LTspice Question

Get the plot showing amplitude and phase.

Bring the cursor over the lets say V(n002) at the top of the plot

Right click and change what is plotted.

re(v(n002)) plots the in phase part of a voltage im(v(n002)) plots the quaderature part.

To get both, you need to get the raw signal displayed again and edit the second one.

Next you double (left) click the scale down the right side and tell it not to plot phase.

Hi Ken, I don't quite have it, I'm plotting (v(in)) which is the voltage out of the signal generator (with internal resistance) feeding a filter. The plot shows voltage amplitude and phase. If I right click and change (v(in)) to re(v(in)) then I still need to do something with im(v(in)). I either have improper syntax or I'm putting it in the wrong place. If I get it right, what is the label on the left y axis? It seems to be voltage I was looking for ohms. Maybe I need to plot something different from the start? Thanks for working me through this, Mike

Yes... OK, there are many ways you could do this. Here's just one.

I'm assuming you will do an AC analysis, to get the response versus frequency. Excite the circuit with a current source, say I1. It's convenient to set it to AC 1 amp, zero phase, but not necessary. Have that current flowing INTO the circuit. Then the voltage at the input divided by the current will be a complex quantity equal to the net impedance. If you made the current = 1A, you don't even have to divide by it, but there's an advantage to do so... The real part of that will be resistance, and the imaginary part will be reactance. Let's say you labeled the input node "in" Then to start, plot V(in). The default is a dB plot. Now right-click on the label at the top. A box will come up; change the plotted quantity to re(V(in)/I(I1)), and "ok" that. Now put your cursor over the plot, and right click. Select "manual limits". Set the left vertical axis to linear. You should now have a plot of the real part of the input impedance, displayed in ohms. But wait! If you change the plotted quantity back to just V(in)/I(I1), it will plot the real part (resistance) on the left axis, and the imaginary part (reactance) on the right axis.

Dat help any?

Cheers, Tom

Thanks Tom, That info got me almost all the way there. The last step, has no effect.

For the last step I did as follows, Right click >> Is it possible to graph R and X of a circuit?

Left double-click on the Y-axis numbers on the graph. The "Representation" selecto box that shows Bode, change that to Cartesian. You will see your results in real an imaginary components. You can get you plot to display |Z| and phase if you select Linear for the Bode plot.

back

Notice that what I said to plot at first was re(V(in)/I(I1)). That plots just the resistance; the dashed line, corresponding to the right vertical axis, is zero. Drop the "re" part (and the parens if you wish) and you display the resistance on the left axis, with the solid line, and the reactance on the right, with the dashed line. Cool thing is that the units are displayed as ohms; it knows that voltage divided by current is ohms. And there's an "i" in the label for the reactance, as there should be.

Cheers, Tom

Stop at this point and admire the trace.

Now click on the same node again.

Look at the two traces. You now have the "re(V(in))" and just "V(in)" in the top of the graph.

Now, move the mouse over the "V(in)" label at the top of the graph. Click and edit like you did before.

You missed a step.

You will only be able to plot voltages or currents. If you force a current of 1A through an impedance, you get a voltage equal to that impedance so this is how you can get impedances to plot.

Z = V/I

If you want you can use 1mA instead and divide by the 1mA. You could use the 1uA or 1pA or 1fA or 1aA or ......

Not true! If you plot V(in)/I(source) -- that is, a function which is a voltage divided by a current -- it will display in ohms, complete with the Omega symbol. That way, you can use any current you want (not that it makes a difference for an AC analysis, which is taken to be linear around an operating point -- but it's nice for transient analyses), and properly display the result in ohms. This works for quite a few other units, too: volts*current is power, etc.

Cheers, Tom