Generate pwm using ic 555: The simple method

Method 1; the simple

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  The simple 555 circuit.

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  The differences in C and output. Note that more than 50% pwm is not obtained

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  The surface and contour plot of the limits of the R₁ and R₂ values.

The capacitor is charging through $R_{1}$ and $R_{2}$ but discharges only through $R_{2}$ using IC 555. Thus $R_{2}$ should be a potentiometer.

The $0.01\mu$F capacitor is to ensure that the CTRL and GND stays on the same voltage level. The CTRL pin of the 555 is to level out any fluctuations in the power supply voltage that might affect the operation.

${\begin{aligned}T_{\text{on}}&=0.693(R_{1}+R_{2})C\\T_{\text{off}}&=0.693R_{2}C\\f&={\frac {1.44}{(R_{1}+2R_{2})C}}\end{aligned}}$

and

${\begin{aligned}{\text{pwm}}\%&={\frac {T_{\text{on}}}{T_{\text{on}}+T_{\text{off}}}}\\&={\frac {0.693(R_{1}+R_{2})C}{0.693(R_{1}+R_{2})C+0.693R_{2}C}}\\&={\frac {0.693C}{0.693C}}{\frac {R_{1}+R_{2}}{R_{1}+2R_{2}}}\\&={\frac {R_{1}+R_{2}}{R_{1}+2R_{2}}}\\\end{aligned}}$

The limits of ${\text{pwm}}\%$

${\begin{aligned}\lim _{R_{2}\to 0}{\text{pwm}}\%&=\lim _{R_{2}\to 0}{\frac {R_{1}+R_{2}}{R_{1}+2R_{2}}}\to {\frac {R_{1}+0}{R_{1}+0}}=1\\\lim _{R_{2}\to \infty }{\text{pwm}}\%&=\lim _{R_{2}\to \infty }{\frac {R_{1}+R_{2}}{R_{1}+2R_{2}}}=\lim _{R_{2}\to \infty }{\frac {R_{2}({\frac {R_{1}}{R_{2}}}+1)}{R_{2}({\frac {R_{1}}{R_{2}}}+2)}}\to {\frac {1}{2}}\\\end{aligned}}$

But note that if $R_{1}$ cannot be zero because then the discharge pin 7 would be shorted to Vcc.

See the attached python file which generates the surface and contour plot shown above: File:555limits.zip. It is a zipped python file.

File:555 simple 300 300.circuitjs.txt is the Falstad file to simulate the circuit. The image above is generated using Falstad circuit simulator.

Caption text

$R_{1}$	$R_{2}$	$C$	$T_{\text{on}}$	$T_{\text{off}}$	pwm%	$f$
300	300	300 nF	0.000125	0.0000624	0.67	5333
300	300	30 nF	0.0000125	0.00000624	0.67	53333
300	300	3 uF	0.00125	0.000624	0.67	533
3000	300	300 nF	0.000686	0.0000624	0.92	1333
300	3000	300 nF	0.000686	0.000624	0.52	762
300	30000	300 nF	0.00630	0.00624	0.50	80

See the file File:555 simple calculations.ods for interactive Libre Office file.

As the resistance of $R_{1}$ becomes much larger than $R_{2}$, the duty cycle increases towards unity (100%) as $R_{2}$ approaches zero. If the resistance of $R_{2}$ increases with respect to $R_{1}$, the duty cycle approaches 50% (or 1:1). It seems to be impossible to obtain less than 50% duty cycle with this system; Thus, only about half range can be achieved.

How to get the pwm% smaller than 50%. Add a diode to short $R_{2}$ during the charge cycle.