Help designing load transient driver

Thread Starter

newbie217

Joined Apr 12, 2009
52
Hello,

In my quest to improve my very weak electronics skills, I thought I would give myself a project to force me to be more hands on.

I want to create a working prototype of a load transient driver using mostly discretes (transistors, MOSFET's, capacitors, resistors, etc), and a few or no IC's. The requirements would be:

Input is driven by function generator (0 to 10 V)
frequency is not too important. Low frequency is fine. Under 1 kHz.
output load is capacitive (up to 10 uF)
Would like to be able to drive this 10 uF load from 0 to 5 V in 25 us or less. My calculations show I would need:

i = C dv/dt

i= ((10u)*(5 V)) / 25 us = 2 A

No limitations on any VCC or VEE rails. Would like the output slew rates to sort of track the input so that the function generator can fine-adjust as needed.

I have some basic understanding of certain transistor topologies: common emitter, emitter followers, push-pull output stages, etc.

My initials thoughts are that after the function generator, I would needs lots of current gain. An emitter follower is the first thing that comes to mind. Or maybe use multiple stages to first boost up the gain... Current mirrors seem to make sense from an IC design point of view... not sure how good they would be when used as discretes / transistor arrays? I often see push-pull output stages used to drive low resistance loads... In this still a good idea w/ a capacitive load at low frequency? The impedance of 10 uF at 1 kHz (max frequency) is ~ 16 ohms.

Basically, I just want to brainstorm ideas, and learn the thought process from the more experienced experts on this forum. How would you tackle this problem?

Thanks!
 

Hi-Z

Joined Jul 31, 2011
158
On the face of it, this looks rasonably straightforward: an arrangement similar to a class-B audio power amplifier (but with unity voltage gain and dc-coupled) would seem appropriate. An op-amp driving push-pull complementary emitter-followers might seem to be the obvious thing.

However, emitter-followers (and source-followers) are notorious for oscillating when they're driving a capacitive load. Now, you may have heard of electrostatic loudspeakers; these look like capacitors to a power amplifier, and in the early days of amplifiers, this presented problems (not only with large currents, but frequency instability (oscillation) also). So much so that a company like Quad designed their amplifier specifically to drive their electrstatic speakers (which looked like 2uF), and warned against use of "unsuitable" amplifiers.

However, I'm digressing a little - these oscillations were associated with the overall closed-loop amplifier, not in the emitter-followers at the output. But I've successfully used pretty standard modern audio amplifiers with my Quad esl's with no problems, and this implies it's possible to drive capacitive loads with a fairly standard class (A)B amplifier (with emitter-followers at the output).

I think the trick is to swamp the capacitance with shunt resistance and to decouple the capacitive load from the amplifier. I think most amplifiers nowadays have a Zobel network at the output, which is a resistor and capacitor in series, placed across the output terminals. Common values might be 8 ohms and 1uF. Speakers are naturally decoupled in a way, because of the cabling. This could be emulated by winding a few turns of wire, using a power resistor (say, 8 ohms again) as the former (I'm sure I remember this as being fairly common practice in audio amps anyway).

So, the overall circuit? I'm going to avoid posting an image (as usual!), so you could base it on this:

http://www.allaboutcircuits.com/vol_6/chpt_6/10.html

but note that for unity gain, the feedback would be direct to the inverting input (no volume pot), and the non-inverting input would connect to the function generator (also, you should connect it to 0V via a 1k resistor for when the function generator is disconnected).

And don't forget the Zobel R-C across the output, and the home-made inductor in series with the load.

I hope the above proves to be oscillation-free - it's certainly the first thing to look for on initial power-up! (If it does oscillate, there are more complex circuits which could re-use these components, so come back to us.)

I've just thought: it might be a good idea to bias the amplifier's output stage to provide a bit of load current when there's no input, and you do this by connect the output to a pulldown resistor (say, 1k down to negative rail).
 

Hi-Z

Joined Jul 31, 2011
158
Regarding the power rails: you don't want these to be unnecessarily high, in order to avoid the possibility of second-breakdown (aka punch-through) in the output transistors. I think I'd limit them to about +10V/-5V, if that's convenient.

Also, the "few turns" inductor should be a couple of dozen turns of single-strand insulated copper wire on an 8 ohm, several watt resistor, with the resistor connected to the ends of the coil.

And on second thoughts, make the Zobel capacitor 2.2 uF...
 
