Thermal rollover

Thermal rollover is the departure of a laser diode's light–current (L–I) curve from a straight line at high current: the slope decreases, the output reaches a maximum, and further current lowers the output. The cause is self-heating: the electrical power that does not leave as light heats the active region, the threshold current rises and the slope efficiency falls with temperature, and above some current each added milliampere produces more heat than light. In the ridge-laser model below the output peaks with the junction about 90 K above the heat sink; an 850 nm oxide-confined VCSEL rolls over at around 10 mA.

Junction temperature

The temperature rise of the active region above the heat sink is the dissipated power times the thermal resistance $R_\text{th}$:

$$\Delta T = R_\text{th}\,(I V - P_\text{out}),$$

where $P_\text{out}$ is the total optical power leaving the chip. For a ridge laser at 150 mA and 1.60 V that emits 65.6 mW from its two facets, the dissipated power is 0.174 W, and with $R_\text{th}$ = 60 K/W the junction runs 10.5 K above the heat sink. Since the threshold rises as $\exp(\Delta T/T_0)$, with a characteristic temperature $T_0$ = 60 K that is already a 19% increase in threshold current.

A worked model

Solving the heat balance and the temperature-dependent L–I curve together locates the peak. Take a laser with a threshold of 10 mA and a slope of 0.25 W/A per facet at the heat-sink temperature, $T_0$ = 60 K, a slope temperature $T_1$ = 200 K, a forward voltage $V = 0.85\ \text{V} + I \times 5\ \Omega$, and $R_\text{th}$ = 60 K/W. Solving self-consistently for the front-facet power:

The local slope falls from 0.23 W/A at 100 mA to 0.13 W/A at 300 mA and 0.06 W/A at 400 mA, and the output peaks at 72 mW near 490 mA, where the threshold has risen to 43 mA. Halving $R_\text{th}$ to 30 K/W moves the peak to 108 mW at 722 mA; halving the series resistance to 2.5 Ω instead gives 100 mW at 678 mA. In each case the peak falls where the junction is about 90 K above the heat sink. The exponential $T_0$ law is an empirical fit, so the model is only a sketch.

Mechanisms

Several temperature-activated losses add up. Auger recombination grows with temperature and with the carrier density, raising the current needed to hold threshold. Carriers escape over the heterobarriers of the separate-confinement layers more readily as the Fermi tails broaden, and this leakage current produces no light. Intervalence-band and free-carrier absorption raise the internal loss, lowering both the slope and the gain margin. In lasers whose wavelength is pinned by a grating or a short cavity, the gain peak moves at 0.4–0.5 nm/K at 1550 nm while a DFB mode moves at about 0.09 nm/K, so the gain at the lasing wavelength drops as the two separate: 16–21 nm over a 50 K rise (wavelength temperature coefficient).

VCSELs

A VCSEL has a small active volume, a current path through resistive doped mirrors and a heat source a few micrometers across, so its thermal resistance is of order a few kelvin per milliwatt. At 10 mA, 2.2 V and 4 mW out, 18 mW is dissipated and with 2.5 K/mW the active region is 45 K above the heat sink; the gain peak and the cavity resonance at 850 nm separate by 0.2 nm/K, 9 nm over that rise. VCSELs are therefore designed with the gain peak on the short-wavelength side of the resonance at room temperature, and their thermal resistance, about 40 times the 60 K/W of the ridge laser above, makes them roll over at currents of order 10 mA.

Measurement and pitfalls

Rollover is read from a CW L–I sweep at a controlled heat-sink temperature. Comparing it with a short-pulse sweep at low duty cycle separates heating from intrinsic effects: the pulsed curve stays straight to much higher current (Pulsed vs CW LIV measurement). The junction temperature itself comes from the wavelength-shift or forward-voltage method (Measuring laser diode junction temperature). A rollover current means little without the heat-sink temperature and mounting, which set $R_\text{th}$, and a slope efficiency fitted into the curved region is understated.

Common questions

What causes thermal rollover in a laser diode?

Self-heating: the dissipated power raises the junction temperature, which raises the threshold and lowers the slope until added current gives less light.

How is thermal rollover reduced?

By lowering the thermal and series resistances, using active regions with higher $T_0$, and choosing the gain-to-mode detuning for the hottest operating point.

References: L. A. Coldren, S. W. Corzine, and M. L. Mašanović, Diode Lasers and Photonic Integrated Circuits, 2nd ed. (Wiley, 2012). G. P. Agrawal and N. K. Dutta, Semiconductor Lasers, 2nd ed. (Van Nostrand Reinhold, 1993). R. Michalzik (ed.), VCSELs: Fundamentals, Technology and Applications of Vertical-Cavity Surface-Emitting Lasers (Springer, 2013).