I have a synchronous 4-bit up/down counter, and in the data sheet it stands that there are 4 pins where I can enter a 4 bit preset number, but it doesn't say what this number represents. I believe that it is the first number in the counting sequence (e.g. if preset is 0001, the counter will start from 0010), am I right?
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3 Answers
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The 74HC192/3 have asynchronous load. As soon as** you assert the /LOAD input, the outputs will change to reflect the 4-bit input you have applied to the PRESET inputs, and the outputs stay there as long as /LOAD is low regardless of clock or anything else (except RESET which overrides the /LOAD).
When you remove the /LOAD (and RESET is not active) then the counter may count up or down from the preset number, if it receives a rising clock edge on one of the clock inputs.
** after some tens of nanoseconds with a 5V supply
answered May 14, 2014 at 17:34
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As you can see on page 8 of the datasheet you link, the preset bits are set in the internal counting register according to their name, i.e. P0 will be stored in Q0, P1 in Q1 and so on. Note that you will need an active low strobe on \$\overline{\text{PL}}\$. After the parallel load normal counting is resumed, so the starting number is what you set, then at the next clock positive edge the number will be increased. Again, refer to page 8.
e.g. if you preset 0001 the counter will store 0001 and hold it until the next clock rising edge, then it will go to 0010 and so on. That is true assuming that the counter is in UP mode, i.e. you are clocking it from CPU.
answered May 14, 2014 at 17:27
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You should read the datasheet first. It clearly says what preset is.
Present is the initial count value. If preset is 0001, counter will start from 0001. It will become 0010 (if up counting) or 0000 (if down counting) only in the next clock.
