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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

cIRCUIT

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  • 2
    \$\begingroup\$ "Now, the typical gain of this opamp is 130" No it is not, please read carefully, the gain is 131 dB. That is 3.16 Million times, not 130 times. That makes sense since it is a precision opamp. I'm too lazy to read a netlist and imagine what your schematic looks like. So include a schematic first than someone might answer. \$\endgroup\$ – Bimpelrekkie Oct 14 '15 at 14:39
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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.

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