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How can I build a simple counter with a HC393 chip?

Discussion in 'General Electronics Discussion' started by mohrr, Jun 20, 2012.

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  1. mohrr


    Jun 20, 2012
    Hi, I am going to be using a counter to take a 5 kHz pulse and bring it down to ~19.5 Hz. I am completely clueless how to begin constructing this circuit, other than the fact that I know that I need to set the clears to zero (ground). Thanks!
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    So you've already worked out that you want to divide the 5 kHz frequency by 256? That will give you an output frequency of 19.53125 Hz.
    The HC393 contains two 4-bit counters, each containing four cascaded divide-by-two stages. The clock frequency is divided by 2 at the QA output, divided by 4 at the QB output, divided by 8 at the QC output and divided by 16 at the QD output. Look at the waveform diagrams in the data sheet and you will see how this works.
    As you said, you need to tie both CLEAR inputs low, because you don't want to disturb the normal "ripple carry" behaviour of the counters. You need to clock the first counter from your 5 kHz signal, clock the second counter from the last output of the first counter, and take the output from the last output of the second counter. Here's a connection list:
    pin 1 from your 5 kHz source frequency
    pins 2, 7 and 12 grounded
    pins 3, 4, 5, 9, 10, 11 not connected
    pin 6 connected to pin 13
    pin 8 is your output frequency
    pin 14 to +5V with decoupling capacitor to ground.
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