Replacing power and hd led?

[TheKeith]

You should either know the voltage and current specifications of the LEDs
you are replacing (and try to match that), or know the MB source voltage
and dropping resistor value.

I'm not sure how to acquire the specs of the led's I'm replacing--my guess
is that they are standard 3mm led's. I made a little tester with first 4x
AAs', and I guess I destroyed one of the leds since it doesn't work anymore
(didn't blow up though). Then I changed the tester to use 2x AA's, and that
seems to work fine; the led lights fine and it doesn't get too warm. Now
these are rechargeables so the voltage for these two batteries is less than
3V.

I checked the source voltage of the power led by touching the leads of my
multimeter to the pins on the mobo--I get around 4.97 V. This seems a bit
high to me since almost all the standard led's I've seen have been roughly
3.5 V, give or take .3 (but again this number is not too reliable since it
can mean different things depending on manufacturer, as I learned from this
thread). I don't believe there is any current limiting resistors in the
wires that go from the pins to the led, so am I to correctly assume that
this extra voltage is not enough to damage the leds?

 

[ric]

Another HAM radio wanker who's electronics knowledge fits on a postage
stamp.
Bill, please learn the difference between current and voltage. Also
read up about potential differential. If you connect the -ve side of
the LED to -15V and the +ve side to +15v you have a PD of 30V which, if
the current isn't too high, is still within the limits of an LED.

Who is the wanker? Without a current limiting resistor in your above
circuit, the current through the LED will be the Isc of the lessor
supply. Maybe *YOU* ought to learn about maximum current flow. Putting
a LED across any voltage without a current limiting resistor will result
in a worthless LED. Your electronics theories in this and other NGs are
laughable, at best.

 

[ric]

I checked the source voltage of the power led by touching the leads of my
multimeter to the pins on the mobo--I get around 4.97 V. This seems a bit
high to me since almost all the standard led's I've seen have been roughly
3.5 V, give or take .3 (but again this number is not too reliable since it
can mean different things depending on manufacturer, as I learned from this
thread). I don't believe there is any current limiting resistors in the
wires that go from the pins to the led, so am I to correctly assume that
this extra voltage is not enough to damage the leds?

The source voltage on the MB is likely 5v. Your reading was a bit less
because you were reading the voltage on half of a two impedance voltage
divider (the multi-megohm impedance of your meter, and the very low in
comparison impedance [around 300-500 ohms] of the MB dropping resistor.)
Once the forward drop of your LED is put into the circuit, the voltage
at those same two pins will drop to approximately the same as the LED's
forward drop (with the remainder across the MB's dropping resistor.)
This drop across the MB's dropping resistor is what determines the LED's
forward current.

 

[Conor]

Who is the wanker?

You ARE.

Without a current limiting resistor in your above
circuit, the current through the LED will be the Isc of the lessor
supply. Maybe *YOU* ought to learn about maximum current flow. Putting
a LED across any voltage without a current limiting resistor will result
in a worthless LED. Your electronics theories in this and other NGs are
laughable, at best.

Err I KNOW THIS. WE're arguing about VOLTAGE not current. The OPs
question was also about connecting to a motherboard header which
already HAS the current limiting in place.

If you're going to butt into an argument first, at least make sure
you've followed it from the start.

 

[Bill Turner]

Another HAM radio wanker who's electronics knowledge fits on a postage
stamp.
Bill, please learn the difference between current and voltage. Also
read up about potential differential. If you connect the -ve side of
the LED to -15V and the +ve side to +15v you have a PD of 30V which, if
the current isn't too high, is still within the limits of an LED.

_________________________________________________________

Jeez, Conor, I don't know whether to laugh or cry.

First of all what the h*** is the "e" in "ve"? Second of all, what the
h*** is PD?

PD *might* mean Power Dissipation, but in the context above it would be
nonsense.

And this statement: "you have a PD of 30V which, if the current isn't
too high, is still within the limits of an LED." Is also utter
nonsense. LED's don't work that way. The *source* voltage could be
100,000,000,000 volts, and if the current is properly limited, the LED
would work just fine. IOW, the "limits of an LED" have NOTHING to do
with the original source voltage PROVIDED the current limiting is done
correctly.

Please tell me which brand of breakfast cereal your education came in.
I never want to buy that brand again.

