My analysis process was a bit different. By observation, there is 13.1V across the 13.7k resistor, so you can calculate the current in the two resistors (ignoring the base current). Then, calculate the IR drop across the 4.32k, and add 0.6 to that.
@@nickcaruso There are many ways to skin the cat. Alan looks at it like a EE and just sees currents. I tried (try) to teach at a more simple level and reduce the problem to something everyone is probably familiar with, voltage divider.
@@IMSAIGuy gotcha. I am (technically at least) an engineer and I am curious about ways of thinking about problems. Such was the intent of the question.
I am a fan of your clips; however, recently you have not published any new clips. Since you were in Tektronix, may be you know who design this circuit and ask him to give a presentation.
@@nickcaruso you might want to think about what the tek designer had to do. +5 at the top .6 in the middle, -12.5 at the bottom. now choose the resistors to use.
Here are a few points: 1- The diode is there to make sure if there is dip in the output voltage -- as a result of transient --, it won't go below - 0.7V , that will save your C789 (39uF Cap), without that your Cap will get reversed biased. 2- you can calculate the current through R783 is equal to ( 0.6 - (-12.5))/13.7K 3. 0.6V at the base of Q780 is our kinda reference in this feedback loop (Variation in hfe of transistors or load current does not change this base voltage, only temperature will) 4. If you open the feedback loop and do loop analyzes, you will realize R788 (which you taken off is there to stabilize the feedback loop) 5. Your open loop gain of the feedback loop is gm(of Q780) * R780 * (R783/(R783+R782)) 6. Your dominate pole (1st pole) in this feedback loop is1/{2*PI* [(R788+ 1/gm(of Q787))*C789]} 7. gm pf any BJT transistor is Ic/VT, VT is about 0.026 V & Ic is the collector current 8. Your first zero of feedback loop is 1/(2*PI*Resr*C789); Resr is the esr resistance of C789 9. Your parasitic pole of the loop is 1/(2*PI*R780* C_parasitic of the Base Q787) diffusion cap of Q787 C~Cb=tf*gm & tf =W^2/(2*Dn) W is effective base & thickness Dn is Diffusivity 10. Q784 & R788 (which you eliminated) act as short circuit protection. 11. The advantage of this Voltage Source over regular LDO: A. It has its own voltage reference. B. Does not require opamp C. Maybe can be designed to be faster ( higher ZGF and still stable) D. Lower over all transistor count. F. Maybe can be can be designed to have lower noise (Bandgap reference noise and opamp noise is eliminated, and we can optimize this type of voltage source for low noise) Anyways, interesting circuit thanks for pointing it out.
@@johnhargenrader8972 Simple: While the voltage on Base-Emitter of Q784 is less than 0.65V , the transistor is off, when the voltage reaches 0.65V It turns on. The Base-Emitter voltage of Q784 is almost equal to R788* Current of Q787. So if R788 is 1.5Ohm then it takes 0.65(V) /1.5 (Ohm) = 426 (mA) to turn Q784 on. Below this current it is as if the transistor does not exist, above this current Q784 would start conducting shot circuit the Base to Emitter of Q787. So the Maximum output current is about 400mA from 12V Supply. ( Assuming the output is short circuited) The Maximum Supply current ( i.e. 12.5 V supply) is 12.5(V) /470 (Ohm) = 26.5 (mA) ( Assuming the output is short circuited)
The Diode also protects Q780 against avalanche breakdown of its BE-diode. This is also the point nobody is thinking about for the typical 2-transistor astable multivibrator design. It should not be operated this way with operating voltages higher than 6 V. Typical avalanche breakdown voltages are between 7 and 12 V for BJT. Especially if you use large capacitors, delivering enough charge, you can easily destroy the BE-diode of the transistors.
@@gkdresden My main concern about this design is short circuit output to -12.5 V If there is a short from output to -12.5 the Diode D788 basically is going to fry, first. Then C789 is going to blow up. This is just beginning of chain reaction that will be very destructive. Therefore, the circuit that used as a load should have no path to -12.5 V.
Brilliant analysis of the circuit! Many thanks. Please do more like this. These are the basic "building blocks" of electronics, and once the student understands HOW they work, they can interconnect each and build larger more complex systems.
I like how in that era it was "we need a voltage regulator here, let's design one with transistors". Please keep this kind of content going, it's always good practice to reason about circuits.
many years ago one voltage regulator ic is cost more expensive than 3-5 transistors. Especially if you need a stabilizer with a current of 20-50-100 mA.
I think the main point of this regulator is that it is based on a separate voltage stabilized reference (The -12.5 V supply). The 0.6 volt drop will drift slightly with temperature (maybe about 0.05V) but the current through the 13.7K will be relatively fixed at 13.1V/13.7K or about 1ma. The current in the base of Q780 will be small (assuming reasonably high beta), so the voltage output will be about 4.32V+0.6V or about 5V. The current limiter will not be thermally stable, but this is just a protection feature and doesn't need to be precise.
@@plainedgedsaw1694 It's -2mV/°C, but it would take a 50°C rise in temperature to decrease the 13.1V across R783 by 0.1V. That's less than 0.02% per °C, which is effectively negligible.
Old Tek manuals are one of the best places to learn good electronic design. Sometimes they over complicated things a lot but usually for good reason. The best one I ever read was the Tek 650 video monitor schematic. The circuit design is excellent, but they also put humorous little drawings in for different circuits, i.e. near the back porch generator circuit was a drawing of a generator sitting on a back porch. I seem to recall there were about a half dozen of these which always cracked me up. The reason the voltage calculated low was that the base of Q780 will bleed a tiny bit of current through R782. Knowing Tek, I bet this was all highly calculated for all temperature and current ranges. The stuff they built from this era generally had very robust designs. For those wondering how it regulates so well, I was going to go through it but there is a great explanation in two comments from Rex Schneider near the bottom of the thread. It is that -12/5V supply that is critical to the proper operation. Great video, I grew up with this stuff, so it is nice to see someone explaining how these circuits work. Still very valuable knowledge to have these days.
Very interesting. With your analysis and explanation it all sounds obvious. But it would have taken me a long time to fully figure it out. Great content!
Did anyone asked why the R783 goes to -12.5V instead of GND? And why R780 goes to +12.5V instead of collector of Q787? There is a reason for this. It is kind of sequencer and protection. If, for some reason, -12.5V line fails (which is a huge disaster for circuits that era) then the lower side of that divider goes up which opens Q780 to saturation which cuts off the 5V line. If, for some reason, +12.5V line fails, then there is nothing, that could open the Q787 transistor, which cuts off the 5V line. There must be -12.5V and +12.5V line present before the 5V line is turned on and if some of them fails, then the 5V line is immediately turned off.
Hi, R783 goes to -12.5V because this circuit is more an amplifier rather than a regulator. It is an inverting amplifier whose input signal is -12.5V. If you connect R783 to gnd (and lower its value to produce 5V nominal at the output), then you realize a Vbe multiplier. The output will vary proportionally to the Vbe (which is -2mV/CentDegree) and thus… a lot. They must have a good regulated -12.5V. To produce 5V they just… amplify it. Clever!
