I designed an opamp using discrete components (no specific purpose).

This is my circuit: circuit diagram

Real circuit: real circuit

I test the opamp by simple voltage follower circuit. Everything seems right.

Yellow: input signal. Blue: output signal. They are almost identical. Therefore, they look like single sine wave. follower waveform

However, when I connected a large resistor (470kohm) between non-inverting input and my signal source, the output amplitude decreased (reasonable behavior due to not-large-enough input impedance) and considerable phase shift happened (this is what I am curious about). source resistor circuit

Yellow: input. Blue: output. source resistor waveform

XY mode view. About 26 degrees of phase shift. xy mode

Could anyone explain to me what cause the phase shift? According to discrete transistors pico-farad level junction capacitance, this should not be the main reason for the phase shift?

By the way, is there any recommended book to read if I want to learn how to design opamp in transistor level?

  • 1
    \$\begingroup\$ Books on op-amp design tend to be aimed at CMOS IC designers. I would suggest the app notes from National, TI (including Burr-Brown) and Linear Technology (anything by Bob Pease or Jim Williams), there some good insights into the innards there and on topics such as compensation. Hans Camenzind's book on analog IC design is free for download. More aimed at audio, but Douglas Self's books are interesting. \$\endgroup\$ May 2, 2023 at 2:47
  • 2
    \$\begingroup\$ This is a well thought, well written and well documented question. Good job OP! \$\endgroup\$ May 2, 2023 at 15:15

1 Answer 1


Around 10pF of capacitance would cause that amount of phase shift with a 470kΩ resistor @16kHz, which is not implausible when you include the breadboard capacitance.

There's also Miller capacitance since the collector voltages are changing. You could reduce that with a cascode configuration.

  • 4
    \$\begingroup\$ In case you missed it, those breadboards have roughly 5pF of capacitance between adjoining sets of contacts; more if they're backed by a sheet of metal. They're great things for finding out what a circuit will do if you take their quirks into account -- but you have to know them to effectively use them to their limits. \$\endgroup\$
    – TimWescott
    May 1, 2023 at 17:26
  • \$\begingroup\$ Your frequency is 20kHz with enough breadboard capacitance to cause reduced gain and phase shift. A breadboard with wires and contact strips all over the place will work much better at 200Hz. \$\endgroup\$
    – Audioguru
    May 2, 2023 at 1:00
  • \$\begingroup\$ Thanks. I didn't noticed that 10pF capacitor has close to 1Megohm impedance at 16kHz, which is competible to 470kohm. Also Miller effect and parasitic start to kick in. Cascode is considered. However, I usually use 5V supply (phone charger). Cascode will consume even more voltage headroom. I will put it to a perfboard later. \$\endgroup\$
    – Willis Lin
    May 2, 2023 at 3:20

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