I'm using a 27MHz oscillator and trying to measure its voltage level and waveform with oscilloscope. But I have no idea how to match the impedance of oscillator and oscilloscope, since the data sheet of the oscillator doesn't specify its output impedance. Could anyone please tell me which values I may assume?


2 Answers 2


Matched impedance is only necessary in certain specific situations. Typically, when you have a very fast driver and you wish to maximize power transfer or minimize reflections (or both!). And in these cases, the driver typically has a very low output impedance (a few ohms) and a series resistor chosen so that the output impedance is 50Ω.

27MHz is fast, but not really fast. Use a standard scope probe set to 10x (so it's a 10MΩ/~10pF load) and you should be able to probe the signal just fine. This way, your scope loads the output as little as possible. At these frequencies, loop area is also important, so use a ground clip and make sure you have a good ground (probing technique is important when you want accurate scope traces!).

At 27MHz, the capacitive load of the oscilloscope probe will be the dominant impedance, which as W5VO points out will be around 500Ω. If you just want to look at levels/frequency, this won't be an issue (the oscillator will drive a 15pF load), but if you are debugging signal integrity issues, this capacitive loading will change your signal enough that passive probes are completely out of the question; an active probe is necessary for these situations.

Finally, if you decide that you need a load, use a load the simulates the input impedance of the circuit that the oscillator will be driving. Typically, you would just ballpark it, and use, say, a 1kΩ/10kΩ resistor as a dummy load.

  • 1
    \$\begingroup\$ It's also worth mentioning that the capacitive load of an oscilloscope probe (x10) will be in the ballpark of 500Ω at 27MHz - the probe may cause too much loading. \$\endgroup\$
    – W5VO
    Apr 27, 2016 at 4:58
  • \$\begingroup\$ @W5VO Thanks for pointing that out, I neglected to explain that. Edited. \$\endgroup\$
    – uint128_t
    Apr 27, 2016 at 5:07

Comon value for digital Output is 50E, but that might be just the DC part. The AC impedance depends on the frequence and the internal circuit.


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