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I'm trying to obtain the output resistance (resistance from collector node) of the following circuit

enter image description here

By inspection, it should be the output resistance of a common emitter amplifier with degeneration

$$R_o = r_o(1+g_mR_1)$$

This transistor has

$$I_c = 20 mA $$ $$V_A = 36$$ $$r_o = \frac{36 V}{22mA} \approx 2k\Omega$$ $$ g_m = \frac{20 mA}{25 mV} \approx 720 m\Omega^{-1}$$

So its output resistance should be

$$R_o = 2k\Omega(1+720 m\Omega^{-1}250\Omega) \approx 360 k\Omega$$

However, the simulations are giving strange results. I tried simulating a small sinusoidal voltage at the collector and measuring the current as to get the resistance of the node, but trying to get the voltage amplitude below 1 V yields very strange results. Here are some captures of the voltage at the collector, which should be sinusoidal, but it isn't. The frequency of operation is 1 kHz

1000 mV ampltiude:

enter image description here

100 mV amplitude:

enter image description here

10 mV amplitude:

enter image description here

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  • \$\begingroup\$ Try setting the maximum timestep to \$10\:\text{ns}\$ and see if that helps with the \$10\:\text{mV}\$ and \$100\:\text{mV}\$ amplitudes. \$\endgroup\$
    – jonk
    Commented Apr 14, 2019 at 3:26
  • \$\begingroup\$ Open control panel and add a couple zeros in front of the default value of "relative tolerance". Your changes to the collector current are really tiny (like in the ppm range). \$\endgroup\$ Commented Apr 14, 2019 at 3:47
  • \$\begingroup\$ Your analysis is a small signal one, so you better switch LTSpice to frequency domain simulation. The latter is exactly what you've done manually, linearization around bias point and analysis. Time domain.TRAN is going to give weird results in a bad conditioned problem like your one. \$\endgroup\$
    – carloc
    Commented Apr 14, 2019 at 7:32
  • \$\begingroup\$ @jonk, your answer is right, it was solved by changing the maximum timestep! \$\endgroup\$
    – MPA95
    Commented Apr 15, 2019 at 1:42

1 Answer 1

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When needing high dynamic range outputs in .TRAN, the .opt plotwinsize=0 card should almost always be added to the schematic. That disables waveform compression which is, by default 300 points (you can check that by R-Click on the waveform window and select View > Mark data points). What you have there is a mV ranged signal on top of V worth of DC.

The downside is (may be) that the .raw file will get larger, in which case the .save card can be used.

If that doesn't help, try what @jonk, and then @SpehroPefhany said, in that order.

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  • \$\begingroup\$ @jonk comment solved it! it was a matter of timestep... \$\endgroup\$
    – MPA95
    Commented Apr 15, 2019 at 1:42

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