I have to admit the person that picked those values did a cute little trick where they did the whole thing as close to 1V=1KR in the divider. Not perfectly, obviously they had to stick to a e series for the resistors (E96 in this case, or 1% tolerance resistors). Still, good on that person for not picking really out there values. edit: It makes life so much easier for the technician. I saw you had alan there in the comments looking at it all as currents as an ee would do. Once you hand it to the tech, you want something you can quickly model in your head
transistor Q780 base bias is 0.8v, not 0.6v. the diode D766 keeps the transistors Q780, Q784 active during overload, which ensures reliable shutdown of the output voltage and its easy restoration back to 5 V when overload is off.
Very helpful!! I’m restoring a Heathkit H89 with a faulty power supply section on the video board and was struggling to understand the theory of operation. This is the most helpful explanation I’ve found. It’s almost exactly the same circuit except the Heathkit has a zener reference and an extra transistor.
Thanks for presenting this piece of old school electronics, very interesting. I think of it this way: it is an opamp in inverting configuration whose gain is -R782/R783 and whose input referred offset is the Vbe of Q780. They use their negative rail as a input to this inverting amp.
Back in the days, in the 60's when I was a student, we would check out the service manuals from HP and Tek. There would be a very good explanation as to how the circuits worked. You don't get that anymore. HP even made it more and more difficult to get diagrams. I always preferred Tek over HP. I think the Tek stuff had better human engineering too. Gee, why didn't they just use a 7805, they didn't exist them. They did have op amps though. Don't think there were 723's either. Thanks, very good. Jim
110 ohm resistor also provides a min load so as the circuit start up voltage can come up , as well as providing a time constant for the circuit while the diode keep it form going blow ground .
Stumbled across your channel, what a blast from the past! Good Stuff. I had to go back through your videos to find out why your "IMSAI GUY", I had an almost identical setup with the addition of two 4Mb hard drives. Lots of FUN, got me through collage and my first couple of jobs. CP/M, Wordstar, dBase, and Fortran. OH My, that was a long time ago.
I clicked immediately on video just because I recognized the Tektronix font and style (layout) on the thumbnail :) . Even if I had repaired Tek scopes for many years, I enjoyed the video and explanation, just for fun. Nice job you do teaching the thru art of creative design (not just ... put an IC and few capacitors around and write some soft) from the times when in design was plenty of room for ideas. The smaller the bricks the more fun for the architect :)
@@caffesenza5836 A design done for production is going to have about 10 times as much current flowing in the resistor divider than flows in the base of the transistor. You do this because the HFE(current gain) of transistors varies a lot from unit to unit. For this reason, a first cut at working out what it does would just ignore the base current.
Your numerical analysis is correct, however I would argue that you should use the general expression for the output voltage because it reveals subtleties that may be significant. Here goes: Call the diode voltage Vd. The -12.5 voltage is actually the"reference voltage" for this generator, so call it Vref. I prefer to think of Vref in its absolute value, namely "12.5V" dropping the negative sign, which will be taken into account shortly. Shorten R782 and R783 to R1 and R2 respectively. Call the output voltage Vo. The equality that inverting gain element Q780 forces is: (Vo - Vd)/R1 = (Vref + Vd)/R2 Solving for Vo gives you Vo = Vref*R1/R2 + Vd*(1+R1/R2) Plugging in values and assuming Vd=0.6V, and Vref = 12.5V, you get Vo = 4.73V, as you calculate at 8:23 The interesting part of this general formula is that it reveals that this voltage has a temperature coefficient to it due to the diode's presence. It is common knowledge that "a diode," namely the transistor's Vbe, has a temperature coefficient of -1.6mV/°C. Therefore as temperature increases, Vo will decrease with it by a factor of (1+R1/R2). The great unknown is "what is the temperature coefficient of Vref, the -12.5V supply? We don't know. What you really want to do is to take the derivative of the equation with respect to temperature, d/dT. If the -12.5V supply has any temperature dependence to it, it will be attenuated by R1/R2. Q780 is the "amplifier" in this circuit that in today's world would be replaced with an op amp running in its inverting mode and its "+" terminal connected to Vd instead of ground. I miss the days of transistor design. It was a great deal of fun.
I obtained the result in a different way. 13.1 volts across 13.7k R783 resistor implies 4.13 volts across 4.32k resistor R782, given equal current (Ohm's law). 4.13 plus 0.6 volt results in 4.73 volts. That would be the output. However, there is some additional small current flowing through resistor R782 to the base of Q780. Depending on this current the voltage drop on R782 will be slightly higher than 4.13 volts so we can probably have exactly 5.0 volts or even slightly higher voltage.
The 110 ohm (R789) must be a startup load as well, as bleed resistors would be more like 1Mohm? That was a really neat circuit though, thanks for the explanation
Thanks! I have a similar voltage regulating circuit in a telephone power supply from the 70s. The voltage divider can be set manually to 3 V, 4.5 V and 6 V. The power supply must be very silent to not make any noise in the telephone receiver. It has that 3rd transistor to keep the current below 300 mA. Thanks for the calculations, very helpful! I suspect your power supply is very silent as well? (probably depends on filter capacitors and a good transformer)
Great video. I think the 110 ohm resistor is to put a minimal load on the output as well as cap bleed off, 110 is rather low for a bleed, unless it has to do it before other supplies bleed off. I have a 80's 2 channel dual trace Tektronix scope that was putting out way over 40VDC out of the preregulator and blowing the fuse. The 40V is provided to a wild push pull oscillator with some control transistors and an optoisolator to a Transistor regulated supply to drive the 120VAC rectified voltage through a Triac to get the 40VDC. A nasty closed loop. The push pull oscillator puts a square wave on the circuitry power supply transformer to regulate most of its output voltages all at once. The manual had a good theory of operation, but I was not able to verify or find a replacement opto isolator. I found an open resistor and a shorted diode in the push pull circuit but it was no help. The transistors are old Tecktronix part numbers and there were not enough waveforms/voltage notes for them. I ended up isolating all the circuity (lifting leads) from the 120V input power switch all the way to the 40V trace. I tried to find an ac/dc converter (or small ac/dc and dc/dc converters) but they would not fit inside by their mech specs. I bought a cheap adjustable 45V 10A supply and strapped it to the outside of the case. It works but it is clunky. I got the scope for free, I bet it is a transistor or two from working properly. If I post the power supply schematic and my notes, could you take a look?
Very nice analysis! I'm a very young engineer and I'm not used to reasoning with BJTs so much, so it's a very nice practice! I want to make one remark though, that in today's world you would much rather have an integrated regulator than discrete BJTs not only because of cost (a SOT-23 LDO is probably cheaper than 3 discrete BJTs) but also because of assembly cost (extra SMD reels that you have to put into the pick and place machine, marginally longer pick and place time, etc) and because of board space savings. So it's not out of laziness that we don't do this anymore, but because designs have become so much more miniaturized and dense.
No voltage reference (Zener diode, TL431 etc.) but it works nicely. I wonder about the ripple voltage. While the voltage drop is 7V, the current limiter will keep the dissipated power down to 2.8W. Not bad for a heatsinked TO-66.
The controlled voltage opposes changes in the -12.5 V supply; they track each other such that the base voltage remains 0.6 V and very little current flows into the base. Tracking power supplies utilize similar tricks, but the resistors would be almost equal in this design.
If they dropped the value of the 13.7K resistor, and added a couple of diodes in series, there would be some temperature compensation in the voltage output, as long as the control transistor and diodes could be close to the same temperature.
