# Low voltage op amp to control high voltage series pass transistor

1. Sep 11, 2013

### d.arbitman

I am in the middle of designing a low drift and low ripple power supply for myself. To control the output voltage, which ranges from 0 to 20V, I am using an op amp whose output is fed into a Darlington pair of power bjts. The output of the Darlington pair is both fed back to the op amp's inverting input (i.e. the op amp is configured as a voltage follower) as well as to the load. The desired output voltage is present on the non-inverting input to the op amp as a reference. I have a transformer that produces +/- 40V rails so I am using those to power the op amp. I have a high voltage op amp but recently, the op amp was fried even though it can be powered from +/- 45V. As a result, I have decided to use a low voltage op amp to control the +/- 40V rails. The problem is, I have no idea how to go about doing this. I have tried using single transistor amplifiers to translate 0 -> 5V to 0 -> 20V but have had no luck. So the question, how do I use a low voltage op amp that is powered from +/- 5V to control +/- 40V rails so that my output ranges from 0 to 20V?

2. Sep 11, 2013

### pullmanwa

3. Sep 11, 2013

### d.arbitman

Uploaded. As you can see, the op amp is powered from +/- 40V but I don't want to use a high voltage op amp like I'm doing at the moment.

#### Attached Files:

• ###### Schematic.PNG
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4. Sep 11, 2013

### pullmanwa

I have a question if you don't mind.

1. where is the output voltage taken across from? I'm asking where is your ground reference for the output voltage?

** The + input to the opamp is missing --- that will cause a problem

5. Sep 11, 2013

### meBigGuy

That's a pretty primitive schematic, but I see what you are trying to do.

You need an amplifier with gain since you want to go from say a 1V swing to a 40V swing.

Here is a circuit that does what you want and does it well.

http://circuits.linear.com/High_Speed_Amplifiers--all-439

Once you understand it, you can simplify it.

The simplest approach is to just drive into a single stage common emitter amp. But that dosn't have much drive. Next simplest is to follow that with an emitter follower. Don't forget to deal with the common-emitter inversion and divide down the emitter follower output for the feedback.

Last edited: Sep 11, 2013
6. Sep 11, 2013

### d.arbitman

I know. That's the input voltage and assumed to be in the range of 0 to 20V. The ground reference is labeled on the schematic (for lack of a better term...the triangle connected to the power supplies is ground).

I have configured the op amp to behave as a voltage follower.

7. Sep 11, 2013

### d.arbitman

That op amp is supplied from +/- 125V which is what I want to stay away from. I want to use a low voltage op amp to do what I want. I not only want the input to be scaled down but also the rails to be, say +/- 5V but I still want it to be able to drive a series pass transistor whose rail is at +40V

I have attached a complete schematic of what I have.

R1 is the load resistance. U1 and its associated resistors mimic a 50V/V current sense amplifier.

#### Attached Files:

• ###### Schematic.jpg
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8. Sep 11, 2013

### meBigGuy

No it isn't. Notice the zener diodes that drop the voltage for the op amp. Or supply the opamp from different supplies.

9. Sep 11, 2013

### d.arbitman

Oh you're absolutely right.

10. Sep 11, 2013

### meBigGuy

OK --- You just need to drive the opamp into a common emitter stage with a pullup to 40V. Follow the common emitter stage with your darlington. Don't forget the inversion of the common emitter stage, which changes things a bit (notice the feedback to the + input in the 125V circuit). You can make U2 inverting, or whatever.

11. Sep 11, 2013

### d.arbitman

I have no idea what's going on in that circuit. I figure the 2n3440 and 2n5415 furthest to the right are the series pass transistors and they are the ones that supply the current. What's the point of the 27R resistors and the 1k that are connected to them? What's the point of the 2n222 and 2n2907 transistors? Why are the diodes 1n4148 connected across the base of the 2n3440 series pass transistor to the output?

How do I go about understanding what that circuit does?

Last edited: Sep 11, 2013
12. Sep 12, 2013

### meBigGuy

The circuit is a bit fancy for what you want to do. It provides high power symmetrical drive with current limiting.

The 27R resistors are for current limiting sense. When the voltage across them reaches 0.7V the 2907 and 2222 transistors turn on, reducing the drive to the output transistor.

See if you can decipher what I said in post 10.

13. Sep 12, 2013

### meBigGuy

describing the circuit a bit more for you. Think of it as two common emitter amps followed by 2 emitter followers.

If you look at the circuit as a big X it is easier to conceptualize what it basically does. The bottom left 3440 is a common emitter amplifier, and the top right 3440 is sort of its emitter follower. And the 2 5415's operate the same way. In combination they are very symmetrical, which you really don't need.

Look at T1 and T5 in this design. It has a double inversion so maybe that will make things easier. Just drive into T1. Read the wikipeda page on common emitter amplifiers to get it biased right.
http://www.bogobit.de/bogobox/

14. Sep 12, 2013

### d.arbitman

What do the 1k resistors do? What is the purpose of the first transistor stage whose base is connected to 1M resistors? How does that first stage work? I figure the diodes prevent crossover distortion, but what about everything else?

As far as post 10 is concerned I don't know which circuit you are referring to when you say drive the op amp into a common emitter stage.

What do you mean by pullup to 40V?

In a common emitter configuration, the output is taken at the collector and the output is inverted. As a result I should make U2 in my schematic inverting? It seems like I would be introducing more parts than I already have to do the same thing.

