Pass Transistor Driver Problems

[amspire]

I agree with Kleinstein. The positive supply for the opamps is too low for the current circuit. The opamp output has to be above 28.7 volts to even turn on the output transistors with a 25V output. Opamps can easily have at least a half volt drop or more on the output, so the maximum base current the opamp can supply will be under 10mA. Maybe much less. If you try a opamp positive voltage of 35 or 40, it will probably work.

[ZeTeX]

My guess would be the diodes D3 and D4 - but the simulation will show for sure.

Yes, it is those 2 diodes, The circuit is working correctly now, Thanks.

[ZeTeX]

I think I finished designing the regulator, it seems to be stable in simulation, no osciliations.

Is there anything that stands out as "Not good"?
Any improvement that could be made, or any feedback about the circuit?
Thanks.

[Kleinstein]

The LT1037 is a good choice for U2, but not for the other OPs. It's not unity gain stable and could thus cause trouble in real life, even if it work in the simulation.

C5 makes the voltage regulation relatively slow - so likely a smaller Value with a resistor (e.g. 1 K)  in series is a better choice.

The output stage should have a relatively small resistor (e.g. 20 Ohms range)  from the emitter of Q2 to the anode of D5. This resistor helps to speed up the transistor stage.
There definitely should be a diode that prevents a to large base to emitter voltage for the output stage. A fast transient could damage Q7 otherwise. This also helps a little to turn down the output - not much, but a little.

With C1 and C2 one should check the type. Usually there should be one with very low ESR (e.g. 100-500 nF ceramic / film type) and one with some ESR (e.g. low ESR electrolytic). Something like 0.1 Ohms or 1 Ohms of ESR can make quite a difference here. I would guess C1 should be somewhat larger (more like 100 µF) to get a better pulse response and to reach the 0.1-0.5 Ohms range for the ESR.

For simulation I would replace V4 and the switch with current source element - that's easier and allows direct changes in load current and run AC simulation to get the output impedance. The curve for the output impedance gives a hint if the supply is stable with any "passive" load.

[ZeTeX]

The LT1037 is a good choice for U2, but not for the other OPs. It's not unity gain stable and could thus cause trouble in real life, even if it work in the simulation.

C5 makes the voltage regulation relatively slow - so likely a smaller Value with a resistor (e.g. 1 K)  in series is a better choice. C

The output stage should have a relatively small resistor (e.g. 20 Ohms range)  from the emitter of Q2 to the anode of D5. This resistor helps to speed up the transistor stage. C
There definitely should be a diode that prevents a to large base to emitter voltage for the output stage. A fast transient could damage Q7 otherwise. This also helps a little to turn down the output - not much, but a little. C

With C1 and C2 one should check the type. Usually there should be one with very low ESR (e.g. 100-500 nF ceramic / film type) and one with some ESR (e.g. low ESR electrolytic). Something like 0.1 Ohms or 1 Ohms of ESR can make quite a difference here. I would guess C1 should be somewhat larger (more like 100 µF) to get a better pulse response and to reach the 0.1-0.5 Ohms range for the ESR. C

For simulation I would replace V4 and the switch with current source element - that's easier and allows direct changes in load current and run AC simulation to get the output impedance. The curve for the output impedance gives a hint if the supply is stable with any "passive" load.

Thanks for commenting and helping, you are helping a lot and I appreciate it.

the OPs are not going to be LT1037, the OPs are going to be "OP275" (DS: http://www.analog.com/media/en/technical-documentation/data-sheets/OP275.pdf).

I've added base emitter diodes + emitter to D5 anode resistor, that does not seems to make things faster but not worse actually, but I have nott tested it much.

I've played with C2 & C1 ESR and capacitance a lot, I figured out that with 4.7uF 0.5Ohm capacitor and 100nF at 0.01Ohm I get the supply most stable. reason for the low capacitance is because with high capacitance the current limiting takes a lot of time to start limiting. lower capacitance - faster current limiting, too low makes oscillations.

One thing I cant understand is using current source as a load, its a current source, it does not sink current, how would that even work? is there any tutorial about this that I can read?

I've tried plotting the output impedance by this tutorial:
http://www.rocklinger.se/Elektronik/Simulering/output_imp/output_impedance_ltspice_en.html

It seems about 0-1ohm depending on the load. (simulation syntax is .ac oct 100 1 1000000)

[Kleinstein]

The current source is also working as a current sink - it just set the current, not matter what. So if the circuit can't provide the current you might see negative voltages.

