The Solar-Powered Park Activation Power Budget
You are activating a park on battery and solar power. Your HF transceiver produces 20 W RF output on SSB and is 60% efficient at 12.0 V DC. On receive it draws 0.5 A. You plan to operate for 4 hours with an estimated 25% transmit duty cycle. Your battery is an 18 Ah LiFePO4 pack, which you do not want to discharge below 20% state of charge. Your solar panel is rated 100 W, and you assume 70% real-world output for 5 equivalent peak sun hours; the charge controller is 90% efficient. The radio is rated for 11.5–14.5 V DC. The weather forecast is clear. You also have a 50 W dummy load and a power meter.
1. Calculate the DC current on transmit and the average current over the 4-hour session.
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First find the DC input power on transmit. If the transmitter is 60% efficient and produces 20 W RF, the DC input is 20 W / 0.60 = 33.3 W. At 12.0 V, the transmit current is 33.3 W / 12.0 V = 2.78 A. On receive, the radio draws 0.5 A. With a 25% transmit duty cycle, the average current is (0.25 × 2.78 A) + (0.75 × 0.5 A) = 0.695 A + 0.375 A = 1.07 A. Over 4 hours, the energy used is about 1.07 A × 4 h = 4.3 Ah. This is an estimate because actual SSB duty cycle, receive current, and voltage vary, but it is a reasonable planning figure.
2. Is the 18 Ah battery adequate with the 20% reserve? Show your reasoning.
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The usable capacity is 18 Ah × 0.80 = 14.4 Ah if you stop at 20% state of charge. The estimated 4-hour load is about 4.3 Ah, which is only about 24% of the battery's nominal capacity and well within the 14.4 Ah usable limit. Even if the duty cycle were 50% instead of 25%, the average current would be (0.5 × 2.78) + (0.5 × 0.5) = 1.64 A, and the 4-hour load would be 6.6 Ah-still less than half the usable capacity. The battery is therefore adequate for this operating plan, provided it starts fully charged and the radio's low-voltage cutoff is set appropriately for LiFePO4. Keep in mind that cold temperatures reduce capacity and that the radio may draw more current on peaks than the average calculation shows.
3. Will the 100 W solar panel replace the energy used during the session?
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The panel's realistic output is 100 W × 0.70 = 70 W. Over 5 equivalent peak sun hours, that is 70 W × 5 h = 350 Wh. After the 90% charge-controller efficiency, about 315 Wh reaches the battery. At 12 V, that is 315 Wh / 12 V = 26.3 Ah of charge-far more than the roughly 4.3 Ah used by the radio. So the panel should replace the energy used and recharge the battery, assuming good orientation, no shading, and moderate temperatures. If you operate while charging, the controller must be able to supply the radio load and charge the battery simultaneously; most small controllers can, but check the voltage and current ratings. The battery still acts as a buffer during transmit peaks.