# How do you solve this opamp circuit?

This is the schematic of a (log) Opamp amplifier. The objective is to calculate the V_out.

simulate this circuit – Schematic created using CircuitLab

To solve this figure I can apply the formula for log (op)amplifier i.e. $$V_{out}= -V_T\space ln(\frac{V_{in}}{I_S\space R_1})$$

The temperature is not given. The npn transistor material is Si and op-amp is IDEAL.

$$\text{How do you solve this circuit for } V_{out} \text{ without using the formula }$$

UPDATE

The circuit simulation( with circuitLAB) gives

V_out = -15volt

It seems, that in simulation, the transistor is being treated as an Open Switch. Should it be ?

• It's not clear what you are asking. You have an equation for Vout, why don't you just plug in the variables on the right hand side and do the math? Jan 3, 2014 at 19:27
• @JoeHass I can't use the formula as I don't know the Operating Temperature. I need an alternate analysis. I've edited the question to be more clear.
– vyi
Jan 3, 2014 at 19:29
• What is your application? Could you substitute in temperature ranges to get upper/lower bounds on the output voltage? Alternatively, could you use heating/cooling systems to get a regulated operation temperature? Jan 3, 2014 at 19:34
• @helloworld922 I don't have an application, I just want to find out. This is a question from an exam I took and I did this one wrong( just like the simulator :p). So, I'm just looking for a correct analysis.
– vyi
Jan 3, 2014 at 19:47
• This is a homework and here, you have to read this. en.wikipedia.org/wiki/Log_amplifier This is a common configuration used in industry circuits. Without this there would be no electronic engineering. Jan 4, 2014 at 16:06

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.

• What should be approximate value for V_out? Near -0.46 volt (for room temperature 300K, Vt=26mV, Is=10^(-10) ) OR -15 volt as given by the simulator.
– vyi
Jan 4, 2014 at 13:46
• I don't know the forward current gain (alpha) or emitter reverse saturation current (Ies) of your transistor, and Is (I sub s) in your circuit is the product of those two values. Are those, or Is, given in the problem? Rough estimates for me give around -0.56 volts for 5V in. Real World you use a more complicated compensating circuit and you measure inputs over several decades like 1mV to 100V, plot their logs and plot the outputs. Fit straight lines and find a scale factor and offset. Jan 4, 2014 at 18:37
• That's what I am asking, an approximation and assurance. You agree that the voltage will be in a .3-.6 volt range.
– vyi
Jan 5, 2014 at 8:08

Whatever the formula you use, in this circuit the output is always -0.7V and it won't change with increase in the input voltage.Because the base is at 0 potential and there is diode from base to emmitter which has a drop of 0.7 volt and therefore the emitter is always at -0.7V.

• No, you are using a very crude approximation for the forward voltage of the base-emitter junction. I will grant you that the change will be small in this case but that's the idea for a log amplifier. The voltage will change, and whether the magnitude of the change is significant is up to the user. Jan 4, 2014 at 14:18
• That is a very rough approximation. Jan 4, 2014 at 14:18
• Yes Joe and Stephen.. I agree with you both. My point is just to show that it is nothing but a normal diode voltage formula with which you have to juggle. Jan 4, 2014 at 15:05
• So you are saying that the input voltage and opamp will have no effect whatsoever, but, the diode junction barrier potential will be determining/dominating factor. That doesn't seems rational.
– vyi
Jan 5, 2014 at 7:34