? Do switching power supplies sink current into the + terminal? I am only a hobbyist, so it would take a great deal of research to try to figure this out, and the odds are good I would come up with the wrong answer!
@@claude77573 It likely can not sink current. Also, if you're going to replicate this experiment, you'll have to check the specs on your supply to see if the outputs are truly floating from ground.
Excellent, i can see that I was the last person to comment. I'm understanding more now and noticed the ring of two as an example of a current source. It would be great to see you do a video on that.
11:45 I'm trying to process this concept. Learning and noticing this current source and current mirror in modern hi-fi solid state amplifiers. I keep rewatching and I still get stuck at why the gain would be different if the load was simply resistive. I'm very slow and just trying to move up comprehension. Thanks
In the example cited, if you used a 190k ohm resistor, the current would have be very low to avoid saturating the transistor, which would then result in a lower gm factor - so the overall gain would be lower (gm*Rc). The current source load gives you very high impedance (Rc) at higher currents which gives you higher gm, so the gain can be very high. In fact, in most cases, the gain when using an active load like a current source is usually so high that they're only used with additional negative feedback to set the overall circuit gain. The benefit of having a very high gain inside of a feedback amplifier is much lower distortion.
It's interesting to see that the output impedance of the 2N2222A of 190K across the range 2V to 10V when sinking around 0.5mA as calculated at 11:34 is almost the same as the output impedance of 192K across the range 5V to 10V, which is given by 5V/(419-396)μA. That's useful because some devices have a "knee" at lower collector voltages where the incremental output impedance drops off. It's also worth noting that the 2N2222A datasheet gives a wide range of output impedances at Ic=1mA and Vce=10V of between 30K and 200K. At Ic=10mA, Vce=10V, that falls to between 5K and 40K, meaning that the 2N2222A is not so useful as a constant current supply at higher currents.
Hah! There's actually a mistake in this video when it comes to measuring and calculating the output impedance! It was discovered after the fact, and I posted a followup video. Note that the output impedance is much higher.... th-cam.com/video/JbCI4Lsnqho/w-d-xo.html
@@w2aew I noticed the discrepancy between the apparent emitter current of 384μA and the 419μA collector current, but I just surmised that the 1K emitter resistor was high by about 8%. The actual value of that resistor would have been the first thing I'd measure. It's been so long since I used an analogue meter, its loading wouldn't have occurred to me. So, the interesting point now is that the 2N2222A seemed to have an output impedance far in excess of the values in its datasheet, but I then realised that the emitter resistor effectively provides negative feedback in a common base configuration, stabilising the collector current against variations due to changes in Vce. Something to explore.
It is in the linear region. The 1k resistor helps to set the current based on the resistor divider at the base and the base-emitter drop. It makes the value of current mostly independent of transistor beta.
When a BJT is in linear region, is Vbe (base-emitter drop) a constant value for different collector current values? If the emitter resistor is removed, and use a pot to replace 2.2K. Can I tune the pot to get a adjustable current source? Thank you.
In the linear region, the Vbe is nearly constant, but will vary a little. Your suggestion would make it very difficult to adjust the current source to a particular value - very unstable and will vary with temperature. Much better to leave the degeneration resistor (1k in emitter) in the circuit, and then adjust the base voltage with a pot.
Is there a way to temperature compensate a current mirror used as to measure high side current? The current mirror I am using had a pnp and npn matched pair. The issue i see is the offset and slope changes as a function of temperature. Is there a way to compensate it?
Brilliant video with much for me to learn...including from the comments! I'm working through building these circuits to solidify my learning but am not sure how to set up a floating power supply. Would you be able to provide a brief explanation on how you set that up please?
I love these fundamental electronics videos. Can I ask you how to match transistors used in a current mirror? Is it ok to compare the measured Hfe? Or are there other parameters needed.. (like thermal coefficients..) ? Thanks in advance.
Nice video - I always forget how useful these are outside of ICs. Have you done a video on the "long tailed pair" before? Would make a good follow up ,,,
Do you have a video you'd recommend for the variable, floating power supply? I'm confused by how its set up. 1) I normally see supplies hooked up with the negative to the ground. 2) It seems to be putting its positive out on the +10V rail. In my mind that would cause a short between the variable supply's voltage (assuming its any value other than 10V) and the 10V rail. But this must not be the case here. Thanks for any insight
Thanks, that helps with the floating aspect. So the last thing I'm struggling with may just be me not understanding notation. At 8:07 you are talking about the variable supply. When I looked at the diagram I assumed + terminal of the variable is going in opposition to the + rail. Does the symbol not represent the direction of the voltage? (By which I mean the - variable terminal is at 10V and the variable + terminal goes down toward the ammeter)
The variable power supply basically sets the voltage that is generated between the + and - terminals of the supply. Since the + end of the supply is connected to the +10V supply rail, the - end of the supply is XX volts below +10V. For example, if the variable supply is set to 3V, then the - end of the supply will be at 7V in this circuit (10-3).
It's a wonderfull video. I hope to see another video about current mirrors using Mosfet, wilson,wildar, wildson modificate, active cascode.... thanks for yours videos
WOW! Amazing video! thank you very much for the detailed explanation and for showing the actual use of a current generator. A question I've been asking myself for a long time was "what are current sources used for?"
You do an incredible job of walking a dumbo like me through this subject. One question........ you said a couple of times that the constant current source can improve the gain of a transistor, the reason isn’t so obvious to me.
A current source is a very high impedance - so when it is used as the load in the collector of an amplifier (instead of RC), the gain will be very high (since gain is gm*RC)
I came across your video while searching for a solution to my problem. I built a circuit exactly like what you have on top right hand side of frame, you're pointing at it at 4:53 I've selected 5V as supply voltage. The resistor is 50 Ohms. I plan on charging a 20Ah NiMH cell connected to the output (via a schottky diode to prevent back flow when the source is off). It provides about 310mA, but quickly runs up and gets into a thermal run away. The current increases until the output transistor is destroyed. I've tried with 2x 2N2222, then went beefy, and used 2x TIP42C NPN transistors. Same effect. Both transistors are matched. Tried a resistor as load instead of battery. Same result. Then I installed a heat sink on the output transistor, and it works fine now! Why is this happening on only 300mA draw? How can I stabilize it? Would a MOSFET with its negative temp coefficient stop the thermal runaway? Do they even make current mirrors using MOSFETs?
Quick example: for a common emitter amp without emitter degeneration, the small signal gain is given by gm*RL. Since gm is a function of collector current, the maximum gain is limited to typically well under 100x. Since a current source can be used as an active load (instead of a passive load like a fixed resistor RL), the output impedance of the current source becomes the load (RL_equivalent). Thus, for the same collector current (same gm), the gm*RL_equivalent product can be very large, leading to very high gain. This is how op amps achieve extremely high open loop gain.
Would it be possible to demonstrate dependent voltage/current sources? Also, are you planning on doing a BJT differential amplifier with a current source?
