That amplifier has one MRF455 transistor, which is good for about 60 watts. The amplifier is a class C amplifier meaning that for calculation purposes, it's about 60% efficient. That means that about 60% of the power consumption will be used to amplify the incoming singal, but the other 40% will be wasted as heat. To figure out how much current it draws, use Ohm's law:
60 watts / 14 volts (typical car charging voltage) / 60% efficiency = 7 amps current draw.
A 10 amp fuse is what you need.
If you're using anything but a PEP wattmeter, you won't see true PEP, and there's no such thing as "peak" RF power. A PEP wattmeter either has internal batteries or an external power supply (wall wart) for powering the additional active circuitry it needs. Some meters have the wall wart, but it's just for powering the lights. Don't believe all of the meter manufacturers' claims.Thanks. I'm seeing 40 watts on my meter. I often see a lower watt and a peak when amp shopping.
Under what conditions would the peak wattage be seen?
Thanks