linearltspicemeshsimulation
I have the following schematic of the circuit, I wanted to added I2 as a Current Dependant Current Source depending on current on in V1 as three times the value, or, I(I2) = 3I(V1)
I have tried various tutorials, and used circuit element F and current both, its not working with any. There is always an error in circuit element I2
What can be done to achieve this result. Thanks a lot!
DC analysis gives you the initial conditions DC steady state values only. You must do a transient analysis to see how the voltages and currents evolve with time. AC analysis is for small-signal sinusoidal steady state only; it is a frequency domain analysis.
I believe you'll need to use a piece-wise linear (PWL) source for your current ramp.
This is defined as a subcircuit:
.SUBCKT PSMN2R0_30PL DRAIN GATE SOURCE
LTSpice needs this to have somewhat special treatment, so you will need to do the following:
CTRL+Right click on the device and you will get this window:
Now edit the Prefix and Value lines: The prefix for a subckt is 'X'. The model name is precisely as defined in the lib file.
Now click OK. You will need to add a spice directive on your schematic:
.include PSMN2R0_30PL.lib This assumes it is in the same directory as the simulation circuit.
LTSpice should now be happy with the part.
Here is what you should see on the schematic:
You can, of course, add it to the LTSpice model tree, but I find it easier to use this method.
Best Answer
You are using the
currentelement (as seen in with theF2browser), which is a regular current source, as opposed to a behavioural source, which, for current, isbiandbi2. Once you place that, the syntax changes and you can writeI=I(...), as the default placeholder suggests.However, as per a recent answer, the manual warns you about quasi-instantaneous current feedback, and you should use an
Fsource (which is, as you guessed, displayed asfwithF2browser). Once you place that, its value should be changed, just as the manual says -- and for your case, in particular -- toV1 3. Note that the current intoV1goes through the+pin, and exits from the-pin, irrespective of placement, so you should take that into account with a proper sign:V1 -3(if that's the case). See the linked answer for an exemplifying picture.