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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.

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    \$\begingroup\$ You did not draw the MR (MasterReset) on the diagram. Is it not necessary to pull that to ground? Otherwise you counter may randomly reset, not giving you a nice divided by 4 signal? \$\endgroup\$ – Mike de Klerk Oct 7 '13 at 9:42
  • \$\begingroup\$ Hmm. MR is CLRA? Didn't see MR pin on SN74LS393 datasheet. \$\endgroup\$ – Darkkey Oct 7 '13 at 10:02
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    \$\begingroup\$ I used the datasheet from here: pdf.datasheetcatalog.com/datasheet/on_semiconductor/… In that document its pin 12 and its description is "A HIGH signal on MR forces all outputs to the LOW state and prevents counting.". So pulling it to ground to be sure might be good practise? \$\endgroup\$ – Mike de Klerk Oct 7 '13 at 12:09
  • \$\begingroup\$ @MIKE-DE-KLERK You're right, CLRA (or MR on On Semiconductor version of 74LS393) should be grounded. Thanks. \$\endgroup\$ – Darkkey Oct 7 '13 at 12:28
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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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  • \$\begingroup\$ So, I should connect 74HC04 1A and 1Y to crystal pins and on 1Y I'll get crystal generated frequency, right? \$\endgroup\$ – Darkkey Oct 7 '13 at 9:49
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    \$\begingroup\$ Yes - so long as you connect the resistor in parallel and two series capacitors to ground. Sometimes you need to buffer the output of the oscillator (often using another two inverters in series) but if you are feeding it only into the counter IC you should be fine. \$\endgroup\$ – David Oct 7 '13 at 9:52
  • \$\begingroup\$ @Darkkey: No. You need an unbuffered inverter, such as those found in the 74HCU04. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 7 '13 at 10:07
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QA is A/2, QB is A/4. Simply connect the oscillator (not the bare crystal) to A, and QB to the circuit that needs a 460.8kHz input.

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