# Electrical – Why is the signal always entered into minus port in inverting op-amp

inverting-amplifieroperational-amplifier

I think output voltage value is same either if I connect input signal into + port or if I connect input signal into – port. The reason why I am thinking like this is come from

$$V_{out} = -\frac{R_f}{R_{in}} V_-.$$

I understand the right figure's output is surely -10V because
$$V_{out} = -\frac{R_f}{R_{in}} V_{in} = -\frac{10000}{1000}{1V} = -10V$$

However, the left figure's is….. I don't know.

Why does all textbook or other documents use the right circuit instead of left one? $$V_{OUT} = A(V_{IN+} - V_{IN-})$$

Where \$A\$ is the open loop gain, \$V_{IN+}\$ is the non-inverting input voltage and \$V_{IN-}\$ is the inverting input voltage. This is the way opamps are designed and A (the open loop gain) is a large number, typically in the order of 10\$^5\$.

From this we can see that if the '+' is 1mV higher than the '-' input the output will head towards 100 V but get clamped at the positive supply rail. Similarly if the '-' input is 1 mV higher than the '+' input the output will head for -100 V but get clamped by the negative supply rail. Figure 1. Inverting configuration.

If we apply 1 V at (A) to the inverting configuration the '-' input (point C) will start to rise. We can calculate from our formula that this will drive the output (B) negative. If the output overshoots point C will go negative and the output (B) will go positive. The system will settle down when (C) is very close to the non-inverting input, 0 V.

At this stage we have a voltage divider created by R3 and R4. (A) is at 1 V, (C) is at 0 V so (B) must be at -10 V. We have an amplifier with a gain of -10. i.e., It's inverting. Figure 2. Inputs swapped.

In the case of Figure 2 we apply 1 V at (A). This starts to pull the voltage at (B) positive. We can see from the gain equation that the output will rise too. The feedback resistor R2 gives positive feedback which pulls the '+' input higher still. If the op-amp is powered from ±12 V then the output will probably rise to about 10 V. With R2 and R1 providing an 11:1 divider the voltage on the '+' input will be 1/11th of the way between 1 and 10 V (about 1.8 V). \$V_B = (V_C - V_A) \frac {R1}{R1 + R2} + V_1 \$.

If we start to decrease the voltage at (A) nothing happens the output for some time. When (A) is 0 V (B) will be \$\frac {10}{11} \$ keeping the output at +10 V.

It's only when \$V_A\$ gets to -1 V that the output will start to switch negative.

This arrangement is known as a Schmitt trigger and is very useful for giving a clean switching signal for a noisy input, for example.