The voltage output never gets that high. Look at it this way. You apply 1V at the input and the output voltage starts to rise (it has a finite rise time due to the op amp bandwidth). This voltage is reduced by the voltage divider R1 and R2. When the voltage at the junction of R1 and R2 equals 1V, the voltage at the op amp plus and minus terminals are equal and the output voltage stops rising and stays at this level. This occurs when Vin = Vout x R1/(R1+R2) which gives a gain of Vout/Vin = (R1+R2)/R1 (assuming very high op amp open loop gain).Thank you for the reply but I still confuse about the way input see different and amplify it.
This circuit for example.
Let's say if VR1=1/2Vout
When the inputs see a difference, say 1V and 0V, the amp massive gain (say 1000) amplifies the difference(1V).
Now Vout is 1000V, half of this was fed back to the inverting terminal. The inputs will be 1V and 500V, Vi=500-1=499V
Again Vout is 499x1000=499000V. The inputs will be 1V and 499000V.
Why Vout is getting bigger??
If the output voltage would be 1000V (it actually can't for a real OP, as mentioned) then the input voltage would be -499V. You mixed up the gain sign, turning the negative into positive feedback.Now Vout is 1000V, half of this was fed back to the inverting terminal. The inputs will be 1V and 500V, Vi=500-1=499V
Again Vout is 499x1000=499000V. The inputs will be 1V and 499000V.
Not for the circuit shown, which is a non-inverting amp. Both input and output have the same polarity.In addition:
If the output voltage would be 1000V (it actually can't for a real OP, as mentioned) then the input voltage would be -499V. You mixed up the gain sign, turning the negative into positive feedback.
You probably didn't look sharp. Vin is the differential OP input voltage, the large 500V term is the feedback generated part. It's surely opposite to Vout in a negative feedback circuit.Not for the circuit shown, which is a non-inverting amp. Both input and output have the same polarity.
Thank you for the reply but I still confuse about the way input see different and amplify it.
This circuit for example.
Let's say if VR1=1/2Vout
When the inputs see a difference, say 1V and 0V, the amp massive gain (say 1000) amplifies the difference(1V).
Now Vout is 1000V, half of this was fed back to the inverting terminal. The inputs will be 1V and 500V, Vi=500-1=499V
Again Vout is 499x1000=499000V. The inputs will be 1V and 499000V.
Why Vout is getting bigger??
You are the one that "didn't look sharp". :wink: The feed back is negative at the negative input, but the input and output voltages are the same polarity for a non-inverting op amp circuit. Otherwise it wouldn't be "non-inverting".You probably didn't look sharp. Vin is the differential OP input voltage, the large 500V term is the feedback generated part. It's surely opposite to Vout in a negative feedback circuit.
..correctVin is the differential OP input voltage
correctthe large 500V term is the feedback generated part.
..It's surely opposite to Vout in a negative feedback circuit.
Let's say if VR1=1/2Vout
When the inputs see a difference, say 1V and 0V, the amp massive gain (say 1000) amplifies the difference(1V).
Now Vout is 1000V, half of this was fed back to the inverting terminal. The inputs will be 1V and 500V ...
Not only. You are right, that it's absurd to calculate with this voltage levels. With negative feedback in a stable loop, the OP output is only changing continously and the feedback loop will settle to steady state without reaching extreme voltage levels.This is where the confusion comes in.
But you stopped quoting post #4 just before the actual expression, which turns negative into positive feedback. With it's action, the output voltage always latches into saturation.
Power Steering on your car works the same way.
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