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.

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.

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!

but I guess the circuit depends on -12.5V rail be regulated with respect to ground all other things being equal.

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!

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

There is beauty in little things after all.

fun little circuit analysis, thanks for taking us along.

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.

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.

Simple and accurate enough! And repairable.

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.

Tektronix had the best electronic drafting style!

Fun to go through this unusual (to me) circuit. Brings back the fun of electronics to just slowly go through it unscripted!

Thank you so much for explaining this in detail, it helps a lot. Regards Helmut

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.

Fascinating, thank you.

More videos like this please ! awesome way to teach !

Liked that! Nifty little circuit. 🤔👍

“Just a cute little circuit.”

Excellent... I love your RPN calculator!!

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

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.

You're a great teacher

Love that. You explained both current and voltage protection. :-)

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.

very nicely explained.

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.

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?

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 .

Really enjoyed that.

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 :)

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.

Great exercise, THANX

Thanks so much for a simple explanation. Would you be able to explain a regulator that uses zener diode?

Not bad. Informative. Thanks.

The circuit is more properly an inverting "op-amp" rather than a regulator. Its output varies with the negative supply voltage.

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.

Good Stuff! Thx!

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.

I wonder what would be difference in performance compared to some standard regulator like 7805.

thats amazing thankyou.

thank you.

Is the res freqency of all 60 herz transforms, 60 herts, or if thet were would that not develope magnetism

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)

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.

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.

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.

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.

Thanks.

Thank you this great video. Liked it even I'm an electronic nobbie

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

that is cool.

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?

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?

This relies of the -12.5V being stable. i expect that there is a regulator with a reference in that supply.

In 8:20 you subtract 12.5 from 17.23 why you did that ?

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.

The 110ohm resistor might also be minimum load, as without a load on output, the base drive would not have anywhere to flow right?

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

You didn't explain why the Q784 transistor can be dropped.

A bit complicated, but it was the best way of doing it at the time.

I could not believe it when you used a calculator to determine 0•6/1•5!

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.

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

What happen to I x R = V equation

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

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 😀.

Why'd you drop the Q784 transistor?

Kewl

I kept hearing your squeaky chair.

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

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?

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,

🧐Burning 50ma unloaded. Think maybe evidently presumably .6 goes up with current for compensation

This regulator depends on the value of the input voltage... how can we call it regulator

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.

Very nice, but awkward. I prefer set PWM sequence.

Your explanation good ,but the omparision with showing the ground at time 8.06 and comparing the two voltages is not clear

Seems to me R789 forms another voltage divider.

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

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...

Why use -12.5V at the bottom of that divider? One could have used 0V, couldn't you?

Who cares what the can is, what type of transistors are they.

OK

Can i ask why my comment is deleted ?

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.

Transistor cu CPU vaccum switch ALU second generation of computer ic intergrade circuit Memory unit (capacitor restance transformer) Primary memory Secondary memory

At first glance, it looks for me just as variation of "Vbe multiplier" circuit.

I am going crosseyed trying to keep up with your hand waving and constant moving of the schematic!

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.

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

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?

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! :)

Can you not swallow your spit in the mic…