When looking at a schematic of a passive network or active network, how can you tell if the passive network or active circuit is a current source or voltage source? Can passive LCR or LR or LC networks be a current source or voltage source how can you tell?
I am bewildered about how can a transistor still work when the base-collector junction is shorted as it is with the current mirror transistors? Does the transistor with the shorted base-collector junction act as a diode only in such a case or does it still have a current gain?
@@uiticus Yes, it will, but the currents won't necessarily match. We're relying on the Vbe of the two transistors being the same to get the same current in both.
@w2aew for the first example of the "source" PNP, does Re become a fixed voltage? I assume Ve is Vb - .6., and does the fixed voltage change with resistance change of the Re? I'm trying to experiment with a 10k and 1k..
Yes, in each of these examples, the intention is that the voltage across Re is fixed, which results in a fixed emitter current, thus a fixed collector current - provided the transistor is kept out of saturation.
Saturation is not defined by operating at max current. Bipolar transistor saturation is define when the collector voltage gets very close to the emitter voltage, such that the base-collector PN junction gets forward biased (turned on). When this happens, the transistor current gain (beta) drops significantly and the transistor collector-emitter begins to look like a short.
Thank you for that awesome video. I love these back to basics videos. You have mentioned that the gain of an amplifier could be massively increased with a current source. It would be great if you could do a video about that topic to show how that works.
Hi, I did not got the part about why the in the current mirror, if the transistors are matched, they will have both the same collector current. Is it because they both would have the same base current and thus the same collector current? Since those are BJT, then I guess Vce if the leftmost transistor can get tu saturation because then it would not be enough to turn on the both transistors right?
Given matched transistors, they'll have the same collector current because they have the same base and emitter voltages. The leftmost transistor can't saturate because it is connected as a diode (base connected to collector).
***** Thank you so much! Wow, you opened my eyes :P since collector is connected to base, Vce = Vbe, and Vce_sat < Vbe so it wont saturate. Taking a 2N3904 as example, Vbe ~ 0.7V and Vce_sat ~ 0.3V. Collector current would be I_c = (Vin - Vbe)/R where Vin: source voltage, Vbe: base - emmiter voltage, and R the resistance in the collector. Is this correct? If I may, one more question: How would you compute base current? With the gain beta @ I_c since it is not saturated? And I forgot to say, Nice video as always!
W2aew, Current source circuits were mostly used for headphone amplifiers cause there was no standard output impedance for headphones? A Constant current source amplifier circuit "can drive any load" impedance that is why they used them for headphones?
Because with the positive lead in the upper left jack, the two ranges that the meter can be in for DC AMPS are the 4mA and 40mA range. Only the 4mA range gives you three digits beyond the decimal point (1uA resolution). The 40mA range give you 10uA resolution. In order for the reading to be in mA with 3 digits beyond the decimal point, it would have to be on the 4A scale, which would require that you use the 10A jack on the lower left corner.
Could you possibly consider a video on different types of oscillator.Specifically why most designs don't oscillate as expected,or work every time in a simulation program but when constructed never do ?
If I connected the positive lead of the 260 to the other side of the ammeter, then the ammeter would have only "seen" the current source and the current reading would be correct. Then, all I'd have to do is measure and account for the burden voltage of the ammeter.
Sorry Alan I should have commented on the other video. Could you have placed a high value resistor between the probe and ground? Mimicking the digital meter?
A floating or isolated power supply is one that generates the desired voltage between its positive and negative terminals, but neither of these terminals is connected to ground.
Thanks Again.. I like back to basics. One question though, Transistor matching is that part number to part number or actually test each transistor of a type for its characteristics.
It really goes beyond just matching part numbers. Where circuits like these are used extensively, and rely on device matching, is typically in integrated circuit design where the transistors are extremely well matched in gain, leakage, temperature, etc. In these cases, device matching is taken advantage of in many circuit types, not just current sources and mirrors. When device matching isn't as practical, then techniques like using emitter degeneration are often used. If you're going to rely on matching for the circuit performance that you need, you should match items like gain/beta, leakage currents, Vbe, etc.
It's funny, I was just recently playing around with a discreet current mirror. You can still get dual matched transistors these days. I was using the BCM847, which is a pair of matched NPN transistors. The PNP version is the BCM856. Unfortunately they tend to be in really tiny packages. For the prototype I hand matched some discreet transistors for gain and glued them together for some thermal matching. I also used some emitter degeneration like w2aew says and this helped a lot. Before doing that my 1ma mirror drifted up to almost 10ma as one side warmed up! You'd be surprised at how poorly transistors will match even one after the other on a paper tape strip. 5%-10% from one to the next is what I saw.
Paul Moir Thermal matching is very important for accurate mirroring. The -2mV/degC change in Vbe can dramatically affect the accuracy/tracking of each side, especially when you consider that each device will likely be dissipating different power.
Hello, nice video ! how to do to have a constant 1amp 14V power supply to charge a battery. WIth classical PSUs, when the battery reaches 13.5v, it takes ages to go to 14v because it draws very few current. Pushing 1Amp all the way will make it charge better and faster. Thanks.
Different battery chemistries have different charge circuit requirements. Lead acid batteries need a constant voltage float charger, while NiCad generally use a constant current charge. Lithium and LiFePO have other requirements. Using the proper charger circuit for your particular battery chemistry will give you the best results, and safest results.
Precesion current source based on simple zener? You can make much better using combination of the Widlar mirror and peaking current source. 2 BJT + 3 resistors, no zener, great stabililty, etc.
Good video, however, in the measure experiment, I don’t understand why you haven’t connected directly a voltage source between collector and ground to fix the voltage of current source.
hi Alan great video but i have a question. i didn't get why you put a 470 ohm resistance in parallel with supply. will you please shed some light on that
slew rate is function of Capacity and Current if and opamp use a bigger current of polarization (this current normaly is obtained of a current mirror) you will improve the slew rate.
I'm not understanding why a piece of equipment has chassis leakage current. What is causing the chassis leakage current or where is it coming from? The Chassis ground I though should be a zero volts and zero amps no leakage.
Something is not right with the second solution of current source (with 2 diodes in series) : I've made the same circuit with : +9V 2N2222A (NPN) 2 * 1N4007 in series with 1k resistor R emiiter : 220 ohms I measure a current less than 1 mA (~0.5 mA) with my DMM. When replacing 2 diodes with a 470 ohms resistor : ~40 mA Which makes sense : there's almost no current flowing into the base: diodes act as a short between the base and GND. Maybe is this made on purpose to have small current source...And if yes, how do we calculate the current that' flows into the base/diodes ? Am I right ?