Nice schematics! Especially interesting because it contains only NPN transistors, even diode is possible to replace with base-collector shortened transistor. Can be used in hack diy or in production - no jokes some times is much cheaper to use only same transistor four times to reduce the costs.
Your 4.7V output would be off slightly because the base current was ignored. But, I would have done the same thing, close enough. :) Haven't thought about if adding in the base current would lower or raise the output voltage. Wonder how stable is this over temperature?
The added base current would shift the output voltage up a little, having the same effect as reducing the value of the 13.7K resistor, but the effect is probably pretty negligible. As for temperature stability, it's probably not great, but the fact that the entire design is dependent on the integrity of the -12.5V rail, it's probably not anyone's biggest concern.
The temperature dependance would be something like 1+4.32/13.7 times the Vbe tempco (2mV/Cent Degree). So it is just 30% more than the variation of the Vbe.
Good video, thanks. Doesn't this circuit rely on the fact the input voltage is regulated (the -12.5V rail) so it works as a regulator for varying loads currents but not for varying input voltages.
That first part of the circuit looks like a fold back current limiter to me... it's probably doing that, which make me think that this probably only regulates properly when it is under load and the no load open circuit voltage probably won't make much sense in respect to the required 5V
I think too low negative rail (like -12.5v) is counterproductive from precision point of view. Also its too dependent on Vbe (of Q780), i.e. Vbe temperature change is not compensated. I would do it better 😀.
You can look at gain and phase shift, and plot the results on a Bode Plot. In a nutshell, you don't want gain and positive feedback at the same time (unless you are making an oscillator). That's the 70,000 ft. quick answer. Perhaps you could analyze it using a Laplace transform, but not sure about that if there's something nonlinear. Nowadays, you would just do the Bode analysis w/a simulator. The 39uF cap likely helps 'helps' slow down the loop so you never have positive feedback w/gain (just an educated guess).
@@IMSAIGuy Thanks for the article! I have wanted a reasonable length article explaining how transfer functions relate to stability. I will take some time to read this, hopefully it helps provide me a framework for thinking about stability.
@@Mark-hb5zf yes i know, but how to do that without directly going to simulation or measurements, is there no "old" way. I the problem I have is I know the control theory and i know circuit and also in the link you can see that it does not really bridge them both. How would you go about designing, if you for example had thought of this circuit, How would know? Also, where would you go to event it from being unstable, where is the place to stabilize it. not sure if you understand what i mean
@@IMSAIGuy depends, also in that article the whole circuit side is missing, i find that always problematic, the control theory is not that complicated, but making a discrete circuit that does work is a different story and is almost never addressed. That was more my questing. If you had designed that supply, how would you have looked at the circuit to determent if it was good in terms of stability. Sure, you can put in a restitor and make a bode plot. But that cant be the only way to at least get a understanding where such a circuit will have critical points. I would be really interesting to look maybe at this circuit and try to do the exact thing. Hope you understand what i meant.
Can one repurpose or adapt this circuit if there is no negative rail? I read the comments that referencing it to ground is not a good idea, but I would think there is enough voltage headroom to do something else.
The divider chain of R782/R783 provides a minimal load of 1mA. The base drive without any more output current is sunk by Q780. At no-load, that is about 13mA, assuming the 470R resistor is tied to something around 12V as suggested in the video. Using 110R to sink 45mA as a minimum load gets the pass transistor into its preferred working region, rather than being just barely switched on. I assume 110R was chosen because that's the smallest value of resistor that keeps its dissipation below 1/4 W on a 5V supply.
Voltage and current control using only 3 active discrete silicon, that's impresive thinking about how long ago that circuit is. I won't like that -12.5 is referenced, I'll go to ground instead. Thank you so much
If you do that, then your output voltage depends on the Vbe of the particular sample of transistor used for Q780, and the temperature drift of the output voltage will be -0.3% per °C. Using a stable, regulated -12.5V supply line eliminates dependency on the particular transistor's Vbe and reduces the temperature drift by a factor or 13.1/0.6 = 22 times. I think I'll stick with the design as it is, if that's okay.
One thing to keep in mind is that the supply rails take some time to start up. The -12V rail could be more noticeably more negative than the +12V rail is positive during startup. The diode will keep the output from going too negative during that time, and keep from damaging components attached to the output.
@@Cynthia_Cantrell Thank you. That also crossed my mind. A job interview side-tracked me from laying this out the other day and your comment just reminded me to build a split supply for an ANDY learner/trainer board for logic. I’ve got a couple hours before the wife’s home. Time to make some sweet MusICs without blowing my ear drum, or getting yelled at. 😂
So it uses Q780 (emitter diode drop) as the voltage reference, and then 2 precision resistors to get the correct ratio against the 0.6v reference. Maybe they were happy enough with the temperature drift on that particular transistor to choose it... I don't think they were too concerned about nailing the voltage exactly, more like they wanted to make it as consistent as possible. I like the over-current clamp though, it's "cute". Perhaps the -12 v low side on the divider was used as it was harder to find precision resistors in low values at the time, or the selection values were limited? These days, the whole range of resistances seem to have a precision value offering. Cheers,
The -12.5 V supply was regulated, this circuit amounts to a buffer of this regulated voltage. The voltage divider is such that the variations in the 0.6v drop (for example, due to temperature) do not contribute much to the output voltage as most of the voltage drop across R783 comes from the regulated supply.
@@plainedgedsaw1694 the output voltage will follow the -12,5V supply. If the 12.5 volts changes by 10%, then the 5v output of this circuit would also see a 10% change. The -12,5 V is the voltage reference for this circuit.
@@argcargv well, yes, it regulates voltage in relation to that -12.5V rail. But if it did connect to ground, it would still be regulated (no significant drift vs input voltage or load), just to different voltage.
@@plainedgedsaw1694 if it was connected to ground then the output voltage would be very sensitive to the temperature of the transistor, which will be indirectly sensitive to load.
No, this is a linear regulator. It does help to calculate and understand the circuit by analysing the "edge cases" but the transient between are not discrete. A so-called linear region.
I was just trying to invent 3.7V regulator for Lifepo4-charger. I was going to use stack of diodes as voltage reference, but that would have been ugly and embarrassing.
Note that this drifts like hell with temp, try tl431, i heard it can be unstable though. Current limit accuracy doesn't matter as much so the transistor should be fine.
@@TimoNoko tl431 isn't precision component either, it's usually 2% precise. But treating li cell well even if treating it worse won't make it blow up, is good idea i think.
If you felt like I think it would be interesting to show a video actually building an example of this and see if it works nicely or terribly, or how efficient
It was designed for a high end Tektronix curve tracer. $40k in todays money. The one I have regulates perfectly at 5.03 volts and is about 32 years old.
It is not very efficient at all, like many linear regulators, namely because all of the current that is available at the output at 5V first has to drop from 12.5V through the pass transistor and current sense resistor - a drop of 7.5V. Since efficiency = Pout / Pin, and (ignoring the regulator transistor) Pout = 5 * I_load and Pin = 12.5 * I_load, efficiency can not be better than 5/12.5 = 40%.
@@IMSAIGuy No, I haven't seen it. But a 15KW out sounds awfully beefy - the standard house outlet can only supply 1.8KW before its breaker trips. Perhaps it's 150V & 10A out?
