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Here is a picture of an amplifier that I'm simulating using LTSpice. When I insert a sine wave of 0.01 volts and 314Hz as an input, I get this:

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The output is taken from the collector and is oscillating between about 2.65 and 2.35 volts. This would mean that the amplitude of the output is about 150mV. This divided by the input amplitude of 10mV gives a gain of 15.

When I do a Bode plot of its frequency response, using 0.01V again as input, I get this:

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Past 100Hz, I get a gain of about -16.4 decibels! This is nothing like the gain of 15 that is claimed by the linear graph. When I convert -16.4 decibels to a gain ratio, I get something like 0.15. So what am I doing wrong?

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  • \$\begingroup\$ If you're not comfortable with AC analysis's "dbV" scale, you can click on that vertical scale (at the left) and change it to "logarithmic volts" or to "linear volts". \$\endgroup\$
    – glen_geek
    Commented Jul 20, 2019 at 14:27
  • \$\begingroup\$ Plot "V(out)/V(in)" otherwise you are plotting the amplitude in dBV \$\endgroup\$
    – sstobbe
    Commented Jul 20, 2019 at 15:45

2 Answers 2

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LTspice always uses db(1V) for the y-axis of the ac simulation results. So, since you actually used 0.01V as the input signal amplitude the actual gain of the circuit is 100 times greater than the output dB value. If you converted -16.4dB to a gain of 0.15 and you expected a gain of 15 then your simulation is actually spot on.

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  • \$\begingroup\$ I'm quite new to Spice, could you explain what do you mean by db(1V)? \$\endgroup\$
    – S. Rotos
    Commented Jul 20, 2019 at 14:23
  • \$\begingroup\$ @S.Rotos: Spice assumes that the input source that is used in AC analysis has an amplitude of 1V. $$A_v = \frac{V_{out}}{1 V}$$ So, when you start reducing this, it looks like it has a lower gain on the plot. \$\endgroup\$
    – Linkyyy
    Commented Jul 20, 2019 at 16:11
  • \$\begingroup\$ Interesting! Thank you. \$\endgroup\$
    – S. Rotos
    Commented Jul 20, 2019 at 21:13
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You've plotted the power in the output while doing a frequency sweep of the input. This is not frequency response!.

$$ 10^{-16.4/20} = 151 \;\mbox{mV} $$

For frequency response, you should plot the ratio of output power to input power (or equivalently due to how logarithms work, subtract the input power in dB from the output power in dB... 10 mV is -40 dB).

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