Sounds like you’ve wired something wrong. The idea with the two diodes is to place the base voltage 2 diode drops above ground, so that the voltage across the emitter resistor is 1 diode drop, thus the emitter current should be close to .65V/RE
@@w2aew Ok, that makes sense to me. 1) So, the potential between the 2 diodes is theoretically the same as the potential of the emitter (because of the base-emitter diode voltage drop), right ? 2) the current we want at the end (of the current source) is determined by RE, right ? And we don't much care about the other resistor as long as the transistor works in its linear area (the "almost-flat" line, the region where Ic doesn't change much with Vce, the region where the impedance is the biggest), right ?
I would like if u could demonstrate the current source using the PNP transistor. I have to design a 100W 8ohm amplifier that has a current source supplying the long tailed pair.
+Anjana Dabideen The diagrams at the bottom of page 1 of the show notes have some of the basic configurations for a PNP current source: www.qsl.net/w2aew/youtube/transistorcurrentsources.pdf You generally start by knowing what value of current you need, call it IC. It's generally a good idea to have 0.5 to 1V or more dropped across the emitter resistor. The "middle" diagram, the one with the two diodes connected from base to VCC, will put about 0.7V across the emitter resistor. Thus, you can calculate the resistor value as 0.7V/IC. Then calculate the resistor that goes from the base to ground. You want the current in this resistor to be about 10x greater than the base current in the transistor (or more). So, you calculate the base current by taking the IC value divided by the transistors Beta or Hfe. Then, multiply by 10 to get the bias current through the base-gnd resistor, call it Ibias. You then calculate the resistor value as (VCC-1.4)/Ibias. Done.
+Anjana Dabideen Can I ask you for one more explanation. I lack some fundamentals on this topic and i am unsure how to go about choosing the input current and voltage to meet the specifications of the amplifier project i am doing. Seeing that i have to design a 100W 8ohm amplifier what is an acceptable Ic?
Thanks for the video, helped a lot! I just came across another current source configuration called the "wilson current mirror". Could you do video about this one and explain why you would pick a wilson current mirror over the basic two transistor current mirror configuration :-)
+Rolfrolfsen Wilson current mirror provides more accurate current mirroring (less error between input and output current) and a higher output impedance (more ideal current source, less load dependent) than a simple two-transistor mirror using degenerative feedback. Might make for an interesting video in the future.
When engineers say an Op amp is a LOWER Characteristic op amp, what do they mean by "lower characteristic? In OP AMP schematics they often use transistor or FET current sources and current mirrors and differential , Common mode rejection. If you look at common Op Amp schematics for common part number some use differential inputs using transistors while others use FETs inputs plus it looks like they are configured in a current source, current mirror, differential configuration for inputs#1 and inputs#2. If the circuit designed is for an op amp LF348 and LF347 which is JFET inputs, you can't use a replacement like an LM741, TLO84, LM324 because those op amps aren't JFET inputs which I'm not sure why it wont work in circuits as an cross reference part. It would be nice if you made a video lesson about common OP amp part number schematics to go over the FET/transistor configurations inside the op amps of the current sources, current mirrors, etc
Another test : 1 current source "feeding" 1 current mirror with +9V supply. - last example of current source (purists prefer to say it is a current sink, not a current source) : 2 NPN (2N2222A) and 2 * 1k resistors. ---> measured current is ~0.7 mA (collector of second transistor), quite unstable actually (breadboard quality ? I have no clue)... Current is the same if I power that current source with +12V : ~0.7 mA (as expected...) then connecting collector of second resistor directly to collector of the first transistor of the first current mirror example (2 * NPN, 2N2222A, no resistors) Then, connecting 3 LEDs at the collector of the second transistor of the current mirror : they indeed all have the same brightness (same goes whatever the color or size of the LEDs) but the current at the LEDs is about 10 times bigger than the current at the current source.... But I think I made a mistake : if I use a current source with NPNs transistors (sink model), should I use a current mirror that uses PNP transistors (instead of NPNs like I did in my test) ?
Hello w2aew! I have a small question. How does the current mirror function in the LT3750 block diagram in the datasheet? You mentioned that it maintains a steady current, so I assume it is for some measuring function? Here's a link to the datasheet: cds.linear.com/docs/en/datasheet/3750fa.pdf Thanks!
It's not really a current mirror. The diode-connected PNP simply establishes a bias voltage on the other PNP transistor. When the stored energy in the transformer is being transferred to the capacitor, the voltage across the cap is reflected (with 1/N) to the primary and appears at the RVout resistor. Since the bias voltage at the RVout pin (the PNP emitter) is basically fixed, the voltage at the transformer is transformed into a current across the RVout resistor. This same current is sent through resistor RBG. Thus, the voltage across RBG is proportional this current, which is in turn proportional to the voltage across the cap. This voltage is compared against an internal 1.24V reference to detect full charge.
+David Senabre The 470 ohm resistor around the variable power supply ensures that the supply always sources current. Without it, the circuit would try to make it sink current, which it can't do.
+w2aew Why would the variable supply sink current? Doesn't the transistor sink current? Current flow, as I see it, will always go from +10V to the collector. The variable power supply just reduces the 10V subtracting from it (as they are opposite polarity), but why should this invert current flow?
+David Senabre A power supply is designed to source current from its positive terminal, and have it return into the negative terminal. If the 470 ohm resistor wasn't across the variable supply, the collector current would have to be sourced from the negative terminal of the variable supply (thus, the positive terminal would have to sink the current from the main supply).
Umm, so my understanding of output impedance is a bit wrong I suspect. I understand how a constant voltage source has ideally no impedance at all since the voltage is constant, but with constant current sources/drains, I am not sure. I thought the transistor will continually change the impedance of itself (it's saturation) thus causing the voltage drop to change, thus leading to a change in the current. Of course if the transistors gate if connected to a resistor in a way that a negative feedback loop is formed, the transistor will end up regulating current at least to some degree. Can a constant current source really be thought of like a really high voltage with very high output impedance, like a resistor on the output? I did wonder that for a long time, but decided that it is not an accurate to think of constant current sources like that.
For BJTs, the collector current will be *nearly* constant vs. load if: 1) The base-emitter voltage is constant (thus base current is constant) 2) The transistor is not cutoff or saturated. For FETS, the channel can be thought of as a voltage controlled resistor at low V(DS) values (linear region), but behave more like BJTs when V(DS) is raised. This is called the "saturation region" for FETs, which is different than the saturation region for BJTs. Yes, you can model the current source as a very high voltage with a larger series resistor. Of course, this model is only valid over an operating voltage range that applies to the circuit being used.
***** Very interesting! I will have to graph out ohms law with that HV + large resistor with considerably lower voltages and see what I can learn. So If I had a infinite voltage source, with an infinite (this infinite is some factor bigger than the infinite voltage) resistance, then would I have a ideal current source? I will see if I can figure this out with mathematically, I will probably end up having to apply some basic calculus knowledge because surely, when I am dealing with infinities, esp. (x*infinity) / (y * infinity), I will need some concept of limits and things!
