Duty cycle

The duty cycle of a pulsed or modulated signal is the fraction of each period during which it is on. For pulses of duration $\tau$ repeating at a repetition rate $f_\text{rep}$ it is $D = \tau f_\text{rep}$, quoted as a fraction or a percentage. A Q-switched laser emitting 10 ns pulses at 10 kHz has $D = 10^{-4}$ (0.01%); a mode-locked Ti:sapphire oscillator with 100 fs pulses at 80 MHz has $D = 8 \times 10^{-6}$; pulsed characterization of a laser diode with 1 µs pulses at 1 kHz runs at $D = 10^{-3}$ (0.1%); a square wave from an optical chopper has $D = 0.5$.

Average and peak power

Because the source is dark between pulses, the average power a thermal power meter reads is the peak power diluted by the duty cycle:

$$P_\text{avg} = P_\text{peak}\, D .$$

For the 10 ns, 10 kHz laser above, an average of 1 W corresponds to a pulse energy of $1~\text{W}/10~\text{kHz} = 100$ µJ and a peak of $1~\text{W}/10^{-4} = 10$ kW. The relation is exact for rectangular pulses; for real pulse shapes the pulse duration is the FWHM and the peak power carries a shape factor (0.881 for sech², 0.939 for Gaussian), which the peak-power entry derives.

How it is measured

On a fast photodiode and oscilloscope, the duty cycle is the on-time divided by the period, which most oscilloscopes report directly. For optical pulses shorter than the detector response, the duration comes from an autocorrelation or other ultrafast method and the repetition rate from a frequency counter, and the duty cycle is computed. The optical duty cycle of a pulsed laser diode can be shorter than the electrical one set on the driver, since the diode emits only above threshold.

Thermal reasons for low duty cycle

Heat follows the average dissipated power when the pulse period is short compared with the thermal time constant of the device and its mount. A pump diode dissipating 1.1 W in CW operation on a mount with $R_\text{th} = 10$ K/W runs 11 K above its heatsink; at $D = 10^{-3}$ the average rise is about 0.011 K. This is why LIV curves are taken with short pulses at about 0.1% duty cycle to separate the intrinsic characteristic from self-heating, as described in Pulsed versus Continuous-Wave LIV Measurement. Some heating still occurs within each pulse, so the pulse itself is kept short, commonly around 1 µs, and the result is checked by changing the repetition rate.

Quasi-CW (QCW) operation applies the same idea at high power. High-power diode bars and stacks for pumping solid-state lasers are driven with pulses of the order of 100 µs to 1 ms at duty cycles of a few percent; 200 µs at 100 Hz is $D = 2\%$. Each pulse reaches steady optical output while the average heat load stays far below that of CW operation.

Duty cycle of modulated signals

In on-off keying, the NRZ format holds the light on for a whole bit slot during each 1. Return-to-zero formats send each 1 as a pulse shorter than the slot: 50% RZ at 10 Gb/s gives 50 ps pulses in a 100 ps slot, and 33% RZ gives 33 ps pulses. With equal numbers of 1s and 0s, the average power of rectangular 50% RZ is one quarter of the on-level, against one half for NRZ.

A modulated signal measured at a single frequency, as by a lock-in amplifier, responds to the fundamental Fourier component of the waveform. For a rectangular pulse train switching between 0 and $A$ the fundamental amplitude is

$$a_1 = \frac{2A}{\pi}\,\sin(\pi D),$$

which is largest at $D = 0.5$, where it equals 0.637$A$ (0.450$A$ rms). At $D = 0.1$ it falls to 0.197$A$. A chopper calibration therefore does not transfer to a blade of different duty cycle.

Pitfalls

Percent and fraction are easily confused: 0.1% is $10^{-3}$, and an error here is an error of 100 in the inferred peak power. Light emitted between pulses, such as amplified spontaneous emission from a pulsed fiber amplifier or leakage from a Q-switch, adds to the average power without adding to the pulses, so the peak power computed from $P_\text{avg}/D$ is too high. The term also has unrelated uses: for a grating it is the fraction of the period occupied by one material, discussed under filling factor.

Common questions

How do you calculate duty cycle?

Multiply the pulse duration by the repetition rate, or divide the on-time by the period. A 1 µs pulse every 1 ms gives $D = 10^{-3}$, or 0.1%.

How do you get peak power from average power and duty cycle?

Divide the average power by the duty cycle: $P_\text{peak} = P_\text{avg}/D$ for rectangular pulses. For shaped pulses, multiply by the shape factor given in the peak-power entry.

Why do laser diodes have a maximum duty cycle?

The junction temperature follows the average dissipated power, and too high a duty cycle at high peak current raises it until the output falls and the facets or junction degrade. QCW datasheets therefore state a maximum pulse width and a maximum duty cycle.

References: A. E. Siegman, Lasers (University Science Books, 1986); W. Koechner, Solid-State Laser Engineering, 6th ed. (Springer, 2006); B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed. (Wiley, 2019).