Digital Electronics - Interfacing to the Analog World - Discussion
Discussion Forum : Interfacing to the Analog World - General Questions (Q.No. 7)
7.
What is the current in the feedback resistor for the circuit given below?
Discussion:
2 comments Page 1 of 1.
Ravi Bhatti said:
1 decade ago
Its simply a OPAMP adder
Here we can use equation of this circuit to calculate Vout as
vout=-[Rf/R1(V1)+Rf/R2(V2)]
=-[16/8(-5)+16/4(-5)]
=-[-10-20]=30v.
Hence we can calculate feedback current If,
If=Vout/Rf
=(30/16KOHM) Ampere
=1.875 mA.
Here we can use equation of this circuit to calculate Vout as
vout=-[Rf/R1(V1)+Rf/R2(V2)]
=-[16/8(-5)+16/4(-5)]
=-[-10-20]=30v.
Hence we can calculate feedback current If,
If=Vout/Rf
=(30/16KOHM) Ampere
=1.875 mA.
Arnab Bhattacharya said:
10 years ago
From KCL, the current in the feedback resistor is equal to the sum of the currents flowing through the two input branches [Since, the inverting terminal is at a virtual ground].
So, current through feedback resistor = 5/8 + 5/4 = 1.875 mA.
So, current through feedback resistor = 5/8 + 5/4 = 1.875 mA.
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