What should be the potential drop across the Oscillator? I am facing a problem that a simple flashing program also does not work. I checked Oscillator pin and found out that the potential drop across these pins is 2.17 VDC. The Microcontroller has 5VDC. What does this 2.17VDC indicate? Does it mean Oscillator is not working?

  • \$\begingroup\$ The clock is 'probably' working since your meter is showing something near half Vdd. You need to put an oscilloscope probe on the oscillator output pin to tell for sure. \$\endgroup\$ Apr 26, 2016 at 11:22
  • \$\begingroup\$ Are you able to download your program? If so, then your osci is correct \$\endgroup\$
    – Swanand
    Apr 26, 2016 at 11:29
  • 2
    \$\begingroup\$ If the oscillator is working and you connect a voltmeter to the oscillator pins then it will probably stop oscillating. It will probably start again when you remove the voltmeter probes, but may not (cycle power to restart.) \$\endgroup\$
    – JRE
    Apr 26, 2016 at 12:03

1 Answer 1


What does this 2.17VDC indicate? Does it mean Oscillator is not working?

The diagram below shows a simple inverter powered from a 9V supply. It shows the gate voltage being raised from a little below half supply to a little above half supply: -

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The output changes from about 9V down to about 1V i.e. this is an inverter and quite often features in the heart of a CMOS oscillator. The gain of the circuit is about -18.8 (at the centre area) and, if there was a high value resistor placed from input to output, negative feedback would ensure that the voltage on the input and output settled at around 4.5 volts (half rail).

If this inverter were on a 5V rail the inputs and outputs would settle at about 2.5 volts. Next apply a XTAL and capacitors around this device and you get an oscillator.

Use a multimeter on either input or output and you'll measure the average voltage on those pins i.e. the standing DC voltage. Your meter won't see the high frequency oscillations because it measures DC. If you tried your meter on AC it probably wouldn't work any better because the meter just won't work at several MHz.

I would estimate that your oscillator is working fine based on your crude DC measurement - it certainly give me no cause for concern.


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