Making an inverter from a transistor
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- เผยแพร่เมื่อ 14 มิ.ย. 2024
- For my first foray into transistors, I decided to try recreating the inverted LED from my earlier video.
-- Links
Sully Station Technologies Circuit Library: sullystationtechnologies.com/c...
Electrical Engineering Stack Exchange pulldown discussion: electronics.stackexchange.com...
Transistor datasheet: www.onsemi.com/pdf/datasheet/...
Breadboard kit: amzn.to/3MYwfho
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(Maybe you and/or someone you know will like them!) - วิทยาศาสตร์และเทคโนโลยี
I like how we can also see your thought process in the making of these, it really helps to hear someone else think out loud when you are also learning. Great video!
Yes, I agree with that.
This one was really cool. I got to see the thinking behind these switches.
Limiting resistor 3 and ground act like a 'voltage divider' and when you turn on the base. Collector and emitter are (more) connected to ground. So the voltage divider dumps most of the voltage to ground (because it's very very low resistance rather than zero).
With a ratio of (very low resistance) to whatever the resistance of the limiting resistor is.
And in the initial setup the voltage is below the LED's ability to function. But not the case when you remove the limiting resistor.
Very cool lesson on this one.
In the second circuit, which is "common emitter" switch, I would do this:
Relpace R3 with a short circuit.
Pick the current I wanted through the LED. Let's say 10 mA.
With a RED LED the voltage across the LED when it is lit up to decent brightness will vary according to the specific type, but let's take 1.5 volts as a reasonable approximation.
With the transistor OFF, the voltage across R1, by Kirchhoff's voltage law will be (Vcc - Vled) = 5 - 1.5 = 3.5 volts
By Kirchhoff's current law, with no current through the transistor the current through the LED will equal the current through the resistor.
To determine the resistance we need we use Ohm's law
R = E/i (or sometimes V/i; normally you'd use capital I for current but it becomes ambiuous in a sans font, so I'll use i instead)
R = 3.5 V / 0.01 A = 350 ohms
There are two common values that are close enough to give us a pretty accurate result - 330 ohms and 360 ohms. 330 is generally much more commonly used.
This is something you just learn from experience.
Checking our current we now have
i = E/R = 3.5 V / 330 ohms = 10.6 mA - good enough!
When we turn a common-emitter transistor ON, the voltage between the collector and emitter, at moderate current, will be a fraction of one volt. The transistor is said to "saturate" (assuming we have enough base current). Say it is at the high end - 1 volt. That is still low enough that the LED can't turn on to any appreciable extent because it needs that 1.5 volts, but lower would give us more margin.
Now let's check using the datasheet.
We look at a graph for collector-emitter saturation voltage versus collector current. It's down to about 0.2, way below the 1.5 V reauired for the LED, with a base current of about 0.06 mA. The graphs are normally of "typical" values and real operating values can be quite a bit different in either direction.
We have a 10 k resistor between the base and +5 V. We analyze this exactly as we did for the LED current, but use 0.7 voltas across the B-E junction.
i = 4.3 V / 10k ohms = 0.43 mA
Going back to the graph in the datasheet we find that's lots more than we need to get the CE saturation voltage down to 0.2 V, so we have lots of margin for unit to unit variation and the like.
Transistor ON - anode voltage of LED is pulled so low that the LED can't turn on.
Transistor OFF - current through the lLED is as we determined previously.
I started just like you, poking around with chips learning, then put together a Z80 cpu and some sram, got it to run basic, and now Im up to a 68030 running a GUI multitasking os
It's fun to watch the experimenting, but especially appreciate all the links you included! Thanks!
Hey, looks great. I'll be watching for more of your content.
A good engineer will never leave a transistor base flying in the air. 🙂 R1 can be removed and the collector can be connected to the output of 2 LEDs - to short the LED. It is good to use a "digital" transistor that already has an input resistor inside and a closing resistor on the base.
resistors Limit current and voltage. Each item ( device ) has a tolerance in the schematic. To PROTECT each device you must not exceed the devices tolerances. This way we do not blow the Led or transistor. Hotter electronics get the shorter their life span. So it makes sense to figure out why you use certain resistors to limit current to each item in the circuit.
Maybe add a potentiometer instead of an on off switch to adjust the flow. Now we can see how the Leds brightness works.
NPN the emitter is connected to the NEGATIVE .
PNP the emitter is connected to the POSITIVE>
Resistors must sit on the Positive side only in a Circuit ( if this helps ).
I'm still not sure why at the moment you'd need NPN vs PNP except maybe in a Positively ground circuit vs a negatively grounded circuit.
As you adjusted each resistor you allowed current to find its balance through the circuit. Because the transistor and LED allow current you have met their needs in a manner to allow power to flow through them resulting in the LEDs brightness with each resistors replacement in the circuit.