I have a simple question, but are there any reasons they used -12.5V as a reference point? I mean, the resistor divider could easily reference to GND...
Hi, if you do that you will see your 5V nominal swing a lot with temperature. Like 17mV / Cent Degree. You will turn this thing into a Vbe multiplier, while it is thought as an inverting amplifier with -12.5V well regulated input
It only depends on already regulated -12.5V rail, but not on it's positive rail. I guess they did it this way because it was simpler and cheaper and did the job.
@@yveslesage8525 but why can’t the voltage divider be referenced to 0V instead? Assuming I want to reuse this circuit in a product that doesn’t have negative supplies. Isn’t it a matter of changing the voltage divider so the 0.6V sits at the right point?
@@DafyddRoche If you do that, then your output voltage depends on the Vbe of the particular sample of transistor used for Q780, and the temperature drift of the output voltage will be -0.3% per °C. It's not suitable for production. You might as well just use a 4.3V Zener supplied from the 470R resistor and only keep Q787. Using a stable, regulated -12.5V supply line eliminates dependency on the particular transistor's Vbe and reduces the temperature drift by a factor or 13.1/0.6 = 22 times. If you want a 5V supply in a modern product, just use an LM7805 or one of the common low drop-out regulators if you have less than 7V to supply it from.
@@RexxSchneider Hi Rex. yes - it'd be easier to use a simple regulator! :) But I am curious, or I should say -- I need to understand - isn't ground considered "stable and regulated"? I think I understand that because the the transistors emitter is grounded, we're now depending on 0.6 or 0.65 or 0.7V turn on voltage difference. Why does having a -12.5V make a difference? surely, a Vbe variance in the transistor makes the turn on voltage equally as variable? (Help! I clearly haven't had my AH-HA! moment! )
@@DafyddRoche Ground is stable, of course. But if you are trying to use a reference that's only about 600mV above ground, i.e Vbe of the transistor, then a variation of 60mV for example (perhaps through temperature change or variation in Vbe between samples) will produce a variation of 10% (60mV/600mV) in the current flowing through the resistor divider and a consequent 10% variation in the output voltage. On the other hand, if you tie the lower resistor to -12.5V instead of ground, then you have a total of 13.1V across that resistor and a change of 60mV in Vbe will only create a change in current of 60mV/13100mV = 0.5%, and hence only a 0.5% change in output voltage. Now do you see the difference?
Messy. Vibe maybe anything between 0.6 and 0.7 V. This significantly affects V out. Assume you want 5 volts output, so work backwards to find Vbe for the resistor values given.
Quite obviously, it's not a regulator at all, since output is at the whims of Vbe of Q780, being highly temp dependant. A simple 2.7v zener in it's emitter would've worked wonders.
Transistor cu CPU vaccum switch ALU second generation of computer ic intergrade circuit Memory unit (capacitor restance transformer) Primary memory Secondary memory
A long and messy explanation. As the output is +5V and the negative supply -12.5 you have their sum (17.5V) across series R782 and R783 (18.02K). The Q780 base is at +0.6, or so (current gain/base current plays here) and is limited to negative ~ 0.7V with that diode. remove R782 and the voltage is limited to about 5.6V. Q780 gain (100 -300) is Rb/Re enough to drive the TO-66 guy (low gain, maybe 25). The easiest calculator to use for this work LTspice. (free, and intuitive to use) Works faster than the calculator. The circuit does not have a fixed reference and a differential circuit, output voltage depends on gain of transistors and stability of negative 12.5V. With an old OP-amp circuit and +/- supplies, this is a quick and dirty voltage regulator.
Your analysis is insufficient, which is why you got an error at the end. You forgot the current into the base of Q780, which can be calculated from its collector voltage, R780 and beta. We know the Q780 collector voltage because Q787 is an emitter follower so its base has approximately same level as its emitter, i.e. +V. The R780 is rather smallish and its current flows into Q780 collector, so the Q780 base current will have a contribution into the overall calculation. I.e. it will take higher output voltage to also feed some current into the base, so it will likely reach 5V on the output.
Hmm.. I was enjoying the discussion but then I started to notice just how many "ok?"s there are and now I'm going to have to exit. It's like almost every second word, mmkay?
Back then you would normally use a Zener on the pass tranny base (if not too high power needed), and the same current limit circuit. Not sure why they did it the way you showed??? Love the To-66's look very cool, Nothing more nurdy than a bunch of TO... cans on a heatsink and a bunch of large inductors next to them, ...heaven man! :)
My analysis process was a bit different. By observation, there is 13.1V across the 13.7k resistor, so you can calculate the current in the two resistors (ignoring the base current). Then, calculate the IR drop across the 4.32k, and add 0.6 to that.
can you say what led you to make that observation, or rather why that was a good place to start the analysis?
@@nickcaruso There are many ways to skin the cat. Alan looks at it like a EE and just sees currents. I tried (try) to teach at a more simple level and reduce the problem to something everyone is probably familiar with, voltage divider.
@@IMSAIGuy gotcha. I am (technically at least) an engineer and I am curious about ways of thinking about problems. Such was the intent of the question.
I am a fan of your clips; however, recently you have not published any new clips.
Since you were in Tektronix, may be you know who design this circuit and ask him to give a presentation.
@@nickcaruso you might want to think about what the tek designer had to do. +5 at the top .6 in the middle, -12.5 at the bottom. now choose the resistors to use.
Here are a few points:
1- The diode is there to make sure if there is dip in the output voltage -- as a result of transient --, it won't go below - 0.7V , that will save your C789 (39uF Cap), without that your Cap will get reversed biased.
2- you can calculate the current through R783 is equal to ( 0.6 - (-12.5))/13.7K
3. 0.6V at the base of Q780 is our kinda reference in this feedback loop (Variation in hfe of transistors or load current does not change this base voltage, only temperature will)
4. If you open the feedback loop and do loop analyzes, you will realize R788 (which you taken off is there to stabilize the feedback loop)
5. Your open loop gain of the feedback loop is gm(of Q780) * R780 * (R783/(R783+R782))
6. Your dominate pole (1st pole) in this feedback loop is1/{2*PI* [(R788+ 1/gm(of Q787))*C789]}
7. gm pf any BJT transistor is Ic/VT, VT is about 0.026 V & Ic is the collector current
8. Your first zero of feedback loop is 1/(2*PI*Resr*C789); Resr is the esr resistance of C789
9. Your parasitic pole of the loop is 1/(2*PI*R780* C_parasitic of the Base Q787) diffusion cap of Q787 C~Cb=tf*gm & tf =W^2/(2*Dn) W is effective base & thickness Dn is Diffusivity
10. Q784 & R788 (which you eliminated) act as short circuit protection.
11. The advantage of this Voltage Source over regular LDO:
A. It has its own voltage reference.
B. Does not require opamp
C. Maybe can be designed to be faster ( higher ZGF and still stable)
D. Lower over all transistor count.
F. Maybe can be can be designed to have lower noise (Bandgap reference noise and opamp noise is eliminated, and we can optimize this type of voltage source for low noise)
Anyways, interesting circuit thanks for pointing it out.
Ok. Now explain the Crowbar current limiter you removed.
@@johnhargenrader8972
Simple: While the voltage on Base-Emitter of Q784 is less than 0.65V , the transistor is off, when the voltage reaches 0.65V It turns on.
