To calculate the transfer function of a current measuring circuit.

Hello all,

So this is my circuit,


Which is basically a current sense differential amplifier with a gain of 22.

The output of the circuit is as expected, (I get a DC sinewave within 3.3 peak with Vrms and Vpp varing according to the current of 0-6 Amps.).

I understood R1 R2 R3 R5 and R6 are used for the differential amplier configuration, R4 I guess is a current limiting resistor because voltage at the shunt is superimposed with 1.65V to shift it into positive cycle.

R7 and R8 act as resistor divider to generate 1.65V and the differential circuit itself is offseted by 1.65V.

I don't if my above stated understanding of the circuit is entirely correct. Please correct me where ever I am wrong.

Also, Will the resistor divider netwrok affect the gain? It does actually when I simulate the circuit in multisim. The gain changes by a small value about 22.019 for 1K and if R7 and R8 are changed to 10K the gain becomes 22.031 .

Why is this happening? I do not undertsand how to proceed with the transfer funtion calculation.

How do I find a definite formula for gain w.r.t R7 and R8, also R4 I don't know whether that would come in the picture too.

All the help is highly appreciated.

For the differential circuit to work properly, the junction of R4 and R6 must have zero impedance to ground but instead, it has the equivalent resistance of R7 and R8 in parallel along with the resistance of R4. So generating the exact transfer function is complex due to that factor.

I don't see the need for R4 so I would remove that from the circuit. Then I would change the value of R6 so the value of R6 in series with R7||R8 equals 220k (or equal to the value of R5). Then the gain becomes simply R5 /R2.

crutschow said:

I don't see the need for R4 so I would remove that from the circuit. Then I would change the value of R6 so the value of R6 in series with R7||R8 equals 220k (or equal to the value of R5). Then the gain becomes simply R5 /R2.

Click to expand...

Thank you for your reply.

Even if current flowing through the shunt is 6 Amps no limiting resistor (R4) is required right?