The LM350 adjustable voltage regulator is capable of producing 3 amps out. Am I correct that in order to get that amount of amps out, you would have to feed it 3 amps in (yes...at a voltage at least equal to the desired output voltage plus dropout)?
Here's my reasoning: Lets assume I wanted 6 volts out of the LM350 at 3 amps.... that's 18 watts. Lets further assume that voltage in to the LM350 was 9 volts. Heat dissipation would be (9-6)*3 = 9 watts. 18 watts output plus 9 watts of heat would equal 27 watts. 27 watts divided by input voltage of 9 volts would equal 3 amps. Am I thinking about this correctly?
Here's my reasoning: Lets assume I wanted 6 volts out of the LM350 at 3 amps.... that's 18 watts. Lets further assume that voltage in to the LM350 was 9 volts. Heat dissipation would be (9-6)*3 = 9 watts. 18 watts output plus 9 watts of heat would equal 27 watts. 27 watts divided by input voltage of 9 volts would equal 3 amps. Am I thinking about this correctly?