# Regarding positive feedback and negative feedback of OP-AMP

## My background knowledge

This picture is ideal OP-AMP:

where $$Vo=A(V_+ - V_-)...(0)$$

When I configure the non-inverting amplifier mode:

$$(V_i-V_o\frac{R_1}{R_f+R_1})A = V_o....(1)$$

$$V_iA=V_o+(\frac{R_1}{R_f+R_1})V_oA=V_o(1+A\frac{R_1}{R_f+R_1})....(2)$$

$$\frac{V_o}{V_i}=\frac{A}{1+A\frac{R_1}{R_1+R_f}}=\frac{1}{\frac{1}{A}+\frac{R_1}{R_1+R_f}}....(3)$$

when

$$A>>\infty$$ (ideal OP-AMP)

then

$$\frac{V_o}{V_i}=\frac{1}{\frac{R_1}{R_1+R_f}}=\frac{R_1+R_f}{R_1}=1+\frac{R_f}{R_1}$$

$$\frac{V_o}{V_i}$$

would converge to

$$1+\frac{R_f}{R_1}$$ (constant)

when I configure OP-AMP as negative feedback

So there is stable.

## My question

If I configure OP-AMP as positive feedback with same circuit and same OP-AMP model

To calculate Vo/Vi,

I just swap $$Vi$$ and $$V_o\frac{R_1}{R_f+R_1}$$ from the equation (1)

Because this equation

$$Vo=A(V_+ - V_-)...(0)$$

is the same whether I configure OP-AMP as positive feedback or negative feedback.

So:

$$(V_o\frac{R_1}{R_f+R_1}-V_i)A = V_o....(4)$$

$$V_iA=(\frac{R_1}{R_f+R_1})V_oA-V_o=V_o(A\frac{R_1}{R_f+R_1}-1)....(5)$$

$$\frac{V_o}{V_i}=\frac{A}{A\frac{R_1}{R_1+R_f}-1}=\frac{1}{\frac{R_1}{R_1+R_f}-\frac{1}{A}}....(6)$$

If

$$A>>\infty$$ (ideal OP-AMP)

then

$$\frac{V_o}{V_i}=1+\frac{R_f}{R_1}$$

Above equation is the same to the result of negative feedback.

-- The constant result

Should positive configuration be oscillation??

Why does I get the constant gain?

So when I configure OP-AMP as positive feedback,

it's stable and linear operation same to negative feedback?

• See this post electronics.stackexchange.com/questions/573990/… Commented Apr 27, 2023 at 16:05
• You should check the sign of input (V+ - V-) and the sign of output Vout. Commented Apr 27, 2023 at 16:11
• I've read here. I surprised A(open loop gain) of OP-AMP is positive! Commented Apr 27, 2023 at 16:47
• There is a "negative" sign before the "A". Commented Apr 27, 2023 at 16:51
• Thank you for your mention. But I still not understand why negative before A. I read here early, and here model is positive before A. Commented Apr 27, 2023 at 17:08

Curlywei - in you calculation, you have assumed that the differential voltage between both opamp input terminals approaches zero (for an ideal amplifier). Please note that such an assumption is allowed for negative feedback only.

Any real circuit with a positive DC feedback factor >1 will bring the circuit into saturation.

Let me add that from the math point of view you did make a sign error only (the result will be a negative gain value). This is a clear indication that something went wrong.

However, it is interesting to know that a DC anylysis of any circuit simulation program will arrive at the same result (stable bias point). We need a TRAN simulation (time domain) to reveal the instability.

• The problem of sign I discovered and I'll fix later. I'm surprised A(open loop gain) of OP-AMP is positive! Commented Apr 27, 2023 at 16:49

Your limit process as a mathematical operation is right in both cases. You assumed that for certain Vi there's a certain Vo. You calculated also the assumed Vo=GVi, where G is 1+Rf/R1. Absolutely right calculation. In both cases.

Unfortunately what you did does not present what happens in practical circuits. The Vo doesn't follow immediately the changes of the voltage difference of the inputs. In any practical circuit there's always some slowness. There's a little signal propagation delay, maybe well below a nanosecond. But Vo changes gradually, because some capacitors (intentionally inserted or parasitic) must get charged or discharged. To model the slowness you should at least imagine that the dependent voltage source is controlled through a RC lowpass filter. That's the simplest possible, but still practically useful slowness model.

The usual non-inverting amp (=the working one) finds the balance. The input voltage difference charges the RC filter towards the wanted voltage, no matter how big the gain A (which is assumed to be > 0) happens to be. Sooner or later the Vo settles to the right value which no more charges nor discharges the RC filter. The found Vo is equal with GVi accurately enough for all practical purposes even if A is finite, but high enough, maybe 10000.

Because the found Vo stays if Vi doesn't change and the balance is found often soon enough, we often skip the whole dynamic balance search process and write Vo=GVi. The balance searching process cannot be skipped if we want to know how our circuit performs with signals which have fast transients.

As already said by others, you can use time domain (=transient) circuit analysis programs to visualize the balance searching process and how the circuit performs with different input signals. The free to use Circuit Lab in this site is a perfect one for a start. To stay in truth any realism needs that internal voltages of the opamp model are limited to some practical operating voltage bounds. Otherwise the said RC-filter could be charged or discharged with no speed limit.

The amp with swapped inputs (=the positive feedback version) does the opposite. A slightest error in the output voltage generates a drive to the RC filter which charges or discharges it more off. Having some error generates more error. If you happened in some way succeed to force Vo=GVi, a slightest amount of noise - maybe only one electron - would collapse the balance and the error grows as big as the available operating voltage allows. Trying to put a pyramid to stand on its tip has the same instability problem. Theoretical balance exists as a possibility, but a slightest displacement kicks it totally off. The displacement grows until the physical limit is reached.

Of course, mathematicians have explored at least 100 years the control theory and found theoretical tools (=the stability theory) for deciding if a feedback circuit finds a balance. Learn it from textbooks.