If the circuit only works with a small output capacitor this is a bad sign. An external load might add capacitance and the supply should work in this case too. It's also hard to find a capacitor 5 µF with a fitting ESR. The speed of current limiting is more limited by the output transistors/regulator part - not by the capacitor. As the regulator is rather slow, a large capacitor might be needed to prevent large transients on load changes. So even 100 µF might be to little.

The OP275 is a low noise JFET OP. Such a simple supply does not need this. So I would suggest the OP37/LT1037 (essentially the same) for U2 as this OP should be fast and low drift / offset. The other OPs are less critical, except for a high supply voltage. Sp I guess there a simple types.

[ZeTeX]

The current source is also working as a current sink - it just set the current, not matter what. So if the circuit can't provide the current you might see negative voltages. T

If the circuit only works with a small output capacitor this is a bad sign. An external load might add capacitance and the supply should work in this case too. It's also hard to find a capacitor 5 µF with a fitting ESR. The speed of current limiting is more limited by the output transistors/regulator part - not by the capacitor. As the regulator is rather slow, a large capacitor might be needed to prevent large transients on load changes. So even 100 µF might be to little.

The OP275 is a low noise JFET OP. Such a simple supply does not need this. So I would suggest the OP37/LT1037 (essentially the same) for U2 as this OP should be fast and low drift / offset. The other OPs are less critical, except for a high supply voltage. Sp I guess there a simple types.

Didnt know about the current soruce as current sink, learned something new- going to test it.

the circuit can work with high output capacitance, it is just that the current limiting then becomes slow as the capacitor has alot of charge left, for example, 10,000µF capacitor, with C1 set to 100u, the current limiting takes about 200uS to get to the set current, while with C1 set to 4.7µF, it takes about 20uS to get to the set current.

I have already OP275 from a different project - better to use them instead of ordering new OPs.

[ZeTeX]

I'm just posting the same schematic just with the fixed cc indicator + the things "Kleinstein" pointed out.
As I said previously I dont want to make C1 any larger because it will slow the current limting.
Any suggestion will be appreciated

[StillTrying]

The inputs to the U7 and U8 LED indicators can't be right.
Surely each op-amps inputs should go to both 22ks, with +/- reversed for one, so that it's output is inverted

[Kleinstein]

If you really insist in such a small output capacitor, it gets a little more tricky: In this case the regulator would need to be really fast unless you can accept a considerable drop  / overshoot  in voltage on fast load changes.

A faster regulator should also use faster output transistors. At least the middle one can be rather easy changed to a faster one, as it does not need that much power. For getting the right compensation you would essentially start over for this - a faster compensation also needs a more accurate choice for C5,C6, R16 and so on.

[T3sl4co1l]

If you really insist in such a small output capacitor, it gets a little more tricky: In this case the regulator would need to be really fast unless you can accept a considerable drop  / overshoot  in voltage on fast load changes.

I never got the fascination with huge caps. The leads you put on outside add more than enough to ruin it.

Big caps are only ever a cheap hack of a fix for a slow regulator design.

My bench supply is little more than a 20V, 10A audio amplifier; it can deliver surges up to 40A, as fast as the cables can handle it.  During such a surge, the waveform is stable at both ends of the cable -- it's not dropping out at all.

A faster regulator also has less windup between voltage and current modes.  I still wouldn't suggest shorting it out into LEDs, but at least the LEDs won't be discharging a big stupid cap, when you do this accidentally.

Tim

[Kleinstein]

There is not only the possibly current pulse from the capacitor, but there is also a delay from a slow regulator reacting to a dropping voltage / short. This is especially by this type of regulator shown here. So even it there is only a 5 µF cap at the output, one a short, a regulator might still behave like a 1000 µF capacitor.  So it is not only the capacitor that is actually at the output. Current regulation in principle behaves like a capacitive source - the faster the smaller the "simulated" capacitance. Windup of the regulator can add an additional current peak - so for fast reaction a different type of regulation (e.g. shunt on the high side and floating current regulator).

Usually an output capacitance in the 100 µF range is not bad. Many commercial supplies have more - sometimes in the 1000 µF range. Getting really fast, so you can use a rather small capacitance also makes the circuit layout sensitive.

[T3sl4co1l]

Yup.  Also, in case it wasn't clear: my bench supply doesn't have current limiting, so it just keeps belting out the amps until something else gives. (I think it's actually hFE or PS limited around 10A.)

