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# [SOLVED]Why resistor is used @ feedback?

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#### 0by1

##### Junior Member level 2
In case of voltage follower, why resistor is used @ feedback??
Is it for current limiting???

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I think at voltage follower you no need to use resistance.
Basically, yes. In some some cases it may be recommended or even required. It's e.g. necessary for current-feedback (CFB) mode operational amplifiers.

Where did you see it?

0by1

### 0by1

Points: 2
Voltage follower is a positive feedback amplifier, the gain of 1.
Gain of non-inverting closed loop OP-AMP configuration can be approxmated as A = 1+(Rf/Rin) where Rf is the feedback resistor and Rin is the input resistor.
Since you need unity gain for a voltage follower, you just make Rin = infinite. Whatever be the value of Rf, you will get the second term zero, hence the gain would be 1.
As far as I understand, there is no role of external resistor current limiter since OP-AMP itself having very high input impedance by birth.
However, I would like to hear comments from experts...

karthickb3e

### karthickb3e

Points: 2
I saw this in some analog circuits... can you explain me little about current-feedback (CFB) mode operational amplifiers. As amplifiers have very high Rin, how feedback current is going to effect gain??

Basically, yes. In some some cases it may be recommended or even required. It's e.g. necessary for current-feedback (CFB) mode operational amplifiers.

Where did you see it?
Dear FVM
Thank you very much for your suggestion.
But i see that case in many buffer without resistance.
would you explain that how calculate R in "CFB" and amount of it's resistance

V
Points: 2
CFB is a special case, I just wanted to mention it, because resistors are needed there. You'll find recommendations for feedback R values in the CFB datasheets. CFB OPs have low impedance -ve inputs, so the resistor is effectively setting the loop gain.

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Points: 2

### jay1691

Points: 2
I saw this in some analog circuits... can you explain me little about current-feedback (CFB) mode operational amplifiers. As amplifiers have very high Rin, how feedback current is going to effect gain??

As long as the opamp can be considered as ideal, the feedback resistor is not necessary for unity gain applications because therre is absolutely no current flowing into the idealized input (input impedance infinity).
However, in real applications it can be beneficial to have such a resistor to compensate for a (rather small) voltage drop across the source resistance (not shown in the simple diagram) in front of the non-inv. input that sometimes cannot be neglected. This method is known as "offset current compensation".

Regarding the current-feedback amp (CFA): It has a low-resistance inverting input (current input) and the current flowing into this input is mirrorred and transferred into a voltage (internally). That means, the input circuitry of a CFA is completely different if compared with the conventional opamp. In this context, it is to be noted that the open loop "gain" of the CFA is given in "ohms" and, therefore is called "transfer impedance" (because the input current is transferred to an output voltage).

The information as given by FvM (regarding the feedback resistor) becomes clear if you compare the closed loop gain of both amplifier types. For comparing purposes the gain is written as a product consisting of (a) the idealized and frequency-independen gain factor Go and (b) a real frequency-dependent error factor E(jw) :

Closed-loop Gain G=Go*E(w) with Go=1+R2/R1 for both amplifiers.

* voltage opamp: 1/E(jw)=1+1/(Go/A(jw) with A(jw): opamp open loop gain
* CFA: 1/E(jw)=1+1/(R2/Z(jw) with Z(jw): CFA open loop transfer impedance.

Advantage of CFA: Z(jw) must not be frequency compensated because R2 always can be chosen - independent on the desired cloosed-loop gain - in such a way that no stability problems arise. For this purpose, the manufacturers specify a suitable (minimum) value for this feedback resistor.

---------- Post added at 14:34 ---------- Previous post was at 13:12 ----------

Remark 1: perhaps I should mention that the CFA calculation assumes that R1>>rin (rin: inv. input resistance, in reality app. 50 ohms or less).
Remark 2: Of course, one also can compare the loop gain of both opamp types with feedback:
*opamp: LG=-k*A(jw) with k=r1/(R1+R2)
*CFA: LG=-R2*Z(jw)

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### helpmejerry

Points: 2
V
Points: 2
Thank you FvM and LvW for your wonderful explanation

Dear LvW, from your answer I have come to conclusion that FB resistor in voltage follower is used for "offset current compensation".

In above figure to compensate offset current from/to +terminal and -terminal, I should have

Rin=~Rf

so that very small voltage drop will be there across both resistors, which cancels each to nullify "offset voltage" @ o/p.
am I right?

Endolith

V
Points: 2

### Endolith

Points: 2
I think you will sometimes see it with very high speed voltage feedback opamps where the feedback resistor interacts with the input capacitance and affects the frequency response.

Keith

Endolith

V
Points: 2

### Endolith

Points: 2
Offset compensation is another possible purpose of a feedback resistor with buffer amplifiers. But it's only reasonable for bipolar OPs without input current compensation. Many modern OPs have input current compensation and don't get any advantage from compensation resistors. Nevertherless are some designer still placing these resistors because the read it in a text book.

For low noise and/or high speed designs, an offset compensation resistor should be bypassed with a capacitor. Besides adding noise, the resistor creates a pole with the OP's input capacitance and reduces the feedback phase margin.

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V
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### Endolith

Points: 2
Hi Guys,

The Advantage of Resistor (Rf) in feedback path is as follows,

1) If input signal is Large scale (> 1V), Pulsating at HIGH frequency, and if NO Rf is used, you can see, Output voltage waveform is distorted during rising edge.

This is because, During fast feedthrough of output to Inverting terminal of Op-amp, in input protection diodes, short the output to input and short circuit protection will be enabled internally.

2) With Rf> 500 Ohm, output is capable of handling current requirement and smooth transition will occur.

3) If RF is larger (>2k), a pole will be created with Rf and Amplifiers input capacitance ( few pF). This creates additional phase shift and reduces Phase Margin(PM). But a small capacitor in parallel with Rf will eliminates this issue.

Conclusion: Rf used in voltage follower in application where, input is > 1V and pulsed at High frequency to get Smooth transition at output at input rise time and fall time.

Regards
Anand

Endolith

V
Points: 2

### Endolith

Points: 2
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