Electronic – Op-Amp gain calculation

The answer is actually pretty simple :

Voltage difference between \$ V_+\$ and \$ V_-\$ is zero. No current flows into the opamps input. Output voltage is the sum of opamp input voltage and voltage across resistor R2. Also, output voltage is dependent on R1.

*NOTE: Ix is the current source between the two input pins. If your source is 0.5Ios, simply substitute the Ix with 0 5Ios. Thanks goes to Andy_aka who caught my nomenclature mistake.*

Thus:

\$ V_+ = R_3 * I_{X}\$

\$ V_- = V_+\$

\$ I_1 = V_+\ / R_1 \$

\$ I_2 = I_1 + I_{X}\$

\$ V_{R2} = R_2 * I_2\$

\$ V_O = V_{R2} + V_{R3}\$

or

\$ V_O = I_{X} * (R_3 + R_2 + R_3 * R_2 / R_1)\$

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.