I would like to know what the output impedance of this circuit is (the impedance seen by the node output).
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1\$\begingroup\$ That will be frequency dependent. You can generate that plot over frequency like this: Related: electronics.stackexchange.com/a/653401/237061 \$\endgroup\$– tobaltFeb 14 at 18:38
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\$\begingroup\$ @tobalt With an ideal capacitor at the output, and including the product rule to reach the constant factor of 2 shown, I find \$R_{_\text{OUT}}=2\underset{t\to 0}{\lim}\frac{ t}{C}\$. Which approaches being arbitrarily small. Or about what you said. ;) \$\endgroup\$– periblepsisFeb 14 at 20:11
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1 Answer
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To find impedance in LTspice you can add a current source, right click it and under Advanced set AC Amplitude to 1. Connect it to the point you want to measure, in this case output, and then run an AC analysis. Set the analysis to something like .ac oct 200 1 1E9, the last number is the upper frequency, 1E9 is 1 GHz, you can reduce it if you don't need to see that wide a bandwidth.
Once you run the analysis, click on output to get a waveform, then right click the scale on the left side of the waveform viewer, you should get a popup that will let you change it from dB to Logarithmic, when you change it the scale will read directly in Ohms. You can right click the right hand scale and select Don't Plot Phase to hide the phase graph, sometimes it's hard to tell which line is the magnitude and which is phase, so that hides the phase line.
You may need to disable the pulse generators and make those points a fixed voltage.
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\$\begingroup\$ Thanks for your input. Is this analysis if I am not driving an ac signal? \$\endgroup\$ Feb 14 at 17:53
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\$\begingroup\$ @DimitarZhekov Yes, I don't think you can drive it with a signal and also do the AC analysis. You might try setting each pulse input to either high or low manually and running the analysis for each possible combination. \$\endgroup\$– GodJihyoFeb 14 at 17:59
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\$\begingroup\$ I see... The scale when changed to logarithmic is in mVs. At 200kHz I read a value of 0.249mV which means it is around 0.2 &ohms;. I am quite surprised by that because I thought the output would be a high impedance output. \$\endgroup\$ Feb 14 at 18:05
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1\$\begingroup\$ @DimitarZhekov Sorry, I accidentally left out one step, you have to set the current source's AC Amplitude to 1. I edited the answer to include that. \$\endgroup\$– GodJihyoFeb 14 at 18:16
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1\$\begingroup\$ @DimitarZhekov I'm not sure about matching it, there might be a better way than this, I'll try to look into it when I have more time. \$\endgroup\$– GodJihyoFeb 14 at 18:19