Generate pwm using ic 555: Add a diode

Method 2; Longer low time: Add a Diode

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  The schematics.

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  The simulations.

The added diode $D$ shorts the out the resistor $R_{2}$ during the charging cycle; thus the capacitor charges directly from $R_{1}$ removing $R_{2}$, though a small leakage current will flow through it. The constants are modified due to the voltage drop of $D$.

${\begin{aligned}T_{\text{on}}&=0.8R_{1}C\\T_{\text{off}}&=0.7R_{2}C\\f&={\frac {1.44}{(R_{1}+2R_{2})C}}\end{aligned}}$

$pwm\%={\frac {T_{\text{on}}}{T_{\text{on}}+T_{\text{off}}}}={\frac {0.8R_{1}C}{0.8R_{1}C+0.7R_{2}C}}={\frac {0.8R_{1}}{0.8R_{1}+0.7R_{2}}}$

which simplifies to $pwm\%\approx {\frac {R_{1}}{R_{1}+R_{2}}}$

The simulation file: File:555 diode 300 300.circuitjs.txt