Pass Transistor Driver Problems

[ZeTeX]

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

you did what and what pdf?

[StillTrying]

The same pdf you were trying to get working.
There might be mistakes, LT1006s and LM358s showed the same 1.8k osc.

[ZeTeX]

The same pdf you were trying to get working.
There might be mistakes, LT1006s and LM358s showed the same 1.8k osc.

no problem for me, pretty stable but I'm still working on the schematic, in the current state it doesn't oscillate. notice I used 33pF caps with 1k resistor, I'm surprised it doesn't oscillate.
EDIT: I noticed in your schematic that you didnt provide any ESR to any of the caps, change the ESR of the output cap to 0.5ohm (didnt test any value yet expect 0.5?) and watch the magic.

[Kleinstein]

The circuit is rather stable, at least with ideal parts. The FTZ849 transistor is relatively fast which allows a fast response. 33 pF is a rather small capacitance, but something like 100 pF would not be that unusual for this type of regulator.

In real life inductance of the shunt and not so perfect raw voltage (e.g. ESR/ESL of capacitor) might require a slightly slower circuit.

[StillTrying]

I hate 33p being written as 330. 

Yep, 0.5R esr on the output cap is/was the only solution, - after spending an hour carefully comparing the 3 schematics, - all 3 are the same!
Only did a quick test going into and out of 2A current limit using a triangular wave on the voltage control and 3R on the output - no oscillation at least.

[timsu]

What would be the right way to control the output voltage with a voltage?
I tried applying a voltage to the inverting input of U2, that kind of worked, but that was not linear.

[StillTrying]

On ZeTeX's image above you connect the voltage source across R15. Yes it's non-linear and don't forget the output voltage is sort of inverted.
If varying, you need 10R across the output to discharge the output cap, to see the output.
Here's a 1V to 2V square voltage source.

Edit here's a 10K pot to fiddle with.

[ZeTeX]

On ZeTeX's image above you connect the voltage source across R15. Yes it's non-linear and don't forget the output voltage is sort of inverted.
If varying, you need 10R across the output to discharge the output cap, to see the output.
Here's a 1V to 2V square voltage source.

Edit here's a 10K pot to fiddle with.

but it is linear, it just took me about 15 minutes to get it right.

(the load was 1k)

[Kleinstein]

The more normal way would be having the resistor at inverting input of the OP going to GND, and have the variable reference voltage at resistor R1 (in the last schematics) instead of the fixed reference.

The down side is, that this will not give you an constant current load, but this is relatively easy to add. If separate it could be before the shunt, so that one might get away without the current compensation any more.

[ZeTeX]

The more normal way would be having the resistor at inverting input of the OP going to GND, and have the variable reference voltage at resistor R1 (in the last schematics) instead of the fixed reference.

The down side is, that this will not give you an constant current load, but this is relatively easy to add. If separate it could be before the shunt, so that one might get away without the current compensation any more.

"that this will not give you an constant current load, but this is relatively easy to add. If separate it could be before the shunt, so that one might get away without the current compensation"
Can you explain more? what constant current load?
I did the way you said and it works great.

[StillTrying]

This is where I ended up with this experiment. It's doesn't seem three bad!

Max V and I is around 3.3A at 26.6V, - you can have a bit more of one for a bit less of the other.
The graph shows 2R being put on the output while the output is set at 20V and 1A, the slopes are mostly the output cap discharging and recharging when the 2R is removed.

[ZeTeX]

This is where I ended up with this experiment. It's doesn't seem three bad!

Max V and I is around 3.3A at 26.6V, - you can have a bit more of one for a bit less of the other.
The graph shows 2R being put on the output while the output is set at 20V and 1A, the slopes are mostly the output cap discharging and recharging when the 2R is removed.

Too slow! ~3ms for the current limit to hit is pretty slow, maybe down programmer will fix it! because the regulator is fast, but the caps at the output have some energy left. with only 100nF caps at the output the supply is fast in the nS range, but the op amp oscillate and that's not a good sign.
have not had much time to play with it in the last days but I'm going to play it with today and see.

[Kleinstein]

This type of circuit usually needs more that a 100 nF cap at the output. Depending on the output transistors used and the compensation , the minimum capacitance is something like 10 µF with an ESR in the 0.1-0.5 Ohms range and about 100 nF with very low ESR in parallel. Though you can get the supply stable with such a small capacitance, one might want a little more (like 100 µF) to reduce the drop and overshoot on load changes.  This is still a rather good value.

