# Effects of negative feedback

1. Oct 28, 2005

### jonathansd

Hi there.
i need some help from you guys
can any of you tell me how to prove the effects of negative feedback mathematically?
i need to prove that negative feedback causes
-reduction in gain
-reduction of distortion
-increase of gain stability
-increase in bandwidth
increase in input resistance

Jon

2. Oct 31, 2005

### Staff: Mentor

I don't think some of those are correct, Jon. I'm not aware of a way that negative feedback can be used to increase Zin, for example. And negative feedback on its own isn't going to increase bandwidth.

Reduction in gain is easy, just compare the open loop gain and closed loop gain for an opamp circuit like a follower. Distortion -- depends on the circuit. Gain Stability -- not sure what that means, unless you mean like emitter degeneration. Just use that as the example, if you mean bias point stability instead of Gain Stability.

Where did you get the question from? They must have had something more specific in mind. What textbook are you using?

3. Nov 1, 2005

### Averagesupernova

Berkeman, I would say all of those are true. You question input impedance. Take a simple common emmiter configured transistor amplifier. Suppose it an emitter resistor of 1000 ohms with a capacitor in parallel that presents a low (let's call it zero) impedance to the signal being amplified on the emitter. So, knowing this the input impedance will be R'e + (beta * the emitter resistor). This gives us a Zin of R'e. Remove the capacitor and you are introducing negative feedback because of the emitter resistor. Do the math again and you will find that the Zin has gone up.

Increasing bandwidth? You bet. It is at the expense of lowering the gain, but it does. Suppose you have an amplifier running open loop and due to parasitic capacitance and etc. of the circuit it has a funny frequency response curve. If you introduce some of the output signal back onto the input (inverted of course) you will reduce the gain and assuming that the feedback path has a flat frequency response the output will try to follow input closer than before. Think of the output as being a scaled version of what is on the inverting input. Basic op-amp theory tells us that inverting and non-inverting inputs always try to track each other. By definition you cannot flatten the frequency response without also reducing distortion. Not necessarily ALL types of distortion. I think gain stability means stability of the gain over a temperature range.

4. Nov 2, 2005

### es

http://users.ece.gatech.edu/~mleach/ece3050/notes/feedback/fdbkamps.pdf
I only read the first ten pages but this paper should cover everything you want except maybe bandwidth calculation. But even if it doesn't the technique is the same but you do it in the frequency domain. (If you're not familiar with Laplace transform now is the perfect time to start)

I think jonathansd actually is referring to the "effective bandwidth" of a system. As I recall the effective bandwidth is defined as the position of the 3db point on the frequency axes.

5. Nov 2, 2005

### Staff: Mentor

Interesting points by ASN. But I don't think the increase of Zin by the emitter resistor can be considered due to negative feedback. It seems more like positive feedback, which can definitely be used to increase Zin (like in bootstrap applications). The emitter resistor in the CE amp configuration acts as negative feedback for the bias point, because it decreases the gain for an increase in base bias voltage. But I don't think it acts as negative feedback with respect to an input signal. You would use the collector voltage for negative feedback in the CE amp configuration, not the emitter voltage.

6. Nov 2, 2005

### SGT

In a feedback circuit you can sample the output voltage or the output current and can compare the feedback with the input signal in series or in parallel.
Comparing in series increases the input resistence, while comparing in parallel decreases it.
Sampling the ouput voltage decreases the output resistence, while sampling the output current increases it.
The example of the common emmiter amplifier, suggested by ASN is one of sampling voltage ang comparing in series, so we should expect an increasing of input resistence and a decrease in output resistence.

7. Nov 2, 2005

### Averagesupernova

The emitter resistor acts as negative feedback for the signal too. Especially if you are using it as a DC amplifier. I have seen in several text books that the emitter resistor is considered negative feedback. You talk about taking the signal off of the collector on a common emitter amp for feedback. If we eliminate the capacitor off of the emitter as I described in an earlier post, you can use the emitter as an output as well. Or for that matter, use both the collector and emitter for outputs. This is known as a phase splitter. You also talk about changing the bias point by changing the emitter resistor. You change the base current and emitter current, but NOT the voltage if you are using voltage divider bias. Take a look at an op-amp with bipolar transistor inputs. Negative feedback WILL increase the input impedance. A noninverting amplifier will have a lower input impedance when running in open loop. Done the experiment in school like 15 years ago.

SGT, I don't think the output Z of a common emitter is determined by the emitter resistor. The emitter resistor does have a play in voltage gain though.

8. Nov 3, 2005

### Staff: Mentor

Well, I could be wrong (it wouldn't be the first time), but I don't see how the emitter voltage can be considered *negative* feedback, when the voltage swing is in phase with the input voltage. Negative feedback would generally mean that the signal that is getting fed back to sum with the input is about 180 degrees out of phase with the input signal. That's why I would consider feeding back Vc negative feedback, and feeding back Ve to be positive feedback.

I checked my basic transistor text (Electronic Circuits by Holt), and I couldn't find any reference to negative feedback in the discussions about Re. I did check the feedback chapter, and found the statements that you were referring to, about how voltage sum feedback can increase Zin, and current sum feedback can decrease Zin. But the polarity of the voltage feedback in the examples is positive, so of course that can increase Zin. That's the bootstrap effect that I was referring to.