The Base-Emitter voltage of Q784 is almost equal to R788* Current of Q787. So if R788 is 1.5Ohm then it takes 0.65(V) /1.5 (Ohm) = 426 (mA) to turn Q784 on.
Below this current it is as if the transistor does not exist, above this current Q784 would start conducting shot circuit the Base to Emitter of Q787.
So the Maximum output current is about 400mA from 12V Supply. ( Assuming the output is short circuited)
The Maximum Supply current ( i.e. 12.5 V supply) is 12.5(V) /470 (Ohm) = 26.5 (mA) ( Assuming the output is short circuited)
The Diode also protects Q780 against avalanche breakdown of its BE-diode.
This is also the point nobody is thinking about for the typical 2-transistor astable multivibrator design. It should not be operated this way with operating voltages higher than 6 V. Typical avalanche breakdown voltages are between 7 and 12 V for BJT. Especially if you use large capacitors, delivering enough charge, you can easily destroy the BE-diode of the transistors.
@@gkdresden
My main concern about this design is short circuit output to -12.5 V
If there is a short from output to -12.5 the Diode D788 basically is going to fry, first. Then C789 is going to blow up. This is just beginning of chain reaction that will be very destructive. Therefore, the circuit that used as a load should have no path to -12.5 V.
Brilliant analysis of the circuit! Many thanks. Please do more like this. These are the basic "building blocks" of electronics, and once the student understands HOW they work, they can interconnect each and build larger more complex systems.
You makes it so easy to understand electronics. You guide in a very methodically and calm way.
Thank You!
I like how in that era it was "we need a voltage regulator here, let's design one with transistors". Please keep this kind of content going, it's always good practice to reason about circuits.
many years ago one voltage regulator ic is cost more expensive than 3-5 transistors. Especially if you need a stabilizer with a current of 20-50-100 mA.
@@whatever2144 Good point, there are still applications where this approach makes sense.
@@whatever2144 It's been quite of a while since I've seen
but I guess the circuit depends on -12.5V rail be regulated with respect to ground all other things being equal.
I think the main point of this regulator is that it is based on a separate voltage stabilized reference (The -12.5 V supply). The 0.6 volt drop will drift slightly with temperature (maybe about 0.05V) but the current through the 13.7K will be relatively fixed at 13.1V/13.7K or about 1ma. The current in the base of Q780 will be small (assuming reasonably high beta), so the voltage output will be about 4.32V+0.6V or about 5V. The current limiter will not be thermally stable, but this is just a protection feature and doesn't need to be precise.
Vbe drifts about 2mV/°C i think. It can be used as thermometer.
@@plainedgedsaw1694 It's -2mV/°C, but it would take a 50°C rise in temperature to decrease the 13.1V across R783 by 0.1V. That's less than 0.02% per °C, which is effectively negligible.
Old Tek manuals are one of the best places to learn good electronic design. Sometimes they over complicated things a lot but usually for good reason. The best one I ever read was the Tek 650 video monitor schematic. The circuit design is excellent, but they also put humorous little drawings in for different circuits, i.e. near the back porch generator circuit was a drawing of a generator sitting on a back porch. I seem to recall there were about a half dozen of these which always cracked me up.
The reason the voltage calculated low was that the base of Q780 will bleed a tiny bit of current through R782. Knowing Tek, I bet this was all highly calculated for all temperature and current ranges. The stuff they built from this era generally had very robust designs.
For those wondering how it regulates so well, I was going to go through it but there is a great explanation in two comments from Rex Schneider near the bottom of the thread. It is that -12/5V supply that is critical to the proper operation.
Great video, I grew up with this stuff, so it is nice to see someone explaining how these circuits work. Still very valuable knowledge to have these days.
Very interesting. With your analysis and explanation it all sounds obvious. But it would have taken me a long time to fully figure it out. Great content!
Did anyone asked why the R783 goes to -12.5V instead of GND? And why R780 goes to +12.5V instead of collector of Q787? There is a reason for this.
It is kind of sequencer and protection. If, for some reason, -12.5V line fails (which is a huge disaster for circuits that era) then the lower side of that divider goes up which opens Q780 to saturation which cuts off the 5V line. If, for some reason, +12.5V line fails, then there is nothing, that could open the Q787 transistor, which cuts off the 5V line. There must be -12.5V and +12.5V line present before the 5V line is turned on and if some of them fails, then the 5V line is immediately turned off.
Hi, R783 goes to -12.5V because this circuit is more an amplifier rather than a regulator. It is an inverting amplifier whose input signal is -12.5V. If you connect R783 to gnd (and lower its value to produce 5V nominal at the output), then you realize a Vbe multiplier. The output will vary proportionally to the Vbe (which is -2mV/CentDegree) and thus… a lot. They must have a good regulated -12.5V. To produce 5V they just… amplify it. Clever!
I have to admit the person that picked those values did a cute little trick where they did the whole thing as close to 1V=1KR in the divider. Not perfectly, obviously they had to stick to a e series for the resistors (E96 in this case, or 1% tolerance resistors). Still, good on that person for not picking really out there values.
edit: It makes life so much easier for the technician. I saw you had alan there in the comments looking at it all as currents as an ee would do. Once you hand it to the tech, you want something you can quickly model in your head
Simple and accurate enough! And repairable.
transistor Q780 base bias is 0.8v, not 0.6v. the diode D766 keeps the transistors Q780, Q784 active during overload, which ensures reliable shutdown of the output voltage and its easy restoration back to 5 V when overload is off.
fun little circuit analysis, thanks for taking us along.
Tektronix had the best electronic drafting style!
Yes, the old hand drawn documents from TEK and HP show the pride and craftsmanship they had.
Very helpful!! I’m restoring a Heathkit H89 with a faulty power supply section on the video board and was struggling to understand the theory of operation. This is the most helpful explanation I’ve found. It’s almost exactly the same circuit except the Heathkit has a zener reference and an extra transistor.
There is beauty in little things after all.
Thanks for presenting this piece of old school electronics, very interesting.
I think of it this way: it is an opamp in inverting configuration whose gain is -R782/R783 and whose input referred offset is the Vbe of Q780. They use their negative rail as a input to this inverting amp.
Fun to go through this unusual (to me) circuit. Brings back the fun of electronics to just slowly go through it unscripted!
Back in the days, in the 60's when I was a student, we would check out the service manuals from HP and Tek. There would be a very good explanation as to how the circuits worked. You don't get that anymore. HP even made it more and more difficult to get diagrams. I always preferred Tek over HP. I think the Tek stuff had better human engineering too.
Gee, why didn't they just use a 7805, they didn't exist them. They did have op amps though. Don't think there were 723's either.
Thanks, very good. Jim
110 ohm resistor also provides a min load so as the circuit start up voltage can come up , as well as providing a time constant for the circuit while the diode keep it form going blow ground .
Stumbled across your channel, what a blast from the past! Good Stuff. I had to go back through your videos to find out why your "IMSAI GUY", I had an almost identical setup with the addition of two 4Mb hard drives. Lots of FUN, got me through collage and my first couple of jobs. CP/M, Wordstar, dBase, and Fortran. OH My, that was a long time ago.
th-cam.com/video/DLFIBQ1WlmU/w-d-xo.html
Thank you so much for explaining this in detail, it helps a lot.
Regards Helmut
I clicked immediately on video just because I recognized the Tektronix font and style (layout) on the thumbnail :) .
