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What is the proper way to use an external 10 MHz clock signal for an HP/Agilent signal generator (in my case: 8648C)?

With my Tektronix, Rohde & Schwarz devices, internal/external reference has to be selected which makes sense.

But the manual for 8648C is just totally silent: How to activate it, how to ensure that is is actually used and locked?

Is it activated automatically? On powerup? When the external clock is connected? 5s after it is connected? ... ?

According to the manual:

10 MHz REF INPUT and OUTPUT These connectors provide the input and output ports for the instrument's timebase reference. The instrument will lock to a 2 MHz, 5 MHz, or 10 MHz external reference source connected to the input that is within ±5 ppm. When the internal timebase is being used, the output connector provides a 10 MHz, 1 Vrms level signal.

But with the scope I can see the 10 MHz signal at the REF INPUT always. Sure enough, the manual also doesn't say how to deactivate the "internal timebase".

I am not sure if the device actually uses the external reference because the measurements suggest I have some frequency mismatch.

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1 Answer 1

Reset to default
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This is a classic ILL or Injection Locked Loop so that no enable switch is needed.

The instrument will lock to a 2 MHz, 5 MHz, or 10 MHz external reference source connected to the input that is within ±5 ppm.

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  • \$\begingroup\$ So as soon as I connect an external reference within spec (10 MHz, between 0.5V and 2Vrms into 50 Ohm) and wait for a few ms the instrument will be locked to the external reference? Sinusoidal, rect, triangle ... does not matter? How would one notice is the amplitude is too low or if locking fails for some other reason? \$\endgroup\$
    – divB
    Commented Apr 10, 2018 at 4:33
  • \$\begingroup\$ It actually works now. I would still appreciate clarification on these questions on the answer. \$\endgroup\$
    – divB
    Commented Apr 10, 2018 at 5:49
  • \$\begingroup\$ Is your source accurate and 1V or so? Usually this is 1e-10 or .0001 ppm error SC cut OCXO 10MHz. Suggesting a triangle seems you are using the wrong Ext f \$\endgroup\$
    – D.A.S.
    Commented Apr 10, 2018 at 5:51

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