When we have an inverting amplifer circuit using opamps, we have resistor R1 which between the voltage input and the opamp, then we have resistor Rf which is between the input of the opamp and Vo. If we assume that V- of the opamp and V+ of the opamp are both zero, doesn't this mean that we should have a constant I1 current. For example if R1 is 100 ohm and the input voltage is 5V, we should have a constant current of 0.05A I1?
\$\begingroup\$
\$\endgroup\$
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
2
Yes - you are correct.
There is a constant current of I1=Vin/R1=50mA (for R1=100 Ohms and Vin=5V). This current goes through the feedback resistor Rf (because we are allowed to assume that no current enters the high-resistive opamp input).
Therefore Vout=-I1*Rf=-(Vin/R1)*Rf. This gives us the classical gain expression
Vout/Vin=-Rf/R1 .
-
\$\begingroup\$ So if I am understanding correctly, the opamp does no amplification but is only there to keep V- at 0, since the amplification would be due to the voltage drop across the resister. \$\endgroup\$– suyol854Oct 14, 2016 at 11:33
-
\$\begingroup\$ No - it is not that simple. The voltage drop across Rf is caused by the current that is driven by Vout. And the voltage Vout is caused by the amplifier and a very small voltage between the inverting opamp input and ground. However, this small voltage (µV range) is normally neglected during hand calculations. As another assumption we set the input resistance of the opamp to infinity - and because of Kirchhoffs current law the current through R1 is the same as through Rf. But the driving voltage for this current is Vout. Recommendation: Study negative feedback systems and all the effects. \$\endgroup\$– LvWOct 14, 2016 at 12:48