Log Op Amp Problem - Odd Configuration - Confused! Help! - Page 1
Forum section: electronics / beginners

author: ExtremHybrid
author_rank: Newbie
timestamp: December 13, 2014, 12:03:31 am
content: 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 iD= Is*(e^(VD/VT) - 1) <--- Approximated to: Is*(e^(VD/VT))My CONFUSION is due to the resistorR2.My answer for Vouthas to include the variables vIN, R1, VT, IS,and R2.So far, I have Vout= -VT*ln(vIN/(IS*R1+1)) <--- Approximated to: Vout= -VT*ln(vIN/(IS*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:Vout= -VT*ln((vIN-R2) / (IS*R1))But this is wrong!Intuitively I know thatR2is a total dud and it has no function in an ideal op amp scenario.No current flows throughR2and 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 stateR2as a variable somewhere in my equation for Voutand 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 someonePLEASEexplain to me what I am missing and where this damnR2belongs in my Voutequation.Thank you in advance.

author: AG6QR
author_rank: Frequent Contributor
timestamp: December 13, 2014, 12:25:43 am
content: 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.

author: tggzzz
author_rank: Super Contributor
timestamp: December 13, 2014, 12:45:02 am
content: Quote from: ExtremHybrid on December 13, 2014, 12:03:31 amTypically, 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.

author: dannyf
author_rank: Super Contributor
timestamp: December 13, 2014, 01:49:29 am
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: December 13, 2014, 03:56:03 am
content: Suppose you consider a simplified equivalent circuit for the op amp where there is an input impedanceRinbetween the (+) and (-) inputs of the amplifier, and where the output voltage is given byVout=G Vin=G(V+-V-)?A currentIin=Vin/Rinwill then flow whenever there is a differential voltage between the inputs and the resistorR2will become relevant.

author: tggzzz
author_rank: Super Contributor
timestamp: December 13, 2014, 09:07:30 am
content: Quote from: dannyf on December 13, 2014, 01:49:29 amThey will be at the same potential. However, since therewillbe a common-mode input offset current, itwon'tbe 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.

author: mikerj
author_rank: Super Contributor
timestamp: December 13, 2014, 09:19:27 am
content: Quote from: tggzzz on December 13, 2014, 09:07:30 amIn practice yes. For a homework assignment where no other information is given, it would be reasonable to assume a perfect op-amp.

author: dannyf
author_rank: Super Contributor
timestamp: December 13, 2014, 11:01:04 am
content: 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.

author: tggzzz
author_rank: Super Contributor
timestamp: December 13, 2014, 03:07:31 pm
content: Quote from: mikerj on December 13, 2014, 09:19:27 amThat, 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.