I'm trying to calculate the total resistance seen by the source v1 in this circuit and we have a voltage dependent voltage source in this circuit with a gain of 0.1 (v1) which is the voltage across the resistor (R1).
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When I do the hand calculations we get R(Thevenin) or the total resistance seen by the circuit as 605K ohms. However when I simulate this in LTSPICE (with 1v voltage source) to measure the total current and then calculate the resistance as 1/current I never get that correct resistance I should get. Not even close. I tried multiple different ways but nothing is working.
What am I doing wrong here and how do I fix it? Thanks
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4\$\begingroup\$ The original question involves voltage-dependent current source but you placed a voltage-dependent voltage source. You should pick "g" or "g2" from components list in LTSpice. \$\endgroup\$– Rohat KılıçJan 14, 2022 at 7:10
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\$\begingroup\$ You are speaking of a "source v1" - I suppose you mean "source vs"? Because v1 is the voltage developped across R1. Furthermore, as your equivalent diagram resembles a BJT in common emitter configuration (with a emitterresistor RE), why dont you simulate the corresponding real circuit using the detailed BJT model? \$\endgroup\$– LvWJan 14, 2022 at 8:55
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1 Answer
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There are many ways to achieve this in LTspice. What I'll do is step a voltage source from \$0\:\text{V}\$ to \$1\:\text{V}\$, to keep the computation simple. And I'll use the .OP card (not the .TRAN card here.) I can think of other ways, too. Perhaps someone will present a more prosaic approach. But this was easy for me. So here it is:
I've arranged for two values to be "stepped" for \$V_1\$. The .MEAS cards shown will capture the current from \$V_1\$ for each step (by checking for the node voltage.) Then a final .MEAS card will perform the computation. Then I can just use the following select after the run:
To find:
Which is just the result you were looking for.
I'm sure others will come up with still easier/better answers. But this is one such approach.
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.OPis the way to go. The.STEPcould have been avoided by using a single value forV1. As an alternative to the.MEAS, one could use a behavioural source withV=-V(VA)/I(V1), and then reading the voltage at that node. Choices, nothing more. \$\endgroup\$ Jan 14, 2022 at 9:20