How to use LTSpice to calculate input impedance? - Page 1
Forum section: electronics / beginners

author: sean87
author_rank: Regular Contributor
timestamp: February 14, 2012, 02:18:08 pm
content: Hi all,I have the multiple feedback low pas filter to design. I want to know the value of input impedance or a graph of it to see if there is resonance or something. Is it possible to do with spice? If so, how can I do it?

author: Bored@Work
author_rank: Super Contributor
timestamp: February 14, 2012, 02:34:31 pm
content: Z = V / IEither add a trace to the plot with an expression using that formula. Or replace the voltage source with a AC 1 0 current source, then plot the input voltage which is equal to plotting the impedance, under this condition.

author: sean87
author_rank: Regular Contributor
timestamp: February 14, 2012, 03:35:53 pm
content: Thanks, I'll try that. Shouldn't I put a resistor in series with votlage source?

author: jahonen
author_rank: Super Contributor
timestamp: February 14, 2012, 04:40:21 pm
content: No. Just plot the voltage of your input node Vin divided by the current of the voltage source V1.Regards,Janne

author: sean87
author_rank: Regular Contributor
timestamp: February 14, 2012, 08:14:51 pm
content: Thanks guys, can you please verify if I am on the right track? Here is the result I got, I will really appriciate if you can tell me what is useful in this plot and what does it mean?! Sorry for being a total noobHere is the circuit I used:

author: Rerouter
author_rank: Super Contributor
timestamp: February 15, 2012, 10:27:53 am
content: it is a frequency vs amplitude graph, the dotted line is the phase of the signal, in short you have a bandpass filter made for roughly 300Hz to 7Khz, with 1Khz as your center frequency, (effective range of a filter is to 63.2% or something close to that)

author: ejeffrey
author_rank: Super Contributor
timestamp: February 15, 2012, 11:28:57 am
content: Yes, the only wrinkle is that you have plotted the result in dB. The formula works out to 20*log10(Z/ 1 ohm). 60 dB is 1 kohm, 64.5 dB in ~1700 ohm. So your input impedance is 1000 ohm at high and low frequency with a peak to 1700 ohm in the middle.At low frequency the capacitors carry no current and therefore R3 carries no current. The node between R1 and R3 is then a virtual ground. The input impedance is given by R1 into the virtual ground. At high frequency, the capacitors are short circuits. In particular, C2 has low impedance to ground, so again the input sees R1 to ground for an impedance of 1 kohm. In between, some current flows through C1, and therefore R3, meaning the node between R1 and R3 is no longer a good virtual ground and the impedance rises. This gives the slight rise in input impedance before C2 causes it to fall off.

author: Spiro
author_rank: Contributor
timestamp: February 15, 2012, 02:16:25 pm
content: Wait a minute. Phase is -180 at ~1kHz. Isn't that negative resistance?

author: ejeffrey
author_rank: Super Contributor
timestamp: February 15, 2012, 02:38:43 pm
content: No, it just means the sense of I(V1) is flipped from your expectation.

author: jahonen
author_rank: Super Contributor
timestamp: February 15, 2012, 03:06:13 pm
content: It is easy to change the scale to linear, just click on the plot window where the y-axis labels are and change the selection from decibel to linear. And yes, voltage source current direction is defined such as that positive current is into voltage source, thus unexpected 180° phase shift. That can be fixed by just putting a - into plot formula.Regards,Janne

author: Spiro
author_rank: Contributor
timestamp: February 15, 2012, 03:13:56 pm
content: I saw that arrow on current clamp icon is opposite of expected current direction. I thought that spicewill automatically orient currents of sources.Trap for young players eeeeyy