I found this circuit online and made it and it works.

It uses an LM358 op amp to amplify an oscillation from a 555 in astable mode.


simulate this circuit – Schematic created using CircuitLab

  1. How were the values of R2 and R3 arrived at? I know they should be equal but why 100k?

  2. What function does the capacitor serve? And how was its value arrived at? Is there a formula for these things or just trial and error?

  • \$\begingroup\$ Why do you need to amplify since what you get is just 0V and 5V? \$\endgroup\$
    – Jason Han
    Aug 2, 2017 at 23:39
  • \$\begingroup\$ I'm not sure what you are asking. The 555 will actually be multiple 555s and the output signal is only around 3V. \$\endgroup\$
    – thatsagoal
    Aug 3, 2017 at 9:37

2 Answers 2


The job of R2 and R3 is to create a voltage that is roughly half the supply. Lower values give you lower impedance of the output voltage, but also increase current. In this case, the voltage only needs to drive a opamp input, which is high impedance. The 50 kΩ output of the divider is plenty low enough, and 45 µA of current was deemed acceptable.

C1 reduces noise on this voltage reference. The noise can come from the power supply directly, and can be picked up from other signals by stray capacitive coupling. With 100 nF and the 50 kΩ impedance, the low pass filter rolloff frequency is 32 Hz.

The choice of 100 nF was probably a knee-jerk reaction. That seems to be a common value to grab as a substitute for actually thinking about the tradeoffs. I would probably have used 1 µF. The downside of too much capacitance is that the circuit takes too long to settle after initial turn-on. With 3.2 Hz low pass filter, the settling should still be fast enough on a human scale. If this circuit is switched by machine, which then wants it to be ready after a short time, then 100 nF could be better. That doesn't seem to be what is going on there, though.

  • \$\begingroup\$ I think the current increase is a minor concern when choosing the lower bound for R2 and R3. The major one is that one has to increase C1 accordingly to get the filter right. \$\endgroup\$ Aug 2, 2017 at 12:49

The values of R2 and R3 are chosen so that the unwanted voltage \$V=G*I_{in}*R/2\$ (Where \$I_{in}\$ the maximum input current of the opamp and \$G\$ is the closed-loop gain) remains acceptably low. This voltage will appear as error on the output.

C1 is chosen so that the low-pass filter formed by R2||R3 and C1 provides the desired PSRR in the target frequency range.


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