No, and there's a few billion bits of experimental proof out there showing that this is entirely incorrect.
1 nanosec is the time of a full cycle of 1 gigahz oscillation.
For example, a JK latch specified at 20 MHz clock has a transition time - propagation delay - of 90 ns. I assume a fast amp features a propagation delay or transition time in the range of 5 microseconds
from time zero when a forward voltage is applied to a fast diode to the time when current flows through the diode is an interval of 50 nsec
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No hahfran that is not correct.
The delay around the forward loop and the feedback loop in a typical audio amplifier is nano seconds - in other words fast enough to ignore. So, feedback around the loop from + to - input is exceedingly fast. To help you visualize it, although the analogy is a bit tenuous, it's about 500 MHz in an amplifier with a total wiring loop of c. 600 mm (and that's being generous it's probably faster).
However, phase is a separate issue. It lags by a few microseconds. And the exact value is highly dependent upon the amplifier specifics. This means that when the input peaks on say a sine wave on the + input, the feedback to the - input is virtually instantaneous, but, it peaks much later.
How can this be? The electrical feedback is instant but the amplitude peak can be microseconds later. It has to do with the fact that capacitive elements in the forward amplifier path accumulate charge, resulting in delays.
Loop delay (nano seconds ) and phase are separate things - don't confuse them.
Note also that very fast amplifiers with nano second rise times and MHz closed loop band widths can have open loop -3 DB BW's of only a few Hz.
The delay around the forward loop and the feedback loop in a typical audio amplifier is nano seconds - in other words fast enough to ignore. So, feedback around the loop from + to - input is exceedingly fast. To help you visualize it, although the analogy is a bit tenuous, it's about 500 MHz in an amplifier with a total wiring loop of c. 600 mm (and that's being generous it's probably faster).
However, phase is a separate issue. It lags by a few microseconds. And the exact value is highly dependent upon the amplifier specifics. This means that when the input peaks on say a sine wave on the + input, the feedback to the - input is virtually instantaneous, but, it peaks much later.
How can this be? The electrical feedback is instant but the amplitude peak can be microseconds later. It has to do with the fact that capacitive elements in the forward amplifier path accumulate charge, resulting in delays.
Loop delay (nano seconds ) and phase are separate things - don't confuse them.
Note also that very fast amplifiers with nano second rise times and MHz closed loop band widths can have open loop -3 DB BW's of only a few Hz.
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OK, so you're not going to read a basic control theory text and work the problems. It's still my best suggestion, rather than arguing that objects tend to fly away from Earth rather than fall.
Really? I know some amps that have aan open loop cutoff of 10Hz - yes you read that right. The performance with feedback is stellar - pretty much perfect for practical purposes like, say, mixing consoles where they mix our music.
So how does that jive then?
Jan
Ok i'll give it another try. As said- from spec- it takes 50 nsec for onset current flow after forward bias is applied to a fast diode. We can without major error model a BJT as 2 diodes,
thus the propagation delay is 100 nsec. If we have ten BJTs in the signal path, we have 1 µsec propagation delay. We are still in time domain. I go back to the train example. The delay of acceleration is the same whether the locomotive starts from zero speed or from 100 mph speed. We only need to assume the same "energy storages" couple the waggons
I think the problem arises that you argue from frequency domain. I don't. From time domain is my point
thus the propagation delay is 100 nsec. If we have ten BJTs in the signal path, we have 1 µsec propagation delay. We are still in time domain. I go back to the train example. The delay of acceleration is the same whether the locomotive starts from zero speed or from 100 mph speed. We only need to assume the same "energy storages" couple the waggons
I think the problem arises that you argue from frequency domain. I don't. From time domain is my point
that is a different story. That it is different becomes obvious with fast transients then we are at the Low TIM thing. Again your viewpoint is frequency domain. In that domain one can make a slow open loop into a fast closed loop provided the open loop gain is very high. But feedback does not change propagation delay.
So you now agree that your statement 'you need at last 100kHz open loop' was incorrect?
Jan
All the transistors are already on running bias current so this turn on time isnt relavent. And show me how 2 diodes have current gain. And how do transistors in digital ccts switch in the giga hz if it takes them 100ns to turn on?
I'm not really sure where the train analogies are really helping your argument, but they are wrong. Accelerating from a dead stop is slower than a rolling start.
no acceleration = dv/dt = d2s/dt² . Only the change of velocity in time matters not its value. And that is just my argument.
Choosing a selective part of a larger picture to prove your point is nonsense. Where does that formula account for the depression of wheels sinking into tracks when stopped? What about backlash in the whole drivetrain and coupling mechanisms? This all effects acceleration.
