Low voltage op amp to control high voltage series pass transistor

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meBigGuy

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Think of the voltage on the base of Q1 as effectively being across the 330 ohm emitter resistor thereby modulating the Q1-Q2 current. The same thing is happening with Q2. If Q2 is "more off" the the voltage at the collector of Q1 will be high, and vice versa. Note that there is no dot between the diodes. The diodes just sorta parallel the two Vbe diodes on the output devices. The 1M feedback resistors kind of clean things up.
 
Think of the voltage on the base of Q1 as effectively being across the 330 ohm emitter resistor thereby modulating the Q1-Q2 current. The same thing is happening with Q2. If Q2 is "more off" the the voltage a the collector of Q1 will be high, and vice versa. Note that there is no dot between the diodes. The diodes just sorta parallel the two Vbe diodes on the output devices. The 1M feedback resistors kind of clean things up.
"The local 1M-50k feedback pairs set stage gain at 20, allowing ±10V LT1055 drives to cause full 120V output swing." Page 6 of that PDF.

What I noticed through simulation is that changing the 1M to 100k lowers the output when the input voltage remains the same. The question is, why and how. Notice that the voltage at the base didn't change much. The voltage at the collector changed by an order of magnitude.
 

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meBigGuy

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OK. I had it wrong (well, the 1M is critical to setting the gain --- the rest explained the mechanism).

The 50K - 1M combination is exactly like the source and feedback resistors in an inverting opamp. Current into the Q1 base node through the 50K must be pulled out through the 1M. Reducing the 1M to 100K reduces the gain by 10 since a 10 times lower voltage will create the current.

I learned something too.
 
OK. I had it wrong (well, the 1M is critical to setting the gain --- the rest explained the mechanism).

The 50K - 1M combination is exactly like the source and feedback resistors in an inverting opamp. Current into the Q1 base node through the 50K must be pulled out through the 1M. Reducing the 1M to 100K reduces the gain by 10 since a 10 times lower voltage will create the current.

I learned something too.
I drew that analog with an inverting op amp too, but still doesn't explain it. Notice in my attached simulation pictures that the current draw increase is through the 510ohm resistor. The current out of the base increased by 1uA when the 1M resistor was changed to 100k


Current into the Q1 base node through the 50K must be pulled out through the 1M. Reducing the 1M to 100K reduces the gain by 10 since a 10 times lower voltage will create the current.
I don't understand what that means.
 

meBigGuy

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The opamp model explains it, but there are other bias currents and finite open loop gain. Notice that the total gain didn't change by exactly 10. Basically the current through the 50K and the current through the 1M will be comparable. Take the 1M out and measure the open loop gain. I'm guessing it will be low enough to account for the discrepency. (model an opamp with an open loop gain of 10 and see what happens).

If there was infinite open loop gain and no input bias currents the current through the 50K and the current through the 1M would be equal and opposite.
 
The opamp model explains it, but there are other bias currents and finite open loop gain. Notice that the total gain didn't change by exactly 10. Basically the current through the 50K and the current through the 1M will be comparable. Take the 1M out and measure the open loop gain. I'm guessing it will be low enough to account for the discrepency. (model an opamp with an open loop gain of 10 and see what happens).
I noticed that the total gain did not change by the same ratio that the resistance changed. I am assuming that is due to the nonzero current change in the base of Q2 (Q1 in LT's schematic) in my schematic as well as the non-linearity of transistors. (Q1 in LT's schematic).

How are the currents through the 50k and 1M comparable?

I attached a picture of a schematic that ought to simulate the behavior of the 1M-50k gain stage. Hopefully it's an accurate model of the gain stage.

So if we set Rg = 50k and Rf = 1M, we know that the output voltage must be higher than if Rf was less 1M due to KCL.

I am having trouble applying this to the actual circuit because of the increase in current through the 510ohm resistor.
 

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jim hardy

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Thanks Guys ! I was thinking about this while installing a new front door on the house today.

You have verified what I had come to think - the 1meg::50K set gain of output stages to 20, so we have two amplifiers in cascade.

And your simulation confirmed something I was trying to reason out:
Observe that of the ~2ma flowing down through R8, hardly any goes into Q5's base.
None should go into Q3 since it's only got 31mv Vbe
which means it has to flow down through D3 and D4 where it robs base drive from Q6.
So as top half of circuit pulls output up, bottom half pushes up due to mismatched base drive currents .
That makes gain of the booster stage quite high, so that wrapping R6 around the booster loop gives a nicely controlled voltage gain. Same way a self-saturating magamp from the 1940's works.

That is what I was trying to reason out when I gave up last night.
So D3 and D4 are NOT current limiters like I initially thought, instead they are part of the biasing.
What REALLY had me going was the dot between D3 & D4 that appeared to tie that point to output.
I couldn't make the circuit go with that dot there.
But it is NOT there in the appnote page 7 !

Ahhh so my alleged brain isn't totally fried. Whew !!!!

I learned a couple things today too -
1. How to hang a door and true up the jamb (it's sure nice having retired carpenter friends)
2. How a nice voltage booster really works (it's sure nice having bright young friends who like electronics)

Thanks Again Guys ! You really made my day !

And congratulations on your victory figuring this thing out.....

old jim
 
And congratulations on your victory figuring this thing out.....
I still can't figure out how that inner gain loop works. Mainly how the ratio of 1M:50k controls the ratio of total output swing to the op amp output swing.
 

meBigGuy

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I think I am out of my element here. I don't fully understand the 510 ohm resistor and its sizing. I'll have to think about it. It's more confusing because there are two sides playing against each other. Have you played with changing the 510R to see what that does.
 
I think I am out of my element here. I don't fully understand the 510 ohm resistor and its sizing. I'll have to think about it. It's more confusing because there are two sides playing against each other. Have you played with changing the 510R to see what that does.
Changing the 510 changes the bias voltage at the base of Q1/Q2, which changes the voltage drop across the 330 and as a result changes base drive to the output transistors. Although I tried and nothing major happened.

I am mainly interested how changing the 1M to 100k changes the maximum output voltage swing, specifically reduces it. In general, how changing the 1M resistor change the maximum output voltage swing.
 

meBigGuy

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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.
 
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.
 

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meBigGuy

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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

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

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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.
 
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.
 
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.
 

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meBigGuy

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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 amout as the 50K current did, therefore a 1M/50K gain (almost).
 
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

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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.
 
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

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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.
 
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

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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/
 

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