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I'm going through the Op Amp section in The Art of Electronics and I'm having trouble understanding their derivation of the gain for the following circuit:

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So by the properties of Op Amps \$ v_A=v_{in} \$, now the way I understand this circuit is that since the potential at A is \$ v_{in}\$ and \$ R_1 \$ is connected to the ground there will be a current flowing through \$ R_1 \$ towards the ground. Since another property of Op Amps is that the current going into them is essentially 0 we know that this current must also pass through \$ R_2 \$ so we can find \$ v_{out}\$ and hence the gain by using \$ v_{out}=v_{in}+iR_2\$ where \$i=\frac{v_{in}}{R_1}\$.

However in the textbook, it is explained as "But \$v_a\$ comes from a voltage divider: \$v_a=\frac{v_{out}R_1}{R_1+R_2}\$", I'm just having trouble seeing why this is valid, this comes from current being \$i=\frac{v_{out}}{R_1+R_2}\$ and then going from the ground to point A through \$R_1\$, I suppose I am confused by the fact that this finds a different current through the resistors than my method and it still works.

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  • \$\begingroup\$ Your equation for i in the first paragraph isn't correct. The one in the textbook is correct. All the current from the output to ground flows through R1 and R2, and there is no current into or out of A at all. \$\endgroup\$ – user207421 Apr 3 '16 at 6:26
  • \$\begingroup\$ @EJP I actually believe my equation for \$i\$ in the first paragraph is still correct it's just that it's described in terms of \$v_{in}\$ instead of \$v_{out}\$, I think I get why now since when using the input voltage to find current it is just 1 series resistor and when using the output voltage it travels through both \$R_1\$ and \$R_2\$ and I suppose \$v_a\$ is in parallel with \$R_1\$ \$\endgroup\$ – Craig Apr 3 '16 at 6:39
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Both solutions are equivalent. They just look different because one current is in terms of vin and the other in terms of vout.

If you substitute (from the gain) \$V_{in} = V_{out}\frac{R1}{R1+R2}\$ you get the equation in the book.

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    \$\begingroup\$ Craig - a short explanation in words: It is correct that the voltage at node A is (nearly) identical to Vin - however, the voltage at node A is not a source! Instead, this voltage is produced by Vout according to a simple voltage divider. \$\endgroup\$ – LvW Apr 3 '16 at 8:18

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