# Electronic – Calculating current in the inverting op-amp

currentinverting-amplifieroperational-amplifier

Okay, so I know the "correct" way to calculate current.

Let's first analyze the circuit:

KVL left side:

\$-V_{in} + i_1 \cdot R_{in} + 0 = 0 \$

\$V_{in} = i_1 \cdot R_{in} \$

KVL right side:

\$V_{out} = -R_f \cdot i_f \$

\$i_1 = i_f = i \$

Gain is:

\$\frac{V_{out}}{V_{in}} = \frac{-R_f \cdot i}{i \cdot R_1} = \frac{-R_f}{R_{in}} \$

\$V_{out} = -18.6 V \$

\$i = \frac{V_{out}}{-16k} = 1.1625 mA \$

How come, applying reasoning that since there's only one voltage supply, the current through \$R_{in}\$ and \$R_f\$ is just voltage divided by equivalent resistance?

\$R_{in}\$ and \$R_f\$ are in series

\$i = \frac{V_{in}}{R_{in} + R_f} = \frac{7}{6k + 16k} = 0.318 mA\$

What's wrong with this reasoning? Why is the current not correct?

You are correct in assuming that \$R_{in}\$ and \$R_f\$ are in series but the potential difference between them is not \$V_{in}\$ but \$V_{in} - V_{out}\$. So instead of the current being (as you suggest): $$i = \frac{V_{in}}{R_{in}+R_f}$$ It is: $$i = \frac{V_{in} - V_{out}}{R_{in}+R_f}$$
Another required equation would be: $$V_{in} - iR_{in} = AV_{out}$$ Solving these equations you can calculate the gain of the amplifier and if you take the limit as \$A -> \infty\$ you will get the same result as you calculated before.
Thus you can see that you could have saved all these calculations by assuming the gain is \$\infty\$ in the first place and potential difference between its inputs is zero like you did in your initial calculations with virtual ground.