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Thread Starter

newbie217

Joined Apr 12, 2009
52
Thanks Hi-Z!

This a rough first draft I put together to visualize what I am trying to do. For now, I just picked a random op amp and hooked it up in unity gain configuration. The output of the op amp is connected to 4x push-pull transistors which connect to the 10 uF load.

I just picked the default transistors, but from the simulation can see the power dissipation exceeds the 500 mW rating for 2N2222 and 400 mW rating for 2N2907. Any recommendations on what output transistors to use?

Also, this rough draft doesn't take into account for stability, oscillations. What are some techniques, methods used to address this? The waveforms are already exhibiting some forms of ringing... Would the next step be to look at the phase margin plots? Do I just connect an AC source to the input and remove the pulse (function generator)? Run AC Sweep and do Bode Plot for VOUT?

I also used GND for the VEE rail and notice that the output starts rising from 0.5 V. I am not getting full swing from 0 V to 5 V exactly.
 

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Hi-Z

Joined Jul 31, 2011
158
Just a few random points:

I re-read your opening post, and spotted the bit about "no ics". Well, I agree there's a lot to be said for avoiding this "luxury" if you're trying to improve your electronics design skills (and I fully approve of what you're trying to do here!). I'll have to think about how to point you in the right direction for a fully discrete solution...

You haven't mentioned anything about the negative-going edge - I'm assuming that the output will be expected to go from +5V to 0V in 25us. If this isn't the case, then the design might be simplified.

Regarding the circuit, the intention is to take the feedback from the overall output rather than the op-amp's output. Assuming supplies of +10V/-5V, then the op-amp will endeavour to make the overall output follow the input - which can require a bit of agility on the part of the op-amp. I see you have access to simulation, and this will be a very useful tool for you to get a feel for what happens at the op-amp output, as well as the overall output.

Note that the article I linked to has a diode in the transistors' base circuit. This is included to limit the op-amp's output excursion necessary when the the transistors switch over the conduction duties. The closer the transistors are to conducting both at once (i.e. the bases would be "separated" by 2*Vbe), the less excursion necessary at the op-amp output. This is useful, because if you look at a simulation which includes a switch in transistor conduction, you will see inevitable overshoot as the op-amp moves at its top slew-rate capability, in order to try to switch instantaneously.

It's for these reasons you would want an op-amp with fast slew-rate, and you'd likewise also want fast output transistors. As I say, the simulations will be very instructive here. Of course, the situation can be improved greatly by going to a class AB design, where both transistors do conduct at the same time, but this is a fairly undesirable complication, as it requires very careful bias arrangements.

I wouldn't worry about loop stability - the op-amp is compensated internally for well-behaved operation at unity gain, and the output transistors aren't contributing to the open-loop gain (they're emitter followers after all). When I talked about oscillation before, I was referring to the possibility of the emitter-follwers oscillating of their own accord. The resistor/capacitor/inductor I mentioned would be there to prevent this. (Note that I rather doubt that simulation would reveal this mode of oscillation - it's more a real-life phenomenon.)

Regarding output transistors, these will have to be in the form of darlington pairs, as the power devices required won't have enough current gain on their own. Have a look here to see what I mean:

http://www.rose-hulman.edu/class/ee/hudson/ece351/labs/ECE351Lab5.pdf

Note that you'll now need more diodes in the base circuit - for class B operation 3 would be appropriate. (Note that you won't need/want emitter resistors for the power devices for class B operation.)

Anyway, do have a bit of fun simulating the above, but make sure you include the output stage in the feedback loop, and do use a negative supply.
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
Much thanks again for that very informative post! I have a feeling this won't be such an easy task, and I will be making many mistakes in the process. But I'll keep persisting :)

Here is a revised schematic modified to better 'follow' the one in the audio for a Class B audio amp. The feedback is now taken from the overall output and fed back to the unity gain op amp.

I'm noticing that there is a lot of ripple on the output now, as the op amp's output swings.

I will need to study the details of the Darlington configuration more... Btw, is the 0.22 uF capacitor still needed for this design? And what is the 10k resistor for?

Also, you mentioned the need for fast output transistors in addition to a fast op-amp (high slew rate capability). Would the TIP 41 / Tip 42 pair work? They are rated for high current at 6.0 A. I'm not sure how to make of the abs max rating for power dissipation. The first line says 65 W and the second line says 2 W.... That's a pretty big difference there. Does this mean anything in regards to either a single application or for continuous, constant power dissipation?
 

