OpAmp Slew

[tggzzz]

Quote from: T3sl4co1l on Today at 12:10:56 AM
    Yes, that's the right idea.  If the stuff around the op-amp is unavoidably slow, you have no choice but to slow down the amp itself to compensate.

And slowing down the amp won't compensate for slow components outside the loop.

Listen to T3sl4co1l, he's right. If the other components inside the op-amp's feedback loop are very slow then it will oscillate, unless you slow the op-amp down to compensate.

No.

You do have to control (= "slow down") the loop gain  so that there is adequate phase margin at all frequencies. But there should be little or no relationship between the loop gain and the opamp gain; if there is then the circuit's stability may be affected by manufacturing variations in the opamp.  In any sane circuit the loop gain is determined by passive components, not the opamp.

[T3sl4co1l]

Gain is a function of frequency.  Loop stability is a function of GBW -- which if it's wildly wrong, WILL cause problems -- but DC gain is generally designed out so its variation is avoided.

Or put another way, GBW is generally more consistent in manufacturing variation than DC gain, which is as it should be.

Tim

[LvW]

No.

You do have to control (= "slow down") the loop gain  so that there is adequate phase margin at all frequencies. But there should be little or no relationship between the loop gain and the opamp gain; if there is then the circuit's stability may be affected by manufacturing variations in the opamp.  In any sane circuit the loop gain is determined by passive components, not the opamp.

Two comments from my side:
1.) ("...phase margin at all frequencies..")
The phase margin is defined for one single frequency only - the frequency where the loop gain crosses the 0dB-line.

2.) (..."should be liittle...relationship between the loop gain and the opamp gain"...)
The loop gain is defined as opamp gain multiplied by the feedback factor - thus, there is a relationship which cannot be avoided. That means, the loop gain is NOT determined by passive componenets only. 

[tggzzz]

Everybody reading this thread is almost certainly living in a house containing feedback loops in which the "opamp" has much higher bandwidth and slew rate than the system being controlled. Read on for justification...

Can I suggest everybody finds out and understands understands how the delightfully named "bang-bang control systems" work.  In these the active gain element has a much higher bandwidth and must have a much higher slew rate (the original point) than the system being controlled.

Bang-bang control systems can be highly advantageous in high power systems, since the high power element is either fully on or fully off. They work extremely well and are widespread. For analysis and loop stability purposes the active gain element is modelled as a comparator, which is effectively identical to a high gain, high slew rate, open-loop opamp (i.e. without negative feedback around the opamp). This is not the slightest problem (either for analysis or for implementation), provided the loop gain is controlled.

So where are they in your house? An immersion heater and a refridgerator. The active element is the thermostat, which probably has a bandwidth of a few Hz - obviously much higher (10* 100*?) than the bandwidth of the heating/cooling element. And the thermostat's slew rate is not far off infinite - it often generates RF energy in the form of EMI!

I hope that dispells the false notion that the active gain element (e.g. an opamp) must have a bandwidth and slew rate that is "similar" to that of the control elements or the loop.

[tggzzz]

No.

You do have to control (= "slow down") the loop gain  so that there is adequate phase margin at all frequencies. But there should be little or no relationship between the loop gain and the opamp gain; if there is then the circuit's stability may be affected by manufacturing variations in the opamp.  In any sane circuit the loop gain is determined by passive components, not the opamp.

Two comments from my side:
1.) ("...phase margin at all frequencies..")
The phase margin is defined for one single frequency only - the frequency where the loop gain crosses the 0dB-line.

Agreed, provided the loop gain only crosses the 0dB line once My error.

2.) (..."should be liittle...relationship between the loop gain and the opamp gain"...)
The loop gain is defined as opamp gain multiplied by the feedback factor - thus, there is a relationship which cannot be avoided. That means, the loop gain is NOT determined by passive componenets only.

What is the "gain" of the bi-metallic switching element in a thermostat used in a bang-bang temperature-control  control system? Whatever it is, it is irrelevant provided it is high enough. Similarly the slew rate must be high enough. To make the example more electronic than electrical, simply replace the bi-metallic switch with a thermocouple and comparator (or opamp without negative feedback).

To the best of my knowledge fridges and hot water tanks have stable temperatures.

[LvW]

What is the "gain" of the bi-metallic switching element in a thermostat used in a bang-bang temperature-control  control system? Whatever it is, it is irrelevant provided it is high enough.

To my knowledge the term "gain" involves an output-to-input ratio which is defined for equal signal forms at the output and the input. With other words: The term "gain" applies to a linear system only - and, hence, not to a bang-bang controller.

[tggzzz]

What is the "gain" of the bi-metallic switching element in a thermostat used in a bang-bang temperature-control  control system? Whatever it is, it is irrelevant provided it is high enough.

To my knowledge the term "gain" involves an output-to-input ratio which is defined for equal signal forms at the output and the input.

Well, it is peripheral to the main point I've been making, but: no, not at all.

Easy example: a transconductance amplifier has an input voltage and an output current.  Imagine a pressure control system: the input would be pressure and the output (say) voltage, so the gain would be measured in V/Pa.

I put the word "gain" in quotes to indicate that it is not a simple definition. But then it doesn't need to be.

With other words: The term "gain" applies to a linear system only - and, hence, not to a bang-bang controller.

No. In practice the gain (and hence the stability analysis) would be based on the mean output, even though the output is never at the mean.

Think about how you would measure and/or analyse, for example, a class-D amplifier.

[David Hess]

Why is the gate resistor so big?  I would guess you don't even need one in this circuit.

