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LM350 DC Voltage Regulator Power Requirements

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tosborn

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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?
 
You have the power calculation right. 9W of dissipation means that the regulator must be bolted to about 100 square inches of aluminum...

Minor nit: any three legged regulator obeys Kirchoff's Current law:
the Σ of currents in the three legs = 0

So technically, if the load draws 3A from the output pin, the current into the input pin is the sum of the output current and the current that flows out the ADJ pin. Also, since it is an adjustable regulator, there are two resistors to set the output voltage. There is some additional current that flows from the output pin down the voltage divider.
 
3A output. 9V input and 6V output.

The load heats with 3A x 6V= 18W.
The regulator heats with 3A x 3V= 9W.

What is the question?
 
The question was "Am I correct that in order to get that amount of amps out (3 amps), you would have to feed it 3 amps in?" Mike confirmed that my thinking was correct.
 
You do not "feed amps in". The regulator takes as many amps as it needs.
If its output load draws 3.0Amps then the input will be about 3.01A.
With no load the regulator draws only 0.01A.
 
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