Even if I had repaired Tek scopes for many years, I enjoyed the video and explanation, just for fun.
Nice job you do teaching the thru art of creative design (not just ... put an IC and few capacitors around and write some soft) from the times when in design was plenty of room for ideas.
The smaller the bricks the more fun for the architect :)
The circuit is more properly an inverting "op-amp" rather than a regulator. Its output varies with the negative supply voltage.
Yes, if one does the math this way it is very easy to figure out the nominal output voltage, dependance on the Vbe and on the base current
@@caffesenza5836 A design done for production is going to have about 10 times as much current flowing in the resistor divider than flows in the base of the transistor. You do this because the HFE(current gain) of transistors varies a lot from unit to unit. For this reason, a first cut at working out what it does would just ignore the base current.
Excellent... I love your RPN calculator!!
th-cam.com/video/pq8XAnIu2T4/w-d-xo.html
Your numerical analysis is correct, however I would argue that you should use the general expression for the output voltage because it reveals subtleties that may be significant. Here goes:
Call the diode voltage Vd. The -12.5 voltage is actually the"reference voltage" for this generator, so call it Vref. I prefer to think of Vref in its absolute value, namely "12.5V" dropping the negative sign, which will be taken into account shortly. Shorten R782 and R783 to R1 and R2 respectively. Call the output voltage Vo. The equality that inverting gain element Q780 forces is:
(Vo - Vd)/R1 = (Vref + Vd)/R2 Solving for Vo gives you
Vo = Vref*R1/R2 + Vd*(1+R1/R2)
Plugging in values and assuming Vd=0.6V, and Vref = 12.5V, you get Vo = 4.73V, as you calculate at 8:23
The interesting part of this general formula is that it reveals that this voltage has a temperature coefficient to it due to the diode's presence. It is common knowledge that "a diode," namely the transistor's Vbe, has a temperature coefficient of -1.6mV/°C. Therefore as temperature increases, Vo will decrease with it by a factor of (1+R1/R2). The great unknown is "what is the temperature coefficient of Vref, the -12.5V supply? We don't know.
What you really want to do is to take the derivative of the equation with respect to temperature, d/dT. If the -12.5V supply has any temperature dependence to it, it will be attenuated by R1/R2. Q780 is the "amplifier" in this circuit that in today's world would be replaced with an op amp running in its inverting mode and its "+" terminal connected to Vd instead of ground.
I miss the days of transistor design. It was a great deal of fun.
I obtained the result in a different way. 13.1 volts across 13.7k R783 resistor implies 4.13 volts across 4.32k resistor R782, given equal current (Ohm's law). 4.13 plus 0.6 volt results in 4.73 volts. That would be the output. However, there is some additional small current flowing through resistor R782 to the base of Q780. Depending on this current the voltage drop on R782 will be slightly higher than 4.13 volts so we can probably have exactly 5.0 volts or even slightly higher voltage.
The 110 ohm (R789) must be a startup load as well, as bleed resistors would be more like 1Mohm?
That was a really neat circuit though, thanks for the explanation
it might be there to keep a little heat in Q787 for stability as well
@@emiliaolfelt6370 Yep, that's valid
Thanks! I have a similar voltage regulating circuit in a telephone power supply from the 70s. The voltage divider can be set manually to 3 V, 4.5 V and 6 V. The power supply must be very silent to not make any noise in the telephone receiver. It has that 3rd transistor to keep the current below 300 mA. Thanks for the calculations, very helpful! I suspect your power supply is very silent as well? (probably depends on filter capacitors and a good transformer)
Great video. I think the 110 ohm resistor is to put a minimal load on the output as well as cap bleed off, 110 is rather low for a bleed, unless it has to do it before other supplies bleed off.
I have a 80's 2 channel dual trace Tektronix scope that was putting out way over 40VDC out of the preregulator and blowing the fuse. The 40V is provided to a wild push pull oscillator with some control transistors and an optoisolator to a Transistor regulated supply to drive the 120VAC rectified voltage through a Triac to get the 40VDC. A nasty closed loop. The push pull oscillator puts a square wave on the circuitry power supply transformer to regulate most of its output voltages all at once. The manual had a good theory of operation, but I was not able to verify or find a replacement opto isolator. I found an open resistor and a shorted diode in the push pull circuit but it was no help. The transistors are old Tecktronix part numbers and there were not enough waveforms/voltage notes for them. I ended up isolating all the circuity (lifting leads) from the 120V input power switch all the way to the 40V trace. I tried to find an ac/dc converter (or small ac/dc and dc/dc converters) but they would not fit inside by their mech specs. I bought a cheap adjustable 45V 10A supply and strapped it to the outside of the case. It works but it is clunky. I got the scope for free, I bet it is a transistor or two from working properly. If I post the power supply schematic and my notes, could you take a look?
Love that. You explained both current and voltage protection. :-)
“Just a cute little circuit.”
Very nice analysis! I'm a very young engineer and I'm not used to reasoning with BJTs so much, so it's a very nice practice! I want to make one remark though, that in today's world you would much rather have an integrated regulator than discrete BJTs not only because of cost (a SOT-23 LDO is probably cheaper than 3 discrete BJTs) but also because of assembly cost (extra SMD reels that you have to put into the pick and place machine, marginally longer pick and place time, etc) and because of board space savings. So it's not out of laziness that we don't do this anymore, but because designs have become so much more miniaturized and dense.
the pass transistor is a TO-66 can, good for 5A. your little SOT-23 will loose that race.
preaching to the choir, cap'
No voltage reference (Zener diode, TL431 etc.) but it works nicely. I wonder about the ripple voltage.
While the voltage drop is 7V, the current limiter will keep the dissipated power down to 2.8W. Not bad for a heatsinked TO-66.
You're a great teacher
Liked that! Nifty little circuit. 🤔👍
The controlled voltage opposes changes in the -12.5 V supply; they track each other such that the base voltage remains 0.6 V and very little current flows into the base. Tracking power supplies utilize similar tricks, but the resistors would be almost equal in this design.
If they dropped the value of the 13.7K resistor, and added a couple of diodes in series, there would be some temperature compensation in the voltage output, as long as the control transistor and diodes could be close to the same temperature.
That's a great idea, very simple, very effective, very old school.
More videos like this please ! awesome way to teach !
Fascinating, thank you.
Thanks so much for a simple explanation. Would you be able to explain a regulator that uses zener diode?
You didn't explain why the Q784 transistor can be dropped.
Nice schematics! Especially interesting because it contains only NPN transistors, even diode is possible to replace with base-collector shortened transistor. Can be used in hack diy or in production - no jokes some times is much cheaper to use only same transistor four times to reduce the costs.
@@harvey66616 agree, NPN
Your 4.7V output would be off slightly because the base current was ignored. But, I would have done the same thing, close enough. :)
Haven't thought about if adding in the base current would lower or raise the output voltage.
Wonder how stable is this over temperature?
The added base current would shift the output voltage up a little, having the same effect as reducing the value of the 13.7K resistor, but the effect is probably pretty negligible.
As for temperature stability, it's probably not great, but the fact that the entire design is dependent on the integrity of the -12.5V rail, it's probably not anyone's biggest concern.
The temperature dependance would be something like 1+4.32/13.7 times the Vbe tempco (2mV/Cent Degree). So it is just 30% more than the variation of the Vbe.
very nicely explained.
