Author Topic: Feedback resistor  (Read 1217 times)

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Offline Sam Longman

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Feedback resistor
« on: April 24, 2017, 08:18:05 am »
I'm struggling to understand how the feedback resistor in a transistor amp works on the physical level (amplifying an AC signal). I've tried googling it but I've not found anything on it. It's probably a dumb question, which might be why other people don't ask it. Anyway, here goes:

1) Why does the feedback resistor not lead to continual feedback and reduction in the input signal to the point where the input is zero? ie if it reduces the input by 10% of the output, why does the new (lower) output not reduce the input again by another 10% of the (new) output and so on round and round until you have nothing left?  Is it because that 10% gets smaller and smaller each time the electricity goes round....

or is it because the input (before the feedback) is always at the original level, so really the feedback is reducing the signal a fraction of time later....however, once the output has reduced, surely it won't reduce the input by as much as before so the output would shoot up again?

(I must be thinking about this all wrong, but you can see how I'm going around in circles trying to understand it. Should I be thinking about the electron flow instead of conventional current but as it's AC I guess the electrons are going back and forth in both directions!!!)

2) would you not get a spike in voltage the very first time the electricity goes through the circuit before the feedback resistor has a chance to reduce the input? or is it because the signal always starts at zero for at least a fraction of time when the signal is first turned on so as the signal voltage climbs it is held down by the feedback from the signal a fraction of time just before the current time)?

3) does reducing the input voltage with the feedback (which I get you have to do or the gain would be too high) then lead to a lower signal to noise ratio? Why don't you attenuate the signal before it goes on the op amp with a voltage divider instead of using a feedback resistor? Is it because that would be top noisy? (I want to build audio equipment (as well as robotics) btw which is why I'm interested in noise... I have Douglas Self's book on Small Signal Audio Design. I'm just trying to learn some basics first).

Thanks,
Sam.
 

Online Rerouter

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Re: Feedback resistor
« Reply #1 on: April 24, 2017, 09:43:11 am »
I'm struggling to understand how the feedback resistor in a transistor amp works on the physical level (amplifying an AC signal). I've tried googling it but I've not found anything on it. It's probably a dumb question, which might be why other people don't ask it. Anyway, here goes:

1) Why does the feedback resistor not lead to continual feedback and reduction in the input signal to the point where the input is zero? ie if it reduces the input by 10% of the output, why does the new (lower) output not reduce the input again by another 10% of the (new) output and so on round and round until you have nothing left?  Is it because that 10% gets smaller and smaller each time the electricity goes round....

or is it because the input (before the feedback) is always at the original level, so really the feedback is reducing the signal a fraction of time later....however, once the output has reduced, surely it won't reduce the input by as much as before so the output would shoot up again?

(I must be thinking about this all wrong, but you can see how I'm going around in circles trying to understand it. Should I be thinking about the electron flow instead of conventional current but as it's AC I guess the electrons are going back and forth in both directions!!!)

2) would you not get a spike in voltage the very first time the electricity goes through the circuit before the feedback resistor has a chance to reduce the input? or is it because the signal always starts at zero for at least a fraction of time when the signal is first turned on so as the signal voltage climbs it is held down by the feedback from the signal a fraction of time just before the current time)?

3) does reducing the input voltage with the feedback (which I get you have to do or the gain would be too high) then lead to a lower signal to noise ratio? Why don't you attenuate the signal before it goes on the op amp with a voltage divider instead of using a feedback resistor? Is it because that would be top noisy? (I want to build audio equipment (as well as robotics) btw which is why I'm interested in noise... I have Douglas Self's book on Small Signal Audio Design. I'm just trying to learn some basics first).

Thanks,
Sam.

You havent given an exact toplogy, but i'll assume something close to this
https://www.electronicspoint.com/tutorialimages/Semi/03313.png

ok, working on that,

1. an NPN transistor amplifier inverts its output signal, so an increase in input amplitude results in a swing on the output that reaches closer to ground, this lower output voltage produces less feedback current to the npn base, and thus reduces how hard the transistor is turned on a little, making it reach a little less towards ground, this process keeps going continuously, tugging how hard the output is switched on away from the voltage rails, and limiting the signal gain (amplification)

the NPN has a very high gain, the feedback resistor just reduces it so your operating in a usable amount of gain,

2. When power is applied with no input signal, the resistances of the various parts of the circuit are limited in how fast they can charge due to tiny parasitic capacitors and inductors, meaning its not exactly immediate, but what you tend to see is the output changes from resting at 0V (before the coupling capacitor) and with power the transistor charges up to its bias point pretty quickly, making the output partly on, this is the thud you hear when you apply power on some amplifiers,

3. You in general want to amplify your signal before any noise is added, and regulate the gain or attenuate it back down after, as all noise is amplified with the signal, so starting with the lowest noise, strongest signal level is ideal.