Jan's question goes right back to Otala's "transient intermodulation distortion" issue. But the simple answer does not seem to me to be clear cut. As Jan says, if you take a standard amp (with we assume Miller compensation) you can at least in simulation see the differential input voltage rise with frequency. The problem this causes is that any increase in base-to-base voltage implies increasing non-linearity and therefore increasing distortion, whether or not the amplifier internally clips. Now this increase in distortion may be low until the b-2-b voltage gets high enough to cause significant non-linearity, which we might say would be 120mV which would cause the currents in the transistors to vary by 10:1.
Peter Baxandall attempted to answer Jan's question by measuring the "peak program". He proposed a simple C-R differentiating circuit which he used to monitor real time music signals. The peak voltage gives a measure of the transients.
So I think the question is "what distortion do you get at frequency f (in open loop)" for a given b-2-b voltage. This could be used to set the limit and if the input signal is band-width limited, falling by the same rate above this as the amplifier compensation falls, then the input stage should not overload.
As for transit times most simulations of most power amps I have run show a delay of 250nS, even for "fast" designs. The issue is complicated by the fact that frequency compensation components add to the loop delays, which without would have been much faster. I assume that the compensation components remain in place when considering "open loop".
I hope this clarifies the question a little: if the input voltage could cause the b -2-b voltage to exceed (define the level which is acceptable) then higher distortion will result, possibly with complete cut-off. Measuring the open loop gain should show the same thing as the simulations: clearly a lower input voltage would have to be used at low frequencies. My answer follows from the above: with Miller feedback you could also add global phase lead (compensation capacitor across the feedback resistor) which will act to reduce the differential voltage, and having determined the maximum acceptable differential voltage you should be able to set the input bandwidth and then the amp should never overload. (Unless there is an exceptional unusual input!)
Peter Baxandall attempted to answer Jan's question by measuring the "peak program". He proposed a simple C-R differentiating circuit which he used to monitor real time music signals. The peak voltage gives a measure of the transients.
So I think the question is "what distortion do you get at frequency f (in open loop)" for a given b-2-b voltage. This could be used to set the limit and if the input signal is band-width limited, falling by the same rate above this as the amplifier compensation falls, then the input stage should not overload.
As for transit times most simulations of most power amps I have run show a delay of 250nS, even for "fast" designs. The issue is complicated by the fact that frequency compensation components add to the loop delays, which without would have been much faster. I assume that the compensation components remain in place when considering "open loop".
I hope this clarifies the question a little: if the input voltage could cause the b -2-b voltage to exceed (define the level which is acceptable) then higher distortion will result, possibly with complete cut-off. Measuring the open loop gain should show the same thing as the simulations: clearly a lower input voltage would have to be used at low frequencies. My answer follows from the above: with Miller feedback you could also add global phase lead (compensation capacitor across the feedback resistor) which will act to reduce the differential voltage, and having determined the maximum acceptable differential voltage you should be able to set the input bandwidth and then the amp should never overload. (Unless there is an exceptional unusual input!)
Jan, I think things are not so simple. An exemple is : feedback cann't solve the crossover distortion problem. I am working on this subject for years. With poor efficiency... if you want ! Feedback cann't solve problem occuring faster than feedback can act. Think twice, and carefully... Feedback operates "a posteriori". Dazzling evidence ! I beg your pardon for my poor english and/or disastrous education
Jan, I think things are not so simple. An exemple is : feedback cann't solve the crossover distortion problem. I am working on this subject for years. With poor efficiency... if you want ! Feedback cann't solve problem occuring faster than feedback can act. Think twice, and carefully... Feedback operates "a posteriori". Dazzling evidence ! I beg your pardon for my poor english and/or disastrous education
Ooops after all.... yes feedback corrects errors always too late. It matters only what too late is. And this must be discussed in time domain, not in frequency domain, that is plain wrong. In frequency domain one assumes steady state signals -continuous- but real music makes its impression from transient states and there time matters. In a simulation, you will choose rectangle as signal.
When you do the sim with NFB and without, if the models are correct ( they are not but...) there should be no difference in delay time or propagation time. Cannot be.
Jan, I think things are not so simple. An exemple is : feedback cann't solve the crossover distortion problem. I am working on this subject for years. With poor efficiency... if you want ! Feedback cann't solve problem occuring faster than feedback can act. Think twice, and carefully... Feedback operates "a posteriori". Dazzling evidence ! I beg your pardon for my poor english and/or disastrous education
I don't agree. Feedback cannot solve the xover problem (for class B) because during xover, the loop gain is essentially non existing. It is the opposite of clipping, if you will - at clipping, feedback is inoperative because there is no loop gain anymore.
If you have a proper biased class AB the xover distortion is decreased by feedback, just as any 'error' is decreased by the difference between open loop and closed loop gain.
Furthermore, the harmonic components in xover distortion are pretty high order, and at those frequencies the loop gain is anyhow low, so that again means less correction than at lower frequencies.
Bottom line: the feedback principle still works as advertised.
Jan
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