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I want to generate a PWM signal from a rectified sinus wave similar to the image below using 555 timer IC.

enter image description here

I have designed the circuit below for this.

enter image description here

I want to get a duty cycle of 80% at the peak of the sine wave, and 0% (5% if not possible) at the zero crossings.

How do I choose R16 and R17 resistor values accordingly? How do I do the calculations?


Note: Period of the sinusoidal is 50 Hz. And period of PWM the is

\$ \tau = (R_{11} + 2R_{12}) C_{11} \ln(2) = (101k\Omega) \times (2.2nF) \times \ln(2) = 154 \mu s \implies \text{f} = 6.49 kHz \$ .

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  • \$\begingroup\$ What's the period or frequency of IC11's output? (Save respondents from having to figure that out before working on your actual question.) \$\endgroup\$ – gwideman Apr 9 '14 at 5:12
  • \$\begingroup\$ @gwideman I've added it as a note. \$\endgroup\$ – hkBattousai Apr 9 '14 at 5:48
  • \$\begingroup\$ Also, I note that VCC is 15V. \$\endgroup\$ – gwideman Apr 9 '14 at 5:53
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    \$\begingroup\$ Mind if I ask: why do you need such a circuit? There must be a better way to do what you are doing. \$\endgroup\$ – Vladimir Cravero Apr 9 '14 at 8:38
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    \$\begingroup\$ Note that with the circuit as drawn, the control voltage to duty cycle mapping will be nonlinear. To linearize it, R15 should be replaced by a current source. \$\endgroup\$ – Dave Tweed Apr 9 '14 at 12:17
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The highest voltage into R16 will be at the peak = (20V - 0.6V) = 19.4V

The lowest threshold will be zero. Unfortunately the threshold will be zero the entire time that the sine is < 0.6V. So there will be a portion of the wave that the PWM output is off, or undefined. To solve that, you might want to add some additional current into R17 (an additional pullup resistor).

But solving with what we have, we want Vthreshmax such that pulse length is ~120us. (80% duty cycle)

Vctl = Vcc * (1- exp(-t/RC) )

Vcc = 15V, t = 120us, R = 10kohm, C = 10nF,

Result: required Vctl = 10.5V

R17/(R16+R17) = 10.5/19.4 = 0.54

R17 = 0.54 R16 + 0.54 R17

0.46 R17 = 0.54 R16

R16 = 0.46/0.54 R17 = 0.85 R17

So, if R17 = 100k, R16 = 85k

Please check the math :-)

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