op amp logarithmic amplifier

[socratidion]

I'm puzzling over the kind of logarithmic amplifier circuit as shown in http://en.wikipedia.org/wiki/Log_amplifier – the second one, that is, the one using a BJT.

I get that the output of the op amp must be -V_BE. But I now don't see why I_C is V_in/R1. Left to my own devices I concluded that I_C=(V_in+V_BE)/R1. That is, the current starting at V_in has to go all the way to the op amp output, so the input voltage has to be measured with respect to that. Everywhere I've looked tells me I'm wrong, but I think I just need someone to tell me what my daft mistake is.

[nfmax]

No mistake, you just need to apply the First Rule of OPAMPS: with negative feedback (and provided the output is not clipped to either supply rail), the voltage on the - input is the same as the voltage on the + input. So the voltage across R1 must be just Vin, and the input current Vin/R1. By the Second Rule of OPAMPS (no current flows into or out of either input) this current has to flow into the collector of the transistor.

[tron9000]

ok so this diagram:


to compliment nfmax's response. An op-amp with feed-back will drive it's output so its inputs are equal. Given that the non-inverting input is at 0V so must the inverting input. If the inverting input is at 0V (virtual earth) then the current flowing through R1 is V_in / R1.

assuming op-amp has infinite input resistance, we can assume no current flows into the inverting input. Therefore (and remembering kirchoff's current law) the curretn flowing through your transistor I_C = I_R1 = V_in/R1

[LvW]

"The voltage on the - input is the same as the voltage on the + input."
......................
"An op-amp with feed-back will drive it's output so its inputs are equal."

According to my experience, somebody could ask: How can a differential input of 0 V cause a finite output voltage?
Therefore, my additional remark: Both statements above are true for IDEAL opamps only (with infinite open-loop gain Ao).
In reality (for real opamps), there remains always a finite diff. voltage between both opamp inputs (Vdiff=Vout/Ao).
Because, usually,  Ao is very large (1E4...1E6) this diff. voltage is very small (µV range) and can be neglected in most cases during calculations.
This leads to the above statements (principle of virtual ground for ideal opamps) . 

[socratidion]

Thank you for the replies. I think, after agonised pondering, that the part I tripped up on was the transistor. Now after staring at the schematic until my head ached, I think I get it.

Correct me if I'm wrong. Between the input and the op amp's output there is a voltage of V_in + V_BE, but as far as the current is concerned, I only care about the voltage difference between the input and the op amp inverting input, which is just V_in (since the inverting input is held at 0V), so the current is set at V_in/R1. I think I was kind of ignoring the transistor, doing the calculation as if it weren't there. I couldn't imagine what a fixed current would do when it came out of the emitter and encountered -V_BE volts. To be honest I still can't: flow, I suppose.

[LvW]

Perhaps some additional explanations may help:
1.) Replacing the transistor by a resistor R2 gives a classical inverting amplifier. In this case, this circuit with negative feedback can be seen as a series connection of the two resistors, which are connected to two voltage sources with different polarities: R1 to a positive (input) voltage and R2 to a negatiuve (output) voltage. And the output voltage is adjusting itself to a value which results in a zero voltage BETWEEN both resistors (at the inv. opamp input node).
2.) Now - the same principle applies to the given circuit: R2 is replaced by a transistor with a non-linear voltage-current characteristic. Hence, the output voltage is adjusting itself based on an exponential law.