How to calculate the output resistance of a opamp circuit? - Page 1
Forum section: electronics / beginners

author: ShowKemp
author_rank: Contributor
timestamp: October 04, 2017, 12:41:20 am
content: Hi,I'm reading a book about OpAmps and simulating the exercises with LTSpice. Here's the exercise and circuit i'm working with:QuoteThe circuit on the book:The circuit built in LTSpice (i have not PSpice so i'm using LTSpice):The results based on a Ideal OpAmp are:Voltage Gain = -R4/R3 = -100k/1k = -100Input resistance = R3 = 1kOutput resistance = 0Results according to the book:Voltage gain = -9.995E+01Input resistance = 1.001E+03Output resistance = 3.787E-02To get the results on the simulation i run an AC Analysis and plot traces with algebraic expressions:Voltage Gain: V(o)/V(i) = -99.949463 (Close to ideal)Input resistance: V(i)/I(R3) = 1.0005004K (Close to ideal)I don't know how to the output resistance, i tried calculating V(o)/I(R2), but got ~100k, this isn't close to 0.So my newbie question is how can i calculate the output resistance of the circuit?Regards

author: ShowKemp
author_rank: Contributor
timestamp: October 04, 2017, 01:12:12 am
content: After further reading i "think" i found the way to calculate the output resistance:Ro = Closed-loop output resistancero = Open-loop output resistance (75ohms)L = Loop Gaina = Open-loop Gain (200,000 V/V or 200 V/uV)Based on the LTSpice schematic:L = a / (1 + (R4/R3))L = 200,000 / (1 + (100k/1k)) = 1980.198Ro = ro / 1 + LRo = 75ohms / (1 + 1980.198)Ro = 0.0379 ohms ~ 3.787E-02 (Close to the book result)Still need to get more familiar with the "OpAmp circuit" terms.Any suggestions on circuit simulations with LTSpice? is it possible to get all this data with one spice command?

author: strawberry
author_rank: Super Contributor
timestamp: October 04, 2017, 01:20:04 am
content: when you take Vgain=99,95 then it will be the same

author: Audioguru
author_rank: Super Contributor
timestamp: October 04, 2017, 01:46:46 am
content: The open loop gain of an opamp is not a fixed number, there is a range of gain, some are high and others even with the same part number are less.The open loop gain of an opamp is high at DC and frequencies below 10Hz then it drops at higher frequencies due to its frequency compensation capacitor. Therefore the output impedance increases at higher frequencies.

author: Zero999
author_rank: Super Contributor
timestamp: October 04, 2017, 08:59:34 am
content: The calculations look right to me. As mentioned above, the open loop gain varies. Also note that this only applies when the op-amp is working in its linear region and with negative feedback. For example, if too much current is drawn, the op-amp's output will either saturate or the current limiting protection will kick in. It will no longer have any negative feedback and the output impedance will rise dramatically.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 04, 2017, 09:07:42 am
content: Yes, this is only correct at DC, in the middle of the range where it's linear. But for that condition, it is correct.Tim

author: danadak
author_rank: Super Contributor
timestamp: October 04, 2017, 09:30:09 am
content: This might help -http://www.ti.com/lit/an/slyt677/slyt677.pdfAlso the Laplace analysis for a single pole comped OpAmp pretty easy. Two pole a little more tedious,use of partial fraction expansion reveals equivalent lumped model characteristics (which yield easyinsight into Zo behavior).Regards, Dana.

author: Cerebus
author_rank: Super Contributor
timestamp: October 04, 2017, 03:19:20 pm
content: Quote from: ShowKemp on October 04, 2017, 01:12:12 amThe DC transfer function simulation will deliver you a transfer function (gain), input and output impedances:Code:.tf V(O) V1

author: Kleinstein
author_rank: Super Contributor
timestamp: October 04, 2017, 05:33:37 pm
content: One way to "measure" the output impedance of a circuit simulated in Spice is adding a current source to the point of interest. The AC simulation with the extra current as source gives the response to output current and thus the output impedance. An offset current might upset the simple V(out)/I(out) formula. In the example I(out) would be from both resistors, not just the 75 Ohms.

author: ShowKemp
author_rank: Contributor
timestamp: October 05, 2017, 04:50:33 pm
content: Quote from: Cerebus on October 04, 2017, 03:19:20 pmThanks, i didn't knew that command, just tested it and the result seems to be the expected: