# High Voltage Series Pass Transistor - Low Voltage Control

I am having some difficulties getting my output stage to work correctly. Here is some info:

I have a +48VDC source with a series pass transistor to drive a load between +45V and 0V. The control system uses op amps and can have an output voltage of anything I wish really. (+5/-5, +5/GND, +12/-12, +12/GND). Its going to depend on my output stage.

Devices I tried:
D44VH10 (Vce=80V, Vbe=7V, Ic=15A)
TIP142 (Vce=100V, Vbe=5V, Ic=10A)
2N6039 (Vce=80V, Vbe=5V, Ic=4A)

What I am having problems with is actually getting my output stage to go from +45V to 0V.

I have tried a few combinations of NPN pass transistors controlled by a PNP in a darlington type setup and this meets all my requirements. However it requires my op amps to run at +48V which they will happily do while on fire. If I lower my op amp's output swing to +12/GND then the load only gets around 26V maximum.

I then tried a PNP pass transistor with an NPN control in a similar darlington setup all controlled by an NPN to ground. This worked and allowed my to use +5/GND as my output swing but even if I applied 48V and 1A to my control NPN my load never received more than 22V.

I then tried the same as above but with a MOSFET to ground and I got a 27V at my load. I modified again an used the MOSFET to drive a photo transistor which in turn drove an NPN series pass transistor from the 48V source. I managed to get my 45V out to the load this way but the temperature variations and unreliability of the photo transistor package does not meet my requirements and worst of all OFF was a state passing around 7mV and 2mA.

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So, I am missing something here. What is the correct way to drive an output stage from a lower voltage control system?

The whole reason I am doing this is because the only op amp that can 40V+ available to me cost $13/piece. If I can get my control circuity to a lesser value I can get them for$1.

• You are missing something. But, so are we. What you are trying to achieve can be done easily and cheaply with lowish voltage opamps. But, we don't know what it is :-). Your description is much better than many, but it seems something you wantthat is not quite usual, or what you have tried should allow it. You say 0-45 V. relative o what>? Is that 0-45 across the load and the load can be referenced to V+. Or must load be ground connected at one end. What IS the load. Is it resistive, or a motor or ... . what resistance and power rating? Tell us everything and you'll get an excellent answer. – Russell McMahon Oct 12 '11 at 3:46
• What drive frequency / speed do you need. Are signals AC or analog across the range or do you want to set a level and leave for some while. eg do you want a say 30V sinewave or just an eg 43 V level. – Russell McMahon Oct 12 '11 at 3:53
• @Russell Here is some more info... The load circuit goes like this: Rectified +48VDC -> Series pass transistor -> 0.1/5W resister -> LOAD -> 0.1/5W current sense resistor -> Rectified 0V and grounded... The load only requires the ability to be varied to an arbitrary point between 0-45VDC (relative to ground, the 0V point after the rectifier) – uMinded Oct 12 '11 at 4:02
• I am designing for two different systems. One will be resistive only and another will be for driving a range of stepper motors for testing. For the inductive load I am simply putting 4-5 pass transistors and have everything clamped and protected with diodes and massive source caps for smooth ripple. – uMinded Oct 12 '11 at 4:05
• Control is a proportionally small control voltage of (Preferably a single supply op amp at +5V/GND. Possibly +5/-5 to help it insure its good and off considering op amp offset) – uMinded Oct 12 '11 at 4:09

There are some open questions, but I'll take a stab at answering. I'll assume you want voltage control of a load that must be ground referenced, the voltage range is 0-45V, a 48V supply is available, the maximum current is 1A, and the control input is a voltage from 0-5V.

Here is a circuit that fullfills the requirements I stated:

This is similar in idea to Russell's circuit with a few key differences. Q2 is a controlled current sink linear with the opamp output voltage in the range of about 600mV to 5V. This current variably turns on PNP transistor Q1. The opamp output from about 600mV to 5V maps linearly to the load current, which should help stability. The compensation cap C2 working against R2//R5 provides a means to add additional stability as needed. C2 shouldn't need to be more than a few 10s of pF.

With 5V on the base of Q2, the emitter will be about 4.3V, so Q2 will sink 70 mA. Assuming the power transistor Q1 has a gain of at least 15 (in the plausible range for this type of transistor), the load current can be up to 1A.

R2 and R5 divide the load voltage into the 0-5V range the opamp can handle. Since stuff happens, you want to make sure all is OK with the full 48V at P1. This 48V divided by R2 and R5 becomes 4.75V into the opamp. That's close enough to 5V to use most of the range but still leave a little margin.