15. Sep 12, 2013

### meBigGuy

Explaining basic transistor circuits is a bit beyond what can be done here. I can point you to a topology that can do what you want and you have to take it from there. I suggested a common emitter followed by a darlington, but two common emitters is a better idea.

Again, look at T1 and T5 in http://www.bogobit.de/bogobox/ That's a basic topolgy for what you want to do.

Don't forget that you need a voltage divider in your feedback loop to keep high voltage from frying the opamp.

I'm not sure whether you will run into stability problems with the additional gain. You may want to rolloff U2.

16. Sep 12, 2013

### jim hardy

17. Sep 12, 2013

### meBigGuy

That's what I posted. But for his circuit the T1/T5 combo in http://www.bogobit.de/bogobox/ is probably adequate. What do you think? I need to get a schematic tool. Too hard to describe circuits.

18. Sep 12, 2013

### d.arbitman

I rebuilt the circuit in that same way...op amp drives an NPN which sinks the base current from a PNP via its collector. I ran some simulations and boy that PNP will dissipate a ton of power.

Thank you for helping by the way.

I have attached my latest schematic.

P.S. I tried loading the circuit that you showed me from LT, the +/- 120V output, with a 10$\Omega$ load and it failed miserably.

#### Attached Files:

• ###### Schematic.jpg
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Last edited: Sep 12, 2013
19. Sep 12, 2013

### jim hardy

Well so it is exact same link ! Sorry - I only saw bogobox....

in his schematic
he needs a current source to turn on his Q1, and a pull-down transistor controlled by the opamp to turn Q1 off.

Establish some base drive to Q1, enough to make rated current
and steal that base drive away to control output voltage.

As you said he should figure out why that Linear Technology circuit works.

Also as you said, it is difficult to paint a good mental picture with just words.

old jim

20. Sep 12, 2013

### jim hardy

Well

120 volts across ten ohms is twelve amps.
Did you notice that the 27 ohm resistor and 1N4148 diode form a ~25 milliamp current limiter on the output stage?
How's it do below 1/4 volt on that ten ohm load?

When you can describe in words how a circuit works you are coming to understand it.

21. Sep 12, 2013

### d.arbitman

Wait a minute. If the 1n4148 AND THE 27ohm form the current limiter, then what do the 2n2222 and the 2n2907 do?

22. Sep 12, 2013

### jim hardy

You got me good !

They make a better one.

23. Sep 12, 2013

### d.arbitman

What do the 510 and 330 ohm resistors do, why are they chosen so small?

Is the smaller of the two values connected to the emitter of the PNP rather than the larger in order to forward bias the transistor?

24. Sep 12, 2013

### jim hardy

They're to make the transistor mimic the current flowing through the 50K ohm resistor. That transistor is a very simple(if admittedly imprecise) controlled current source.
I did arithmetic assuming opamp is at zero out and get two milliamps through that transistor.
That is the base drive current available to the final transistors, but some of it is diverted by your current limiters..

The basis of this thing is a balancing act with two controlled current sources driven by opamp through those 50K resistors.

See how cleverly they've separated the high voltage from opamp?
50K resistors and high voltage transistors.
The output is "felt" through the 100K resistor and balanced against input via its 10K

Also observe the driver stages invert polarity, that's why the LT 1055's + and - inputs are backward from your normal inverting opamp configuration.

I'm still trying to figure out the biasing at full output.
Perhaps you or me-big-guy will beat me to it ...

Here's as far as I got;
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http://circuits.linear.com/img/439_circuit_1.jpg

The objective is to control ~ 120 volts with only a ~12 volt "handle".
We have to do that because
The opamp is powered by those zener sources, look at their datasheet they're 15 volts.
So the LT1055opamp can only move his pin6 up and down about 12 volts.
We'll call that point in the circuit 'our 12 volt handle'.....

Let us look at top half of circuit when handle is at zero volts.
The majority of the voltage drop between power supply and our "handle" is across the 50K resistors.
Current through the 50K resistor is around ~(125 ) / 50.5K = 2.475ma.
Voltage across the 510 = 2.475ma X 510 ohms = 1.262 volts
If Veb for the 2N5415 is 0.6V, the there must be (1.262 - 0.6) = 0.662V across the 330 ohm resistor. That'd be 2 milliamps through it.

That 2 ma multiplied by 2N3440's probable hfe of ~ 30, would cause 60 ma to flow out of its emitter but your current limiters limit it to ~25 ma.

By symmetry, the bottom half of the circuit is doing the exact same thing, except of course current is in other direction - leavingthe output terminal instead of entering it...
So the two 25ma currents cancel and output is zero.

Now let us push our 'handle' down to -12 volts.
Current through the 50K resistor is now ~(137) / 50.5K = 2.712 ma
Voltage across the 510 = 2.673ma X 510 ohms = 1.383 volt
voltage across 330 = (1.383-0.6) = 0.783 v
that'd be 2.374 ma through it

So - lowering our 'handle ' twelve volts increased the base drive current for top 2N3440 by 374microamps.

By symmetry, bottom output transistor should see close to same decrease in base drive.

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I've not yet considered the 1 meg resistors because I think they're negative feedback,,

it's late and i'm pooped. I hope you guys have it explained in the morning !

old jim

Last edited: Sep 13, 2013
25. Sep 13, 2013

### d.arbitman

Last edited: Sep 13, 2013