Yours are the best electronics instructional videos on the internet. If I magically had access to these videos back when I was a kid I'd be a retired physicist by now with a 60 year hobby in amateur radio. But better late than never. You're filling in many of the gaps in my Swiss cheese of electronics knowledge. Thank you.
When I left college as an electronic engineer I built thyristors and breakover diodes in a semiconductor plant. Your videos and explanations are reminiscent of the Forrest Mims books back in the day. Excellent and well put together videos!
I learned a lot from the Forrest Mims books!
@@w2aew I still have the books! Bought from Radio Shack in the 70s and diligently studied. I remember my interview at the semiconductor plant in 1982. My boss to be gave me a sketch pad and asked me random questions like "draw me a single transistor class A amplifier circuit with example biasing resistor values." That "back of the envelope" method helped to get me the job and I hope your videos help get new kids coming through a grip on electronics. Breaking things down into smaller parts helps to get a handle on the common techniques.
@@w2aew books by R.M. Marston were also essential reading.
The best explanation about electronics I've seen. Wished I had a teacher like you. I wish all the success you deserve.
same
It still nice to see the old analog Simpson meter. Still in use . Good video well explained!
I've been a student of EE for 25 years and your style is the most impressive I've seen so far! I love the simply laid out notes and common themes. I'm on the binge and I want a vid specifically detailing the inside of an opamp from you. All the others I watch aren't cutting it!
I recently retired and now have time to learn about electronics. You are a good teacher, and I appreciate your videos. Thanks for all your hard work. 73!
This channel is fantastic! The explanations of important fundamentals are clear and the practical applications are always mentioned. I follow along and experiment with each topic covered. Thank you so much for your effort and I wish you all the best!
Best practical electronics & education channel out there. Hands down.
Professor, so much thankful for your videos. I never never have had the opportunity of learning with such a perfectly explanations. THANKS.
Your VOM is in parallel with the current source and this causes a reading error.
I guess it is a 20kohm/V meter, so in the 10V range it has an impedance to 200k, that matches the impedance you calculated.
Yikes!! You are absolutely right!! Although I grew up using analog meters like the 260, how quickly we can forget about the meter's impedance after getting spoiled by the 10Mohm input impedance of modern DMMs. Ugh, what a bonehead error! I'll have to do a follow up video (with egg on my face). Thank you for keeping me honest!
As a subscriber of w2aew's videos I'd like to say kudos for the sharp eye and a professional and polite comment on what was done wrong. You are the antithesis of stereotypical youtube comments. Thank you.
***** This is good stuff! You, one of the gods of youtube electronics, made a simple to overlook mistake and from just a comment made a whole video explaining the error and give credit to the one who found it. I'm impressed at the level of sportsmanship/professionalism here. Thanks for this. This type of discourse is REAL education.
Agreed!
When real professionals meet they usually behave like gentlemen. But you are right, this kind of chivalry is becoming the exception in YT comments.
The HP RPN calculator is a really nice touch. I am still using my trusty HP35s.
Valued info! Thanks! I love how the majority of your projects involve less than a $5 of a handful of parts.
Brilliant. Thanks for explaining these mysterious little circuit blocks!! Now I see how they help increase the gain in audio stuff.
the slope of the IV characteristics is 1/R @ 6 min. So for a ideal constant current source, output impedance is infinite (1/infinite is 0 which means flat line).
Love this video, looking to understand transistors since ~1 year and making slooooowly progress.
Time really flies when watching your videos. Thanks for the great explanation.
@ 12:00 mark, I am interested to know why using the current sink with an impedance of 190k is an advantage over simply using a 190k resistor in the amplifier application you mention.
Because, in order to use a 190k resistor, the collector current would have to be very low in order to not saturate the transistor. The low collector current will lower the transconductance (gm), and therefore lower the gain...
Thanks for the video.
(My Fluke 87vMax multimeter is back for a safety recall, if I ever see my meter again, if they repair it. Check yours?)
This setup gave me some issues because I was down to my intrepid Cen-Tech P37772 meter with the patent infringment yellow bumper, and worn selector switch, which also doesn't do microamps. So I needed to scale the current up the setup. I also used two power supplies to keep my bias divider in one place, but that might be cheating.
What I found: I was displeased with the linearity (low input-impedence)
Why I found it: likely because I was against the compliance voltage for my altered setup.
What made recovered the fun: I replaced R2, with a diode, which brought up the bias voltage, and linearized the setup better. Also, placing 1 though 5 LEDs in the current source with very little change in apparent brightness made it fun, once again, all the way past 12 volts. They didn't smoke at 15.
What opportunity I found: By watching the LED brightness when changing the voltage, I could see the limits of the voltage compliance. Oddly, the 2n222 handled the current quite well. Apparently, some have a max continous current of 600mA, others 800mA (likely with some kind of clip-on heatsink.)
What opportunity I missed: while I practiced calculating the input impedance a few the circuit, I should have for this setup, but didn't.
Thanks, again for the videos.
Great video. Earlier today I built most of the circuits you showed, my favorite was the voltage divider biased BJT with two diodes in place of R2 in the divider. This circuit was remarkably stable, even when I changed the supply voltage or dropped in transistors with different betas. One configuration I tested, a 500 uA sink, showed less than 10 uA change with a 100% swing on the supply voltage. Remarkable.
Yes, this is exactly what you'd expect - nice job. Note that I made a mistake with my use of the old Simpson 260. It's input impedance was too low to be used to measure the voltage on the current source output. See my next video for details...
I really like your back to basics videos, I always learn something new. Very well done too!
Cheers Alan, simple, straightforward, and very useful! Back to my bench!
One of the best channels to refresh and learn electronics ... As always , Excellent work W2
Just been building your second "sink" circuit - 0.6V across a 120 ohm emitter Resistor with a view to create a 5mA constant sink. That particular circuit was in an exam paper , GCE A level - June 1981 AEB Electronic Systems Paper 1. Testing the sink current in an actual circuit I'm getting about 10mA. The paper also asks for graphs of collector current and output voltage when you charge a 1000uF capacitor.
Well, something sounds "off" if you're measuring 0.6V cross that emitter resistor, but also seeing 10mA current.
@@w2aew I thought it could be my "random" silicon npn transistor - I used a C4881 (TO-220) which measured 171 beta on my Peak tester. It appears to have survived being soldered on to a strip-board. The bias config is 2 diodes in series with a 300-ohm 1-watt resistor - dropping 15V - 1.2V.
I will measure again - with a load
@@armandine2 Always a good idea to backup your measurements with a second method. For example, if you're measuring 10mA collector current, double check that by measuring the voltage across the emitter resistor, as well as the resistor value - all should lead to the same conclusion - if not, then something is wrong.
@@armandine2 Make sure that load doesn't saturate the transistor.
11:33 .... re: Using the current sink as a load for an amplifier. Dumb question: Is this a good strategy for a LNA block that must amplify very low-V (uV) raw sensor output (assuming the opamp ckt is well designed for noise)?