You will have to think about the power dissipation of Q1 carefully. It could be quite a lot depending on what current your load really draws. Worst case the load voltage is half the supply, so 24V, and drawing 1A. That puts 24W on Q1, which is quite a lot. If your load really can draw up to 1A, then Q1 probably should be a TO-3 with a good heat sink and forced air cooling. If that's too much, you need to consider switching topologies to accomplish what you are doing. 24W is not trivial to deal with.

Q2 could also get toasty, but nowhere near as bad as Q1. At the maximum of 5V on it's base, it will drop about 43V at 70mA, which is 3W. That's not too hard to deal with, like a TO-220 with a small heat sink. Of course if your load doesn't really need 1A this all scales down linearly.

## Oops:

I updated the schematic to get rid extra resistor in series with the opamp negative input. The circuit evolved as I was drawing it and I didn't notice this resistor was no longer needed when the circuit was posted originally. The description has been updated accordingly.

• Thanks for the great breakdown. I'm not missing anything component wise, one of my designs was exactly like your example I only had a different R4 value. I still get my odd behavior though so maybe I am misunderstanding the data sheets. For my example I am using a TIP32 pass transistor. Ratings are Vce=100V, Vbe=5V, Ic=3A, according to the graph Hfe is 100 from Ib=1-100mA. However on that graph it also says that Hfe curve at Vce=4V In real life though my gain is not 100 as expected its only 18.5! and my drop across Vce=3.3V – uMinded Oct 13 '11 at 4:14
• My strange behaviour was casued bu underestimating the gain. As a result I would only get an output of like 26V and my op amp would have 5V on its non inverting input and like 2.6V on its inverting yet would not be driving Q2 with full output voltage. I assumed it was a problem with the control circuity. Not until you mentioned "Q1 has a gain of at least 15 (in the plausible range for this type of transistor)" did I actually measure both currents and do the math... Only way to learn I guess. Thanks Guys! – uMinded Oct 13 '11 at 4:19
• Quick question to double check. For example if the data sheet says Vce=100V, Ic=5A, P=30W @ 25deg that means any combination of voltage and current not exceeding those values (6V@5A, 300mA@100V, etc) will cause that package to only be 25deg? – uMinded Oct 13 '11 at 4:25
• @uMinded: No, that probably means the transistor can dissipate up to 30W if you hold its case at 25 degC. That's unlikely to be possible, especially at 30W. The datasheet should give you maximum die temperature and thermal resistance to case. With the heat sink spec you should be able to estimate case to air thermal resistance. With the total die to air resistance you can calculate how much power the transistor can safely handle in your particular circumstance. – Olin Lathrop Oct 13 '11 at 12:14
• @uMinded: Yes, the gain of bipolar transistors can be significantly less near the maximum current limit than at the highest gain operating point. This is why the gain of power transistors in particular is usually specified at several operating points. – Olin Lathrop Oct 13 '11 at 12:17

This circuit appears to do what you want.

This is a cut and paste markup of a circuit copied from the internet from here - not ideal but good enough for illustrative purposes. (That webpage is not very useful).

The opamp is shown running from 4V. Somewhat more is useful but not much. Depends on output swing of opamp used relative to +ve rail. eg an LM324 {datasheet} would be happier with V+ >= 6V. 9V to 12V would be fine.

The output voltage (load voltage) is divided by N:1 by two resistors to match the desired control voltage level. eg if you make a 20:1 divider ( eg 19K:1k) then for Vout = 50V, Vcontrol ("Volage") = 2.5V as 50/20 = 2.5.

For a given current at 50V, The MOSFET must dissipate whatever power is not dissipated in the load.

eg if 100 mA is taken at 20V in the load.
Load power = 20V x 100 mA = 2 Watt.
MOSFET power = V x I = (50-20) x 100 mA = 3 Watt

At higher currents substantial heat sinking would be required.

For

Maximum desired voltage = Vmax
Maximum desired current = Imax

Maximum MOSFET dissipation is when Vout = 0 = V x I
= Vmax x Imax.

• Note that the MOSFET in the schematic is placed upside-down! Source and drain should be swapped. – stevenvh Oct 12 '11 at 7:09
• @stevenh - Yes, thanks. That's what I get for copying diagrams from the net. I noted it at the time but then forgot to change it. – Russell McMahon Oct 12 '11 at 7:49