I would like to discuss the following circuit described in the attachment below. Please do not post anything in relation to any other op amp typologies such as the differential, inverting ect… configurations… I want to only discuss the configuration shown in the attachment.
My question is, by applying 1Vdc at the inverting input, what rf resistor would one need in order to get -5.96 Vdc at the output. We don’t know the current nor the rf resistor.
I know rf’s value, since I built the circuit and measured vout. But what I don’t understand is how to calculate rf? :oops:
It would be very appreciated if one can show the calculations… :mrgreen:
But I am curious as to why this circuit behaves the way it does. You may try this circuit yourself and replace rf with values from 10 to 1K and the op amp will output a useful voltage range between aprox +6Vdc to -14 Vdc respectively
But I haven’t a clue on whats actually happening ?? :?
Bof, I can simply forget about it and stay ignorant, or I keep on trying until its clear!!! :roll:
The op amp is current limiting, trying to bring the (-) input to 0Vdc. Since the current limit is a function of the particular op amp you are using, the design does not produce predictable results.
That circuit actually does have a use, if the input to the op amp (-) terminal is a current instead of a fixed voltage. Then it becomes a current-to-voltage converter. This circuit is typically used to convert the output of a photodiode or phototube, which generates a current proportional to the light intensity.
That circuit actually does have a use, if the input to the op amp (-) terminal is a current instead of a fixed voltage. Then it becomes a current-to-voltage converter
Yes, that’s exactly what I learned yesterday
These op amps … is there anything they can’t do!!! :mrgreen:
The output depends not only on Rf but also on the impedance or series resistance on the 1 VDC voltage source. So the equivalent series resistance of the voltage source then becomes Rin which along with Rf determines the gain.