The limited response time of a CV supply looks like inductance (i.e., the error amp can't regulate fast enough and the output dips a little, then recovers); while, that of a CC supply looks capacitive (voltage doesn't change immediately in response to current, but takes some catching up).

The interaction of the two (CV to CC or vice versa) typically looks like diode reverse recovery: upon passing the threshold condition, it stays in whatever mode it was, until after that time has passed.

The mechanism for this, in most circuits, is integrator windup: you have two independent integrators (normally, an integrator is an inverting amp with a C across it; in this case, it's the op-amp's intrinsic integrating behavior (dominant pole compensation) acting as a smaller C, in parallel with the R+C network explicitly used), and while one of them is inactive, it saturates to the opposite rail.  The time taken to go from rail to setpoint is the recovery time, which varies with output level and load, but is on the order of the loop time constant (~ms for a switching converter -- awful!).

Unfortunately, as transconductance amps are rarely used, it's very difficult to avoid this in a practical circuit.

Some solutions include:
https://www.eevblog.com/forum/projects/limiting-op-amp-output/msg450476/#msg450476 and
https://www.eevblog.com/forum/projects/limiting-op-amp-output/msg732595/#msg732595

but these are more applicable to voltage --> current --> Gm amp --> output loops (the loops are stacked), not merging two loops into one.

Tim

[ZeTeX]

There is not only the possibly current pulse from the capacitor, but there is also a delay from a slow regulator reacting to a dropping voltage / short. This is especially by this type of regulator shown here. So even it there is only a 5 µF cap at the output, one a short, a regulator might still behave like a 1000 µF capacitor.  So it is not only the capacitor that is actually at the output. Current regulation in principle behaves like a capacitive source - the faster the smaller the "simulated" capacitance. Windup of the regulator can add an additional current peak - so for fast reaction a different type of regulation (e.g. shunt on the high side and floating current regulator).

Usually an output capacitance in the 100 µF range is not bad. Many commercial supplies have more - sometimes in the 1000 µF range. Getting really fast, so you can use a rather small capacitance also makes the circuit layout sensitive.

thanks, I got the psu to float but the current limiting does not work with capacitive load at the output, but does work the resistive load.
C3 get charged to 3V and thats it. it starts with 12A charging current and drops slowly.

[Kleinstein]

Just moving the shunt to the high side is making things only worse - if you move the shunt high side, you would need to change the current regulator too, e.g. make the current regulator floating, possibly with a second isolated supply, a separate reference and so one - so not a small change but more like a new start.

The downside with the old regulator type is, that the current regulator needs to bring the voltage down from a possibly high value. This takes quite some time and thus makes response slow, especially if the current loop is also running into windup, that is the OP going all the way to the positive limit and charge the capacitor all the way. At first the current can rise quite far above the set point.

The old circuit uses the diodes to get the output voltage from the voltage or current regulator which ever is lower. This kind of works, but not very well with critical loads and is slow for current regulation. The alternative way is having a current setting output stage and use the diodes to use the lower current level from both regulator parts. This type of regulation is faster with current limiting, but needs a fast tuning to make the voltage regulation work well. If done right is can give good voltage regulation too. Usually it needs a slightly larger capacitor at the output, but has less virtual capacitance coming from the regulator part - so the overall current pulse on a short is usually smaller.

[ZeTeX]

Just moving the shunt to the high side is making things only worse - if you move the shunt high side, you would need to change the current regulator too, e.g. make the current regulator floating, possibly with a second isolated supply, a separate reference and so one - so not a small change but more like a new start.

The downside with the old regulator type is, that the current regulator needs to bring the voltage down from a possibly high value. This takes quite some time and thus makes response slow, especially if the current loop is also running into windup, that is the OP going all the way to the positive limit and charge the capacitor all the way. At first the current can rise quite far above the set point.

The old circuit uses the diodes to get the output voltage from the voltage or current regulator which ever is lower. This kind of works, but not very well with critical loads and is slow for current regulation. The alternative way is having a current setting output stage and use the diodes to use the lower current level from both regulator parts. This type of regulation is faster with current limiting, but needs a fast tuning to make the voltage regulation work well. If done right is can give good voltage regulation too. Usually it needs a slightly larger capacitor at the output, but has less virtual capacitance coming from the regulator part - so the overall current pulse on a short is usually smaller.

Alright, new start is no good. I'm back to the the previous design..

