Voltage-Summing Current-Sensing Amplifiers

 Give the Following Circuit. This is an operational amplifier. 

And the following will be the circuit that is the equivalent to the op amp. 

The feedback network is the bottom. If there is no current through Ri, nothing goes through Ri, and Vi is very small, and thus the impedance approaches infinity. If the only place for current to go is through Ri, the impedance will be very big as a result. The circuit was within the Norton setup. The calculations will be easier with Rs in Parallel with Ri, because if impedance is infinity, Rs || infinity = Rs. 

It is easier to draw things in big in order to find the correct values. 

We first want to find G0, by doing voltage injection. I assume 

vx = 0. Voltage injection requires putting the voltage source in series with the existing branch. When killing a voltage source, we get a short, and we need to get the original circuit back. 

Current source needs to be in parallel with branch, and voltage in series so we get the original circuit back. 

When vx is 0, this means vi is 0 due to the amplifier configuration. So, Vt at the top is IsRs. This indicates there is no current through Ri and the voltage Vs appears at the top of Vt and the bottom of R1. V' at the bottom of the resistor Ri is the same as Vt at the top since the input voltage = 0. 

If Ii or the input current is zero, this means that R1 and T2 are in series. Because these are in series, I can do a voltage division, and get that v' = vs. As a result, v''/vs = v''/v' which equals (R2 + R1)/R1.

 

v' = v''    r1/(r1 + R2) 

v'' = vy r2/(R0 + RL + r2) 

Vx = -Avi = vx/vy = A Ri/(Ri + Rs) * (r1/(r1 + R2))  r2/(R0 + R2 + r2)

The second step in this is to try to find Tn.  This time I need to find Vx/Vy when I0 = 0.

 

Now we want to make sure that there is no current through RL. 

This indicates that v'' = vy, which shows that Vd = v'' which equals v' R3/(R2 + R3).

vx = -Avi, vi = Ii Ri, where v' = Ii(R1||(R2 + R3)

vx = -A Riv'/(R1||(R2 + R3))

This makes -A Ri(R1 + R2 + R3)/R1R3

The next step is to figure out what Rn is. Here, use the Norton source. If Vi = 0, we have infinite impedence. 

The usual formula for the gain is utilized. Rin = Rin0( (1 + 1/TNin) / (1 + 1/TDin) ).

TN is the loop gain when the output is 0 (Rs is shorted here), and TDin is the loop gain when the input is 0 (Rs is parallel). 

We want to set everything to the equation 1/Td = I0 + I1 Rs where I0 and I1 are constants. 

TNin = Td when Rs = 0, so it just equals to I0.