How did I blow the fuse in my multimeter?

K

kony

Shine a light in a LED, and you *DO* get a voltage out! Yes, free energy
(minus the cost of the LEDs, as mentioned.) Yes, very inefficient, but
they *DO* put out a voltage. Note that no mention was made as to how many
LEDs would be needed to provide useful power.


We seem to be interpreting the question differently... I read it to mean,
powered LEDs, their light shining on a solar cell to generate electricity
from the solar cell.
 
A

Andy Foster

kony said:
We seem to be interpreting the question differently... I read it to mean,
powered LEDs, their light shining on a solar cell to generate electricity
from the solar cell.

That was how it was written.
 
H

half_pint

Raymond Sirois said:
There IS a VERY small voltage drop across a diode, as a result, there
is a small amount of impedance. The power supply will see this as a
load and therefore there will be current applied to the diode...
never speak in absolutes. As I said, LEDs draw an INSIGNIFICANT
amount of current...


You are wrong. They draw a lot of current, thats why a series resistor is
needed to stop them blowing.
 
R

ric

kony said:
We seem to be interpreting the question differently... I read it to mean,
powered LEDs, their light shining on a solar cell to generate electricity
from the solar cell.

Yeah, I guess so. Obviously powered LEDs need to pass enough current
to provide light. But like a diode, they pass whatever current is provided
to them, not draw current on their own. They'd likely not provide enough
light to power a solar cell in any regard.

I once powered a small LCD clock with a series/parallel array of LEDs.
Worked great! But it was quite a bummer to reset the clock every morning
after I turned on the shop's lights. ;)
 
C

Cyde Weys

ChrisJ9876 wrote:

My first thought was that the OP doesn't know the difference between voltage &
current, and therefore shouldn't be messing with a multimeter.

LOL, I do know the difference, I'm just not very handy with a multimeter.
 
C

Cyde Weys

Trent© said:
If yer shorting the resistance-test circuit...and then the meter is
moving...then you blew the resistance fuse.

If you blew the resistance fuse, that means you had it in resistance
mode when you checked the battery.

Having the meter in ohm mode when checking the battery current will
blow the fuse every time.

I think you'll find that the meter has 2 fuses inside.

Nope, there's only one fuse in there. I just checked. Unless one of
the fuses is designed to be inaccessible and is squirrelled away in the
back of the unit (wouldn't make a lot of sense).
 
K

Kmano

Let me try and explain this whole LED thing, and the fuse blowing
though it might be to late for this thread. The amount of current
that is 'drawn' by the LED depends on the LED and the battery.
Formula is this:
A=V/R
Where A is Current, V is Voltage, and R is resistance of the
LED/multimeter. So a 9 V battery with a LED (typically around 1-10
ohms) will yeild a significant amount of current, ~1000 to 9000 mA.
So to prevent this a resister is placed in the circuit to drop the
current. This is also why the multimeter blew the fuse. on a DC Amp
setting the resistance will be small, mine is rated at 25 Ohm. So
the current will be more than the fuse could handle. Some
multimeters have higher ranges, which just means there is more
resistors in the circuit, and in turn a different range displayed.

I hope this helped, I have a problem of my own... My dad, for reasons
I don't know, tried to measure the current on a live 220V circuit and
blew the multimeter without even blowing the fuse. I'm trying to
trace the component that blew and need to find some schematics of a
multimeter, any will do, just to help figure out what some parts do.
if anyone has attempted to fix something like this let me know.
Thanks

==============
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