# Frequency Divider Circuit

Discussion in 'General Electronics Chat' started by nDever, Jan 13, 2011.

1. ### nDever Thread Starter Active Member

Jan 13, 2011
154
4
Hey Guys,

I have a 24 MHz crystal and some 7400 series ICs. I would like to achieve a signal clocked at 2-5 MHz with a 33% duty cycle. Does anyone know of a simple circuit that I could quickly build to achieve this? If you would like to know some of the ICs I have, let me know. I know with certainty that I do not have a frequency divider IC.

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,504
512
From Wiki: "Frequency division: a chain of T flip-flops as described above will also function to divide an input in frequency by 2^n, where n is the number of flip-flops used between the input and the output."
http://en.wikipedia.org/wiki/Flip-flop_(electronics)

3. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,851
966
Divide by eight with flip flops to get 3Mhz.

The duty cycle will depend on the duty cycle of the signal being divided with this method.

4. ### tom66 Senior Member

May 9, 2009
2,613
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Simplest way to do this that I can think of is with a binary counter and a quad-NOR gate, but I can only get 25% duty cycle.

5. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
Regardless of duty cycle of the original signal, any division by the multiples of 2 will result in perfect square wave output.

Divide a square wave by odd divisor often result in non-square wave output waveform.

In your case, you can try a division of 6 to get 4MHz, i.e. divide by two follows by a division of three.

The division by three could give waveform without 50/50 duty cycle. Just a thought though. Please note that 99% of counter circuits out there in the internet would divide by three first then by two to result in a square wave output. That's not you would have wanted.