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.

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  The contour and surface plot of varying resistances.

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  Added the second diode. This will make the output identical if R₁ and R₂ are identical.

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  One more resistance (or potentiometer) to make it more symmetric. This is discussed in more detailed in the next chapter, because the two potentiometers can be easily replaced by a one.

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.

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

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

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

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

The leakage current through R_2 can be eliminated by introducing a second diode D_2 in series with that. This will make the output identical if R₁ and R₂ are identical. See the image above. The corresponding simulation file is File:555 diodes2 300 300.circuitjs.txt. Adding one more resistance (or potentiometer) to make it more symmetric. This is discussed in more detailed in the next chapter, because the two potentiometers can be easily replaced by a one.