Designing a high current (10A) voltage buffer

Discussion in 'Electronic Design' started by Michael, Jul 28, 2007.

1. MichaelGuest

Hey there - so my job doesn't allow me to do much in the way of analog
design, so for fun last night I decided to design a high current
voltage buffer. The idea is that it'd take a 0-10V input, and would
output the same voltage, except with the ability to drive up to about
10A. The design eventually evolved in my head into a PID controller
using op amps driving a P FET. So I scribbled down some notes and
passed out. This morning I drew it out, corrected a couple oversights,
and chose some components. I have posted what I came up with here:

The op-amp choice was very arbitrary - I just wanted a rail to rail
quad op-amp, and the OPA4234 was already in my schematic capture
program's library. I chose the FET a slight bit more carefully - I
wanted a low on resistance FET capable of dissipating a lot of power
while also being able to handle a lot of current. I ended up with the
IRF IRLR9343. It can't completely handle the specs I gave - but it can
get fairly close.

So - in this design R2, R3, R4, and R5 set the PID gains, with P = R2/
R3, I = R2/R4, and D = R2/R5. I put in fairly arbitrary values for R2-
R5 - giving initial gains of 0.5 for all three. This would obviously
need tweaking based on the application. Also, the resistor between the
op-amp and the gate of the FET was a pretty arbitrary value - I put it
in there to protect the op-amp, though it'll also significantly slow
down the circuit.

So - can anybody spot any mistakes? I suspect there are at least a
couple. Also - is there any way to do this on a unipolar supply? Any
way to get rid of an op-amp? My only thought on getting rid of an op-
amp was to make the summer into a combination summer/difference
amplifier so that it'd sum the I and D terms while subtracting the -P
term, but I couldn't find a way to make the math work out cleanly. Any
other comments or suggestions?

Thanks!

-Michael

2. John LarkinGuest

You won't need the derivative term, and you can combine the integral
and proportional into one opamp by using a series r+c as the feedback
element.

Make the gate resistor 33 ohms maybe; it keeps the fet from doing RF
oscillations on its own.

John

3. linnixGuest

Is your supply capable of 10A? Driving the gate of the FET wouldn't
gain anything. The op amp is doing nothing at all. You can't boost
current without L & C.

4. John PopelishGuest

Yes. The gate looks capacitive, so, if the MOSFET is
following only slowly changing voltages and the circuit is
stable, the gate current will always be small. And most
opamps have enough current limit in their output stages to
protect the opamp for quite a while. The only real need for
a gate resistor might involve preventing high frequency (RF)
oscillations in the MOSFET, by killing the Q. But something
in the 10 to 100 ohms range usually works better for that.
A ferrite bead over the source lead is also sometimes used,
but it will probably saturate with 10 amps DC in this
application.
I haven't solved any equations, but it looks functional, if
over complicated. It should be possible to merge all that
gain network (5 opamps and all their resistor and capacitor
input and feedback capacitors) into a single opamp.
Sure, if the input stays between those rails, and there is
enough voltage to drive the MOSFET. A single +12 rail
should do it.
Draw a single opamp follower. Add a MOSFET current booster
to it, an inverting booster, in this case. This extra
inversion will require that you reverse the inputs on the
opamp. Build it, and you will find that it is not stable,
because the internal gain versus phase of the opamp that
made the opamp follower stable did not take into account the
additional gain and phase shift of the MOSFET inverter
inside the feedback loop.

You have to add some lead lag compensation to the feedback
(and possibly to the input side) that approximates what you
are doing with the PID construction, that lets you control
the gain roll off and phase shift so that, at the frequency
where the loop gain falls to 1, the phase shift does not
approach or exceed 180 degrees, so that stable, negative
feedback operation takes place. It can be done the one opamp.

I don't want to hand you a finished answer, right away,
because I think you are having too much fun thinking about
this, right now.

1. The design looks owerweighted. The whole PID regulator can be done in one
opamp unless you need to adjust the parameters independently.

2. You have Q1 in common source mode. Thus the transconductance of Q1 is
included into the loop gain. I don't like this idea because it is a variable
parameter. I would use rather use bipolar Darlington as the emitter
follower.

DSP and Mixed Signal Consultant
www.abvolt.com

6. MooseFETGuest

The Power Supply Rejection Ratio will really suck. You only need one
op-amp to do the whole thing. The differential section won't work.
You don't need a -12V supply. You didn't do anything to control the
gm of the MOSFET section. Your design suggests a 10uF capacitor that
I suspect you will have trouble with. 25nA times 100K puts the offset
voltage at 2.5V. You are putting an heavier load on Vin than you need
to. Your feedback path runs through 3 350KHz op-amps forcing you to
use a low gain crossover frequency. You didn't provide for an anti-
windup circuit. The output pulses high at power up. The resistor
feeding the gate of the MOSFET is too high. You didn't show any
bypass capacitors.