I wonder what would be difference in performance compared to some standard regulator like 7805.
Good video, thanks. Doesn't this circuit rely on the fact the input voltage is regulated (the -12.5V rail) so it works as a regulator for varying loads currents but not for varying input voltages.
Yes, correct
True, but the input voltage (collector of the output BJT) can still vary without producing any significant output drift
Can we replace Q780 with TL431 and get normal voltage regulator, insead of temperature sensor?
In 8:20 you subtract 12.5 from 17.23 why you did that ?
This relies of the -12.5V being stable. i expect that there is a regulator with a reference in that supply.
That first part of the circuit looks like a fold back current limiter to me... it's probably doing that, which make me think that this probably only regulates properly when it is under load and the no load open circuit voltage probably won't make much sense in respect to the required 5V
I think too low negative rail (like -12.5v) is counterproductive from precision point of view. Also its too dependent on Vbe (of Q780), i.e. Vbe temperature change is not compensated. I would do it better 😀.
seems useful nice explanation from circuitry but i mean 7805 just need 12 volts not plus and minus 12 volt supplys to get 5 volts
Is the res freqency of all 60 herz transforms, 60 herts, or if thet were would that not develope magnetism
Not bad. Informative. Thanks.
How would you got about stability of such a circuit, to be sure that it will not oscillate? Would be nice to know how to do that
that is a very advanced topic. www.analog.com/en/technical-articles/power-supply-loop-stability-loop-compensation.html
You can look at gain and phase shift, and plot the results on a Bode Plot. In a nutshell, you don't want gain and positive feedback at the same time (unless you are making an oscillator). That's the 70,000 ft. quick answer. Perhaps you could analyze it using a Laplace transform, but not sure about that if there's something nonlinear. Nowadays, you would just do the Bode analysis w/a simulator. The 39uF cap likely helps 'helps' slow down the loop so you never have positive feedback w/gain (just an educated guess).
@@IMSAIGuy Thanks for the article! I have wanted a reasonable length article explaining how transfer functions relate to stability. I will take some time to read this, hopefully it helps provide me a framework for thinking about stability.
@@Mark-hb5zf yes i know, but how to do that without directly going to simulation or measurements, is there no "old" way. I the problem I have is I know the control theory and i know circuit and also in the link you can see that it does not really bridge them both. How would you go about designing, if you for example had thought of this circuit, How would know? Also, where would you go to event it from being unstable, where is the place to stabilize it. not sure if you understand what i mean
@@IMSAIGuy depends, also in that article the whole circuit side is missing, i find that always problematic, the control theory is not that complicated, but making a discrete circuit that does work is a different story and is almost never addressed. That was more my questing. If you had designed that supply, how would you have looked at the circuit to determent if it was good in terms of stability. Sure, you can put in a restitor and make a bode plot. But that cant be the only way to at least get a understanding where such a circuit will have critical points. I would be really interesting to look maybe at this circuit and try to do the exact thing. Hope you understand what i meant.
Can one repurpose or adapt this circuit if there is no negative rail?
I read the comments that referencing it to ground is not a good idea, but I would think there is enough voltage headroom to do something else.
The 110ohm resistor might also be minimum load, as without a load on output, the base drive would not have anywhere to flow right?
Maybe! You do need some load to get regulation, as the transistors are set up in push, not pull
The divider chain of R782/R783 provides a minimal load of 1mA. The base drive without any more output current is sunk by Q780. At no-load, that is about 13mA, assuming the 470R resistor is tied to something around 12V as suggested in the video. Using 110R to sink 45mA as a minimum load gets the pass transistor into its preferred working region, rather than being just barely switched on. I assume 110R was chosen because that's the smallest value of resistor that keeps its dissipation below 1/4 W on a 5V supply.
Voltage and current control using only 3 active discrete silicon, that's impresive thinking about how long ago that circuit is. I won't like that -12.5 is referenced, I'll go to ground instead. Thank you so much
If you do that, then your output voltage depends on the Vbe of the particular sample of transistor used for Q780, and the temperature drift of the output voltage will be -0.3% per °C.
Using a stable, regulated -12.5V supply line eliminates dependency on the particular transistor's Vbe and reduces the temperature drift by a factor or 13.1/0.6 = 22 times.
I think I'll stick with the design as it is, if that's okay.
@@RexxSchneider Sorry I assume the input as unregulated. I agree with you
thank you.
Really enjoyed that.
I’m going to put this circuit on the bench today. I knew that wasn’t a Ziener, and am going to figure out why, and what if?
One thing to keep in mind is that the supply rails take some time to start up. The -12V rail could be more noticeably more negative than the +12V rail is positive during startup. The diode will keep the output from going too negative during that time, and keep from damaging components attached to the output.
@@Cynthia_Cantrell Thank you. That also crossed my mind. A job interview side-tracked me from laying this out the other day and your comment just reminded me to build a split supply for an ANDY learner/trainer board for logic. I’ve got a couple hours before the wife’s home. Time to make some sweet MusICs without blowing my ear drum, or getting yelled at. 😂
I could not believe it when you used a calculator to determine 0•6/1•5!
Yes and he was so surprised to find 0.400 that he verified once more... 😉
So it uses Q780 (emitter diode drop) as the voltage reference, and then 2 precision resistors to get the correct ratio against the 0.6v reference. Maybe they were happy enough with the temperature drift on that particular transistor to choose it... I don't think they were too concerned about nailing the voltage exactly, more like they wanted to make it as consistent as possible.
I like the over-current clamp though, it's "cute".
Perhaps the -12 v low side on the divider was used as it was harder to find precision resistors in low values at the time, or the selection values were limited? These days, the whole range of resistances seem to have a precision value offering.
Cheers,
The -12.5 V supply was regulated, this circuit amounts to a buffer of this regulated voltage. The voltage divider is such that the variations in the 0.6v drop (for example, due to temperature) do not contribute much to the output voltage as most of the voltage drop across R783 comes from the regulated supply.
@@argcargv this circuit would regulate fine as it is without the -12.5 V, just less precisely.
@@plainedgedsaw1694 the output voltage will follow the -12,5V supply. If the 12.5 volts changes by 10%, then the 5v output of this circuit would also see a 10% change. The -12,5 V is the voltage reference for this circuit.
@@argcargv well, yes, it regulates voltage in relation to that -12.5V rail. But if it did connect to ground, it would still be regulated (no significant drift vs input voltage or load), just to different voltage.
@@plainedgedsaw1694 if it was connected to ground then the output voltage would be very sensitive to the temperature of the transistor, which will be indirectly sensitive to load.
What happen to I x R = V equation
thats amazing thankyou.
wait so this is a voltage divider with a rapidly switching element so that it doesn't waste the current? could this be a way to think about it?
No, this is a linear regulator. It does help to calculate and understand the circuit by analysing the "edge cases" but the transient between are not discrete. A so-called linear region.
A bit complicated, but it was the best way of doing it at the time.
Thanks.
Great exercise, THANX
Your explanation good ,but the omparision with showing the ground at time 8.06 and comparing the two voltages is not clear
I kept hearing your squeaky chair.
Very nice, but awkward. I prefer set PWM sequence.
Good Stuff! Thx!
I was just trying to invent 3.7V regulator for Lifepo4-charger. I was going to use stack of diodes as voltage reference, but that would have been ugly and embarrassing.