I'm aware of the this downside, but I could not think of a way to stop it. "The alternative way is having a current setting output stage and use the diodes to use the lower current level from both regulator parts."
could you explain about the "current setting output stage"? how does it work?

I've seen this schematic : https://www.circuitspecialists.com/pdf/1802X_schematic.pdf
it seems pretty simple psu, similar to mine but floating, is it good?

[StillTrying]

@ ZeTex : Could you upload the Ltspice/.asc file here?

The .asc of his last but one version is on a post above.

This is near enough it, but with the op-amps dragged into better positions - so I can understand it!

[Kleinstein]

The Circuit from the link
 https://www.circuitspecialists.com/pdf/1802X_schematic.pdf
looks good. This is the alternative design with current setting output stage and floating regulator circuit, very much like the older HP power supplies. So if the caps / resistors for compensation are chosen right it provides good performance. Though usually there is one more cap and 2 extra diodes.

The way it is drawn is a little strange - so it may even more difficult to follow.

The 3rd transformer winding is just for the relay to switch between the transformer taps. For a high current version this might to be modified a little (e.g. use 4 taps or use transistors instead of relay).

[ZeTeX]

The Circuit from the link
 https://www.circuitspecialists.com/pdf/1802X_schematic.pdf
looks good. This is the alternative design with current setting output stage and floating regulator circuit, very much like the older HP power supplies. So if the caps / resistors for compensation are chosen right it provides good performance. Though usually there is one more cap and 2 extra diodes.

The way it is drawn is a little strange - so it may even more difficult to follow.

The 3rd transformer winding is just for the relay to switch between the transformer taps. For a high current version this might to be modified a little (e.g. use 4 taps or use transistors instead of relay).

what does this part do?
also they are not using current sense amplifier, I guess because they just use low voltage to adjust. (Isense point goes directly to the current error amplifier.) would not that cause problem because the low voltage adjustment (mV range)?

[Kleinstein]

The circuit part looks like compensation for the current needed for the output divider. So the meter for the current does not include that little current used by the regulator itself.

They do not use an extra amplifier for the current signal, as they don't need to. The output of the regulating OPs should be somewhere in the 0.5-1.5 V range, so not such a big change. Looking only at the current loop, this very much looks like the classical constant current sink.
The circuit with the output stage as an emitter follower needs the OPs to go all the way from near zero to a little more than full voltage. So the current regulating OP may need to work with a much higher gain. In this case the extra amplification of the current signal helps to make the regulator faster - at least if the amplification is fast.

[ZeTeX]

The circuit part looks like compensation for the current needed for the output divider. So the meter for the current does not include that little current used by the regulator itself.

They do not use an extra amplifier for the current signal, as they don't need to. The output of the regulating OPs should be somewhere in the 0.5-1.5 V range, so not such a big change. Looking only at the current loop, this very much looks like the classical constant current sink.
The circuit with the output stage as an emitter follower needs the OPs to go all the way from near zero to a little more than full voltage. So the current regulating OP may need to work with a much higher gain. In this case the extra amplification of the current signal helps to make the regulator faster - at least if the amplification is fast.

They have not specified resistor values, as for now I copied the circuit to ltspice but it doesn't work, I'm guessing because the resistors values are not chosen correctly. 
Whats the problem ?
original circuit: https://www.circuitspecialists.com/pdf/1802X_schematic.pdf

[Kleinstein]

The circuit has resistor value - just the strange coding as used with SMD resistors. So a resistor marked 103 should be 10*10^3 or 10 K.

One also has to be careful abut where ground is connected. I think there is a small mistake in that pin 1 of the connector CON2 -> CN2/CN4 should not go to ground.

[ZeTeX]

The circuit has resistor value - just the strange coding as used with SMD resistors. So a resistor marked 103 should be 10*10^3 or 10 K.

One also has to be careful abut where ground is connected. I think there is a small mistake in that pin 1 of the connector CON2 -> CN2/CN4 should not go to ground.

ok, I will put the correct resistor values and see if it works,
I noticed the mistake - didnt include it in the schematic.

[StillTrying]

I did a LT version of that pdf. Here's a 0-1A-0 load. 

[Kleinstein]

The circuit might need to have some ESR (e.g. 0.1 - 1 Ohms range) for C28 (the output capacitor). Normally electrolytic caps have this, but a simulation be default assumes 0 ESR.

Usually there is also a capacitor in parallel with parts of R54-R56. This might be needed for extreme loads, like a perfect current sink often used in simulation.