Taking your example of an inverting amplifier -- the input impedance would be Ri+Rf with no negative feedback (assuming a low input impedance for the stage that the output of the opamp is driving), and Ri with negative feedback. How does the negative feedback increase the input resistance? It sure looks like it lowers it using that example.

9. Nov 3, 2005

### Averagesupernova

Berkeman I was just thumbing through Malvino's Electronic Principles Third edition. Page 507 (chapter 16) describes what I was referring to about the emitter resistor. They have a common emitter NPN transistor amplifier. It is set up with voltage divider bias on the base, a collector resistor and 2 emitter resistors in series. The lower emitter resistor is bypassed with a capacitor. The upper one is the one considered the feedback resistor. Also known as a swamping resistor. They call this configuration non-inverting current feedback. BUT, the effect is negative feedback. They also show some other configurations of discreet feedback schemes. They show a resistor going from the collector back to the base and it is described as inverting voltage feedback. It would lower the input impedance. They show 2 other schemes but they both involve a second transistor stage. They are: Inverting current feedback (lower input impedance) and Non-inverting voltage feedback which a second inverting stage feeds the signal on its collector back to the first stage emitter. This configuration raises input impedance.

They also describe in another area of the book how negative feedback increases the bandwidth of a circuit. I thought I covered that well enough earlier.

Elaborate please? Not sure what Ri and Rf are.

10. Nov 4, 2005

### SGT

Remember that this is a series comparison. The input signal is from base to ground, but the effective signal is from base to emmiter, that is, the difference between the input signal and the feedback signal (the emmiter to ground voltage).
If you sample the collector signal and feed it back to the base through a resistor, you have parallel comparison: the input current and the feedback current sum up at the input node, so you must have opposite phases in order to have negative feedback.
The opposite is true for the bootstrat configuration you mentioned. Since you have parallel comparison and in phase currents, you have positive feedback.
In this case there is no effective difference. The output resistence is divided by $$1 + \beta$$, but since it is infinite, dividing it by a finite value makes it still infinite.

11. Nov 15, 2005

### cyeokpeng

Quick math proof:

1. Reduction in Gain
Closed-loop gain = Open-loop gain / (1 + Open-loop gain*B)
where B = feedback factor > 0
Can be derived from Transfer Function analysis
as you can se fro the relation, the denominator > 1, so Closed-loop gain
is effectively reduced by negative feedback!

2. Reduction in Linear Distortion
Consider the voltage transfer characteristics of both open-loop and
closed-loop amplifier(with B=0.01).
Assume open-loop voltage transfer characteristics to be piecewise linear.
Open-loop gain varies from 1000 to 100 to 0.
Plot the open-loop voltage transfer characteristics vo/vi on graph paper.
Then apply negative feedback B = 0.01.
Closed-loop gain 1 = 1000 / (1000 + 1000*0.01) = 90.9
Closed loop gain 2 = 100 / (100 + 100*0.01) = 50
Closed-loop gain varies from 90.9 to 50 to 0.
Plot the closed-loop voltage transfer characteristics vo/vi on graph paper.
From the plot, you can see that the slope=gain change is gradual for
closed-loop voltage TF compared to open-loop TF.
So, linear distortion is effectively reduced by negative feedback!

12. Nov 15, 2005

### cyeokpeng

Quick Math Proof:

3. Increase in gain stability / Gain Desensitivity
Consider closed-loop gain = open-loop gain / (1 + open-loop gain*B)
Af = A / (1 + AB)
Applying calculus approximation method,
d(Af) = dA/(1+AB)^2
Divide by Af,
d(Af)/Af = (dA/A) * (1/(1+AB) only accurate for small dA/A
As can be seen from above, a significantly large change in open-loop gain (dA/A) will cause little change in closed-loop gain (d(Af)/Af), due to the division of negative feedback densensitivity factor (1+AB) >> 1.
Thereby, the gain is stabilized by applying negative feedback!

4. Increase in BW
Consider high-frequency amplifier TF with dominant pole Wh.
A = Ao / (1+(s/Wh))
Applying the formula of negative feedback,
Af = A/(1+AB)
Af = (Ao/(1+AoB)) * (1/(1+(s/(Wh*(1+AoB)))))
From the above, we can see that
Aof = 1st term = Ao/(1+AoB)
Whf = last term = Wh*(1+AoB) -----(1)
From equation (1), we see that the high frequency 3-dB point is increased by a factor of (1+AoB) with negative feedback applied!

Uising the same transfer function analysis for low-frequency point, we arrive
Wlf = Wl / (1+AoB)
---effectively reduced by factor (1+AoB) with negative feedback applied!

So overall, BW = Whf - Wlf is effectively increased due to negative feedback!

13. Nov 15, 2005

### cyeokpeng

An additional property of negative feedback you never mentioned is NOISE REDUCTION performance.

5. Increase in input resistance
This is not necessary true!
It depends on the actual feedback topology you employed in the amplifier circuit.
Basically, you need to apply voltage feedback to input signal (Series mixing) so that the input resistance of the feedback amplifier is effectively increased by a factor of (1+AB)!
If you apply current feedback to input (Shunt mixing), the input resistance is effectively reduced by factor (1+AB)!

Hope you have a hapy time reading!