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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). [Original Circuit1 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. LTSPICE Circuit What am I doing wrong here and how do I fix it? Thanks

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    \$\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\$ Commented Jan 14, 2022 at 7:10
  • \$\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\$
    – LvW
    Commented Jan 14, 2022 at 8:55

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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:

enter image description here

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:

enter image description here

To find:

enter image description here

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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  • \$\begingroup\$ +1 .OP is the way to go. The .STEP could have been avoided by using a single value for V1. As an alternative to the .MEAS, one could use a behavioural source with V=-V(VA)/I(V1), and then reading the voltage at that node. Choices, nothing more. \$\endgroup\$ Commented Jan 14, 2022 at 9:20

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