Electronic – Analyze op amp circuit

You have some boundary conditions: In your circuit the collector voltage is zero because the inverting input is a virtual ground. Also in the transdiode configuration, Ic=Ie, and Ve=Vout.

Use a reasonable approximation to the Ebers-Moll equation for currents that are greater than a uA. $$I_{c}\; =\; I_{s}\; e^{-\frac{qV_{E}}{kT}}$$ Is is emitter reverse saturation current. You solve for a variable in the exponent of an exponential by using its inverse, the natural log. $$\ln \left( \frac{I_{c}}{I_{s}} \right)\; =\; \frac{qV_{E}}{kT}$$ and solve for V $$V_{E}\; =\; -\frac{kT}{q}\; \ln \left( \frac{I_{c}}{I_{s}} \right)$$ This is V across the transistor junction and because the virtual ground at the inverting input is zero, this is V on the output, and is negative (inverted of course).

In your circuit, current into the virtual ground must equal current out, so current through the transistor is the same as current through the resistor. This means $$I_{c}\; =\; \frac{V_{in}}{R_{1}}$$ and substituting $$V_{E}\; =\; V_{o\; }=\; -\frac{kT}{q}\; \ln \left( \frac{V_{in}}{I_{s}R_{1}} \right)$$ kT/q has units of volts so for simplicity, replace with a scaling voltage in front. $$V_{o\; }=\; -V_{t}\; \ln \left( \frac{V_{in}}{I_{s}R_{1}} \right)$$

This is the first time I have used LaTex so I was a little distracted. Hope it makes sense. There are approximations and simplifications I can expand on.

Feed back will try and make the two inputs the same. So the inverting input is also at zero volts. The negative voltage (CV-) is sucking 5/3k = 1.66mA from that node. The only place that current can come from is the opamp output. You need 1.66 mA flowing through R3... and that gives 1.66V on the opamp output. (No current can come from R5 because both sides are at zero volts... that is only true when Vin is zero.)