It's usually not that good to set the voltage by adjusting the feedback divider, as this also makes the loop gain dependent on the setting. So it's better to adjust the ref. voltage instead.

If one wants it even faster, I would chose a true two quadrant output stage, which allows the output transistors to work at a minimum bias level under all conditions. But this type needs some extra effort and has higher standby power consumption / heat production. A fast regulator also gets sensitive to the layout - not a surprise if you go for an output impedance (mainly inductance in the 100 nH range)  comparable to a 10 cm piece of thick (e.g. 100mm²) wire.

[timsu]

"that this will not give you an constant current load, but this is relatively easy to add. If separate it could be before the shunt, so that one might get away without the current compensation"
Can you explain more? what constant current load?
I did the way you said and it works great.

I'm not quite sure, but if I understood it right, but the current regulations seems a bit off.
The current through the 0.1R shunt resistor is not exactly the same as through the load (it seems to be around 1-2mA off).
I'm not quite sure where it comes from and it is also just a very small error.

Also the forward voltage drop in  diodes  can quite differ if I read the datasheets correctly.
Will this become an accuracy problem?

[StillTrying]

Too slow! ~3ms for the current limit to hit is pretty slow,

I agree, could do it 30 times faster in firmware!

I got it down to 0.6ms by reducing the output caps to 47u+4.7u+0.1u all with 0.5 ESR, but then the output rises by 0.5V for 20+ms when the load is removed.
I've got it down to 1ms ATM with 110uf output cap for a 1A load, and 0.2V output rise. I don't think that's too bad considering the rubbish components I'm using - LM324 & 2N3055.

It's usually not that good to set the voltage by adjusting the feedback divider, as this also makes the loop gain dependent on the setting. So it's better to adjust the ref. voltage instead.

I agree, even on the current control op-amp where the reference is simply applied to the +ve input, there's changes in the bias currents especially while it's being used as a comparator, I'm surprised it's simulating as good as it is.

[ZeTeX]

Too slow! ~3ms for the current limit to hit is pretty slow,

I agree, could do it 30 times faster in firmware!

I got it down to 0.6ms by reducing the output caps to 47u+4.7u+0.1u all with 0.5 ESR, but then the output rises by 0.5V for 20+ms when the load is removed.
I've got it down to 1ms ATM with 110uf output cap for a 1A load, and 0.2V output rise. I don't think that's too bad considering the rubbish components I'm using - LM324 & 2N3055.

It's usually not that good to set the voltage by adjusting the feedback divider, as this also makes the loop gain dependent on the setting. So it's better to adjust the ref. voltage instead.

I agree, even on the current control op-amp where the reference is simply applied to the +ve input, there's changes in the bias currents especially while it's being used as a comparator, I'm surprised it's simulating as good as it is.

I'm trying to implement a down programmer that will sink current, that way the output caps can get discharged fast, it seems to work, but the ESR of the output caps play a big rule here, The down programmer works very good with the output cap at ESR of 0.01ohm, but the circuit osciliate when the down programmer is off, but with 0.5ohm ESR the down programmer is a little bit slower but the circuit is stable.

[Kleinstein]

The current through the shunt is not exactly the same as the output current. There is also the base current for the output stage and the current for the voltage divider flowing through the shunt.  The current for the voltage divider was compensated in the original circuit with the one extra OP.

If you need an accurate current, one should use a separate shunt on the other side of the GND connection and than have a separate emitter resistor. You need emitter resistors anyway, if more than one transistor in parallel is to be used.

Just as a note: essentially the same circuit also works with a MOSFET instead of the darlington transistor. This might be attractive if high speed at relatively low output voltage is needed. However MOSFETs don't work that well at relatively low current.

If you want a down-programmer to make it a limited 2 (or more limited even 4) Quadrant supply, one could use a PNP transistor at the + 5 V auxiliary supply. Just to give an Idea I add an LTspice file. There are also a few modifications for slightly faster crossover from CV to CC mode. Accurate current regulation and measurement  would need an extra shunt. V5 could be used to adjust the standing current or choose class B operation.