One of the reasons you may chose to use a transistor as a switch in this way, rather than just using the switch is mainly to do with the current flow. In the transistor circuit most of the current to the device (LED in this case) is flowing through the transistor instead of the switch.
In your case it really makes no odds, as both the switch and the transistor will have no problems handling the current pulled by the LED.
However, lets assume your LED is a strip of LED's pulling significantly more current and your switch is the output pin of another IC (chip) or similar. In this case the switch on its own will not be able to handle the current required for the LED strip, so you put a transistor in the middle to handle the current and the IC can control the transistor.
Look up using resistors as voltage dividers.
@0:19- Transistors are analog switches
The schematic isn't about how to turn an LED on and off. It's a lesson to show how common emitter transistors are used in circuits to control the flow of current (like a switch does).
Pull up or pull down resistor is needed when circuit will be operating in a RF noisy area. The base without (pull up or down resistor), acts like an antenna. RF can trigger the transistor inadvertently, without the extra resistor or special shielding. For this simple discussion on transistor biasing, the extra resistor isn't needed.
R1 and R3 together create a voltage divider circuit. R3 assures the current is correct for the LED (too much will smoke it). R1 is required to make sure the transistor always goes completely off when the base is open.
Inverter circuit- R1 and R3 again create a divider circuit. R1 keeps current flow down to the transistor's published specifications. R3 protects the LED from over-current damage. In this circuit, R3's value is determined by the value of R1 (for total amount of resistance against VCC). Eliminating R1 puts the LED in danger, and eliminates the divider circuit required to operate the transistor properly. With only a 220 Ohm resistor between the LED and ground (the path of least resistance), virtually no current makes it through the 10K Ohm resistor to the transistor base (start the timer before the LED smokes itself).
Reading the "Concept of Operation" above the schematics will answer most of your questions.
Transistors can be used as digital switches. Some are more suited to this than others. Some transistors are better used in an analogue fashion. Frequency response, gain, collector current, everything in general..... even the style of the case......... Some transistors come out in different cases. Then we get into Common Emitter, Common Collector and Common Base method of connection ( or circuits)...... each method is a practical study into how semiconductors work. Then there is the configuration.... normal amplifier, driver stage, power output......... class A , B C etc etc...... maybe darlington connected, maybe complementary symmetry, it depends on what you want as a designer......... switching or analogue................ allow a voltage drop of 0.5 to 0.7 Volts over a silicon semiconductor junction, whether a transistor or a diode. ) 0.2 or 0.3 V if Germanium junction semiconductor. Then you have polarity...... NPN or PNP....Have fun, check twice before connecting it all up..... semiconductors like transistors and diodes self-destruct silently and extremely fast. Reverse connections or over current is fatal to your semi conductor. Accidentally shorting base to collector with your screwdriver or multimeter is another good way to destroy things. Ruined transistors test shorted, or open, with a multimeter. Conductive reading should appear as a typical junction. If zero ohms or open..... it is to be thrown in the bin!!!
As to being an 'inverter', some circuits invert the signal output. Like high in... low out. That's it. It's not an inverter circuit like a microwave or air conditioning or generator inverter......... that's a totally different thing.
When you spend a few years gaining competency with diodes, transistors......... then you can get into the world or FET's and IC's !!!! Mind you, these days you may get FETs and IC's first.....
E=MC2
In a project as simple as this, the placement of the limiting resistor doesn't really matter.
However you do change the radio properties (yes radio signals are naturally generated in a projects) when you do this. Before the led its basically a filter, radio noise stops here (especially if there's a capacitor). After the led its a band pass filter, a specific range of noise (or signal) is allowed to flow to the rest of the circuit.
As you get more amd more complex with transistor based circuits this needs to be a forethought, as npn vs pnp may change the radio properties of the signal, meaning the oscillating wave may or may not give a "correct" result ("correct" being the desired/anticipated result).
Also combining npn and pnp transistors for the sake of the same result can improve performance, as you can have 2 or more math or logic executions within the same signal.
This circuit couldn't have less to do with RF effects. I'm not saying they aren't present, but they are negligible and irrelevant. Everything that happened in this video can and should be analyzed as DC and steady state.
@@whatelseison8970
I agree with you. RF is negligbile in this project. That's why I said "In this project, the placement of the limiting resistor doesn't matter."
Loved the vid! Super cool to see your thought process and sources.
My thought for the R1: it could be in parallel to the LED to reduce the amount of current flowing through the LED. Maybe this is due to the max rated current the LED can handle
Bi-junction transistors do not require a pull down resistor because they switch based on current flow across the base, not the presence of voltage on the gate like a mosfet.
Just keep experimenting until you learn something. Forrest Mims has a series of books on electronics that should be easy to read. Most of his books were published by Radio Shack at one time.
cool vidoes man
Also, BJT transistors are current driven, whereas MOSFET transistors are voltage driven.
