I am trying to use a 555 timer in my circuit to produce a signal (rather generally) above 100kHz. I have been successful with this goal, but have had an issue where the voltage measured on my positive and GND rails drops significantly whenever there is a change.

I am using a bench supply that is set to supply 9v at a maximum of 80mA, but the whole circuit only draws 19mA when operational.

555 Timer This photo shows the 555 timer, with a 470Ω resistor connected to pins 7 and 8, a 10kΩ resistor connected to pins 6 and 7, a 330pF capacitor from pin 6 to GND, and a 1nF capacitor from 5 to GND.

Waveform measurement of supply voltage (yellow) and the 555 timer output voltage (blue) The image above shows the measurement of the supply voltage in yellow and the 555 timer output voltage in blue.

Waveform Here is a zoomed in version of the second photo.

This voltage drop is causing the other ICs in my circuit to not work properly.

My guesses is that there is a capacitor somewhere that is too large or small, that this is a property of the power supply or 555 timer, or that the sudden change of voltage requires a lot of current to drive that I somehow cannot detect with my power supply.

Is there any way that I can minimize the voltage drop of my source?

  • 6
    \$\begingroup\$ Put a 10uF capacitor on the supply rail. \$\endgroup\$
    – earl
    Commented Dec 1, 2023 at 17:39
  • \$\begingroup\$ That works! How come that fixes this problem? \$\endgroup\$ Commented Dec 1, 2023 at 17:42
  • 1
    \$\begingroup\$ Large inductance between the power supply and breadboard from interconnecting wiring. \$\endgroup\$
    – vir
    Commented Dec 1, 2023 at 17:51

2 Answers 2


The problem is most likely the lack of a power supply decoupling capacitor. When using a breadboard to prototype analog circuits I always put a 10uF capacitor on the supply rail.


The root problem is not external inductance, it is an internal design flaw.

The original bipolar 555 design is notorious for output stage cross-conduction. This is a condition where both output transistors, the one for pulling the output up and the one for pulling it down, are conducting at the same time. IOW, one turns on before the other one is off. This produces a spike in the current being drawn by the device, an almost-dead-short across the power rails that sometimes draws over 300 mA. At high operating frequencies, the cross-conduction time becomes a significant percentage of the period of one cycle, increasing greatly the average current draw of the circuit.

The solution is a larger-than-normal decoupling capacitor, something like a 10 uF to 47 uF electrolytic in parallel with a 0.1 uF ceramic.

  • 9
    \$\begingroup\$ Yes +1. Another solution is to toss the bipolar NE555 in the e-waste bin and use a CMOS variant such as TLC555. Then 100nF bypass is fine. Also much better for OP's 100kHz frequency, which is at the edge of what the bipolar part can do. \$\endgroup\$ Commented Dec 1, 2023 at 18:36
  • 3
    \$\begingroup\$ IF - he is not relying on the bipolar 555's extra-beefy output stage. \$\endgroup\$
    – AnalogKid
    Commented Dec 1, 2023 at 20:18

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