Other than this, its a great design.

7. MooseFETGuest

Its easier to just add a small source resistor to the MOSFET. You
would still get nearly the full swing but have a gm that is well
enough controlled.

8. MichaelGuest

What is Q? I'm not familiar with that term.
I had that suspicion - I always tend to use way more op-amps than are
really needed.
But the P term will initially dip below zero.
An op-amp follower - as in a buffer? (ie
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampvar2.html)

An inverting MOSFET current booster is what I drew, right? Or are you
referencing something else?

I don't follow what the feedback system would be in this case. Can you
elaborate on that?
Ah - lead/lag design - fun stuff. How do you implement a lead or lag
controller with an op-amp? I've only done lead lag design with DSP
systems - never analog.
By the way John - I'm not sure if you realized this or not - but this
is Michael - the robot guy. Good to see you!

Thanks,

-Michael

9. MichaelGuest

Why skip the D? I mean, it doesn't hurt anything - when properly
tuned, it only helps...

On another note - something's been bothering me - would it be possible
to make just a proportional controller? I feel like if I set the I and
D gain terms to zero this thing would oscillate terribly, but looking
back through my control systems textbook P controllers are shown to be
stable, though they can have a steady state error.

So - where'd you get 33 from? That was one of those values that I just
didn't have a clue about...

Thanks!

-Michael

10. MichaelGuest

In my theoretical design - yes, the supply is infinite. I don't follow
what you're saying about the op-amp not doing anything?

11. MichaelGuest

In my theoretical design - yes, the supply is infinite. I don't follow
what you're saying about the op-amp not doing anything?

12. John PopelishGuest

Quality factor of a resonant system. High Q implies sharper
resonance and large stored energy, compared to losses per
cycle. The gain and feedback capacitance from drain to gate
can produce a resonant system that can self excite with some
gate driver source impedances (oscillate).

http://en.wikipedia.org/wiki/Q-factor

(snip)
No problem, if it helps you deal with all aspects of the
problem, conceptually. But then you get down to the
engineering aspect of merging all those features into a
reasonable circuit.
Under what situation?

(snip)
Yes. The simplest opamp amplifier. ~infinite current gain,
voltage gain = 1.
Your MOSFET amplifier is exactly what I am talking about.
The simplest idea is to switch the opamp inputs and take the
voltage feedback from the drain, instead of from the opamp
output. But that will almost certainly turn out to be unstable
Lead/lag refers to the phase shift of a given network. You
also get a frequency response that comes along with it.
I did not realize. Nice to be thinking with you again.

The final feedback must, at DC have a gain of 1 from the
MOSFET output back to the effective inverting input, just as
it does with the simple 1 opamp follower (which has that
gain at all frequencies, not only at DC).

But the feedback network (that can include signals from both
the opamp output and the MOSFET output) can have other gains
and phase shifts at other frequencies.

A very simple, though, perhaps sub optimal (not the highest
frequency response) version would be a resistor from MOSFET
output back to + and a capacitor from the opamp output back
to +. This pair has a perfect gain of 1 at DC from the
MOSFET output (since the capacitor has infinite impedance at
DC). But above some frequency, where the capacitor
impedance approaches the resistor resistance, the capacitor
takes over as the effective feedback, and the MOSFET output
no longer follows the input, but the opamp is kept in a
stable feedback situation.

There are many variations on this theme that also include
components connected to ground to make the feedback network
some sort of voltage divider, and series RC loads on the
output to narrow the range of gains and phase shifts
possible as the load impedance changes.

If the actual load is fixed, things get a lot simpler. If
the load can vary anywhere between infinity and 1 ohm, you
have a lot more cases to deal with. This is one reason why
a big follower booster is often used (emitter or source
follower configuration). That configuration varies in its
gain and phase shift a lot less as the load impedance varies.

13. MichaelGuest

I was assuming that the op-amp and the fet were on separate 12V
supplies, if that is what you were referring to. Sorry that that
I found one PID controller implemented on a single op-amp, but it
didn't give a nice way of tuning - if you changed any component you'd
change all three gain terms. Can you suggest a better method that
allows individual tuning of gains?
I think I do with how the circuit is currently set up. The P section
will be generating a negative voltage. Unless I completely change the
circuit - I believe it is necessary.
How would you suggest changing the circuit so that it is controlled?
What's so hard about sourcing a 10µF cap? I don't follow. I didn't
want the resistor to be any larger (being that the input impedance of
the op-amp is in the Mega ohm range) - hence the large cap. I suppose
I could compensate for that when setting the gain - but that's a
little ugly.
Can you elaborate on this? I don't know what you're referring to.
I agree completely. I'll change R6 through R9 to 100Ks.
Like I said - the op-amp choice was arbitrary. I am really quite
confused when it comes to choosing op-amps - when is fast too fast? I
always hear warnings that really fast op-amps can cause really bad
oscillations.
I am not familiar with this term. Can you elaborate on it?
I'm aware of that one. I don't agree that it is absolutely a bad thing
- I mean in some applications the output being low on startup could
cause failures. How would one change the circuit so that you could
control the initial Vout?
Another arbitrary value choice. How does one choose this value? I'm
thinking that you would look at the gate capacitance and then choose a
resistor such that the time constant (RC) had a period of half of the
frequency you want the circuit to run at. You would also have to
verify that that won't be asking the op-amp to drive too much current.
I guess I always feel those are just assumed
Hey now, I am stuck just doing embedded design at work - I'm rusty on
all this analog stuff

Thanks, I really appreciate it.

