The duty cycle of a pulse wave is the ratio of the time it spends high to the total duration of the high and low state. We calculated these durations in step 5, and we can combine them to calculate the duty cycle of the 555:

**duty cycle = (time spent high) / (total time duration of high and low states)**

substitute the equations from step 5 to get:

**duty cycle = (0.7*(R _{A} + R_{B})*C) / (0.7*(R_{A}+2*R_{B})*C**

**)**

this simplifies to:

**duty cycle = (R**

_{A}+ R_{B}) / (R_{A}+ 2*R_{B})In the equation above, when R

_{A}is much larger than R

_{B}(you can ignore the R

_{B}terms) you end up with a duty cycle ~= 1 and when R

_{B}is much larger than R

_{A}(you can ignore the R

_{A}terms) you get a duty cycle =~ 1/2. So the limits of the duty cycle with the circuit shown in fig 2 are 50% to 100%. If you wanted to get a duty cycle that was less than 50%, you have to use a circuit like the one shown in fig 1. In this circuit a diode bypasses R

_{B}during the charging phase of the 555 (while the output is held high). So how does this affect the durations of the high and low phases of the output?

The duration of low output remains:

**t = 0.7*R**

_{B}*C secondsthis happens when the capacitor is discharging, so current is flowing from the capacitor, through R

_{B}(in the upwards direction in fig 1), and into the discharge pin of the 555. This is the opposite direction of current flow that the diode will accept, so no current flows through the diode. During this time, the circuit in figure 1 is functionally equivalent to the circuit in fig 2.

The duration of high output does change, most notably the R

_{B}contribution goes away because it is being bypassed by the diode. In this case the capacitor is being charged so current is flowing from the power supply Vcc, through R

_{A}(in the downward direction in the schematic), and through the diode to the capacitor. Current will not flow through R

_{B}because the path through the diode is the path of lease resistance; the diode is essentially acting as a wire across R

_{B}.

Previously the duration of high output was:

**t = 0.7*(R**

_{A}+R_{B})*C secondswe cannot simply remove R

_{B}from the equation because we need to account for a slight voltage drop (about 0.7V for silicon diodes) across the diode. We calculated the general form of the duration of high output in step 5. I've reproduced it below:

**t = -(R**

_{A}+R_{B})*C*ln[1/3*Vcc/(Vcc - V_{0})]we should subtract the voltage drop from the diode (Vd) from both instances of Vcc in this equation and remove the contribution from R

_{B}

**t = -R**

_{A}*C*ln[(1/3*Vcc-Vd)/(Vcc - Vd - V_{0})]as in step 4, the initial voltage V

_{0}equals 1/3Vcc

**t = -R**

_{A}*C*ln[(1/3*Vcc - Vd)/(Vcc - Vd- 1/3Vcc)]**t = -R**

_{A}*C*ln[(1/3*Vcc - Vd)/(2/3*Vcc - Vd)]**t = -R**

_{A}*C*ln[(Vcc - 3*Vd)/(2*Vcc - 3*Vd)]so the the duration of the high output is now

**t = R**

_{A}*C*ln[(2*Vcc - 3*Vd)/**(Vcc - 3*Vd)**

**]**

Notice how there is no R

_{B}dependance. Also notice how the voltage drop across the diode and the supply voltage have an effect on the equation.

It's good to note here that you can also use monostable mode with an external trigger to create a PWM signal of duty cycles between 0 and 100%. I explained how to do this at the end of step 4. Even more info about PWM with the 555 timer can be found on the datasheet.