Great lesson on current sources for tabletop circuits. My area of application is industrial and automotive where temperatures change from -20C to 70C at best. My designs have to cope with -40C to 80C. These basic current controllers won’t be useable without temperature compensation. Just consider the 0.6v Vbe reference voltage these circuits use, over a small 50C temperature range the voltage will change by 2mV per degree, that’s 100mV change for a 600mV reference! Even the self heating can render the current out of range. Throw in todays trend to run everything from a single battery or a USB voltage and you suddenly have a much more interesting design.
I consider any circuit design to be incomplete if it doesn’t have some temperature compensation or a good reason not to use it. I’m sure you have a few design options for temperature correction, perhaps an update video one day.
Fantastic presentation Allen. I really enjoy your presentations and get allot of inside from them.
Gosh I wish you were my electronics teacher on my graduation. Thank you for the excellent videos.
Another really great back to basics video, thanks! Also liked the HP 15C calculator.
Great video. Just correction on current changed is in milliamp, but Allen kept say Microamp.
Really well done videos! I do have a question....At about 16:06 in the video you say that you take the first current source output and tie it to the PNP current mirror input. This happens to be the emitter on the PNP....my question is : What did you tie the collectors of the PNP mirror to?
Take a careful look at the schematic at 15:47 - the input of the PNP current mirror is the base-collection junction labeled 'IN". The emitters are tied to the positive power supply.
@@w2aew ...my apologies......In prior video you always had a number (ie +10V, etc) labeled and I didn't recognize the 'circle' without the label....my bad....thanks
This is very helpful for some working I'm doing on pulse magnet motors and what to do with the excess energy output.
I just noticed you also provide a pdf of the ckts - SWEET!
thanks!
Can you explain why it was necessary to put a 470 ohm resistor across the outputs of the power supply to keep it sourcing current? Since a bench power supply typically has readouts for volts and current, I suppose one could "adjust" those readings to compensate for the resistor, correct?
In this setup, the power-supply would have to "sink" current (current flowing into the + terminal), and linear power supplies like this can't sink current. So, by placing a small resistor across the supply, the output will always "source" current and force the set voltage across the resistor. There is no adjustment to any readings necessary since we're not measuring the output current of the supply.
@@w2aew Thank you for your quick reply! I have a switching power supply. Can I ask a follow-up question? Do switching power supplies sink current into the + terminal? I am only a hobbyist, so it would take a great deal of research to try to figure this out, and the odds are good I would come up with the wrong answer!
@@claude77573 It likely can not sink current. Also, if you're going to replicate this experiment, you'll have to check the specs on your supply to see if the outputs are truly floating from ground.
Very nicely explained. Thank you very much.
Excellent, i can see that I was the last person to comment. I'm understanding more now and noticed the ring of two as an example of a current source. It would be great to see you do a video on that.
Nicely explained, although I'll have to replay it a few times for my mind to take it all in!
I very much enjoyed this. These have always been a little fuzzy for me but this cleared it up nicely.
11:45 I'm trying to process this concept. Learning and noticing this current source and current mirror in modern hi-fi solid state amplifiers. I keep rewatching and I still get stuck at why the gain would be different if the load was simply resistive. I'm very slow and just trying to move up comprehension. Thanks
In the example cited, if you used a 190k ohm resistor, the current would have be very low to avoid saturating the transistor, which would then result in a lower gm factor - so the overall gain would be lower (gm*Rc). The current source load gives you very high impedance (Rc) at higher currents which gives you higher gm, so the gain can be very high. In fact, in most cases, the gain when using an active load like a current source is usually so high that they're only used with additional negative feedback to set the overall circuit gain. The benefit of having a very high gain inside of a feedback amplifier is much lower distortion.
It's interesting to see that the output impedance of the 2N2222A of 190K across the range 2V to 10V when sinking around 0.5mA as calculated at 11:34 is almost the same as the output impedance of 192K across the range 5V to 10V, which is given by 5V/(419-396)μA. That's useful because some devices have a "knee" at lower collector voltages where the incremental output impedance drops off. It's also worth noting that the 2N2222A datasheet gives a wide range of output impedances at Ic=1mA and Vce=10V of between 30K and 200K. At Ic=10mA, Vce=10V, that falls to between 5K and 40K, meaning that the 2N2222A is not so useful as a constant current supply at higher currents.
Hah! There's actually a mistake in this video when it comes to measuring and calculating the output impedance! It was discovered after the fact, and I posted a followup video. Note that the output impedance is much higher.... th-cam.com/video/JbCI4Lsnqho/w-d-xo.html
@@w2aew I noticed the discrepancy between the apparent emitter current of 384μA and the 419μA collector current, but I just surmised that the 1K emitter resistor was high by about 8%. The actual value of that resistor would have been the first thing I'd measure. It's been so long since I used an analogue meter, its loading wouldn't have occurred to me. So, the interesting point now is that the 2N2222A seemed to have an output impedance far in excess of the values in its datasheet, but I then realised that the emitter resistor effectively provides negative feedback in a common base configuration, stabilising the collector current against variations due to changes in Vce. Something to explore.
Hi Pro. At 9:28, is the BJT in saturation or linear region? Why there is a 1.0K resistor at emitter?
Thank you for these knowledgeable videos.
It is in the linear region. The 1k resistor helps to set the current based on the resistor divider at the base and the base-emitter drop. It makes the value of current mostly independent of transistor beta.
When a BJT is in linear region, is Vbe (base-emitter drop) a constant value for different collector current values?
If the emitter resistor is removed, and use a pot to replace 2.2K. Can I tune the pot to get a adjustable current source?
Thank you.
In the linear region, the Vbe is nearly constant, but will vary a little. Your suggestion would make it very difficult to adjust the current source to a particular value - very unstable and will vary with temperature. Much better to leave the degeneration resistor (1k in emitter) in the circuit, and then adjust the base voltage with a pot.
Is there a way to temperature compensate a current mirror used as to measure high side current? The current mirror I am using had a pnp and npn matched pair. The issue i see is the offset and slope changes as a function of temperature. Is there a way to compensate it?
Brilliant video with much for me to learn...including from the comments! I'm working through building these circuits to solidify my learning but am not sure how to set up a floating power supply. Would you be able to provide a brief explanation on how you set that up please?
As usual, clear and informative video.
Thank you.
Thanks Alan - a good tutorial on a subject I'm a bit fuzzy on. I'm going to have to watch this again. Cheers!!
Great video about current soutces.
I love these fundamental electronics videos. Can I ask you how to match transistors used in a current mirror? Is it ok to compare the measured Hfe? Or are there other parameters needed.. (like thermal coefficients..) ? Thanks in advance.
Most important would be to match Vbe, and then beta.
Thanks. Makes sense though if the bases are connected together in the current mirror circuit, the Vbe should be close.