With op amps, you might notice the divider is from the output, and connecting to the the alternate pin to the input signal, so any noise from the feedback network or output gets partially cancelled out by feedback action.
« Last Edit: April 24, 2017, 09:49:36 am by Rerouter »
 

Offline Sam Longman

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Re: Feedback resistor
« Reply #2 on: April 24, 2017, 10:33:34 am »
Thanks for your quick reply.

Yes, the circuit you've shown is the right one.

Question one I still need to do some thinking about. I'm guessing there's an element of the system being in balance. The output making the input lower, which in turn makes the output lower, which reduces the amount the feedback reduces the original signal, which in turn means the output tries to go up again, but again this pushes the input down again.

Thanks.
« Last Edit: April 24, 2017, 10:35:34 am by Sam Longman »
 

Online Rerouter

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Re: Feedback resistor
« Reply #3 on: April 24, 2017, 11:23:04 am »
To try and better explain it, lets begin with no feedback resistor, a tiny input signal slams and saturates against either rail, resulting in more like a square wave, this is because a small signal NPN likely has a current gain between 70-300 (Its "beta") this is clearly not ideal,

At stead state as the output will be resting somewhere close to center of the supply rails, and the input will be resting somewhere above where the tranistor begins conducting by the biasing resistors, placing it somewhere around lets say 0.5V on the base pin, and by adjusting this bias and adjusting how much its turned on, you can adjust where the output rests, generally centered between the 2 supply rails

The biasing resistors are feeding a tiny amount of current into the base of the transistor to turn it on, and as the output is at a higher voltage than the input, the feedback resistor is also summed into this, adding a very tiny amount of current, but at steady state you have already corrected this out.

When you have a signal coming in, you actually have a divider between the signal impedance, and the feedback resistor, the lower the value of that feedback resistor, the more influence it has over the input signal, as its inverted to the input signal this reduces the input amplitude, and reduces the effective gain of the amplifier.

Note this is more or less happening all at once, with some feedback the corrections ideally converge, but if you do something to add significantly more phase offset between the output of the amplifier (180 degrees) and its input, say bring it closer to 360 degrees, you can end up with an oscillator, where your feedback action applies in the wrong direction. this plays in to control loop stability and bode plots, handy tools, but depends on how far you want to dive down the rabbit hole for a transistor amplifier (you want a gain of less than 1 before non ideals add up to more than 180 degrees of additional phase shift, but for such a circuit, its likely in the MHz, not Khz)
 

Offline Sam Longman

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Re: Feedback resistor
« Reply #4 on: April 24, 2017, 11:30:01 am »
Excellent. I think I get it now. Thank you very much.
 

Offline danadak

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Re: Feedback resistor
« Reply #5 on: April 25, 2017, 12:43:57 pm »
You can build control loops that drive signals to 0, like a control
loop to null out the offset of DC amplifiers, like OpAmp offset
correction loops. Often referred to as servo circuits.

Or build them to force a constant peak amplitude, or RMS amplitude,
or even average amplitude. No end to the fascinating methodologies
for signal path control. And this does not discuss frequency shaping
possibilities. Or phase control, energy or power control......


Regards, Dana.
Love Cypress PSOC, ATTiny, Bit Slice, OpAmps, Oscilloscopes, and Analog Gurus like Pease, Miller, Widlar, Dobkin, obsessed with being an engineer
 

Offline David Hess

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Re: Feedback resistor
« Reply #6 on: April 25, 2017, 11:30:01 pm »
1) Why does the feedback resistor not lead to continual feedback and reduction in the input signal to the point where the input is zero?

For an ideal amplifier, it does.  Finite open loop gain prevents restoring the summing point to zero.

Quote
2) would you not get a spike in voltage the very first time the electricity goes through the circuit before the feedback resistor has a chance to reduce the input?

You do get a voltage spike before the amplifier can react.

Quote
3) does reducing the input voltage with the feedback (which I get you have to do or the gain would be too high) then lead to a lower signal to noise ratio? Why don't you attenuate the signal before it goes on the op amp with a voltage divider instead of using a feedback resistor?

Feedback lowers the gain, increases the linearity, and lowers the noise.  If the signal was attenuated and no feedback is used, (1) then the equivalent input noise would be amplified by the open loop gain.  If the signal is larger and feedback is used, then the signal *and* the noise is lowered by the feedback.

Some amplifiers do operate with no feedback but they have a limited input voltage range to keep linearity under control.  Operational transconductance amplifiers like the still produced LM13700 operate this way and are often used as electronic gain controls in audio applications.  If you add an external feedback resistor to them, then it defeats the purpose of using them.

(1) Current feedback operational amplifiers sort of operate like this.  A capacitor cannot be placed across their feedback resistor to limit bandwidth and noise unless a sacrifice is made to the god of clever circuits so in order to limit bandwidth and noise beyond the selection of the feedback resistor which also controls the frequency response, the signal has to instead be filtered at the input and output to control noise.
 


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