555 Timer with Diode — Duty Cycle
Calculate 555 astable frequency and duty cycle with optional diode for duty cycles below 50%.
Component Values
Results
Duty cycle = Ra / (Ra + Rb) with diode
555 Timer Astable with Diode
In the standard 555 astable circuit, the capacitor charges through Ra + Rb and discharges only through Rb. This means tHigh is always longer than tLow, and the duty cycle is always above 50% — you cannot go below 50% without modification.
Adding a diode in parallel with Rb (cathode toward pin 7) creates a bypass path during charging. The capacitor now charges through Ra only and discharges through Rb only. This makes tHigh and tLow independent: tH = 0.693 × Ra × C and tL = 0.693 × Rb × C. The duty cycle becomes Ra/(Ra+Rb), freely settable from near 0% to near 100%.
To get exactly 50% duty cycle with a diode, set Ra = Rb. To get a duty cycle below 50%, make Ra < Rb. Note that the diode forward voltage (about 0.7 V) introduces a small timing error at low supply voltages. Use a Schottky diode for better accuracy.
With Diode
tH = 0.693 × Ra × C
tL = 0.693 × Rb × C
D = Ra / (Ra + Rb)Without Diode
tH = 0.693 × (Ra + Rb) × C
tL = 0.693 × Rb × C
D > 50% alwaysKey Points
- Without diode: duty cycle always > 50%
- With diode: duty cycle = Ra / (Ra + Rb), any value possible
- To get 50% duty cycle: set Ra = Rb
- Use 1N4148 or Schottky diode for best accuracy
Applications
- PWM dimmer with precise duty cycle control
- Clock generation for digital circuits
- Audio tone generation at specific duty cycle
- Motor speed control with adjustable on/off ratio
Practical Examples
Generate a 50% duty cycle 1 kHz square wave using a 555 + diode circuit with C = 100 nF.
Rb = 1.44 / (2 × 1000 × 100e-9) = 7.2 kΩ → use 6.8 kΩ standard value
Symmetric 10 Hz LED flash with C = 10 µF for visible equal on/off times. Ra kept small (1 kΩ) for near-50% duty.
Rb = 1.44 / (2 × 10 × 10e-6) = 7.2 kΩ · ton = toff = 72 ms
Did you know? Adding a diode to the 555 astable circuit to achieve duty cycles below 50% was a common hack in 1970s electronics magazines. With the standard circuit, the minimum duty cycle is set by the ratio Ra/(Ra+2Rb), which approaches 0% only as Rb → ∞ (impractical). The diode bypasses Rb on the charge path.