The second resistor in the led path is to also discharged the small capacitors in the chip
I think the LED is still on a little bit because without the resistor in series, the resistance of the path with the LED is low enough to allow current to flow but still many current runs away to the transistors' path, so the LED is dimmer than before. And as you have replaced the 10K resistor, I think there is also with the resistor of the LED path enough current for to flow in both directions and turn the LED on. I don't know if this is right, and sorry for possibly bad English.
8:50 This discussion isn't about some sort pull-up/-dowm problem what you have. You must know, that B-E has some capacity, and when you OFF signal, then transistor is still ON short amout of time. If you grounded [B]ase you can shorten this time. ()
hi sir i want to ask a question sir
in ds2501 transistor there is data ground pin and here is emiter base collector
can we make npn transistor like ds2501 memory chip ic using arduino ?
thnx sir
2:32 It's definitely NOT silly....You cannot limitlessly switch current through a switch so a transistor is usefull in that case to switch higher loads. Also, if the base current must come from a different (less "powerful" or lower voltage) source to switch something, you'd use a transistor.
0:30 Nope....If anything, they're analog switches....Providing current (or voltage, depending on type) gradually switches them/ makes them amplify/conduct...
When "switching" they're more like a tap.
NP or PN junctions drop 0.5v approx. so when you short R3 and press SW1, Q1 will drop 1.0v (C to E) which is plenty of voltage for D1 to be on. D1 should still be fairly bright.
All devices like this drop the voltage by .7 volts.
@@garycook5125 No. A silicon PN junction drops about 0.7V in forward bias but germanium diodes will drop something like 0.4 and Schottky diodes, which have a metal-semiconductor junction drop as little as 0.2V. LED's on the other hand use more exotic junctions, usually based on III-V compounds like gallium arsenide, indium phosphide, gallium nitride, etc. and have a forward voltage drop related to their color. Red LED's have a forward drop of about 2.1V, green are around 3, and blue 3.5. UV LED's are higher still.
Typically, switching transistors have a different parameter called saturation voltage, which will be the minimum voltage from emitter to collector when turned on fully and can be anywhere from 1 to 2V and is dependent on temperature and collector current in addition to the material properties.
@@whatelseison8970The video is about transistors, so that is what I'm talking about.
The resistor R1 will ensure the transistor to fully turn on. Without the R1 the transistor will work as an Emitter follower circuit and act as a variable resistor. With a led as load it's no big deal but with heavy loads on the emitter the transistor will overheat since it can't turn on fully. Without the R1 the transistor will not reach full saturation basically al low voltages also the applyed voltage to the R3 will nevver be Vcc but instead Vcc-BE-LedVforward. Measure the voltage on the emitter with R1 and without the R1 and you will understand why it is there. Can it work with out it? Absolutely but you loose voltage stability. With R1 the voltage on the emitter can be calculated Vcc-BE otherwise it will be Vcc-BE-LED_Vf. Since leds haves different Vf values so will your output.
An emitter follower will never saturate the transistor and adding an extra load between the emitter and ground helps nothing. It is just a power waster.
8:50 That is not a "battle"about a "pull-down"in this schedule, this discussion involves microcontroller in/output...You only need a resistor to make current flow....thus it is pointless to place a "pull-down" Only in case of "switch-bouncing" it would make sense...Nothing needs to be pulled down here.
Pulling the base down can reduce current in the collector-emitter path by providing an alternative path for collector-base leakage current. With modern silicon transistors operating at moderate temperature CB leakage current usually isn't a problem but it can be at elevated temperature. The magnitude of the leakage approximately doubles for every 10 °C increase in temperature and the leakage current is multiplied by the current gain of the transistor. If you use a Darlington pair that doesn't have internal resistors you really do need to pull the base down.
Pulling the base down also increases the speed with which the transistor turns off by "sucking" carriers out of the base region (otherwise they persist until they recombine). With something like an indicator LED this is completely irrelevant.
When a transistor is being driven by a typical "push-pull" ("totem pole") digital output a pull-down is of virtually no value except to clearly determine the state of the transistor if the digital device is powered down on in reset and outputs are floating. Whether you really need to be concerned with that depends on what you are driving. If it is the cycle release for a 100 ton punch press you better be darned sure behavior is very well defined in all circumstances.
*_You've been burned by a bad datasheet!_*
With the flat side facing toward you and the pins down, the pins are EBC, left to right. I think your transistor was in the circuit backwards. Many transistors will work to some extent that way but the current gain is dramatically reduced. That is consistent with everything observed.
You really need to measure voltages at all the nodes to get a better understanding of what is going on in these circuits.
Many years ago, Motorola, from which ON Semi descended, sent out unsolicited samples of one of their 3-terminal regulators in the same package as this transistor. The datasheet they supplied with it had the wrong pinout.