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My Equation of the circuit using the node \$V_1\$ is

$$ \frac {V_1-V_o}{(1-x)R_p} + \frac {V_1-0}{xR_p+R_3} - \frac {V_1-0}{R_2}=0 $$

I would like to know how to eliminate or substitute \$V1\$ from the equation (if it's right in the first place!) so that I can combine with the equation \$V_{\text{in}}=-\frac {R_2}{R_1} \cdot V_o\$ (so that I have \$ \frac {V_o}{V_{\text{in}}}\$ and calculate for values of \$x=1\$ and \$x=0.3\$.

Hope this makes sense.


2 Answers 2


Your equation for Vin is wrong. The same current flows through R1 and R2, hence express V1 in terms of Vin.

Also, the last term in the node V1 equation is wrong: \$\Sigma\$currents away from node \$=0\$

  • \$\begingroup\$ Ok I think the last term in the equation is (V1-Vin)/(R2/(R2+R1)). V1 in terms of Vin is VinR2/(R2+R1)=V1. If we replace V1 in the nodal equation with "VinR2/(R2+R1)=V1" can we then transpose for Vo/Vin? \$\endgroup\$
    – AbramsM101
    Feb 4, 2016 at 23:23
  • \$\begingroup\$ No, to both. You haven't summed the currents AWAY from node V1 properly. And, use Ohm's law on R1, and then on R2; in other words, do nodal analysis on the node between R1 and R2. \$\endgroup\$
    – Chu
    Feb 4, 2016 at 23:31
  • \$\begingroup\$ well there is zero p.d across R1 and R2 so for that we have V1=-IR1 and V1=-IR2 \$\endgroup\$
    – AbramsM101
    Feb 4, 2016 at 23:39
  • \$\begingroup\$ The node between R1 and R2 is at 0V, therefore there can't be zero across these resistors in general. Use Ohm's law; there's no current flowing into the op amp. \$\endgroup\$
    – Chu
    Feb 4, 2016 at 23:43
  • \$\begingroup\$ On the node between R1 and R2 I have 0-Vin/R1 and 0-V1/R2 \$\endgroup\$
    – AbramsM101
    Feb 4, 2016 at 23:54

The easiest way to think about this is by considering that the op-amp's real output is at the point called V1. The gain equation with V1 as the output is: -

V1/Vin = -R2/R1 i.e. as per a normal op-amp - if you think about this it'll eventually make sense.

If you want the gain at the real op-amp output Vo, then work out what the attenuation is between Vo and V1 is. If (say) it is 5 then then....

Vo/Vin = -5xR2/R1

  • \$\begingroup\$ Why does your Vo/Vin equation disregard R3 and the pot? \$\endgroup\$
    – AbramsM101
    Feb 5, 2016 at 0:10
  • \$\begingroup\$ It doesn't - I said if the effect of the attenuation is 5:1 i.e. Vo/V1 = 5, then Vo/Vin = -5x R2/R1. The attenuation of "5" is the effect of R3 and the pot. \$\endgroup\$
    – Andy aka
    Feb 5, 2016 at 9:26

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