Boltar says that is deliberate as a test.  The large gate resistance is adding so much lag when combined with the large capacitance of the pass transistor that even the relatively slow LT1013 is oscillating.  This may be corrected in the feedback network and like T3sl4co1l says, a more complex feedback network will work even better to correct the problem.

If the opamp bandwidth is too low then probably the phase will be such that the phase margin is eroded, often to the point at which oscillation occurs. In addition, the gain will be inaccurate since there will be insufficient feedback at the frequency of interest.

You have this backwards.  For a given amount of external phase lag before the gain drops to unity, a slower operational amplifier will be more stable because the external added lag is a smaller fraction of the operational amplifier's own lag.  You can compensate for excessive phase lag externally in the feedback loop or use a slower operational amplifier.  Operational amplifiers which support external compensation, external overcompensation, or programmable supply current can be useful in these situations because their GBW may be controlled directly.

[T3sl4co1l]

  Bang-bang controllers are inherently and intentionally unstable -- oscillating -- the original point was stable, continuous systems (like op-amps)!

Tim

[T3sl4co1l]

Think about how you would measure and/or analyse, for example, a class-D amplifier.

Also a good example, because average-control methods are standard, i.e., treating the clock frequency as a dominant pole and working below that.  e.g. http://seventransistorlabs.com/ClassD1/index.html

Tim

[DeweyOxberger]

Why is the gate resistor so big?  I would guess you don't even need one in this circuit.

1) Op amps generally don't like a capacitive load.  The FET's gate is capacitive. It will cause instability in the op amp behavior.
2) As the resistance goes to zero the output noise of the op amp feeds into the gate.  You'd like a low pass filter there to help reduce overall noise.  So some R to work against the gate C is usually nice.  The downside is the output voltage slew range of the op amp increases and you risk hitting your slew rate if the resistance gets too large (the op amp will bang rail to rail).

[LvW]

Easy example: a transconductance amplifier has an input voltage and an output current.  Imagine a pressure control system: the input would be pressure and the output (say) voltage, so the gain would be measured in V/Pa.

@Tggzzz: If you carefully read again my post#30 you will notice that I spoke about equal waveforms - not about equal quantities (voltages, currents, ...).
I suppose that an OTA gives a sinus output (current) for a sinus input (voltage), does it not?

[tggzzz]

Easy example: a transconductance amplifier has an input voltage and an output current.  Imagine a pressure control system: the input would be pressure and the output (say) voltage, so the gain would be measured in V/Pa.

@Tggzzz: If you carefully read again my post#30 you will notice that I spoke about equal waveforms - not about equal quantities (voltages, currents, ...).
I suppose that an OTA gives a sinus output (current) for a sinus input (voltage), does it not?

Actually you didn't, you used the phrase "equal signal forms", which doesn't mean much to me. Your clarification makes sense.

Of course an OTA (T=transconductance or transimpedance ) has a linear relationship between input and output, at least until clipping occurs. At which point the input-vs-output transfer curve looks similar to the output of a high-gain comparator or open-loop opamp

[Zero999]

Why is the gate resistor so big?  I would guess you don't even need one in this circuit.

1) Op amps generally don't like a capacitive load.  The FET's gate is capacitive. It will cause instability in the op amp behavior.
2) As the resistance goes to zero the output noise of the op amp feeds into the gate.  You'd like a low pass filter there to help reduce overall noise.  So some R to work against the gate C is usually nice.  The downside is the output voltage slew range of the op amp increases and you risk hitting your slew rate if the resistance gets too large (the op amp will bang rail to rail).

1) True but don't forget that there's a resistor in series with the source terminal which will reduce the gate capacitance seen by the op-amp.

2) I thought that op-amps were generally more noisy at lower frequencies so a low pass filter won't have much effect there.

[David Hess]

2) I thought that op-amps were generally more noisy at lower frequencies so a low pass filter won't have much effect there.

Flicker noise contributes more noise for a given bandwidth and increasing noise at lower frequencies but broadband noise has a lot more bandwidth so it dominates in all but low frequency applications.  A discussion about how to filter out flicker noise would be fascinating but beyond the scope of the discussion here.

[DeweyOxberger]

1) True but don't forget that there's a resistor in series with the source terminal which will reduce the gate capacitance seen by the op-amp.

2) I thought that op-amps were generally more noisy at lower frequencies so a low pass filter won't have much effect there.

1) The output has a finite effective resistance for sure.  It might not be implemented as a series resistance on the output.  It's usually an artifact of the design.  That resistance may not help with stability under a cap load (the devil is in the details).  For many op amps you will need some series resistance on the output.

2) There a many sources of noise.  Don't forget the noise from the op amp power supply (CMRR). The FET has a high gain.  A little R can help.

[atferrari]

Hola Boltar, are you still there? 

He's gone, I think.

[Simon]

Hola Boltar, are you still there? 

He's gone, I think.

I think it got too technical and "anal" about fine detail well out of his understanding. I guess some people didn't notice this was posted under beginners, no one asked for a 400 page technical write up. People generally have to learn in small steps (I do) or it's too much to get ones head around at once.

[tggzzz]

People generally have to learn in small steps (I do) or it's too much to get ones head around at once.

It is an interesting dilemma. Personally, while that's true for me, I also like to lurk on topics I only quarter understand - that way I can begin to map out the areas I don't yet know.

But in this case I'm sure you are right.

[T3sl4co1l]

One of the best professors I ever had was for Statistical Mechanics.  A tough physics course to begin with, but he added so much more, from some history in calculus to special functions and quantum mechanics.  Testing was on a very loose curve -- you were expected to retain a little bit of everything, and due to the sheer quantity, that was enough.

Tim