Nice video - I always forget how useful these are outside of ICs. Have you done a video on the "long tailed pair" before? Would make a good follow up ,,,
Yeah, I was thinking of a video on a diff pair, then maybe even using some current source active loads to make a crude discrete op amp.
*****
Sounds great - diff amps are important, and also fun!
Really clearly explained, again! Thank you, love to see this continue on into a diff amp too.
Do you have a video you'd recommend for the variable, floating power supply? I'm confused by how its set up.
1) I normally see supplies hooked up with the negative to the ground.
2) It seems to be putting its positive out on the +10V rail. In my mind that would cause a short between the variable supply's voltage (assuming its any value other than 10V) and the 10V rail. But this must not be the case here.
Thanks for any insight
Please see this video, it might help to clear this up for you:
th-cam.com/video/mJL11UF1arQ/w-d-xo.html
Thanks, that helps with the floating aspect.
So the last thing I'm struggling with may just be me not understanding notation. At 8:07 you are talking about the variable supply. When I looked at the diagram I assumed + terminal of the variable is going in opposition to the + rail. Does the symbol not represent the direction of the voltage? (By which I mean the - variable terminal is at 10V and the variable + terminal goes down toward the ammeter)
The variable power supply basically sets the voltage that is generated between the + and - terminals of the supply. Since the + end of the supply is connected to the +10V supply rail, the - end of the supply is XX volts below +10V. For example, if the variable supply is set to 3V, then the - end of the supply will be at 7V in this circuit (10-3).
It's a wonderfull video.
I hope to see another video about current mirrors using Mosfet, wilson,wildar, wildson modificate, active cascode....
thanks for yours videos
I'm glad you liked it. I'll keep these topics in mind.
WOW! Amazing video! thank you very much for the detailed explanation and for showing the actual use of a current generator. A question I've been asking myself for a long time was "what are current sources used for?"
You do an incredible job of walking a dumbo like me through this subject. One question........ you said a couple of times that the constant current source can improve the gain of a transistor, the reason isn’t so obvious to me.
A current source is a very high impedance - so when it is used as the load in the collector of an amplifier (instead of RC), the gain will be very high (since gain is gm*RC)
Thank you hit the nail on the head I was looking for something on current I like the way you present your vid's.
Wow!! Great explanation
really love these back to basics videos.
I came across your video while searching for a solution to my problem.
I built a circuit exactly like what you have on top right hand side of frame, you're pointing at it at 4:53
I've selected 5V as supply voltage. The resistor is 50 Ohms. I plan on charging a 20Ah NiMH cell connected to the output (via a schottky diode to prevent back flow when the source is off). It provides about 310mA, but quickly runs up and gets into a thermal run away. The current increases until the output transistor is destroyed.
I've tried with 2x 2N2222, then went beefy, and used 2x TIP42C NPN transistors. Same effect. Both transistors are matched.
Tried a resistor as load instead of battery. Same result.
Then I installed a heat sink on the output transistor, and it works fine now!
Why is this happening on only 300mA draw? How can I stabilize it?
Would a MOSFET with its negative temp coefficient stop the thermal runaway? Do they even make current mirrors using MOSFETs?
How are current sources used to increase the gain of an amplifier when used instead of a resistive load? I didn't get that part...
Quick example: for a common emitter amp without emitter degeneration, the small signal gain is given by gm*RL. Since gm is a function of collector current, the maximum gain is limited to typically well under 100x. Since a current source can be used as an active load (instead of a passive load like a fixed resistor RL), the output impedance of the current source becomes the load (RL_equivalent). Thus, for the same collector current (same gm), the gm*RL_equivalent product can be very large, leading to very high gain. This is how op amps achieve extremely high open loop gain.
***** Got it. As always thanks for a great video!
Would it be possible to demonstrate dependent voltage/current sources? Also, are you planning on doing a BJT differential amplifier with a current source?
I do plan on a diff amp, and show how current sources can be used...
When looking at a schematic of a passive network or active network, how can you tell if the passive network or active circuit is a current source or voltage source? Can passive LCR or LR or LC networks be a current source or voltage source how can you tell?
I am bewildered about how can a transistor still work when the base-collector junction is shorted as it is with the current mirror transistors? Does the transistor with the shorted base-collector junction act as a diode only in such a case or does it still have a current gain?
With the base and collector shorted, it is essentially a two-terminal device that looks and acts like a diode.
@@w2aew So can we technically replace that transistor with a diode to make it still work?
@@uiticus Yes, it will, but the currents won't necessarily match. We're relying on the Vbe of the two transistors being the same to get the same current in both.
@w2aew for the first example of the "source" PNP, does Re become a fixed voltage? I assume Ve is Vb - .6., and does the fixed voltage change with resistance change of the Re? I'm trying to experiment with a 10k and 1k..
Yes, in each of these examples, the intention is that the voltage across Re is fixed, which results in a fixed emitter current, thus a fixed collector current - provided the transistor is kept out of saturation.
w2aew so you don't want the PNP running at max current? In this case, does saturation turn off the transistor?
Saturation is not defined by operating at max current. Bipolar transistor saturation is define when the collector voltage gets very close to the emitter voltage, such that the base-collector PN junction gets forward biased (turned on). When this happens, the transistor current gain (beta) drops significantly and the transistor collector-emitter begins to look like a short.
Thanks for another great, thorough explanation.
Thank you for that awesome video. I love these back to basics videos. You have mentioned that the gain of an amplifier could be massively increased with a current source. It would be great if you could do a video about that topic to show how that works.
I plan on doing something on this topic soon.
Hi, I did not got the part about why the in the current mirror, if the transistors are matched, they will have both the same collector current. Is it because they both would have the same base current and thus the same collector current? Since those are BJT, then I guess Vce if the leftmost transistor can get tu saturation because then it would not be enough to turn on the both transistors right?
Given matched transistors, they'll have the same collector current because they have the same base and emitter voltages. The leftmost transistor can't saturate because it is connected as a diode (base connected to collector).
***** Thank you so much! Wow, you opened my eyes :P since collector is connected to base, Vce = Vbe, and Vce_sat < Vbe so it wont saturate. Taking a 2N3904 as example, Vbe ~ 0.7V and Vce_sat ~ 0.3V. Collector current would be I_c = (Vin - Vbe)/R where Vin: source voltage, Vbe: base - emmiter voltage, and R the resistance in the collector. Is this correct? If I may, one more question: How would you compute base current? With the gain beta @ I_c since it is not saturated? And I forgot to say, Nice video as always!
Yes, you have the collector current computation correct. Since the transistor isn't saturated, base current is just IC/beta.
W2aew, Current source circuits were mostly used for headphone amplifiers cause there was no standard output impedance for headphones? A Constant current source amplifier circuit "can drive any load" impedance that is why they used them for headphones?
How do you know the current is microamps rather than milliamps? Am I missing something looking at the Fluke?
