Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Or is there something about PA's that can somehow force too much current
into a device?

Thomas G. Marshall wrote:

Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Or is there something about PA's that can somehow force too much current
into a device?



The rating on the supply indicates what it can deliver to the device in
terms of current. All this means is a connected device requesting
current can not ask for any more than 800 ma.
The device connected governs how much current will flow, the supply
only indicates the amount it can give with out damage or shut down to
it self.

Voltage of the supply must be close or exact to what the device requires.
In your case you have a 600 ma reserve or power, so you are fine..

--
"I'd rather have a bottle in front of me than a frontal lobotomy"

http://webpages.charter.net/jamie_5"

"Thomas G. Marshall" wrote:

Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Basically that's the idea.


Or is there something about PA's that can somehow force too much current
into a device?

See other posters' warnings about unregulated adaptors having a higher voltage
when only partially loaded.

Graham

On Sun, 20 Jan 2008 15:13:05 GMT, "Thomas G. Marshall"
. com wrote:


Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Or is there something about PA's that can somehow force too much current
into a device?



It may work, and it may not... Some crappy designs deliver the rated
voltage only at (or near) the rated current, and as current drops,
output voltage rises. As such a lightly loaded unit may in fact cause
damage because of overvoltage.

Realize that these cheap wall warts are *not* regulated, and the
output voltage is approxmate at best in most cases.

On Jan 20, 5:13 pm, "Thomas G. Marshall"
. com wrote:
Hugely fundamentally ignorant question.
A very frequently asked one, as well. And very frequently answered.
Others gave real answers. I agree with the correct ones

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Or is there something about PA's that can somehow force too much current
into a device?

Actually, I believe. One kind of adapter that might be able to put
more current into a device (by raising the voltage, of course) than is
"asked for" is a constant current supply (battery charger for variable
number of cells). But it's very unlikely that you have one. The plate
would probably have to be wrong, too.

On Sun, 20 Jan 2008 16:37:03 +0000, Eeyore
wrote:



"Thomas G. Marshall" wrote:

Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Basically that's the idea.


Or is there something about PA's that can somehow force too much current
into a device?

See other posters' warnings about unregulated adaptors having a higher voltage
when only partially loaded.

---
Geez, I guess without your helpful "me too" he'd be a little lost
lamb?

On Sun, 20 Jan 2008 15:13:05 GMT, "Thomas G. Marshall"
. com wrote:


Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of 12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw the
proper 200mA (because the voltages are matched)?

Or is there something about PA's that can somehow force too much current
into a device?

---
Most small unregulated wall-warts and transformers have secondaries
wound with wire with a resistance which will result in the specified
nominal output voltage being obtained with the specified load
connected to the output and nominal mains voltage applied to the
input.

That means that with high mains and/or lower than specified output
current the output voltage will be higher than nominal and with low
mains and/or higher than the specified output current the output will
be lower than nominal.

The reason for this is that the resistance of the secondary looks like
it's in series with the secondary, so as current into the load
increases the voltage dropped across the secondary resistance also
increases, decreasing the voltage available to the load, like this:

View in Courier:

Vs Vl
/ /
MAINS---+ +--[Rs]--+
| | |
P||S |
R||E [Rl]
I||C |
| | --Il |
MAINS---+ +--------+

Typical regulation of small transformers is about 30% no-load to full
load, so if we assume that's true for the transformer in your
wall-wart, Vs will be 1.3 times Vl, or 15.6V, when Vl is at 12V.

With Il = 800mA, Rl will be equal to:

VL 12V
Rl = ---- = ------ = 15 ohms
Il 0.8A

and Rs will be equal to:

Vs - Vl 15.6V - 12V
Rs = --------- = ------------- = 4.5 ohms
Il 0.8A

So, to simplify things, we now have:


15.6V E1
|
[4.5R] R1
|
+-- E2
|
[15R] R2
|
0V

Your load is going to draw 200mA at 12V, so its resistance under those
conditions will be:

E1 12V
R2 = ---- = ------ = 60 ohms
Il 0.2A

and the circuit now looks like:

15.6V E1
|
[4.5R] R1
|
+-- E2
|
[60R] R2
|
0V

so the current in the circuit will be:

E1 15.6V
Il = --------- = ------------- = 0.242A
R1 + R2 4.5R + 60R

and the voltage across your device will be:

E2 = Il * R2 = 0.242A * 60R = 14.52V

Here's a data sheet for a typical smallish wall-wart which will
illustrate the principle:

http://204.202.11.159/tamuracorp/cli...830AS12080.pdf

JF

John Fields wrote:

lots of good stuff snipped


Here's a data sheet for a typical smallish wall-wart which will
illustrate the principle:

http://204.202.11.159/tamuracorp/cli...830AS12080.pdf

JF

_Nice_ link.

On Wed, 23 Jan 2008 23:50:35 GMT, ehsjr
wrote:

John Fields wrote:

lots of good stuff snipped


Here's a data sheet for a typical smallish wall-wart which will
illustrate the principle:

http://204.202.11.159/tamuracorp/cli...830AS12080.pdf

JF

_Nice_ link.

---
Yeah, they've got great data sheets!

Little problem with that one though, they've got the width of the
mains prongs dimensioned out as 2.5" instead of 0.25"!

I just emailed them about it, so we'll see what happens. I've done
business with them before and they've always been decent, so I'll bet
they email me back with something like, "Oops... Thanks!" :-)

JF

"Thomas G. Marshall" . com
wrote in message news:5yJkj.1858$XI6.1836@trndny04...

Hugely fundamentally ignorant question.

If I have an electronics device that claims to take a power adapter of
12V
200mA, and I plug in one rated at 12V 800mA, will the device merely draw
the
proper 200mA (because the voltages are matched)?

Or is there something about PA's that can somehow force too much current
into a device?




Yes IF the power are really what the label said.

How do you know the powers are matched? Not all adapters telling your the
truth. Use a meter, do not trust the cheap wall-adapters made by the
Chinese.

Jack...