0
\$\begingroup\$

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

\$\endgroup\$
2
  • 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\$ 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
    Jan 14, 2022 at 8:55

1 Answer 1

2
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\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\$ Jan 14, 2022 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.