Author Topic: Crystal oscillator startup time  (Read 456 times)

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Online HwAoRrDkTopic starter

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Crystal oscillator startup time
« on: January 14, 2024, 09:57:00 pm »
If a crystal oscillator - for example a common 4-pin 3225-package 32.768 kHz oscillator - has specified a startup time of a few milliseconds, what exactly defines "startup"?

What I mean is, is that the time it takes from power-up (i.e. power applied to VDD pin) for the output to be stable? Or is that the time it takes the output to be stable after the 'enable' pin is activated? None of the datasheets I've seen specify exactly.

I suspect it's the former, but I'm not sure.

Also, another question: what would be the point of having an AC-coupling cap, for instance 100 nF, in series with an oscillator's output? I have seen a reference design with this. It would seem to serve no purpose to me. :-//
« Last Edit: January 15, 2024, 02:03:01 pm by HwAoRrDk »
 

Offline MarkT

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Re: Crystal oscillator startup time
« Reply #1 on: January 14, 2024, 10:29:07 pm »
Can you provide the part number?
 

Online langwadt

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Re: Crystal oscillator startup time
« Reply #2 on: January 14, 2024, 10:35:50 pm »
I'd expect it to be time from enable, it takes time to get the xtal oscillating

The series cap might be for when driving an input meant for an xtal and not an oscillator?
 

Online HwAoRrDkTopic starter

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Re: Crystal oscillator startup time
« Reply #3 on: January 14, 2024, 11:07:58 pm »
I'd expect it to be time from enable, it takes time to get the xtal oscillating

I was thinking that the enable pin would simply be gating the buffer on the output, and that would have no effect on the internal crystal circuit. You think the opposite?

The series cap might be for when driving an input meant for an xtal and not an oscillator?

I believe it's going to an input that is just meant to take a ~50% duty cycle square wave 3.3V clock signal.
 


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