CW Power
Why the Continuous-Wave Rating Is the Hardest Case
A CW signal is an unmodulated carrier held at constant amplitude, so the device dissipates a fixed amount of heat every microsecond with no idle period in which to recover. This makes the CW power figure the conservative thermal limit on any datasheet. Under pulsed operation a part gets cooling intervals between pulses, allowing instantaneous power far above the CW number; under CW there is no such relief, and the steady-state junction temperature must stay below its rated maximum at all times. For this reason a transistor rated 50 W CW may comfortably survive several hundred watts of peak power at a low duty cycle.
The governing physics is thermal, not electrical. The maximum heat a die can shed is the allowable temperature rise divided by the total junction-to-case thermal resistance. Once that dissipation ceiling is known, the deliverable RF output follows from the power-added efficiency: the lower the efficiency, the more of the DC input becomes waste heat, and the less CW output the device can sustain before reaching its temperature limit. Modern GaN-on-SiC devices push CW capability higher than older GaAs or LDMOS parts precisely because SiC spreads heat better and tolerates junction temperatures above 200 degrees C.
Engineers specifying components for a transmitter chain treat the CW rating as a budget. They sum the worst-case dissipation across the line-up, confirm each stage stays within its CW limit at the highest expected baseplate temperature, and then derate for reliability so the mean time to failure meets the program target. Getting this wrong shows up as thermal runaway, gain compression drift, or outright device burnout.
From Thermal Limit to Deliverable CW Output
Pdiss(max) = (Tj(max) − Tcase) / Rth(j-c)
Average power from peak and duty cycle:
Pavg = Ppeak × D (CW is the case D = 1)
Deliverable CW output from efficiency:
Pout ≈ Pdiss × η / (1 − η)
Power in dBm:
P(dBm) = 10·log10(PmW), 0 dBm = 1 mW, 30 dBm = 1 W
Where Tj(max) = max junction temperature, Rth(j-c) = junction-to-case thermal resistance (°C/W), D = duty cycle, η = drain/PAE efficiency. Example: Tj(max) = 225°C, Tcase = 50°C, Rth = 2°C/W → Pdiss = 87.5 W; at η = 0.60 → Pout ≈ 130 W CW.
CW Power Across Device Technologies
| Technology | Typical CW Output | Max Tj | Rth(j-c) | Useful Frequency | Strength |
|---|---|---|---|---|---|
| GaN-on-SiC HEMT | 50 to 300 W | 225 °C | 0.5 to 2 °C/W | DC to 40 GHz | High CW density, rugged |
| LDMOS | 100 to 1000 W | 200 °C | 0.1 to 0.5 °C/W | DC to 3.5 GHz | Low cost, high power through S-band |
| GaAs pHEMT | 1 to 20 W | 150 °C | 5 to 30 °C/W | 2 to 60 GHz | Low noise, microwave linearity |
| Si BJT | 5 to 50 W | 175 °C | 1 to 5 °C/W | DC to 2 GHz | Mature, inexpensive |
| TWT amplifier | 50 to 3000 W | n/a (collector) | n/a | 2 to 100 GHz | Very high CW at mmWave |
Frequently Asked Questions
How does CW power differ from peak and average power?
CW power is average power at 100% duty cycle, with the carrier on continuously and no modulation envelope. Peak power is the instantaneous pulse maximum and can be far higher: average power equals peak times duty cycle, so 1 kW peak at 10% duty is 100 W average. CW is the most thermally demanding case because the device never gets an off period to cool, so a 50 W CW part may handle several hundred watts peak under low-duty pulsing. Datasheets list CW and peak ratings separately, and CW is almost always the lower number.
How do you calculate the maximum CW power a transistor can dissipate?
Start from the thermal limit: Pdiss(max) = (Tj(max) − Tcase) / Rth(j-c). A GaN HEMT with Tj(max) = 225°C, a 50°C case, and Rth = 2°C/W can shed (225 − 50)/2 = 87.5 W of heat. Deliverable CW output then depends on efficiency: at 60% drain efficiency the heat is 40% of DC input, so the RF output that produces 87.5 W of heat is roughly 130 W. Designers derate further to keep junction temperature low enough for a long mean time to failure.
Why does CW power handling drop as frequency increases?
Two effects compound. First, semiconductor efficiency falls as the operating frequency nears device cutoff, so more DC input becomes heat rather than output for the same delivered power. Second, millimeter-wave devices use smaller gate peripheries that concentrate heat and raise junction-to-case thermal resistance. A GaN part delivering 100 W CW at 3 GHz may be limited to 5 to 10 W CW at 40 GHz, which is why mmWave power is often produced by combining many small devices with active cooling and SiC substrates.