[Kalvin]

This has been quite educating discussion thread. Designing a stable control loop is not easy. But it can be made easier using a simulator and analysing the loop stability with different loads and conditions, thus reducing the guess work.

Edit: You may find the following LTSpice tutorial quite useful for analysing the loop stability:
http://www.linear.com/solutions/4449

For someone wanting to take a bit closer look at the basics of the feedback systems, here is an excellent set of tutorials on feedback systems and stability, breaking the loop etc. :
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-1-general-structure-and-essential-concepts/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-3-improving-noise-linearity-and-impedance/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-4-introduction-to-stability/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-5-gain-margin-and-phase-margin/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-6-new-and-improved-stability-analysis/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-7-frequency-dependent-feedback/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-9-breaking-the-loop/
http://www.allaboutcircuits.com/technical-articles/negative-feedback-part-10-stability-in-the-time-domain/

For those who like to see videos instead, here is a nice introduction to feedback systems by Prof. J. K. Roberge:



A lecture and a demo for a feedback system compensation by Prof. J. K. Roberge:

[ZeTeX]

The current through the shunt is not exactly the same as the output current. There is also the base current for the output stage and the current for the voltage divider flowing through the shunt.  The current for the voltage divider was compensated in the original circuit with the one extra OP.

If you need an accurate current, one should use a separate shunt on the other side of the GND connection and than have a separate emitter resistor. You need emitter resistors anyway, if more than one transistor in parallel is to be used.

Just as a note: essentially the same circuit also works with a MOSFET instead of the darlington transistor. This might be attractive if high speed at relatively low output voltage is needed. However MOSFETs don't work that well at relatively low current.

If you want a down-programmer to make it a limited 2 (or more limited even 4) Quadrant supply, one could use a PNP transistor at the + 5 V auxiliary supply. Just to give an Idea I add an LTspice file. There are also a few modifications for slightly faster crossover from CV to CC mode. Accurate current regulation and measurement  would need an extra shunt. V5 could be used to adjust the standing current or choose class B operation.

I have not found a difference in speed between MOSFET and NPN transistor, Could you explain about the opreation of the down programmer?
I've downloaded your file, but its a little bit messy and I'm having hard time understanding how it works, because copying it directly to the schematic doesn't work.

[Kleinstein]

The down programmer / second quadrant part might be a little difficult to under stand. Sorry for the confusing order, the negative output is on the upper side of the current source/sink and the positive side is at the bottom, where the ground symbol is. So a little explanation:

The main idea is to use the positive auxillary supply to pull up the negative output side. So using this as a way to allow slightly negative output voltages and have room for the transistors to work even at zero output voltage. The very basic circuit would be just Q3 and a resistor from the +5 to the emitter of Q3 - this would give a constant standing current.

V5 and Q8 check if the controlling voltage for the normal output stage is going to low so that the current is set to zero or very low. At low control voltage more current is flowing through Q8.  From V5 one could adjust the point where the negative side sets in.

Q9 with R16 and R15 works like a crude kind of current mirror and amplification of the current. The voltage amplification from Q8 compensates for the higher value of R15 compared to the R1 (0.1 Ohms shunt). Q10 is a crude, fixed current limiting for the negative side current.

Q3 is than just there to reduce the power for Q8 and thus allow a small and fast type there. As a second effect it prevents output voltages below about  -0.7 V.

At high currents MOSFETs are rather fast compared to power BJTs, especially if it comes to turning off. This is at least true if the circuit is made to have a reasonable strong gate drive. However the downside is that they behave different at low currents, so not so much speed advantage unless a considerable standing current is used. Normally I see not big advantage in using MOSFETs - for this circuit the current reading would be a little more accurate compared to the BJT version.

[ZeTeX]

The down programmer / second quadrant part might be a little difficult to under stand. Sorry for the confusing order, the negative output is on the upper side of the current source/sink and the positive side is at the bottom, where the ground symbol is. So a little explanation:

The main idea is to use the positive auxillary supply to pull up the negative output side. So using this as a way to allow slightly negative output voltages and have room for the transistors to work even at zero output voltage. The very basic circuit would be just Q3 and a resistor from the +5 to the emitter of Q3 - this would give a constant standing current.

V5 and Q8 check if the controlling voltage for the normal output stage is going to low so that the current is set to zero or very low. At low control voltage more current is flowing through Q8.  From V5 one could adjust the point where the negative side sets in.

