Calculate the load and throw in a resistor in series to get the needed voltage drop
For example if the cig lighter voltage is 15.5V, and needed voltage is 13V, and the radar detector draws 300mA, a 8 ohm resistor (rated at 0.75W) wired in series should do the trick.
V=IR so 0.3 amps x 8 ohms = 2.5 volts dropped.
P=IV = 2.5 volts x 0.3 amps = 0.72 wattage resistor required
The caution here is I'm assuming a constant 300mA current draw by the radar detector, but it could be more for startup and less for continuous operation, which means the voltage drop across the resistor and the voltage to the radar detector will fluctuate, as will the power across the resistor, so a higher rated resistor is better.
Please use any of this information at your own risk it might be completely wrong. Also remember, a resistor dissipating energy will heat up a little.
For example if the cig lighter voltage is 15.5V, and needed voltage is 13V, and the radar detector draws 300mA, a 8 ohm resistor (rated at 0.75W) wired in series should do the trick.
V=IR so 0.3 amps x 8 ohms = 2.5 volts dropped.
P=IV = 2.5 volts x 0.3 amps = 0.72 wattage resistor required
The caution here is I'm assuming a constant 300mA current draw by the radar detector, but it could be more for startup and less for continuous operation, which means the voltage drop across the resistor and the voltage to the radar detector will fluctuate, as will the power across the resistor, so a higher rated resistor is better.
Please use any of this information at your own risk it might be completely wrong. Also remember, a resistor dissipating energy will heat up a little.