 

[ric]

Err I KNOW THIS. WE're arguing about VOLTAGE not current.

What does *that* mean? The only place that voltages exists without current
is in an open circuit. Are you saying that this is an open circuit? LEDs
develop their own forward voltage drop. They're diodes for gawd's sake.
The source voltage and current limiting resistor are the important factors
here. I don't care if the source voltage is 1KV. The LED will never see it
if the current limiting resistor is sized right.

The OPs
question was also about connecting to a motherboard header which
already HAS the current limiting in place.

And what MB header has the +15v and -15v that you brought up in

If you're going to butt into an argument first, at least make sure
you've followed it from the start.

But I have. I gave the OP his answer long before you provided your usual
incorrect info. Correcting your posts seems to be a hobby with many
throughout Usenet.

 

[Conor]

_________________________________________________________

Jeez, Conor, I don't know whether to laugh or cry.

First of all what the h*** is the "e" in "ve"? Second of all, what the
h*** is PD?

PD *might* mean Power Dissipation, but in the context above it would be
nonsense.

I'm British so use British terminology.

+ve = positive voltage
-ve = negative voltage
P.D. = Potential Differential.

 

[Conor]

What does *that* mean? The only place that voltages exists without current
is in an open circuit. Are you saying that this is an open circuit? LEDs
develop their own forward voltage drop. They're diodes for gawd's sake.
The source voltage and current limiting resistor are the important factors
here. I don't care if the source voltage is 1KV. The LED will never see it
if the current limiting resistor is sized right.

No shit Eintstein. THe OP was wanting to know if they could safely
connect a LED to the motherboard header. The motherboard header limits
to the current to a safe level. Most LEDs will operate safely at the
levels outputted by the motherboard.

And what MB header has the +15v and -15v that you brought up in


But I have. I gave the OP his answer long before you provided your usual
incorrect info.

Except it isn't. Just a question, where did you learn about
Electronics? I personally did a 2yr course and then worked at Euromax
Electronics in the UK as a bench engineer repairing videogame boards
where the manufacturers regularly scrubbed component identification
numbers/colours off as well as using custom SIPP/DIL packages to try to
prevent cloning.

 

[Bill Turner]

I'm British so use British terminology.

_________________________________________________________

This explains a lot. The joke here in the Colonies goes: Why do the
Brits drink their beer warm? Because Lucas makes their refrigerators.

I'll bet you are the Chief Engineer.

 

[KCOM]

What does *that* mean? The only place that voltages exists without current
is in an open circuit. Are you saying that this is an open circuit? LEDs
develop their own forward voltage drop. They're diodes for gawd's sake.
The source voltage and current limiting resistor are the important factors
here. I don't care if the source voltage is 1KV. The LED will never see it
if the current limiting resistor is sized right.


And what MB header has the +15v and -15v that you brought up in


But I have. I gave the OP his answer long before you provided your usual
incorrect info. Correcting your posts seems to be a hobby with many
throughout Usenet.

Conor is a Gay Porn Star. See?

http://Conor.Turton.of.Hull.UK.swellserver.com/hosted_sites/gay_site.php

He's also a troll.

 

[Conor]

_________________________________________________________

This explains a lot. The joke here in the Colonies goes: Why do the
Brits drink their beer warm? Because Lucas makes their refrigerators.

I'll bet you are the Chief Engineer.

Lucas = automotive electrical and fuel injection manufacturer in the UK
so the humour doesn't carry over. It is probably hilarious to you but
makes no sense from here unless you mean a different Lucas.

 

[ric]

Except it isn't. Just a question, where did you learn about
Electronics?

Asked and answered in another NG thread that you finally bowed out of
after being told by a half dozen posters how wrong you were. (Cal
Berkeley, EE.)

 

[Christopher Pollard]

Lucas = automotive electrical and fuel injection manufacturer in the UK
so the humour doesn't carry over. It is probably hilarious to you but
makes no sense from here unless you mean a different Lucas.

Joseph Lucas, the Prince of Darkness. So named due to the unreliable nature of
vehicle electrical systems long before electronic fuel injection existed.

Oh, and you can't put 30v across an LED unless it's current limited, in which
case the LED ill pull it down to about 1.6V and the limiting resistor will drop
the rest.


Chris Pollard