This is a follow-up question to (this question).

Here is the amplifier section in question:

enter image description here

In the answer, the author (@Brian Onn) gives the following explanation:

"Slow vs. fast is not about the gain of the circuit. The feedback is going to be "fast" when it has high enough bandwidth."

I thought I understood this at the time, but after investigation (and it's hard to find any direct references), I realize I don't. So, the comment warrants its own question.

Here's a brief summary of what I understand so far:

1) The input to the feedback amplifier (meter driver) is the output of the bridge. It'll be a phase altered AC signal of various amplitudes when the bridge is unbalanced and a signal closer to DC when balanced.

2) There are two feedback paths in the amplifier (one for AC and one for DC). Each path effects the gain of the amplifier differently. The design objective (as Brian points out) is to make the tracking "snappier" when the bridge is unbalanced and less so as the bridge approaches a balanced state.

3) Negative feedback reduces the gain and broadens the bandwidth. I understand this well enough.

But, what I don't understand is how a change on the input to the amplifier can result in a slower or faster response (from the output) depending on the bandwidth.

I see difficult mathematical references to step response and slew rate and my feeling is that this is where the answer lies. Can anyone provide a more "intuitive" explanation?

  • \$\begingroup\$ Draw a circuit to explain you thoughts. \$\endgroup\$ – Andy aka Jan 14 '16 at 11:41
  • \$\begingroup\$ I've included the circuit from the original question for convenience. If I understood the answer from the previous question correctly, increased responsiveness is a function of increased bandwidth in general. So, just about any negative feedback amplifier (that reduces gain and widens the bandwidth) would achieve the same outcome. \$\endgroup\$ – Buck8pe Jan 14 '16 at 12:06
  • \$\begingroup\$ I'm not understanding this part of your question: "But, what I don't understand is how a change on the input to the amplifier can result in a slower or faster response (from the output) depending on the bandwidth." \$\endgroup\$ – Andy aka Jan 14 '16 at 15:13
  • \$\begingroup\$ See the response by @FakeMoustache. If I can get an answer regarding "these high frequency signals are needed to change the output of B faster", the picture will be complete. \$\endgroup\$ – Buck8pe Jan 14 '16 at 15:18
  • \$\begingroup\$ I asked that based on that answer - I couldn't understand from your question or the answer what you are saying. In other words, how can a change produce a slower or faster response when bandwidth is fixed for a given amplifier? That's what you appear to be saying and that's the problem I'm having but I'm probably stuck in a loop now and need a hefty prod to read it any differently. \$\endgroup\$ – Andy aka Jan 14 '16 at 15:23

I don't understand is how a change on the input to the amplifier can result in a slower or faster response (from the output) depending on the bandwidth.

What you have to realize is that all bandwidth limited circuits (and that is basically all practical circuits) exhibit delayed response from input to output. A circuit with a very low bandwidth will take longer to respond than a circuit with a high bandwidth. Why ? Because the circuit with the low bandwidth has large time constants internally which delay the signal.

But there is more. Suppose we have two circuits, with the same gain etc, only their bandwidths are different:

circuit A has a bandwidth of 1 kHz

circuit B has a bandwidth of 1 MHz

Now we apply the same sinewave signal of 1 Hz (that is one Hertz) to their inputs.

What will we see at the outputs ?

Now we apply the same sinewave signal of 10 kHz to their inputs.

What will we see now at the outputs ?

Think about it before reading any further !

For the 1 Hz case the outputs of circuits A and B will be identical. Their bandwidths are much larger than 1 Hz so that signal is unaffected, not delayed etc.

But for the 100 kHz case the signal at the output of circuit A will be attenuated significantly but not so for circuit B. Circuit B has enough bandwidth, circuit A does not.

Now take it one step further, we apply a step function to the inputs of both circuits. A step function is a signal that changes it's value suddenly from one moment to the other. If you would perform a fourier transformation on a step function you would find that it is the combination of an infinite number of frequencies. So it contains all ! Including the 1 Hz, 100 kHz, 1 MHz, all of them !

Now what will we see at the outputs of circuits A and B ?

At the output of B we will see all frequencies up to 1 kHz, the rest is attenuated.

At the output of circuit B we will see all frequencies up to 1 MHz, the rest is attenuated.

So the signal from B contains more high-frequency signals, these high frequency signals are needed to change the output of B faster compared to the output of A (which is missing the signals between 1 kHz and 1 MHz when compared to output B). So output B will respond quicker than output A. For a signal to change quickly you need high-frequency components.

Any signal in the time-domain you can think of can be represented also in the frequency domain. Slowly changing signals consist of low frequencies. Quickly changing signals must contain high frequencies, the lower frequencies are optional. This time-domain vs frequency-domain representation of signals is done by fourier transformation. Most of us are familiar with time-domain representations of signals, like on an oscilloscope but you can also measure and represent a signal in the frequency domain using a spectrum analyzer.

Using an inverse fourier transformation you could reconstruct the signal in the time-domain at both ouputs. You would then find that the signal at output B responds earlier and changes faster to the step function at its input. Circuit A is slower so the signal at its output will start later and it will be less steep (less change over time).

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    \$\begingroup\$ "these high frequency signals are needed to change the output of B faster". This is absolutely a key phrase for me. Can you please expand a little more on this? \$\endgroup\$ – Buck8pe Jan 14 '16 at 14:14
  • \$\begingroup\$ In other words, what is it about high frequency signals arriving back at the input that hastens their departure from the output in the time domain? \$\endgroup\$ – Buck8pe Jan 14 '16 at 14:21
  • \$\begingroup\$ I do not understand your last comment, the high frequency signal do not travel back to the input. The signals only travel from input to output. Maybe you're confusing this with feedback. Feedback will not make the high frequency signals travel back to the input. They can't because the circuit attenuates these signals. \$\endgroup\$ – Bimpelrekkie Jan 14 '16 at 15:17
  • \$\begingroup\$ Probably you're confused because feedback can increase bandwidth. But I am not discussing this in my answer ! My answer has nothing to do with feedback. First fully understand my answer, then we can discuss feedback. \$\endgroup\$ – Bimpelrekkie Jan 14 '16 at 15:20
  • \$\begingroup\$ For understanding why feedback can increase bandwidth, read this: en.citizendium.org/wiki/Negative_feedback_amplifier and especially Bandwidth extension and look at the gain vs frequency plot on the right. Blue is no feedback, red is with feedback. The bandwidth is the point where the curve starts to drop. For the red curve this is at a higher frequency but the red curve has less gain at lower frequencies. There you have it: gain was exchanged for bandwidth ! But note how the red curve always remains under the blue curve ! \$\endgroup\$ – Bimpelrekkie Jan 14 '16 at 15:27

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