Note that this drifts like hell with temp, try tl431, i heard it can be unstable though. Current limit accuracy doesn't matter as much so the transistor should be fine.
use a tl431
I already tried with 3.6 Zd temp killed it.
@@plainedgedsaw1694 This is not a precision instrument. Charging current is small and LiFePo4 does not break easily.
@@TimoNoko tl431 isn't precision component either, it's usually 2% precise. But treating li cell well even if treating it worse won't make it blow up, is good idea i think.
If you felt like I think it would be interesting to show a video actually building an example of this and see if it works nicely or terribly, or how efficient
It was designed for a high end Tektronix curve tracer. $40k in todays money. The one I have regulates perfectly at 5.03 volts and is about 32 years old.
It is not very efficient at all, like many linear regulators, namely because all of the current that is available at the output at 5V first has to drop from 12.5V through the pass transistor and current sense resistor - a drop of 7.5V. Since efficiency = Pout / Pin, and (ignoring the regulator transistor) Pout = 5 * I_load and Pin = 12.5 * I_load, efficiency can not be better than 5/12.5 = 40%.
have you seen this thing? 70lbs and a beast. 1500V and 10amps output. no, efficiency was not important
@@IMSAIGuy No, I haven't seen it. But a 15KW out sounds awfully beefy - the standard house outlet can only supply 1.8KW before its breaker trips. Perhaps it's 150V & 10A out?
@@Cynthia_Cantrell 1500v and 10A are subject to 220W max output
Thank you this great video. Liked it even I'm an electronic nobbie
I have a simple question, but are there any reasons they used -12.5V as a reference point? I mean, the resistor divider could easily reference to GND...
Hi, if you do that you will see your 5V nominal swing a lot with temperature. Like 17mV / Cent Degree. You will turn this thing into a Vbe multiplier, while it is thought as an inverting amplifier with -12.5V well regulated input
This regulator depends on the value of the input voltage... how can we call it regulator
It only depends on already regulated -12.5V rail, but not on it's positive rail. I guess they did it this way because it was simpler and cheaper and did the job.
Seems to me R789 forms another voltage divider.
no, the top of the resistor is the +V that is being regulated. that resistor is just a load.
Why use -12.5V at the bottom of that divider? One could have used 0V, couldn't you?
No, -12.5V is clearly the voltage reference in this circuit
@@yveslesage8525 but why can’t the voltage divider be referenced to 0V instead? Assuming I want to reuse this circuit in a product that doesn’t have negative supplies. Isn’t it a matter of changing the voltage divider so the 0.6V sits at the right point?
@@DafyddRoche If you do that, then your output voltage depends on the Vbe of the particular sample of transistor used for Q780, and the temperature drift of the output voltage will be -0.3% per °C. It's not suitable for production. You might as well just use a 4.3V Zener supplied from the 470R resistor and only keep Q787.
Using a stable, regulated -12.5V supply line eliminates dependency on the particular transistor's Vbe and reduces the temperature drift by a factor or 13.1/0.6 = 22 times.
If you want a 5V supply in a modern product, just use an LM7805 or one of the common low drop-out regulators if you have less than 7V to supply it from.
@@RexxSchneider Hi Rex. yes - it'd be easier to use a simple regulator! :) But I am curious, or I should say -- I need to understand - isn't ground considered "stable and regulated"? I think I understand that because the the transistors emitter is grounded, we're now depending on 0.6 or 0.65 or 0.7V turn on voltage difference. Why does having a -12.5V make a difference? surely, a Vbe variance in the transistor makes the turn on voltage equally as variable? (Help! I clearly haven't had my AH-HA! moment! )
@@DafyddRoche Ground is stable, of course. But if you are trying to use a reference that's only about 600mV above ground, i.e Vbe of the transistor, then a variation of 60mV for example (perhaps through temperature change or variation in Vbe between samples) will produce a variation of 10% (60mV/600mV) in the current flowing through the resistor divider and a consequent 10% variation in the output voltage. On the other hand, if you tie the lower resistor to -12.5V instead of ground, then you have a total of 13.1V across that resistor and a change of 60mV in Vbe will only create a change in current of 60mV/13100mV = 0.5%, and hence only a 0.5% change in output voltage. Now do you see the difference?
that is cool.
Who cares what the can is, what type of transistors are they.
Why'd you drop the Q784 transistor?
watch the whole video before asking questions, Nick.
@@nickcaruso Don't be so hard on Nick, Nick. :)
@@Mark-hb5zf 🤣
Messy. Vibe maybe anything between 0.6 and 0.7 V. This significantly affects V out. Assume you want 5 volts output, so work backwards to find Vbe for the resistor values given.
you are correct. 0.6V is not going to be true. They must have designed those resistors for the parts called out, as the output of my unit is 5.03V
Can i ask why my comment is deleted ?
the youtube algorithm didn't like something. it hates links
@@IMSAIGuy ooh, okay. I will keep that in mind. grtz
Quite obviously, it's not a regulator at all, since output is at the whims of Vbe of Q780, being highly temp dependant. A simple 2.7v zener in it's emitter would've worked wonders.
I am going crosseyed trying to keep up with your hand waving and constant moving of the schematic!
Transistor cu CPU vaccum switch
ALU second generation of computer ic intergrade circuit
Memory unit (capacitor restance transformer)
Primary memory
Secondary memory
Kewl
At first glance, it looks for me just as variation of "Vbe multiplier" circuit.
OK
A long and messy explanation.
As the output is +5V and the negative supply -12.5 you have their sum (17.5V) across series R782 and R783 (18.02K).
The Q780 base is at +0.6, or so (current gain/base current plays here) and is limited to negative ~ 0.7V with that diode.
remove R782 and the voltage is limited to about 5.6V.
Q780 gain (100 -300) is Rb/Re enough to drive the TO-66 guy (low gain, maybe 25).
The easiest calculator to use for this work LTspice. (free, and intuitive to use) Works faster than the calculator.
The circuit does not have a fixed reference and a differential circuit, output voltage depends on gain of transistors and stability of negative 12.5V.
With an old OP-amp circuit and +/- supplies, this is a quick and dirty voltage regulator.
this is amazing. thank you
is this ai? why did you choose to say and leave in saying we could v have kkk
wouldn't you rather be within the ccc
Your analysis is insufficient, which is why you got an error at the end. You forgot the current into the base of Q780, which can be calculated from its collector voltage, R780 and beta. We know the Q780 collector voltage because Q787 is an emitter follower so its base has approximately same level as its emitter, i.e. +V. The R780 is rather smallish and its current flows into Q780 collector, so the Q780 base current will have a contribution into the overall calculation. I.e. it will take higher output voltage to also feed some current into the base, so it will likely reach 5V on the output.
Hmm.. I was enjoying the discussion but then I started to notice just how many "ok?"s there are and now I'm going to have to exit. It's like almost every second word, mmkay?
OK
@@IMSAIGuy I miss Pete. He was special.
The very bad side of this schematic is 5V output is dependent on -12V line stability.
Back then you would normally use a Zener on the pass tranny base (if not too high power needed), and the same current limit circuit. Not sure why they did it the way you showed???
Love the To-66's look very cool, Nothing more nurdy than a bunch of TO... cans on a heatsink and a bunch of large inductors next to them, ...heaven man! :)