Log Op Amp Problem - Odd Configuration - Confused! Help!

[ExtremHybrid]

Hello, I have an odd log op amp configuration I cant solve.

I am given the task to form a mathematical expression for this circuit:


Given to me are:
Shockley ideal diode equation  i_D = I_s*(e^(^V_D/V_T) - 1) <--- Approximated to:  I_s*(e^(^V_D/V_T))

My CONFUSION is due to the resistor R₂. 

My answer for V_out has to include the variables v_IN, R₁, V_T, I_S, and R₂.

So far, I have V_out = -V_T*ln(v_IN/(I_S*R₁+1)) <--- Approximated to: V_out = -V_T*ln(v_IN/(I_S*R₁))

And this equation would be correct for your standard logarithmic op amp configuration as shown below:



At some point I came up with the equation:
V_out = -V_T*ln((v_IN-R₂) / (I_S*R₁))

But this is wrong! 

Intuitively I know that R₂ is a total dud and it has no function in an ideal op amp scenario.
No current flows through R₂ and the voltage at the + terminal of the Op Amp is the same as the - terminal, which is GND (i.e 0V) !!!

But the answer to the problem requires that I state R₂ as a variable somewhere in my equation for V_out and I don't know where. If it were a differential amplifier configuration  with a voltage divider connected to the + terminal of the Op Amp, then I would know how to figure it out.

But I am confused and frustrated at this problem.
Intuitively I know the answer, but mathematically I can't express it as required. FML.

Could someone PLEASE explain to me what I am missing and where this damn R₂ belongs in my V_out equation. 

Thank you in advance.

[AG6QR]

Since you understand that the current through R2 is zero, and therefore R2 is irrelevant, I'd say you understand the electronics.  You're just stumped by the test-taking aspect of the homework problem.

If it were me, I'd include a 0 * R2 somewhere in your expression.  That represents the voltage developed across R2 (assuming ideal op-amp with 0 current through its inputs), and thus it represents the voltage at the + input.

Sure, your algebra teacher would tell you to simplify 0 * R2 to just 0.  But if your EE teacher wants to see R2 in there, put it in there correctly.  You might even use an expression for the input current instead of 0, which would make it easier to adapt your expression to real-world op-amps with nonzero current on their inputs.

[tggzzz]

Could someone PLEASE explain to me what I am missing and where this damn R₂ belongs in my V_out equation. 

Typically, but not always, such "irrelevant resistors" can be understood by considering an op-amp that has a non-zero input-offset current.

For more information, see any opamp "cookbook" or detailed application notes.

[dannyf]

The non-inverting terminal and the inverting terminal should be at the same potential: GND.

Thus, the current floating through R1 (and the diode) is V1/R1 - here, V1 is limited to be positive.

The output would be such that the current through the diode is V1/R1, and is negative.

Hope it helps.

[IanB]

Suppose you consider a simplified equivalent circuit for the op amp where there is an input impedance R_in between the (+) and (-) inputs of the amplifier, and where the output voltage is given by V_out = G V_in = G (V⁺ - V^-)?

A current I_in = V_in / R_in will then flow whenever there is a differential voltage between the inputs and the resistor R₂ will become relevant.

[tggzzz]

The non-inverting terminal and the inverting terminal should be at the same potential: GND.

They will be at the same potential. However, since there will be a common-mode input offset current, it won't be GND.

Some opamps have very low common-mode input offset currents, but logarithmic amps will also be operating with very low currents in the feedback loop.

[mikerj]

The non-inverting terminal and the inverting terminal should be at the same potential: GND.

They will be at the same potential. However, since there will be a common-mode input offset current, it won't be GND.

Some opamps have very low common-mode input offset currents, but logarithmic amps will also be operating with very low currents in the feedback loop.

In practice yes.  For a homework assignment where no other information is given, it would be reasonable to assume a perfect op-amp.

[dannyf]

If you were to consider Is and R2, it is equally simple:

the two input terminals are at the same potential. Assuming that Is flows into the opamp on the non-inverting end alone, the non-inverting end would be at -Is * R2.

So the current going through R1 would be (V1+Is*R2) / R1. and the math follows.

You can extend this to a case where Is is present on the inverting input terminal as well.

You can also extend it to cases where input offset voltage exists, etc. But for a typical opamp, it makes minimum difference.

[tggzzz]

For a homework assignment where no other information is given, it would be reasonable to assume a perfect op-amp.

That, of course, depends on what the assignment is designed to teach.

Since the OP was given that circuit, it is plausible that including the common-mode input offset current was part of the assignment.