Because with the positive lead in the upper left jack, the two ranges that the meter can be in for DC AMPS are the 4mA and 40mA range. Only the 4mA range gives you three digits beyond the decimal point (1uA resolution). The 40mA range give you 10uA resolution. In order for the reading to be in mA with 3 digits beyond the decimal point, it would have to be on the 4A scale, which would require that you use the 10A jack on the lower left corner.
@@w2aew Thank you! I'm not familiar with all the various Fluke meters.
So, without the VOM it is actually a higher impedance. Great! It would be nice to see the same video with the VOM in parallel to the voltage source.
Yes - in fact I made a followup video that points out this error in measurement technique...
th-cam.com/video/JbCI4Lsnqho/w-d-xo.html
Could you possibly consider a video on different types of oscillator.Specifically why most designs don't oscillate as expected,or work every time in a simulation program but when constructed never do ?
What would you have done differently to get your Simpson 260 to not effect the readings?
If I connected the positive lead of the 260 to the other side of the ammeter, then the ammeter would have only "seen" the current source and the current reading would be correct. Then, all I'd have to do is measure and account for the burden voltage of the ammeter.
Sorry Alan I should have commented on the other video. Could you have placed a high value resistor between the probe and ground? Mimicking the digital meter?
That would solve the issue of the parallel current path, but the VOM wouldn't indicate the correct voltage.
Great video!, Thanks a lot! I really enjoy your videos.
Can i ask you a question?
In the video at 7:44, what do you mean by "floating power supply"?
A floating or isolated power supply is one that generates the desired voltage between its positive and negative terminals, but neither of these terminals is connected to ground.
Thanks Again.. I like back to basics. One question though, Transistor matching is that part number to part number or actually test each transistor of a type for its characteristics.
It really goes beyond just matching part numbers. Where circuits like these are used extensively, and rely on device matching, is typically in integrated circuit design where the transistors are extremely well matched in gain, leakage, temperature, etc. In these cases, device matching is taken advantage of in many circuit types, not just current sources and mirrors. When device matching isn't as practical, then techniques like using emitter degeneration are often used. If you're going to rely on matching for the circuit performance that you need, you should match items like gain/beta, leakage currents, Vbe, etc.
Thank you for the full answer, previously heard of temperature q point etc effecting device to device.. so your answer makes sense.
It's funny, I was just recently playing around with a discreet current mirror. You can still get dual matched transistors these days. I was using the BCM847, which is a pair of matched NPN transistors. The PNP version is the BCM856. Unfortunately they tend to be in really tiny packages.
For the prototype I hand matched some discreet transistors for gain and glued them together for some thermal matching. I also used some emitter degeneration like w2aew says and this helped a lot. Before doing that my 1ma mirror drifted up to almost 10ma as one side warmed up! You'd be surprised at how poorly transistors will match even one after the other on a paper tape strip. 5%-10% from one to the next is what I saw.
Paul Moir Thermal matching is very important for accurate mirroring. The -2mV/degC change in Vbe can dramatically affect the accuracy/tracking of each side, especially when you consider that each device will likely be dissipating different power.
Hello, nice video ! how to do to have a constant 1amp 14V power supply to charge a battery. WIth classical PSUs, when the battery reaches 13.5v, it takes ages to go to 14v because it draws very few current. Pushing 1Amp all the way will make it charge better and faster. Thanks.
Different battery chemistries have different charge circuit requirements. Lead acid batteries need a constant voltage float charger, while NiCad generally use a constant current charge. Lithium and LiFePO have other requirements. Using the proper charger circuit for your particular battery chemistry will give you the best results, and safest results.
Precesion current source based on simple zener? You can make much better using combination of the Widlar mirror and peaking current source. 2 BJT + 3 resistors, no zener, great stabililty, etc.
Good video, however, in the measure experiment, I don’t understand why you haven’t connected directly a voltage source between collector and ground to fix the voltage of current source.
hi Alan
great video but i have a question.
i didn't get why you put a 470 ohm resistance in parallel with supply.
will you please shed some light on that
Because the power supply can't "sink" current. The 470 ohm resistor ensures that it is always "sourcing" current.
I'm enjoying your show - especially the b.e.e. stuff.
I am curious about something you told about slew rate. Value of current inside op amp current mirrors limit slew rate.
slew rate is function of Capacity and Current
if and opamp use a bigger current of polarization (this current normaly is obtained of a current mirror) you will improve the slew rate.
What is the purpise of mirror circuit in power amp circuit?
Current sources are very useful. THANKS
I'm not understanding why a piece of equipment has chassis leakage current. What is causing the chassis leakage current or where is it coming from? The Chassis ground I though should be a zero volts and zero amps no leakage.
Chapeau Maestro! Very clear and well explained...
Something is not right with the second solution of current source (with 2 diodes in series) :
I've made the same circuit with :
+9V
2N2222A (NPN)
2 * 1N4007 in series with 1k resistor
R emiiter : 220 ohms
I measure a current less than 1 mA (~0.5 mA) with my DMM.
When replacing 2 diodes with a 470 ohms resistor : ~40 mA
Which makes sense : there's almost no current flowing into the base: diodes act as a short between the base and GND.
Maybe is this made on purpose to have small current source...And if yes, how do we calculate the current that' flows into the base/diodes ?
Am I right ?
Sounds like you’ve wired something wrong. The idea with the two diodes is to place the base voltage 2 diode drops above ground, so that the voltage across the emitter resistor is 1 diode drop, thus the emitter current should be close to .65V/RE
@@w2aew Ok, that makes sense to me.
1) So, the potential between the 2 diodes is theoretically the same as the potential of the emitter (because of the base-emitter diode voltage drop), right ?
2) the current we want at the end (of the current source) is determined by RE, right ? And we don't much care about the other resistor as long as the transistor works in its linear area (the "almost-flat" line, the region where Ic doesn't change much with Vce, the region where the impedance is the biggest), right ?
@@gloubiboulgazeblob yes, exactly! You got it.
@@w2aew Thanks !
...and finally I understand current sources. Awesome video. Thank you :)
Thanks,very very nice and simple tutorial,
i hope you go forwards and waiting for more basics videos
+Moon Light Shadow Please let me know what topics you'd like to see in future "basics" videos!
So awesome! Thank you!
very clear explanation, thank you
Thanks for the tutorial.
You mention microamps a few times (as shown on the meter), is that right?
Yes. The meter shows 0.418 milliamperes, which is 418 microamps.
Ahh, I did not notice the probe in the mA socket. The dial confused me a bit :D
Another good (and useful) video Alan. Thank you for prparing it.
How can i make a 4 to 20 milli_ampers current source ? just for testing industrial Analog PLC input or analog sensor.
Here is a good example:
www.edn.com/design/power-management/4371307/Convert-1-to-5V-signal-to-4-to-20-mA-output
wonderful ,thanks a lot.
Can you make any less readble???
calmly taught, thank you.
hi good day I need your help in designing a current source using the diode pair. Can u do a tutorial showing the design and calculations?
