Electronic – How to get 460.8KHz from 1.8432Mhz oscillator using counters

clockcounter

I need to supply 460.8kHz clock to IC (NCR20C12). However, I can only get 1.8432MHz Oscillator. How can generate clock divided by 4 from it? I have a couple of SN74LS393 counters and want to ask, is it possible to divide clock with counter and how to do it? Can't find solution on google, may be because of messing some terminology.

Is the following circuit correct?

schematic

simulate this circuit – Schematic created using CircuitLab

Sorry for my bad english, stupid question and thanks in advance.

Best Answer

You could do it with your SN74LS393 counter chip by using the correct output but you cannot simply connect the crystal like that. You need to turn the output of the crystal into a digital pulse.

You can do this in any number of ways but the most simple is a "Pierce Oscillator":

Pierce oscillator

U1 is an inverter, something like a 74HC04 or you can find tiny SMD parts in SOT/SC-70 cases with a single inverter. C1 and C2 should be chosen according to the capacitive load your crystal expects, check the datasheet. R1 is a feedback resistor, you could start with a high value and work downwards until you find it is stable.

Another common way to do this is using D flip-flops, which is essentially how the internals of your counter chip will be working. The Wikipedia page on frequency dividers illustrates a divide-by-four counter nicely:

Divide by four counter using D flip-flops

I have previously done this using 74HCT74 ICs.

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