Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

50% duty cycle

Status
Not open for further replies.

elnaz

New Member
I have a digital pulse with nonsymmetric duty cycle. How can I make it 50% when it's frequency remains unchanged?
 
what are you using the generate this less than 50% duty signal?

It might be easier to modify that cct to give 50%
 
I am using some kind of delay process to generate this pulse and I can't change it.
I am searching for a logical gate or something to do this automatically.
 
If the frequency of the pulse doesn't vary a tremendous amount, you could feed it to a PLL (no change in frequency) and then use the sine wave output of the VCO to drive a simple squaring circuit (a Schmitt trigger comes to mind here).

Dean
 
Can't use a flip flop because it'll divide the frequency in half and he specified that it had to remain the same.

Dean
 
if the freq doesnt change (much) and neither does the pulse width you could use a schmitt trigger (74HC14 or make one out of a comparator).

With a low-pass filter with the input being the PWM and the output feeding the Schmitt. By putting a diode across the resistor pointing towards the schmitt you can choose the resistor and the capacitor to effectively stretch the pulse to be ~50%

otherwise use a toggle flip-flop to change it output from high-low-high at each leading edge. this will 1/2 your freqency, but there are simple circuits to double frequencies again using XOR gates and capacitors

Why can you change the circuit that makes this sub 50% duty? and why cant you duplicate it to have the 2nd make 50% duty?
 
why is 50% duty nececary? what are you driving?
You can use frequency doubler and then flip-flop to get perfect 50%.
 
panic mode said:
why is 50% duty nececary? what are you driving?
You can use frequency doubler and then flip-flop to get perfect 50%.

err thats what I said!!!
 
What is the frequency, pulse width, and polarity of the pulses? What logic family and supply voltage are you using?
 
I am using the attached circuit to double my VARIABLE frequency,which produces a nonsymmetric pulse based on a not gate's delay.
Because I am feeding some and gates through latches and these are me latches enables,I need it to be perfect 50%,without change in frequency.
I am doing this in Xilinix Foundation's software's schematic editor which provides me with just simple logic gates.(no cap or IC)

Actually I am trying to make LS7084 IC through FPGAs.

But finally I reached my goal through another way with no 50% duty cycle pulse.


(but I still don't know how can I make my pulse symmetric with flip-flop without changing it's frequency :roll: )

Thank you all for your attention
 

Attachments

  • 6easy_4.gif
    6easy_4.gif
    3 KB · Views: 521
Styx said:
panic mode said:
why is 50% duty nececary? what are you driving?
You can use frequency doubler and then flip-flop to get perfect 50%.

err thats what I said!!!


Sorry Styx,

It must have happened at about same time. Just before hitting
submit button I was called away from my desk to a meeting.
I tought I hit it but I gues I missed it. When I came back your post
was there already and I didn't check.
Mea culpa... :roll:
 
is this your output?

Z is the output with a positive edge clock input , with a negetive edge clock invert y and Z.

if you use a one shot you should be able to stretch the output to whatever you want plus you can use a positive or negetive edge to trigger it..

also what i have as Y input will follow F(in) with a negetive edge Clock input
with a positive edge Clock input Y is the oposite of F(in).
the width of Z or F(out) is the propagation delay of the exclusive Nor gate.as you said .
 
there has to be a better way.. anyway this is the other one
 

Attachments

  • image020.jpg
    image020.jpg
    82.7 KB · Views: 484
Yes,we can use two monostables (a positive triggered and a negative triggered) and then add their outputs to get it symmetric.
But as i said before I don't know how can I Make a monostable with basig logic gates I have in simulation editor.
 
Yes,we can use two monostables (a positive triggered and a negative triggered) and then add their outputs to get it symmetric.
But as I said before I don't know how can I Make a monostable with basic logic gates I have in simulation editor.
 
how to use a 74hc175 to make a 50% duty cycle?

i connect the MR pin to vcc(5v) and the d pin to qnot pin .i get the input frequency to clk pin i get the output from q .but it didnt work.help me in that area?
 
Status
Not open for further replies.

New Articles From Microcontroller Tips

Back
Top