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Thread Starter

newbie217

Joined Apr 12, 2009
52
Also, no requirement on the falling edge rate. If the negative edge can also go from 0 to 5 V at 25 us, that would be gravy ;)
 

Hi-Z

Joined Jul 31, 2011
158
Well, as long as neither edge can slew faster than this we should be OK (otherwise we risk excessive currents).

As fo rthe 65W/2W, the first figure is for the transistor case held at 25C - which would require a cool room and a large heatsink; the 2W is for the transistor without heatsink in a room which is at 25C. I shouldn't worry too much - your duty cycle is very low, so your average power dissipation won't be a problem with these supply rails (but the current capability of these transistors is handy).

Talking of power rails, you've still not got yourself a negative supply, which is why the output starts at over 1V.

The 0k resistor is to maintain a bit of current through the diode, which is just acting as a constant-voltage element (because there's always forward current). I'm not sure about the 0.22uF capacitor - I'll have to have a think. You may well be better off without it.

But it's that "ripple" that has me wondering... It looks like the circuit is oscillating (at about 10kHz) when the top transistor starts to conduct. If you could run a simulation without any load other than a 1k resistor (deleting the 10uF, 8R and 2.2uF), that should prove illuminating. (And don't forget a negative supply!)

I would imagine you'd get better simulation results with C2 deleted and a darlington configuration in place.

I'll be interested to see how you get on!
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
Here is the revised circuit adding the Darlington output, and removing the 0.22 uF cap. Also added the negative supply rail.

I tried the simulation w/o the capacitive load, and it simulates fine w/ R = 1k at the output. No oscillations. As soon as I introduce the 10 uF load back, I get the oscillations again.

Still need to play w/ it some more... and this is where my problem solving skills are starting to fall apart... not too familiar w/ how to analyze oscillations, stability. I've read about pole compensation, but don't really know how to attack a stability problem (rather than just inserting random capacitors anywhere and reading a phase margin plot) :confused:
 

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Hi-Z

Joined Jul 31, 2011
158
Firstly, just a couple of points about the circuit: with 4 diodes you run the risk of having both output transistors conducting at the same time, with potential for massive currents and thermal runaway. As a class B amplifier (as opposed to class AB), you want the transistors to be close to conduction, so 3 diodes would be probably about right.

C1 should be across all the diodes, but you could probably omit it anyway. Also, I would have a pulldown resistor at the output, taken to the negative rail; for a -5V negative rail you could try a value of 470 ohms. This will bias the circuit so that the upper transistors are conducting when in the quiescent state.

As for the oscillation, I think it's unlikely that the simulator is modelling the "emitter-follower self-oscillation" mode I mentioned earlier (but it might be). So, it's perhaps more likely that it's an overall feedback loop problem. Now, the 10uF load will look more like 10nF as "seen" by the op-amp's output (because the capacitance is reflected at the input of an emitter-follower in diminished form - by a factor of the combined current gains of the two transistors in the darlington configuration).

I think there are probably quite a few op-amps that can handle a 10nF load in unity-gain configuration, but I don't think the LT1678 is one of them (I came to that conclusion looking at its schematic and overshoot performance). Do you have any other op-amps to try?
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
Yea, I wasn't too sure about using the 4 diodes. Before inserting the output stage into the load circuit, I ran some simulations feeding a sine wave input (no feedback from output). I noticed distortion w/ only using 2 diodes, so I just doubled that number.

As far as op-amps, I am using LTSpice simulator, so I have access to a ton of Linear Tech op-amps.

I found this in a search, but even this model gave me oscillations. I'm running into the oscillations it seems regardless of what op-amp I am using.

http://www.linear.com/docs/Design%20Note/dn136.pdf

Which diode to remove if using three?
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
I tried the 1464 again, and can remove all the oscillation when driving up to about 30 nF at the output. Oscillations will start to show up for values larger than this.
 

Hi-Z

Joined Jul 31, 2011
158
You can remove any of the 3 diodes. As for the oscillation, do you get very different results for the various different op-amps? It could well be that the op-amp models aren't particularly accurate when it comes to loop stability (but I may be wrong here).

The LT1464 seems to have the right idea (for driving capacitive loads), but it's slew rate is fairly ordinary. It seems to me that it ought to be able to drive more than 30nF via the darlingtons - again, are all the op-amps like this?
 
