I am designing a differential amplifier that will basically help me in measuring the current through a sense resistor of 0.015R.
Now the Vout formula is
$$Vout = {(R_f/R_i)*(V_2-V_1)} $$
\$V_2\$ – non inv voltage,
\$V_1\$ – inv voltage.
Based on this (i am using the ISL70218SRH op amp).
Now, the typical gain of this opamp is 130. I want to tune(!!!!) it to give a gain of 100. So Rf = 1000K and Ri = 10K. So gain = (Rf/Ri).
Now – from the Vout formula we can deduce that if the voltage difference (this is the drop across the sense resistor) is 0.03V, then
Vout = 100*0.03V = 3V.Right ?
But, when I do a spice simulation, the Vout is almost 4.8V.
V1 2 0 5
V2 1 0 4
*V2 7 0 3.3
R9 1 2 0.015
*R12 2 0 100000K
R8 1 4 10K
R11 2 3 1K
R7 4 6 166K
R13 3 0 166K
R10 6 0 1K
X1 3 4 1 0 6 ISL70218
.TRAN 1m 300m 1m
*.PLOT TRAN V(3)
.PROBE
Best Answer
V2-V1 in your equation is the voltage across the current sense resistor R9. Rf/Ri is the gain, which (from your schematic) is 166/1. So the output voltage will be 0.03 * 166 which is 4.98V
To set a gain of 100, you can use 1000K resistors in place of the the 166K, and replace the 1K with 10K as you suggest.