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I need help with what is probably a very simple question that I don't know.

When you put 12v at 10 watts and bring it down to 5v with a regulator would the wattage change?

And just say the regulator had no max on what it could output.

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Do you mean a linear voltage regulator, like a 7805? Or a switching regulator? –  David Mar 25 at 8:32

2 Answers 2

The wattage isn't something you inflict on a load but the voltage is. It seems to me that you might be taking the voltage (12V) and reading the label on the power supply (that says 10W) and therefore believing you will be putting 10 watts across to any load connected to it.

This isn't true - you put voltage on a load and the load determines what current is taken by ohms law (I = \$\dfrac{V}{R}\$).

So, if the load resistance is 14.4 ohms, the current will be 0.8333 amps and the power delivered will be 12V x 0.8333A = 10 watts.

If you inserted a 5V regulator between supply and 14.4 ohm load, the load would see 5 volts and take a current of 0.347 amps (ohm's law again) - this would be a load power of 1.736 watts. However, if the voltage regulator were a linear type there would be 7 volts dropped by the regulator at a current of 0.347 amps and this would be a power of 2.43 watts lost in the regulator.

Now the total power consumed is 2.43 watts + 1.736 watts = 4.17 watts (not 10 watts). It's load dependent.

If you used a near-100% efficient switching regulator there would be no power lost in the regulator and the power taken from the power supply would be just that consumed by the load i.e. 1.736 watts.

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I think this answers the OPs misconception very well, but it would be worth noting that a linear regulator from 12V to 5V at 10W would generate a fair amount of heat (see also this question). –  David Mar 25 at 9:23
    
@David I did mention that - "However, if the voltage regulator were a linear type there would be 7 volts dropped by the regulator at a current of 0.347 amps and this would be a power of 2.43 watts lost in the regulator". –  Andy aka Mar 25 at 10:02

Your question is not very clear, but my understanding of your question is this:

"If I have a power supply that provides a Voltage of 12V and can supply a maximum Power (Wattage) of 10W, what Power could be provided to a load if I regulate the Voltage down to 5V?"

The answer simply relates to the efficiency of the 5V regulator.

$$Efficiency = \frac{Power_{out}}{Power_{in}}$$

So

$$Power_{out} = Efficiency \times Power_{in}$$

LINEAR REGULATOR: $$Efficiency = \dfrac{V_{out}}{V_{in}} = \dfrac{5}{12} = 42\%$$

So your 10W rated PSU can supply \$10W \times 0.42 = 4.2W\$ to the load at 5V.

SWITCHING REGULATOR :

Efficiency = 80% (for example)

So your 10W rated PSU can supply \$10W \times 0.8 = 8W\$ to the load at 5V.

You may ask: where does the rest of the power go? It's lost as heat in the regulator. Dissipating 6W in a linear regulator can amount to a lot of heat!

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