Difference between 1/4W and 1/2W resistors?

edfair wrote:

For peace of mind you might want to ask for flame proof resistors.

I take it by now, Man-wai Chang is running for the hills, and not looking back :-)

As Timothy suggests, a little headroom is all you need, for peace of mind.
Selecting the next size up by calculation, is enough. The resistor will end
up stone cold to the touch, as proof it's not headed for flameout.

It takes deliberate carelessness, to create a spectacle. And the arithmetic
in this thread, is enough to prevent such a spectacle.

Paul

On Wed, 16 Mar 2011 10:23:34 -0700, Timothy Daniels rearranged some
electrons to say:

"edfair" wrote:

Most of the stuff I've dealt with used 330 ohm 1/8W resistors with
standard LEDs on 5V supply. Doesn't let thru enough for full brightness
but doesn't lead to premature death.

For peace of mind you might want to ask for flame proof resistors.
Nothing like miscalulating the required value or getting a defective
one and ending up with a candle.

IIRC the LED drop is 1.7V which would put 3.3V across the 330 ohm
resistor.

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating. The
1/4watt resistors mentioned previously in this thread should suffice
quite nicely.

*TimDaniels*

No. 1/8W = 125 mW, not 8 mW.

"david" wrote:
Timothy Daniels wrote:

"edfair" wrote:

Most of the stuff I've dealt with used 330 ohm 1/8W resistors with
standard LEDs on 5V supply. Doesn't let thru enough for full brightness
but doesn't lead to premature death.

For peace of mind you might want to ask for flame proof resistors.
Nothing like miscalulating the required value or getting a defective
one and ending up with a candle.

IIRC the LED drop is 1.7V which would put 3.3V across the 330 ohm
resistor.

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating. The
1/4watt resistors mentioned previously in this thread should suffice
quite nicely.

*TimDaniels*

No. 1/8W = 125 mW, not 8 mW


(.125W)/(.033W) = 3.79 . Review your math.

*TimDaniels*

Timothy Daniels wrote:

[snip]

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating.
The 1/4watt resistors mentioned previously in this thread should
suffice quite nicely.

*TimDaniels*

There's a problem there .033W (33mW) is a lot less than 1/8W (125mW).

--
Mark Lloyd
http://notstupid.us

"Creationist: (1) One who follows creationism. (2) A moron. (3) A person
incapable of doing math. (4) A liar. (5) A very gullible true believer."

On Thu, 17 Mar 2011 08:46:42 -0700, Timothy Daniels rearranged some
electrons to say:

"david" wrote:
Timothy Daniels wrote:

"edfair" wrote:

Most of the stuff I've dealt with used 330 ohm 1/8W resistors with
standard LEDs on 5V supply. Doesn't let thru enough for full
brightness but doesn't lead to premature death.

For peace of mind you might want to ask for flame proof resistors.
Nothing like miscalulating the required value or getting a defective
one and ending up with a candle.

IIRC the LED drop is 1.7V which would put 3.3V across the 330 ohm
resistor.

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating.
The 1/4watt resistors mentioned previously in this thread should
suffice quite nicely.

*TimDaniels*

No. 1/8W = 125 mW, not 8 mW


(.125W)/(.033W) = 3.79 . Review your math.

*TimDaniels*

Review yours. Do I need to remind you of what you wrote?

that works out to .033watts, or about 3.79 times the 1/8watt rating.

33 mW is 26% of the 125mW rating.

You should have said: "A 1/8W resistor has a power dissipation rating
that is 3.79 times greater than the power dissipated by the LED circuit,
so it will be more than adequate for this application".

On Thu, 17 Mar 2011 20:26:57 -0500, Mark Lloyd rearranged some electrons
to say:

Timothy Daniels wrote:

[snip]

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating.
The 1/4watt resistors mentioned previously in this thread should
suffice quite nicely.

*TimDaniels*

There's a problem there .033W (33mW) is a lot less than 1/8W (125mW).

Tim hit the multiply button instead of the divide button on his
calculator.

"david" tried again:
Timothy Daniels wrote:

"david" wrote:
Timothy Daniels wrote:

"edfair" wrote:

Most of the stuff I've dealt with used 330 ohm 1/8W resistors with
standard LEDs on 5V supply. Doesn't let thru enough for full
brightness but doesn't lead to premature death.

For peace of mind you might want to ask for flame proof resistors.
Nothing like miscalulating the required value or getting a defective
one and ending up with a candle.

IIRC the LED drop is 1.7V which would put 3.3V across the 330 ohm
resistor.

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating.
The 1/4watt resistors mentioned previously in this thread should
suffice quite nicely.

*TimDaniels*

No. 1/8W = 125 mW, not 8 mW


(.125W)/(.033W) = 3.79 . Review your math.

*TimDaniels*

Review yours. Do I need to remind you of what you wrote?

that works out to .033watts, or about 3.79 times the 1/8watt rating.

33 mW is 26% of the 125mW rating.

Which says the same thing. Learn some arithmetic.


You should have said: "A 1/8W resistor has a power dissipation rating
that is 3.79 times greater than the power dissipated by the LED circuit,
so it will be more than adequate for this application".


Which is the same thing. Learn some arithmetic.

*TimDaniels*

"Mark Lloyd" wrote:
Timothy Daniels wrote:

[snip]

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating.
The 1/4watt resistors mentioned previously in this thread should
suffice quite nicely.

*TimDaniels*

There's a problem there .033W (33mW) is a lot less than 1/8W (125mW).


Yes, that's the point. A 1/4watt resistor easily dissipates the heat.

*TimDaniels*

Timothy Daniels wrote:



[snip]

that works out to .033watts, or about 3.79 times the 1/8watt rating.

33 mW is 26% of the 125mW rating.

Which says the same thing. Learn some arithmetic.

"about 3.79 times the 1/8watt rating" is about .47375 watts. This is MORE
than 1/4watt.

[snip]

--
Mark Lloyd
http://notstupid.us

"Creationists are the best evidence yet of the absence of intelligent
design"

Timothy Daniels wrote:

"Mark Lloyd" wrote:
Timothy Daniels wrote:

[snip]

Assuming that P=VI (power = voltage x current), or V**2/R,
that works out to .033watts, or about 3.79 times the 1/8watt rating.
The 1/4watt resistors mentioned previously in this thread should
suffice quite nicely.

*TimDaniels*

There's a problem there .033W (33mW) is a lot less than 1/8W (125mW).


Yes, that's the point. A 1/4watt resistor easily dissipates the heat.

*TimDaniels*

Ture, it does. However, the phrase "about 3.79 times the 1/8watt rating" is
contradictory to that. .033 watts is reasonable for a LED. What isn't, is
using almost 4 times that value.

--
Mark Lloyd
http://notstupid.us

"Creationists are the best evidence yet of the absence of intelligent
design"