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Thread Starter

newbie217

Joined Apr 12, 2009
52
Yes, the behavior is the same regardless of what op-amp I use. Basically, I will see oscillations for sure above 30 nF.

LT also has this high slew rate (250 V / us) op-amp that is unity-gain stable, large load capacitance driving. I tried this as well, but the simulation results were much worse (stable to 5 nF).

http://cds.linear.com/docs/Datasheet/1220fb.pdf
 

Hi-Z

Joined Jul 31, 2011
158
Well, it's always a problem when you're trying to get some assurances from a simulation, and you realise that the simulator may not be too trustworthy.

I don't know if you're in a position to easily try out the real thing - access to components, power supplies, 'scopes function generators etc? Sometimes it's better to try things out on the bench, though you have to appoach things a little cautiously if you're concerned about excessive currents, thermal runaway and oscillation.
 

Hi-Z

Joined Jul 31, 2011
158
Regarding the simulation, you might like to try just the op-amp in unity-gain configuration, and with various capacitive loads (this would give an indication of how accurately the op-amps are modelled, since some of them are guaranteed to be able to drive 10nF directly).

I've looked at the minimum current gain for the TIP42, and at 3A collector current this can go as low as 15. Similarly, for a 2N3906, this is specced at 30. This would give a worst-case darlington current gain of 450, which would imply the op-amp could see as much as about 20nF (but mostly it'll be less than this).

But the main implication here is that R2 and R3 need to be resuced in value - I would suggest 750 ohms for supllies of +10/-15V. The reason here is that the base current for the darlingtons needs to be provided via these resistors, and they will have only about 3.5V across them when the output is at maximum excursion (5V and 0V).

I hope you're finding all this instructive!
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
Yes, this is all very informative and much appreciated :) Also makes me realize how much I don't know! But then again, that was the whole purpose of doing this project. I'll give it another go after work today...
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
I simulated the follower w/o the output stage, and it can drive the 10 uF load. Simulation does not show oscillations w/ just the follower, regardless of load capacitance. I will try and tweak the resistor values to see if the results change.
 

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Hi-Z

Joined Jul 31, 2011
158
Well that's interesting. I wonder if the LT1678 works equally well with 10uF? The reason I ask is that it's possible that the op-amp models aren't exact, and I'd be suspicious if the LT1678 doesn't oscillate with that load.

Anyway, it does seem we have a problem - which isn't too surprising, I suppose, since capacitive loads, amplifiers and emitter-followers are pretty notorious when uttered in the same sentence!

I think it would be good to run a simulation of the overall circuit as follows:

Op-amp negative supply pin connected to -5V, not 0V.
No capacitors except the 10uF load (so no 8ohm/2.2uF).
R2, R2 at 750 ohms.
3 diodes (don't care how they're arranged).
With a 100 ohm pulldown to -5V at the output.

I'm not necessarily expecting good results, but at least this would give us an idea of what happens with a bare-bones circuit.

Reasons for oscillations can be tricky to pin down in real-life; it ought to be easier in the simulation environment, but bear in mind that device models may be flawed. As I hinted at earlier, it may be that we need a different approach to the problem, but as this is a somewhat academic educational exercise anyway I suspect setbacks are not too important.
 

Thread Starter

newbie217

Joined Apr 12, 2009
52
So, I tried simulating w/ smaller resistors, and the oscillations will disappear when R = 100 ohms. I tried increasing, but even at 200 ohms, there are oscillations. The supplies are set to 10 V and - 5 V. The 8 ohm + 2.2 u at the output doesn't appear to be influencing the results.

I'm curious as to how to set the bias currents for the base input. As you mentioned the input voltage swings from 0 V to 5 V. For 5 V at the op-amp output, this gives the voltage across the top resistor, VR3 = (VCC - (Op_Out + VD1 + VD2) = (10 - (0.7 + 0.7 + 5) = 3.6 V.

IR3 = 3.6 / 100 = 36 mA

For 200 ohms, IR3 = 3.6 / 200 = 18 mA

How do we know how much current is going to the Base and how much is going through the diodes? How much base current is actually needed? When I simulate the case for R = 200 ohms, everything is basically oscillating, so I'm confused how to ascertain any useful information.

Attached are the revised circuit and waveforms at 100 ohms and 200 ohms. Also, at 100 ohms, the average power dissipation across each resistor is about 440 mW. Does that mean a power resistor is needed?
 

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