+Anjana Dabideen Can you be more specific about what you need? "Designing a current source using the diode pair" doesn't make sense to me.
I would like if u could demonstrate the current source using the PNP transistor. I have to design a 100W 8ohm amplifier that has a current source supplying the long tailed pair.
+Anjana Dabideen The diagrams at the bottom of page 1 of the show notes have some of the basic configurations for a PNP current source:
www.qsl.net/w2aew/youtube/transistorcurrentsources.pdf
You generally start by knowing what value of current you need, call it IC. It's generally a good idea to have 0.5 to 1V or more dropped across the emitter resistor. The "middle" diagram, the one with the two diodes connected from base to VCC, will put about 0.7V across the emitter resistor. Thus, you can calculate the resistor value as 0.7V/IC. Then calculate the resistor that goes from the base to ground. You want the current in this resistor to be about 10x greater than the base current in the transistor (or more). So, you calculate the base current by taking the IC value divided by the transistors Beta or Hfe. Then, multiply by 10 to get the bias current through the base-gnd resistor, call it Ibias. You then calculate the resistor value as (VCC-1.4)/Ibias. Done.
+w2aew Thanks alot for your help.
+Anjana Dabideen Can I ask you for one more explanation. I lack some fundamentals on this topic and i am unsure how to go about choosing the input current and voltage to meet the specifications of the amplifier project i am doing. Seeing that i have to design a 100W 8ohm amplifier what is an acceptable Ic?
Thanks for the video, helped a lot!
I just came across another current source configuration called the "wilson current mirror".
Could you do video about this one and explain why you would pick a wilson current mirror over the basic two transistor current mirror configuration :-)
+Rolfrolfsen Wilson current mirror provides more accurate current mirroring (less error between input and output current) and a higher output impedance (more ideal current source, less load dependent) than a simple two-transistor mirror using degenerative feedback. Might make for an interesting video in the future.
When engineers say an Op amp is a LOWER Characteristic op amp, what do they mean by "lower characteristic? In OP AMP schematics they often use transistor or FET current sources and current mirrors and differential , Common mode rejection. If you look at common Op Amp schematics for common part number some use differential inputs using transistors while others use FETs inputs plus it looks like they are configured in a current source, current mirror, differential configuration for inputs#1 and inputs#2. If the circuit designed is for an op amp LF348 and LF347 which is JFET inputs, you can't use a replacement like an LM741, TLO84, LM324 because those op amps aren't JFET inputs which I'm not sure why it wont work in circuits as an cross reference part. It would be nice if you made a video lesson about common OP amp part number schematics to go over the FET/transistor configurations inside the op amps of the current sources, current mirrors, etc
Another test : 1 current source "feeding" 1 current mirror with +9V supply.
- last example of current source (purists prefer to say it is a current sink, not a current source) : 2 NPN (2N2222A) and 2 * 1k resistors.
---> measured current is ~0.7 mA (collector of second transistor), quite unstable actually (breadboard quality ? I have no clue)...
Current is the same if I power that current source with +12V : ~0.7 mA (as expected...)
then connecting collector of second resistor directly to collector of the first transistor of the first current mirror example (2 * NPN, 2N2222A, no resistors)
Then, connecting 3 LEDs at the collector of the second transistor of the current mirror : they indeed all have the same brightness (same goes whatever the color or size of the LEDs) but the current at the LEDs is about 10 times bigger than the current at the current source....
But I think I made a mistake : if I use a current source with NPNs transistors (sink model), should I use a current mirror that uses PNP transistors (instead of NPNs like I did in my test) ?
Thanks Alan! You're the man!
another extremely informative video. thank you
Hello w2aew!
I have a small question. How does the current mirror function in the LT3750 block diagram in the datasheet?
You mentioned that it maintains a steady current, so I assume it is for some measuring function?
Here's a link to the datasheet: cds.linear.com/docs/en/datasheet/3750fa.pdf
Thanks!
It's not really a current mirror. The diode-connected PNP simply establishes a bias voltage on the other PNP transistor. When the stored energy in the transformer is being transferred to the capacitor, the voltage across the cap is reflected (with 1/N) to the primary and appears at the RVout resistor. Since the bias voltage at the RVout pin (the PNP emitter) is basically fixed, the voltage at the transformer is transformed into a current across the RVout resistor. This same current is sent through resistor RBG. Thus, the voltage across RBG is proportional this current, which is in turn proportional to the voltage across the cap. This voltage is compared against an internal 1.24V reference to detect full charge.
Thank you!
That was very informative.
I don't fully understand why the 470 ohm resistor is required. What would happend if not used?
Great video. Thanks.
+David Senabre The 470 ohm resistor around the variable power supply ensures that the supply always sources current. Without it, the circuit would try to make it sink current, which it can't do.
+w2aew Why would the variable supply sink current? Doesn't the transistor sink current? Current flow, as I see it, will always go from +10V to the collector. The variable power supply just reduces the 10V subtracting from it (as they are opposite polarity), but why should this invert current flow?
+David Senabre A power supply is designed to source current from its positive terminal, and have it return into the negative terminal. If the 470 ohm resistor wasn't across the variable supply, the collector current would have to be sourced from the negative terminal of the variable supply (thus, the positive terminal would have to sink the current from the main supply).
Umm, so my understanding of output impedance is a bit wrong I suspect. I understand how a constant voltage source has ideally no impedance at all since the voltage is constant, but with constant current sources/drains, I am not sure. I thought the transistor will continually change the impedance of itself (it's saturation) thus causing the voltage drop to change, thus leading to a change in the current. Of course if the transistors gate if connected to a resistor in a way that a negative feedback loop is formed, the transistor will end up regulating current at least to some degree.
Can a constant current source really be thought of like a really high voltage with very high output impedance, like a resistor on the output? I did wonder that for a long time, but decided that it is not an accurate to think of constant current sources like that.
For BJTs, the collector current will be *nearly* constant vs. load if:
1) The base-emitter voltage is constant (thus base current is constant)
2) The transistor is not cutoff or saturated.
For FETS, the channel can be thought of as a voltage controlled resistor at low V(DS) values (linear region), but behave more like BJTs when V(DS) is raised. This is called the "saturation region" for FETs, which is different than the saturation region for BJTs.
Yes, you can model the current source as a very high voltage with a larger series resistor. Of course, this model is only valid over an operating voltage range that applies to the circuit being used.
***** Very interesting! I will have to graph out ohms law with that HV + large resistor with considerably lower voltages and see what I can learn.
So If I had a infinite voltage source, with an infinite (this infinite is some factor bigger than the infinite voltage) resistance, then would I have a ideal current source? I will see if I can figure this out with mathematically, I will probably end up having to apply some basic calculus knowledge because surely, when I am dealing with infinities, esp. (x*infinity) / (y * infinity), I will need some concept of limits and things!
***** Can the output impedance be described as di/dv (a derivative )?
Yes, it is dV/dI