For a long time this seemingly simple amplifier circuit has baffled me. I hope to be able to understand it this time.

I will be using the example from the Art of Electronics 3rd editon (page 88).

Here is the circuit in LTSpice. I changed one thing: I added C2 at the output, otherwise the output voltage is between 7 and 17 V at the output and now with the DC being blocked it's +/- 4.5 V (or at least it is in the LTSpice simulation).

For signal-in I set it to a sinewave with an amplitude of 0.5 V. The book says the gain is -10 so this will easily fall between...wait a minute, can the signal-out even go negative with a supply voltage of +20 V?

Anyway, let's test it on the breadboard.

Failure! I think I replicated the example circuit perfectly, but my output signal (yellow) is even smaller then the input (blue). Surely I made a blunder somewhere? Note that when I made the image below I had the supply voltage around 8 V as the 20 V did not seem to make a difference.

This is more like it. All I changed was the transistor. Before, I used a C376 and now a BC548. Unfortunately I cannot seem to find a datasheet for the C376 to find out the cause. I expected it to work as a regular CBE NPN transistor.

What is the gain when I use the supply voltage as in the example from the book (20 V)? About 5.

• On a 1 Vpp signal I measure 888 mV as signal-in and 4.56 V signal-out.
• On a 3 Vpp signal I measure 3 V as signal-in and 14.6 V signal-out.

What is the minimum and maximum voltage when I set my scope to DC coupling? With a signal-in of 1 Vpp the signal-out has a min of 7.6 V and a max of 13 V.

With a DC signal-in of 500 mV (measured at signal-in as 540 mV) the signal-out is an average of 10.3 V.

I am no closer to understanding the circuit, but at least I have a working practical example now.