Negative Impedance Converter and its Sensor Application

Discussion in 'Homework Help' started by mjakov, Oct 16, 2014.

1. mjakov Thread Starter New Member

Feb 13, 2014
20
5
Hi!

How would you find the v-i relationship of the ideal op-amp circuit below for the interval of linear operation ?

Let $V_{out}$ be the voltage between $R_3$ and $R_2$. Then:
$V_{out} = v - R_3 i$
Because of $V^+ = V^-$ and KCL:
$V_{out} - (i + i_0) R_2 = v$
$i_0 + i = \frac{v}{R_1}$
Solving this system of equations we get:
$v = - \frac{R_3}{R_2} R_1 i$

However, according to the book the correct solution should be:
$v = - \frac{R_2}{R_3} R_1 i$

Which of the two solutions is the correct one?

2. Jony130 AAC Fanatic!

Feb 17, 2009
3,962
1,098
If Vin = 1V and R2 = 10K ; R3 = 1K; R1 = 1K.
Vp = Vn = 1V so
IR1 = 1V/1K = 1mA, this current will flow through R2. So voltage at op-amp output is Vout = Vin/R1 *(R1 + R2) = 11V
From this we can find Iin = (Vin - Vout)/R3 = (Vin - Vin/R1 *(R1 + R2))/R3 = -10mA
And after we solve this Iin = (Vin - Vin/R1*(R1 + R2))/R3 for Vin we have
Vin = - R3/R2 * R1*Iin
Vin = - 1K/10K * 1K*-10mA = -0.1*-10V = 1V

Oct 9, 2007
2,285
331
4. WBahn Moderator

Mar 31, 2012
17,788
4,807
Try solving from as different a perspective as you can and see if you get the same answer. For instance, because the two op-amp terminals are the same, you know that whatever voltage your drop across R2 you have to get back across R3. So that means that (I)R3 = - (I+Io)R2. See if you can use that to back your way into a solution.

5. mjakov Thread Starter New Member

Feb 13, 2014
20
5
Thank you all for the helpful replies!

The following circuit is supposed to contain the application of the negative impedance converter for a sensor. Element $v_s$ is the sensor voltage source, $r$ is the internal resistance of the sensor and $R$ is the load resistance. The goal is to make the current $i$ independent of the load resistance. It should be proportional to the voltage source only.

So, here are the equations:
$v_A = v_C$
Using the node voltage method with nodes A and C:
$\frac{v_s - v_A}{r} - \frac{v_A}{R} - \frac{v_A - v_B}{R_3} = 0$
$- \frac{v_C}{R_1} + \frac{v_B - v_C}{R_2} = 0$
Solving this we get:
$v_A = \frac{-v_s}{r \frac{R_2}{R_1 R_3} - 1 - \frac{r}{R}}$
If we choose $R_1, R_2, R_3$ so that
$\frac{R_2}{R_1 R_3} = \frac{1}{R}$, then:
$v_A = v_s$

So this would mean that i does not depend on the internal resistance r of the sensor. But, it does not seem that i is independent of the load R, because
$i = \frac{v_A}{R} = \frac{v_s}{R}$.
Would you agree with this conclusion and is there some other way to make i independent of R ?

Could you recommend some good information source as a reference or tutorial on these kinds of circuits? My textbook on the analysis of electric circuits sometimes omits details such as the names of the circuits like the negative impedance converter above. Could you please give some directions as to what part of electrical engineering does this subject matter belong to?

6. Jony130 AAC Fanatic!

Feb 17, 2009
3,962
1,098
Are you sure that your circuit should not look like this

Then for this circuit try to find IL. And if you set the voltage gain = (1 + R2/R1) = 2V/V and R3 = r then you get IL current independent of the load resistance.

$IL = \frac{R_1 R_3 }{ R_1 R_3 r + R_1 R_3 R - R_2 r R } * Vs$

Because this time the circuit is a howland current source.

See the example here