I am trying to measure the current through the MOSFET switch M1 using low side shunt resistor method (on MCLV-2 dev. board). I can read the signal `Vout' on the analog input of a dsPIC33F MCU. I believe that the capacitors are there to filter high frequency switching noise and clean up the measurement. Converting the ADC reading from MCU to a meaningful current reading in amperes requires solving this circuit. So I used a simplified circuit to apply the circuit theory to solve for the voltages. I arrived at the simplified circuit using the fact that in steady state the capacitor will behave as an open circuit. I came up with the following equations:

Vn = Vp                       (negative feedback)
(V1 - Vn)/2K = (Vn - V3)/30K   (KCL at node Vn with shown current direction)
(Vref - Vp)/30K + (V2 - Vp)/2K = 0      (KCL at node Vp with shown current direction)

My questions are: (1) Is the simplified circuit correct? (2) Are these equations correct? (3) Do I need knowledge of V1' andV2' for solving this circuit? Any other help regarding this will also be greatly appreciated! The schematics of the original and simplified circuit are shown below.


simulate this circuit – Schematic created using CircuitLab


simulate this circuit

  • \$\begingroup\$ Why are you using two current sense resistors of 25 milli ohm? \$\endgroup\$
    – Andy aka
    Apr 22, 2015 at 17:19
  • \$\begingroup\$ @Andy, This is a simplified schematic. The actual circuit is a 3-legged bridge, too large to put here. The lower current sense was added to show that V2 terminal was not hooked to ground. Hope it clarifies:) \$\endgroup\$
    – Adeel
    Apr 22, 2015 at 17:26
  • \$\begingroup\$ How can this be analysed without full knowledge of the shunt? \$\endgroup\$
    – Andy aka
    Apr 22, 2015 at 17:31
  • \$\begingroup\$ Ideally, looks to me like it's a differential amplifier with a gain of -15 times the input current*R1, referenced to Vref, 1.36V. So for a current of 1A there would be 25mV across the sense resistor, and 1.36-.375V = .985V at the output of the op-amp. This of course ignores resistor mismatch CMR errors, offset errors, etc. \$\endgroup\$
    – John D
    Apr 22, 2015 at 17:43
  • \$\begingroup\$ @Andyaka the edited schematic should clarify things. \$\endgroup\$
    – Adeel
    Apr 22, 2015 at 17:58

1 Answer 1


(1) Is the simplified circuit correct?

Yes. It is a valid simplification of your 'original' circuit.

(2) Are these equations correct?

Yes (though they might be easier to solve if you define Vref as 'ground').

(3) Do I need knowledge of V1' and V2' for solving this circuit?

No. The formula for a differential op amp is Vout = (V2-V1)*Rf/Rin, so you only need to know the voltage difference between V1 and V2. This voltage is determined by the resistance of R1 and the current flowing through it.

NOTE: In your circuit current flows from V1 to V2, so Vout will go negative (relative to Vref) for positive current. However the MCLV-2 board's amplifier inputs are connected in reverse compared to your circuit, so their output voltages will not be inverted. Also the reference voltage is 1.65V (3.3V/2) not 1.36V.

  • \$\begingroup\$ Thanks for the correction. (1) Which resistance you are referring to by Rin? (2) Also, with knowledge of Vout, Rf and Rin, current through the shunt resistor R1 is simply (V2-V1)/R1, correct? (3) If you are referring to the amplifier OA1 in the above schematic, it is connected as shown in the MCLV-2 schematic. I did not follow what you meant by are connected in reverse compared to your circuit. Can you please elaborate? \$\endgroup\$
    – Adeel
    Apr 23, 2015 at 8:39
  • 1
    \$\begingroup\$ (1) Rin = 2k in your simplified circuit. (2) The voltage across R1 is V1-V2. In the gain equation it is changed to V2-V1 to match the op amp inputs (output goes positive when V2 > V1). (3) in the MCLV-2 schematic, SHUNT_HI_SUM is below the shunt resistor and SHUNT_HI_x is above it. You may be confused because they appear to be the other way around in sheet 2. \$\endgroup\$ Apr 23, 2015 at 11:55
  • \$\begingroup\$ just out of curiosity, have you worked with an MCLV-2 or answered just by looking at is datasheet? \$\endgroup\$
    – Adeel
    Apr 23, 2015 at 14:28
  • \$\begingroup\$ I just looked at the datasheet. \$\endgroup\$ Apr 23, 2015 at 17:50
  • \$\begingroup\$ Comment on NOTE in the answer: Vout = Vref + (Rf/Rin)*(V2 - V1). So with positive current flow, Vout will go negative only if Vref=0. In the case of MCLV-2, Vref is 1.65V, hence Vout will go negative whenever the term (Rf/Rin)*(V2 - V1) < -1.65V. \$\endgroup\$
    – Adeel
    Apr 25, 2015 at 7:49

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