-Michael

14. MichaelGuest

I was assuming that the op-amp and the fet were on separate 12V
supplies, if that is what you were referring to. Sorry that that
I found one PID controller implemented on a single op-amp, but it
didn't give a nice way of tuning - if you changed any component you'd
change all three gain terms. Can you suggest a better method that
allows individual tuning of gains?
I think I do with how the circuit is currently set up. The P section
will be generating a negative voltage. Unless I completely change the
circuit - I believe it is necessary.
How would you suggest changing the circuit so that it is controlled?
What's so hard about sourcing a 10µF cap? I don't follow. I didn't
want the resistor to be any larger (being that the input impedance of
the op-amp is in the Mega ohm range) - hence the large cap. I suppose
I could compensate for that when setting the gain - but that's a
little ugly.
Can you elaborate on this? I don't know what you're referring to.
I agree completely. I'll change R6 through R9 to 100Ks.
Like I said - the op-amp choice was arbitrary. I am really quite
confused when it comes to choosing op-amps - when is fast too fast? I
always hear warnings that really fast op-amps can cause really bad
oscillations.
I am not familiar with this term. Can you elaborate on it?
I'm aware of that one. I don't agree that it is absolutely a bad thing
- I mean in some applications the output being low on startup could
cause failures. How would one change the circuit so that you could
control the initial Vout?
Another arbitrary value choice. How does one choose this value? I'm
thinking that you would look at the gate capacitance and then choose a
resistor such that the time constant (RC) had a period of half of the
frequency you want the circuit to run at. You would also have to
verify that that won't be asking the op-amp to drive too much current.
I guess I always feel those are just assumed
Hey now, I am stuck just doing embedded design at work - I'm rusty on
all this analog stuff

Thanks, I really appreciate it.

-Michael

I have to disagree. The voltage gain of this stage should be more or less
constant regardless of load. That can be achieved by local voltage feedback,
not by current feedback. Something like R or RC between drain and gate will
probably do.
Actually, it is fun to design RR output stages because you have to optimize
several different things.

DSP and Mixed Signal Consultant
www.abvolt.com

16. John LarkinGuest

It can be a huge noise amplifier, and gets quirky in a lot of
situations.
Sure. But the opamp isn't ideal, and its gain will start to droop at
some frequency, so it's never going to be pure proportional. And a lot
depends on the load... if it's a pure resistor as shown, good, but any
load capacitance complicates life, and the fet itself has capacitance.
So the proportional gain may have to be very low to keep the loop
stable, so then you's need some integral to get accuracy.

This ain't simple. Keep in mind that the transcondance of the fet will
vary with current, so the loop dymamics aren't fixed. The safe thing
to do is to make a *slow* P+I controller, slow as in low KHz loop
bandwidth, if you can tolerate that.

Seems to keep most mosfets from becoming RF oscillators. Bigger values
work too, but may slow down the fet gain and complicate the loop even
more.

John

17. John LarkinGuest

I didn't follow that either!

John

18. D from BCGuest

Maybe it's just me..but don't P Fets suck?..
I'm no pro but on a few occasions have compared N Fet specs and prices
and P Fet specs and prices.. I currently have the belief that N power
Fets generally outperform P power FETs.

Hope you can keep that P Fet junction temperature within safe limits
with 10A max.

D from BC

19. John PopelishGuest

I think N fets make better use of the silicon. In other
words, for a given chip size, you get a higher performance
fet if it is N-channel than if it is P-channel. Higher
performance is defined as
1/(gate capacitance * channel resistance) or some such.

20. MooseFETGuest

This is because holes are so hard to move. Electrons are nice round
little balls but holes have sharp corners all over the place.

The very nice thing about a P-MOSFET is the fact that you can switch
the positive rail with a signal less than that. They make very nice
"you hooked the battery up wrong" diodes and power switches for
circuit sections as a result.

10A * 12V = 120W

If you buy a good one in a TO-218, you can do this with boiling water
cooling. No L-N2 is needed.

If you add a small balasting resistance, you can parallel about 10
MOSFETs to get a more reasonable power in each. I suggest the
balasting resistor because at low currents, the Tc is in the "this
thing runs away" direction.