What Ray says is correct - with a small note.

All power converters are inefficient - some more than others.
At the top end, you can get a DC-DC converter at 95% efficient.
Average is about 80% for an AC-DC converter.
The rest is lost in heat.

Generally, you buy a converter for a voltage out, which means the current out
is going to be less than the "perfect" converter.
(There are constant-current supplies, too
starting to be common for LED supplies - I can forward a link to a webinar
by a company that makes the chips.)

Thus, good engineering practice is to give yourself maybe 50% margin over
max.
An under-powered power supply leads to all sorts of problems,
and the worst ones are when you have a peak load at unexpected times,
because they usually look like something else.
This has happened to me in the past.

I now will put an o-scope probe on the output of the power supply for
new board projects to make sure the power supply is OK.
Note a multi-meter will not work - they average out the peaks.

So - Ray is correct, but keep in mind this also assumes that the power
supply has enough margin at the input side to draw the extra current
if the power supply needs it, and does not exceed th max draw
of the power supply.

For example: if the board needs 5v at 5 amps (W=VA == 5*5 = 25 watts)
you want to spec the power supply at least at 5v at 7.5A = 37.5 watts;
better is 5v at 10A = 50 watts (if you need a healthy margin).
(80% of 50 watts is 40 watts out; OR:

(N watts in) * (80/100) = (50 watts out)
(N watts in) = ((50 watts out) * 100)) / 80
N watts in = 62.5 max draw

If the average draw is 25 watts, then the input draw is ((25 * 100) / 80)
= 31.25

Assume the board draws 5v at 6 amps at some point == 30 watts.
The 80% 5v/10A power supply input at that point draws ((30 * 100) / 80) =
37.5

Now, if the total input draw cannot exceed 50 watts (for whatever reason,
perhaps a dying battery):

(50 watts in) * (80/100) = 40 watts out

then it does not matter what the supply is rated at - there is not enough
power
to feed it at the max draw if the board starts drawing over 5v at 8 amps
(5*8 = 40 watts draw, at the max of the 40 watts you can get out).

A simple analogy: a fresh battery can crank over a car starter;
but run the battery down - the draw demand stays the same, but the supply
is limited to the point it cannot satisfy the draw.

(I hope this illuminates rather than confuses.
I tried to use "draw" in the same way Ray does.
The simple way to think of it is how much you *need*
vs what can be *provided*.
The only difference I am really pointing out is Ray is "assuming" a perfect
example (ie ignore the heat loss) which is great for a classroom example,
but the real world can be messier than that.
I have seen power converter chips that don't work like the spec sheet says
and it can be a real pain, and systems with peak draws that are out of the
blue
because the EE thought about the average, not the peak when *everything*
booted up at the same time. I put together a system with disc drives that
peaked at 18 amps, averaged 11 amps when running, and had to be on a
20 amp aircraft breaker. we *always* started it on ground power....
The inverter and DC-Dc converters could get toasty under the right
conditions.)

... bandit

> > First, "Watts In" _always_ equals "Watts Out". Because Voltage * Current
> > = Watts, voltage and current are both very relevant in this case.
> >
> > When dealing with transformers and transformer-type circuits you always
> > have a trade off between current and voltage. If you up the voltage on
> > the output, you get proportionally less available current. If you design
> > the circuit to provide more current then you have to have proportionally
> > less voltage on the output.
> >
> > This makes sense because "Voltage * Current = Watts" and "Watts In = Watts
> > Out" (minus small heat losses that will generally ignore).
> >
> > So as an example:
> >
> > If I have a transformer circuit that takes in 12V and draws 1A that is 12
> > Watts of power. If the output is say 1000V then the available current
> > can't be more that 0.012A because:
> >
> > Watts In = Watts Out
> >
> > 12V * 1A = 12W (Input)
> >
> > 1000V * 0.012A = 12W (Output)
> >
> > In order for "Watts In" to equal "Watts Out" you have to have a
> > proportional trade off between voltage and current.
> >
> > So no, you can't have more power (watts) output then you have going in.
> > Although, you can have (proportionally) different current and voltage
> > going in versus going out as long as everything still adds up to the same
> > wattage in as going out.
> >
> > The terms "current" and "draw" are not interchangeable.
> >
> > Simply stated, current is a measure of the amount of electrons traveling
> > through a wire (measured in Amps). Draw is a more generic term meaning
> > how much current an electric device uses.
> >
> > Say if a light bulb operates on 1A at 120V, then you would say that it has
> > a 1A draw at 120V, or that it draws 1A. So, "draw" deals with how much
> > current is _consumed_.
> >
> > As such, you wouldn't say that you could "get 200mA worth of draw" out of
> > a power supply. Rather you would say that you could "get 200mA of
> > current" out of a power supply. Here we are dealing with current
> > _provided_ so you wouldn't use the term "draw".
> >
> > Hope that helps.
> >
> > - Ray
> >
> >
> > On Mon, 2 Jan 2012, Morgan Gangwere wrote:
> >
>> >> I've got a simple EE question.
>> >>
>> >> Ended up getting in an argument over power supplies. Namely, over
>> >> current and draw.
>> >>
>> >> I gave the example of [this flyback
>> >> supply](http://www.dos4ever.com/flyback/flyback.html#boost2) and said
>> >> that you'd be able to at least get 200mA worth of draw out of the
>> >> other end, given a 12v@1500mA source.
>> >>
>> >> I was promptly informed that this was impossible because my output was
>> >> more than my input. Their logic is as follows:
>> >>
>> >> Watts = Amps * Volts.
>> >> W_i = ( 1.5 * 12 ) = 19.2W
>> >> W_o = ( .02 * 180 ) = 36.0W
>> >> W_i < W_o, therefore perpetual energy => impossible.
>> >>
>> >>
>> >> I argue that this is perfectly valid given that the input current is
>> >> greater than the output current and that voltages here are irrelevant.
>> >>
>> >> Am I correct in saying this? Or has my fundamental understanding of
>> >> electronics gone down the tosser?
>> >>

So, if we were to wsnt to get the 75mA or so we are looking for we would have to place at least 2A i to the circuit (ignoring any drain from the 555)

75mA is really low power, even assuming a really large % loss from heat, we would need no more than 100 mA

80% of 100 mA = 80 mA, so a crappy wall-wart should produce the 75 mA. Even the crappy wall-warts we have at home are rated to 100 mA - most are at least 200 mA to 500 mA.

Look at your math again - 2A is:

2A = 0.075A * (x / 100)
x = (2A * 100) / 0.075A
x = 200 / 0.075
x = 2666
meaning 2A is 2666% of 75 mA (0.075 * 26 = 1.95; 0.075 * 26.66 = 1.9995)

100mA at 180v is a different story. Wall warts rarely boost. 18w is not a lot, but you can't beat conservation of energy. 3w in, the output voltage will drop to less than 30v (mean) with a draw of 100mA (mean)... assuming the circuit could theoretically drive that.

Yes, Morgan... conservation of energy is your friend. If you can't solve the whole circuit, you can always find limits through physics. Go ahead and ignore the minor inefficiencies of oscillator and other heat loss to get the max theoretical output. 3w in, you'll never see more than 3w out. In practice, expect tops 90% efficiency. 80% is not uncommon when you have inductors (transformers). 90% is not uncommon for good capacitor-only converters.