Q9 with R16 and R15 works like a crude kind of current mirror and amplification of the current. The voltage amplification from Q8 compensates for the higher value of R15 compared to the R1 (0.1 Ohms shunt). Q10 is a crude, fixed current limiting for the negative side current.

Q3 is than just there to reduce the power for Q8 and thus allow a small and fast type there. As a second effect it prevents output voltages below about  -0.7 V.

At high currents MOSFETs are rather fast compared to power BJTs, especially if it comes to turning off. This is at least true if the circuit is made to have a reasonable strong gate drive. However the downside is that they behave different at low currents, so not so much speed advantage unless a considerable standing current is used. Normally I see not big advantage in using MOSFETs - for this circuit the current reading would be a little more accurate compared to the BJT version.

I have a problem where, if I set V5, to higher voltage, then at the output the current get limited to higher value then I set in op amp that limits the current.
Are there any advantages to this circuit then just shorting the output if the current that flowing is higher then the set current like I did before?

[Kleinstein]

The down programmer should operate in a linear mode too, so allowing for 2 quadrant operation and support regulation at very low current. This way you could even do load steps like +1 A to -10 mA and they behave very much like a 1 A to 100 mA step.  Just switching in a resistor will not work well: to much at high voltage and to little at low voltage plus some extra glitches - so I don't think this would be an option at all.

If you set the voltage limit to high (e.g. more than about 2 V for V5 in the circuit) there can be too much standing current. But there is a reasonable large range (e.g. 1.2 - 1.9 V) that should work OK. Even without a standing current and thus in class B mode the dead zone is not that large. One could get away without Q3 and the extra 10 µF capacitor for speed up if you want is simpler.  In a real circuit the voltage should be something like 3 diode drops or a VBE multiplier to compensate for temperature dependence. It is rather similar to setting the bias current in an audio amplifier, though it is less critical here as we are not concerned with crossover distortion.

[ZeTeX]

The down programmer should operate in a linear mode too, so allowing for 2 quadrant operation and support regulation at very low current. This way you could even do load steps like +1 A to -10 mA and they behave very much like a 1 A to 100 mA step.  Just switching in a resistor will not work well: to much at high voltage and to little at low voltage plus some extra glitches - so I don't think this would be an option at all.

If you set the voltage limit to high (e.g. more than about 2 V for V5 in the circuit) there can be too much standing current. But there is a reasonable large range (e.g. 1.2 - 1.9 V) that should work OK. Even without a standing current and thus in class B mode the dead zone is not that large. One could get away without Q3 and the extra 10 µF capacitor for speed up if you want is simpler.  In a real circuit the voltage should be something like 3 diode drops or a VBE multiplier to compensate for temperature dependence. It is rather similar to setting the bias current in an audio amplifier, though it is less critical here as we are not concerned with crossover distortion.

then, maybe not switch just a resistor, what about turning on constant current sink? therefore the drawn current will always be the same, but even if I will just short the output with 0ohm resistor (direct short) it will still work great even at low voltages.
It doesn't matter if I set V5 to 1.2V or 1.9V, the output waveform is the same thing.

I'm also trying to a find a "true" 2 quadrant bench power supply schematics on google and I did had much luck with that.

[Kleinstein]

One can use a constant current sink. It's possible with just the PNP transistor with the base to GND and a resistor from the emitter to +5 V.

However this gets a little wasteful if a higher negative current is desired. So something like 50 or 100 mA might be ok, though already adding some waste heat all the time, even if a large positive current is needed. The maximum current is reduced by the maximum negative current set.  Having a variable negative current might also allow for a slightly faster setting of the regulation as the range with very low current (and thus usually the lowest speed) of the power transistor can be avoided. However this only works if the negative side is faster than the NPN at low currents.

[StillTrying]

This 2 op-amp circuit is always going to have a delay/gap in swapping control from V to I. When the V op-amp output is controlling at around 0.5V, the I op-amp output has to come down from 3.6V to take over, not only is there a gap/delay before it does, but during this time the V op-amp(in linear mode) moves in the wrong direction and overshoots before the I amp gets down to 0.5V.

Here's a messy graph showing the  pass tr (grey) overshooting by 1.5A before the I amp takes over to bring it down to the 0.5A set limit.