I am trying to implement a current output amplifier circuit shown below. Current is measured across sense resistor Rs to adjust input voltage Vin.

enter image description here

Gain: G = -Iout/Vin = R2/(GainIA1 × R1 × RS)

The gain of this circuit is set at 1 with

  • R2 = 1 kΩ
  • R1 = 100 kΩ
  • RS = 0.01 Ω
  • RL is the load.

It has an instrumentation amplifier IA1 on the feedback path for less noise and better resolution. The gain of IA1 is also 1.

If I decide to increase resistance of Rs and decrease resistance of R1 while keeping the gain at 1, how would changing sense resistor value affect my system performance?


1 Answer 1


you formula for gain is not correct,

\$ \frac{- I_{out}}{V_{in}} = \frac{R_2}{R_1*IA1*R_S}\$

for your purpose, you need to change R1

  • \$\begingroup\$ Sorry I updated the question. By changing the resistance of R1 & RS, would that affect the my system performance? Is there any advantage of changing sense resistor value? \$\endgroup\$
    – Linh
    Commented Nov 19, 2018 at 6:45
  • \$\begingroup\$ @Linh, by changing R1, input impedance will be change. if you change Rs, you should aware of power dissipation, \$ P_d = R_s * i^2 \$, \$\endgroup\$
    – M KS
    Commented Nov 19, 2018 at 11:11
  • \$\begingroup\$ What if I keep the same value for R1 and change the value of R2, what should i notice? \$\endgroup\$
    – Linh
    Commented Nov 19, 2018 at 18:06
  • \$\begingroup\$ @Linh, I think R2 could just change the loop gain and this cause instability in the loop. I don't know which value this instability happens. you could simulate this circuit for instability region. \$\endgroup\$
    – M KS
    Commented Nov 19, 2018 at 18:35
  • \$\begingroup\$ Could you give me more detail on how it causes instability in the loop? Isnt that the point of using negative feedback is to improve stability? \$\endgroup\$
    – Linh
    Commented Nov 19, 2018 at 22:06

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