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calculate rise and fall time

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saeed451

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hi . how i can calculate rise and fall time of a pulse with this specification :

f = 1 khz
v = 5 v
duty cycle = 70%
 

Hi,

It can not be calculated.

For an ideal pulse the timeis considered to be zero.
But it depends on the driver capabilities what's the real value.
It is totally independent of frequency, duty cycle and voltage.

Klaus
 

Hi,

It can not be calculated.

For an ideal pulse the timeis considered to be zero.
But it depends on the driver capabilities what's the real value.
It is totally independent of frequency, duty cycle and voltage.

Klaus

ok so what s the simplest circuit for duty cycle 70%?

f= 1khz

how we should to design ?(without555)
 

Hi,

You could use a comparator as analog solution,
or an (8 pin) microcontroller as digital solution.

And many other solutions..

Klaus
 

Hi,

These are just a few, from RC + 3 inverters, single op amp, logic + crystal/logic + RC, two transistor + RC oscillators, the CD4060/HEF4060 + crystal/RC, a VCO IC, ..., ..., and there are plenty more. The "Relaxation Oscillators" pdf has the one using a single op amp and a few discrete components. Not sure why a 555 is not suitable, it's one of the simplest solutions to get a 1kHz, 70% duty cycle and makes it easy to adjust "on" and "off" time.
 

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  • Circuit Techniques for Clock Sources Linear an12fa.pdf
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  • Relaxation Oscillators.pdf
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  • CMOS Oscillators AN-118.pdf
    77.6 KB · Views: 113
Hi,

These are just a few, from RC + 3 inverters, single op amp, logic + crystal/logic + RC, two transistor + RC oscillators, the CD4060/HEF4060 + crystal/RC, a VCO IC, ..., ..., and there are plenty more. The "Relaxation Oscillators" pdf has the one using a single op amp and a few discrete components. Not sure why a 555 is not suitable, it's one of the simplest solutions to get a 1kHz, 70% duty cycle and makes it easy to adjust "on" and "off" time.

so how i can make 70 % or more with op-amp ?

(sry i couldnt understand Pulse Techniques in colleje very well)
 


Make a basic OP/comparator based relaxation oscillator https://en.wikipedia.org/wiki/Relaxation_oscillator#Comparator.E2.80.93based_relaxation_oscillator

Adjust the duty cycle by moving the comparator threshold up or down (connect the resistor in the lower a left not to ground but to an offset voltage).

tnx all . so if i want to design it with 2 tr and resistor and capacitor in base and collector (ASTABLE) i should to first : calculate frequency with ;


so : T1 = ln2 r1c1 = 0.7 ms and T2 = ln2 r2c2 = 0.3 ms (for duty cycle 70%)

and it give f= 1/T = 1khz

correct?
 

Hi,

I don't think that's so, but I don't know what you mean by "ln2". And the "Relaxation Oscillators" pdf puts f = 1/1.4CR, look at the picture below. You could read pages 5 and 6 if you want to do the BJT version rather than an op amp version as described in the post above yours, there are more pleasant versions of the basic BJT one, such as "Variable Mark to space Ratio."

Relaxation.JPG
 

Hi,

I don't think that's so, but I don't know what you mean by "ln2". And the "Relaxation Oscillators" pdf puts f = 1/1.4CR, look at the picture below. You could read pages 5 and 6 if you want to do the BJT version rather than an op amp version as described in the post above yours, there are more pleasant versions of the basic BJT one, such as "Variable Mark to space Ratio."

View attachment 135905

this is correct becuse 1/ln2 = 1.4 and the formule 1/1.4CR in the picture is for duty cycle 50 % with the same R and C in the circuit . ok ?

so i should to writed it T=1.4(C1R2 + C2R3) and by changing c1r2 and c2r3 we can change duty cycle --- F = 216hz >>> T = 4.6ms >>> t1 =3.2 ms >>>t2=1.4 ms >>>> it gives duty cycle 70 %
 

how i can calculate rise and fall time of a pulse with this specification...

Rise and fall times depends both on the driver and the load. You can observe (measure manually) the rise and fall times with a scope.
 

Hi,

Not sure, quick calculations for one transistor as "on" and other as "off," assuming you only use 1 output from the oscillator, not both, might be using BJT1 as MARK and BJT2 as SPACE:

C1 + R2: 220nF + 9275R = 700Hz C2 + R3: 220nF + 21K6 = 300Hz

I did the calculation as 0.7, not 1.4, as it isn't a 50% duty cycle:

Fo BJT1 = 1/(0.7 * 0.00000022 * 9275) = 700Hz
Fo BJT2 = 1/(0.7 * 0.00000022 * 21600) = 300Hz

9275 isn't a standard value, I'm not up on the standard values for decades of 9, but 21600 is more or less a -1% 22k resistor, typical when you measure batches you buy.

You'd best check the calculations are functionally correct and/or works before using them...

The problem with this kind of thing is that average user like myself purchase capacitors that are usually +-5% to +-20%, and resistors +-1% to +-10%, depending on what you're able/willing to spend, and trimpots are okay but drifty with temp/time/slight vibrations of board for something that needs to be 1kHz. You'd want to measure the passive components to a) separate wheat from chaff, and b) compare calculation to performance.
 

Hi,

i don´t think the calculation is correct.

I rather think you should go for
0.7ms high time (= half a period of 714 Hz)
0.3ms low time ( = half a period of 1667 Hz)

Klaus
 
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    d123

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sry i couldnt understand Pulse Techniques in colleje very well...

That depends; if you want accurate frequency and pulse duration (say 1 kHz +/- 1e-3 Hz and pulse width accurate to the third decimal), you have to use digital tools and techniques.

Please let us know your objectives.
 

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