A basic question about DC power conversion

Assuming that there is no other power flow or storage other than needed for the conversion, those values represent the limiting case that a perfectly efficient converter would impose.

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Your question confuses me, so I'm assuming you meant:

I have a 15V power supply which supplies a 5V linear regulator U1.

Linear regulators are series elements (for the most part) and thus pass exactly the same current as the circuit as they draw from the supply (there is some leakage for control, but they try to minimize this.)

Thus, for a linear regulator, the current in (300mA) is equal to the current out (300mA), and the regulator itself dissipates I*V = 300mA *

10V = 3W.

On the other hand, a switch mode power supply will attempt to use other means to minimize the power consumed by the regulator itself. For a perfectly efficient regulator (which does not exist) the loss in the regulator would be 0W. Thus, the total power drawn from the supply would be 300mA * 5V = 1.5W, and you would thus draw 1.5/15 = 100mA.

However, there is always some inefficiency in the regulator, expressed as a percentage. Thus, a 90% efficient regulator would dissipate 10% of the total energy drawn from the supply, and deliver 90% to the load. Consequently, the power consumed would be

Pload = 0.300 * 5 Preg = Pload / 9 = 167mW Ptotal = 1.666W

So the average current from the supply would be

I = 1.666/15 = 111mA

Not unless you neglect power used by the regulator.

--
Regards,
   Robert Monsen

"Your Highness, I have no need of this hypothesis."
     - Pierre Laplace (1749-1827), to Napoleon,
        on why his works on celestial mechanics make no mention of God.

Thanks Larry and Robert. Device D is not a linear regulator. It is a board which has 4 output ports at 5V but needs 15V input power.

--- Here's your situation:

+-----------+ +15V>------| PORT1|------------------------+->5V | PORT2|-----------------+->5V | | PORT3|----------+->5V | | GND>--+----| PORT4|---+->5V | | | | +-----------+ | | | | | [RL4] [RL3] [RL2] [RL1 | | | | | +--------------------+------+------+------+

You said in your first post that the current drawn by the loads on the

5V output ports is 300mA, so the total power taken out of the board would be:

P = IE = 300mA * 5V = 1.5 watts.

Since you'd have to put at least that amount of power into the board to get that much out, with a 1.5 watt input requirement the input current requirement would be:

P 1.5W I = --- = ------ = 0.1A E 15V

So the answer to your first question would be a guarded "yes", subject to the conditions Larry Brasfield explained in his reply to your first post.

-- John Fields

Along with the other asnswers, then provided you have an efficient converter (switch mode supply, for instance), then the answer is the 5V device draws V5/Vsys (where Vsys is your 15V input) , [5/15 * I(15v)] / converter efficiency. For this case, that's 100mA / (n < 1), so it's something above 100mA.

There is literally no such thing as a 100% efficient converter. A typical off the shelf converter for this range will typically have an efficiency ranging from 80-90%. It is possible to get much higher efficiency for a specific system, but off the shelf converters are, by definition, not optimised for a particular problem, except 15V in, 5V out, I(o) >= some value.

For a typical converter then, the current drawn at the input to the converter to feed device 'D' would probably be about 115mA (typ).

Cheers PeteS

On 7 Mar 2005 11:35:55 -0800 in sci.electronics.basics, "aman" wrote msg :

So what you are talking about is a product you have bought that performs the voltage conversion? If you can find out the efficiency of the device, as others have mentioned, you can figure out how much input current it will draw from the total of the output current, by using the equations given.

--
Al Brennan