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Binary Counters

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h3lladvocate

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I'm kind of new at this. I have a 16 pin dual 4 bit counter. I've set up a 555 timer IC so that the output high frequency is about 15Hz. I've connected 5 V to the Vcc on pin 16, the output for the 4 bits each to an LED, so i can see the state of each, pin 8 is connected to 0 V, and pin 2 is connected to the output of the timer. This is what im really not sure about. I've looked up the data sheet for the counter(74HC390) and it only has 3 inputs, a reset, divide by 2 and divide by 5. Im not sure where to connect the output of the 555 timer to. Can anyone help me with my counter. I can't find any tutorials on how to wire one up, and the manufacturer "assumes" you know how to use it...
 
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First of all I hope you are using resistors in series with the LED's on the output pins. For a 5V supply and red,green, or yellow LED's I would recommend a minimum of 150 ohms.
Both counters do have 4 outputs. They are labeled Q0,Q1,Q2,and Q3. The master reset for normal operation is connected to pin 8. If you connect pin 3 to 4 and feed the output from the 555 to pin1 the output from Q3 will be 1/10 the input frequency. If you connect pin 7 to pin 1, and the 555 output to pin 4, then pin 3 will still be 1/10 the input frequency but the waveform will have equal logic 1 and logic 0 levels. Even if you are not using the CPO,CP1 and the MR pins of the other counter these pins should be connected either high(Vcc) or to GND. They should never be left floating.
 
Thanks, my problem was I didn't have the reset connected to pin 8. I guess that would explain why the light were basically randomly flashing. I dont get the whole waveform and different frequency part, could someone explain that?
 
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k7elp60 said:
the waveform will have equal logic 1 and logic 0 levels.

What k7elp60 is saying is that the length of time that the output is high ( 1 ) will be the same as the length of time the output is low ( 0 ) in other words it will have equal duty cycle.

If you feed 10Hz into a divide by 2 then it's output will only change state every second cycle. This means that the output frequency will be 5Hz, or the input divided by 2.

So, duty cycle is the ratio of on time to off time in a single cycle, and frequency is the length of time it takes for one cycle to complete. Therefore the duty cycle can change and the frequency will still be the same.
 
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