# Help analysing a discrete voltage regulator circuit

I need a -140V reg which doesn't exist, so I'm trying to learn how to build simple discrete regulators (starting with positive regs). As I only need a few tens of milliamps with a fairly constant load, I feel this shouldn't be too onerous a task. I've spent some time reading and simulating, but as I often find with transistor circuits I start to get lost about what is controlling what. In this case, I have been looking at the circuit below:

The explanation of this circuit says:

When the output voltage drops, T4 pulls the emitter of T3 lower. This drives T2 harder, which causes the output voltage to rise again. R4 restricts the base current of T2. C1 and C2 have been added to improve the stability of the circuit.

Now, I can see that the idea is you're providing feedback to give the regulator better stability and that they're saying that's coming from R6/R7 back to the pass transistor via T4 and T3.

I'm not understanding how the emitter of the T4 can be pulled low by the R6/R7 potential divider. When I looked at it I thought that T3's base must be at 78V (because of the two Zeners), therefore T4's base must be two Vbe drops below at roughly 76.8V and so that would be the voltage at the R6/R7 junction.

What am I missing here? Why does the voltage at T4's base go up and down and why T3's base voltage not override this due to the Zener voltage which I would have thought is constant and quite "stiff". I assume what must be happening is that as the output drops, T4's base drops and T3's base is dropping, however, how that works when the Zeners would to my limited knowledge be keeping T3's base at 78V I do not understand.

• How much accuracy, output ripple, PSRR, noise, output impedance, etc do you need? What is the load? Tony explained how the circuit works, but you still have to check if it will be a good match for your application. Commented Dec 6, 2017 at 17:32
• Yeah, all those questions are good questions. The load is a small motor that will be driven at a constant speed. Accuracy is not so vital, ripple and noise more so. However, baby steps in understanding the circuit first! :D I think the LM317/137 route will do the trick and provide me a more than ample supply, Commented Dec 6, 2017 at 17:40
• If you need constant speed and the motor's torque load varies, then regulating the speed would be better than regulating the voltage... Also you could use a PWM, but that's a bit more complex. Do you need well controlled speed? Or constant voltage vill be OK? Commented Dec 6, 2017 at 18:20
• I am regulating the speed. That part is the easy bit! :D Commented Dec 6, 2017 at 18:28
• You have one too many i’s in “analysing” in your title. Commented Dec 6, 2017 at 21:49

Think of T4 as an emitter follower buffering the feedback voltage (from R6/R7 voltage divider). The collector resistor is just for current limiting, it does not normally affect the operation.

You can thus think of the feedback $V_O \frac{R_7}{R_6 + R_7} + V_{BE}$ as being applied directly to the emitter of T3.

When the emitter of T3 drops below the zener voltage minus the 0.7V Vbe drop, T3 conducts more and drives the pass transistor harder.

• I'm sure I'm I am being slow, but I don't see how this answers my question about the seeming fight between the Zener voltage on T3's base and the R6/R7 voltage on T4. Sorry that sounds sarcastic, it's not meant to be. Commented Dec 6, 2017 at 16:51
• T3 base is at constant voltage 78V. R6/R7 voltage varies with the output voltage. Negative feedback through the pass transistor T2 balances the voltage at the base of T4 so that it is 78+0.7+0.7 = 79.4V, so the output voltage (due to the 2:1 divider) is about 160V. That won't happen without the whole regulator working. Does that answer your question? Commented Dec 6, 2017 at 17:09
• I think so; as in the comments to Tony's answer, I think I was missing the whole point of the feedback keeping that voltage difference of 1.4V as you have pointed out in the previous comment. Hopefully I'm seeing it now anyway. Commented Dec 6, 2017 at 17:37

Look at T3/T4 as a differential amplifier with a 2 diode drop on the feedback ratio with the Zener voltage as the reference.

Then it will make sense, even though it is a Common Collector driving a Common Base, the voltage error gain is the same and the emitters don't need extra loads as they are share current as a virtual low impedance (base-ESR/hFE) with high collector to emitter gain ratio.

For simplicity I used a PNP series pass, but LDO's use a current source like that. The outputs use a Darlington Emitter Follower with high gain in front, so ignore that.

I just want to show that the impedance and gain of this feedback arrangement is just like a differential amp with offset and fewer parts.

simulate this circuit – Schematic created using CircuitLab

For more precision

Except use a LM137HVQML and complement everything for PNP emitter follower drivers on a negative rail used in this IC.

• I don't understand the bit about looking at it as a differential amplifier, but the LM317 circuit looks interesting. I spent hours and hours Googling how to boost the output voltage of a regulator and the only hits I could find were upping the output current. Commented Dec 6, 2017 at 16:53
• LM317 only pulls up current output, with NPN Darlingtons inside, so you need LM137 but even a SMPS buck regulator is feasible. Commented Dec 6, 2017 at 17:17
• Yep, that bit I do get! Thank you for that apps note, that is excellent. Commented Dec 6, 2017 at 17:18
• I hope you understand the Diff Amp with offset part. If you had a low current source to a 140V transistor Vce in breakdown mode to act as a Zener, you'd have the fix I used in 1976 when my TV Hsweep supply died. But if you imagine a 10uA CC to a 140V Zener being buffered with a Darlington to drive the Vadj point of a 3 terminal regulator. Then it only has to regulate 20V. Commented Dec 6, 2017 at 17:24
• Yes, that makes sense the way you have drawn it I think. You can start to see it looking like the feedback of an audio power amp output going back to the the long tailed pair at its input. I suppose the point is that the voltage between the bases does have to be 1.4V and it's the feedback circuit ensuring that it does stay the same. If I am understanding it correctly now anyway. Commented Dec 6, 2017 at 17:35