Working dual-clock speeds
Dan Michaels email (remove spam text)
Roman Black wrote:
>Alok Dubey wrote:
>> use a 2:1 mux , the PIC line as a select line for the mux and the clock
>> source as the line i/ps to the mux..
>> this is the only soln u can go for if u want i select line
>Hi Alok, no I posted a working solution before for Dan, using
>4 diodes and 3 resistors. No problem there, just that it used
>2 PIC output pins to control it.
>The challenge is to do it with ONE output pin??
Yes that "was" the challenge. However, there is a slight problem
with the original by Roman and also with Rich Otteson's sol'n.
Depends on how fast the clocks are.
Any time you use a diode OR or AND gate, you need a pulldown
[pullup] load resistor to keep the diodes in conduction after
the input signal changes direction. So this brings up several
1 - With Roman's ckt, Rload has to be small enough to pull down
the pin in a timely fashion. At hi-freq, it has to be quite
small --> eg, at 20 Mhz and assuming 10 pF capacitance at the
pin Rload = 1/[2*pi*f*C] = 1/[2*pi*20Mhz*10pF] = 800 ohms.
Not so bad, of course, at lower freq.
2 - Now, further implications are that Rload divides with Rout
at the oscillators, so Rout must be << Rload.
3 - Also, since Rload is greater than Rout, risetimes and falltimes
will be different, so the effective duty-cycle of the clock will
not be 50%.
4 - Similarly, with RichO's ckt, I don't think it will work,
since for the one oscillator you need a "pullup" resistor
at the common output pin for correct operation, while you need
a "pulldown" resistor at the same point for the other oscillator
to work properly.
Unfortunately, these ckts may be "too" simple for good operation.
Probably need an real mux to do it. Roman's sol'n is close, but
you have the problem of how to match Rout vs Rload values, as
- Dan Michaels
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