Low voltage op amp to control high voltage series pass transistor

[meBigGuy]

Good, I'm glad the 510 has minimal impact. The 1M is the negative feedback resistor that sucks or adds current to the base node to control the gain. It does that for sure. Take it out to see the open loop gain.

Normalize the currents to 0V input and look at the deltas. It may make more sense then. The currents for 0 output voltage are the basic bias currents, and they swing relative to that. The open loop gain affects the gain change vs. resistor size.

 

[d.arbitman]

I simplified the circuit. I took out the op amp and feedback resistors because they were unnecessary for understanding what is happening. I also took out the current limiting transistors because there was no load.

So I applied 0V to the input and -4V with a 1M resistor. I attached the results.

 

[meBigGuy]

For example the 0 level current for the 1M resistor is 123 uA, so when it changes to 23uA when the output goes to 100V, that's actually a -100uA change in current. Then look at the delta through the 50K.

 

[meBigGuy]

It seems there is significant offset since you get 43V for 0 input. Not sure where that is coming from.

 

[d.arbitman]

It seems there is significant offset since you get 43V for 0 input. Not sure where that is coming from.

It's using real models as opposed to ideal transistors. NPN and PNP transistors are very different.

For example the 0 level current for the 1M resistor is 123 uA, so when it changes to 23uA when the output goes to 100V, that's actually a -100uA change in current. Then look at the delta through the 50K.

Where do you see 123uA?

 

[meBigGuy]

If the output voltage was 0V (and all should be relative to an OUTPUT of 0V) then there would be 123V or so across the 1M resistor. When the output goes to 100V, there will be 23V or so across the 1M. That means the 1M current changed by -100uA. What do you think the 50K current will change by? Maybe not exactly 100uA, since there is finite open loop gain and input bias currents, but it should be consistant with the gain change you are seeing.

 

[d.arbitman]

If the output voltage was 0V (and all should be relative to an OUTPUT of 0V) then there would be 123V or so across the 1M resistor. When the output goes to 100V, there will be 23V or so across the 1M. That means the 1M current changed by -100uA. What do you think the 50K current will change by? Maybe not exactly 100uA, since there is finite open loop gain and input bias currents, but it should be consistant with the gain change you are seeing.

Oh, you're considering an ideal case; I thought you were talking about my simulation and I couldn't find 123uA anywhere. You were starting to scare me.

According to the simulation pictures I posted. The current decreases by ~60uA through the 1M while the current increases by ~80uA through the 50k which makes sense, since the voltage decreases by 4V.

4V/50k = 80uA


The difference of 20uA is compensated for via the base drive and the current through the 510ohm resistor. I still can't explain the inner loop gain.

 

[d.arbitman]

So if we assume that the input can swing between -10V and +10V, I have simulated the output swing using 1M and 100k resistors. I only changed the resistor values and I want to know why/how the output swing is larger using the 1M resistors.

 

[meBigGuy]

The difference of 20uA is compensated for via the base drive and the current through the 510ohm resistor. I still can't explain the inner loop gain.

You just explained it. The input voltage goes negative causing an increased current to be pulled out of the 50K. That current comes from the 1M resistor. (well, the 1M resistor draws less current because the output voltage went up). The output had to go up enough to change the current by the same amount as the 50K current did, therefore a 1M/50K gain (almost).

 

[d.arbitman]

I finally figured it out. I forgot to take into account the quiescent current and the associated change in said current when the input drops. Because 510 is 1% of 50k, the increase in current through the 510 is negligible, so when the input drops, the current through the 1M has to decrease by roughly the same amount. When you use a lower value resistor the voltage change is not as significant because more quiescent current flows and a small change in current doesn't produce a large change in voltage. However when the resistor is large, the quiescent current is small and a small change in current changes the output voltage by a larger value.

 

[meBigGuy]

Not sure I totally agree with your use of quiescent, but I'm glad you see the 50K/1M currents vs the resistor ratio as setting the gain.

Think of the base of Q1 as a contant voltage point since the Q1 collector current does not change significantly. Or, think of the Q1 current as constant since the voltage at Q1's base doesn't change significantly.

 

[d.arbitman]

Not sure I totally agree with your use of quiescent, but I'm glad you see the 50K/1M currents vs the resistor ratio as setting the gain.

Think of the base of Q1 as a contant voltage point since the Q1 collector current does not change significantly. Or, think of the Q1 current as constant since the voltage at Q1's base doesn't change significantly.

By quiescent, I meant when the input is 0V.

 

[meBigGuy]

As I reread what you said, I think you almost got it. Basically the current change in the 50K needs to be matched to the current change in the 1M. The output goes to whatever voltage is required to do that. The gain is not ideal because the effective impedance at the Q1 base is not infinite, and the open loop gain is not infinite.

The quiescent current has nothing to do with it. It is just quiescent. Maybe it helps you think about it, but it isn't part of the equation.

 

[d.arbitman]

As I reread what you said, I think you almost got it. Basically the current change in the 50K needs to be matched to the current change in the 1M. The output goes to whatever voltage is required to do that. The gain is not ideal because the effective impedance at the Q1 base is not infinite, and the open loop gain is not infinite.

The quiescent current has nothing to do with it. It is just quiescent. Maybe it helps you think about it, but it isn't part of the equation.

It's the change in current from the state when no input is applied. When the in put is -4V, then change in current is 80uA. The drop in voltage at the base is almost negligible (and the additional current through the 510ohm is on the order of 2uA) which is why the additional change in current out of the base is 1uA or so. The rest of the current must be decreased through the 1M which causes less of a voltage drop across it. Since ~77uA through 1M is much greater than through 100k, the output can swing much more.

 

[meBigGuy]

The output "MUST" swing much more as opposed to "can".

Is this the implementation you will use? Now that you understand this circuit, look at the simpler two transistor solution using T1 and T5 in http://www.bogobit.de/bogobox/

 

[d.arbitman]

The output "MUST" swing much more as opposed to "can".

Is this the implementation you will use? Now that you understand this circuit, look at the simpler two transistor solution using T1 and T5 in http://www.bogobit.de/bogobox/

I will most like not be using it. I just HAD to understand it. I am using a NPN transistor that sinks base current from a PNP. The NPN is driven by an op amp. Pretty much how T1 and T5 are configured.

 

[meBigGuy]

Well, I'm glad you persisted because I learned a bunch in the process.

 

[jim hardy]

Well, I'm glad you persisted because I learned a bunch in the process.

Likewise. I'm glad both of you persisted.

 

[d.arbitman]

I'm just glad you guys were very helpful.