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

On Tue, 15 Mar 2011, Man-wai Chang wrote:

It's only the resistance value of the resistor and the load that
determines the current drawn, not the wattage of the resistor.

Then why are resistors (in that shop at least) classified resistors by W?

Because there are plenty of times when more current goes through the
resistor, and it needs to be a higher wattage.

The wattage of a resistor means nothing so long as your demands are
not too much for that wattage resistor. A higher wattage resistor
won't be noticed by the circuit. But you'll end up with a physically
bigger resistor than you need.

Forty years ago, the schematics in the hobby magazines would routinely
specify 1/2w resistors. But it was just routine, for a long time 1/2
resistors were the most commonly available. It was only the transistion
to solid state that 1/4 resistors became common, and while they could be
used due to lower power circuits, the change was really because the 1/4w
resistors were physically smaller.

Michael

"Man-wai Chang" wrote:
It's only the resistance value of the resistor and the load that
determines the current drawn, not the wattage of the resistor.

Then why are resistors (in that shop at least) classified resistors by W?


The wattage rating of a resistor just indicates how much
power it can dissipate as heat without getting so hot that it
changes its resistance. Using a resistor with a higher wattage
rating than is needed just adds useless bulk but is otherwise
harmless.

*TimDaniels*

Man-wai Chang wrote:
So you can choose the correct wattage for it's intended use.

Let me review your replies first.


The resistance and power rating are independently controlled factors.

Resistors come in families, like 10 ohms to 22 megohms at 1/8th watt,
10 ohms to 22 megohms at 1/4 watt, 10 ohms to 22 megohms at 1/2 watt
and so on. Once you get to a certain wattage level, the wattage
rating is harder to predict.

In a DC situation, where the resistor has some voltage across it,
a current flows. We use Ohms law, amongst others.

V = R * I

If you have a 3 volt drop across the resistor, and the resistor is 68 ohms,
then to work out the current:

3V / 68 ohms = I = 0.044 amps or 44 milliamps

The power dissipated by the resistor, given the conditions above,
is as you stated P = VI. To make the latter two forms, you do substitution
of Ohm's law.

P = V * I

= V**2 / R

= (I * R) * I = I**2 * R

You select one of the previous three equation forms, according
to the numbers you have available to work with. For example,
I might select

= (3 volts * 3 volts ) / 68 ohms = 0.132 watts

That amount of watts, is slightly more than 1/8th watt, so you'd need a
higher wattage resistor. Perhaps a 1/4 watt resistor would be a
good choice. The 1/4 watt resistor will dissipate that heat, without
being damaged. A higher wattage resistor will also do the job,
such as a 1/2 watt resistor. As long as the resistor selected,
can handle 0.132 watts minimum, it's good enough.

If you use a resistor in a confined space, where it can't receive
any cooling air, then it might need to be "derated". The following
curve, demonstrates when to derate. At the boiling point of water (100C),
I can use about 60% of the normal power rating of the resistor. Notice
that this particular curve, says the resistor can't really be used
at over 150C ambient.

http://www.interfacebus.com/Resistor-Derating-Graph.jpg

So, say I go to the store, with a piece of paper in my hand
saying "I need 68 ohms at greater than 0.132 watts". I can
buy a 1/4 watt resistor, or a 1/2 watt resistor or a 25 watt
resistor. Any of those will handle the level of power being
dissipated. The smaller resistors will be cheaper and
take up less space, for this kind of application.

+ -
+5V ----- resistor ---- LED ------ Ground

HTH,
Paul

On 3/14/2011 4:52 AM, Man-wai Chang wrote:
20ma * 3V = 60mW so a 250mW resistor is sufficient in this case. You
have to substitute the voltage drop and desired current for the LED that
you want to use.

What if I used a 1/2W resistor instead of a 1/4W (250mW?)? Would it
reduce the current too much?


In your case, yes...

Seriously.. do some search on specifications.



--
The black flies were coming.
(Alastair Mayer in "Small Penalties")

Seriously.. do some search on specifications.

Working on it... slowly!

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used due to lower power circuits, the change was really because the 1/4w
resistors were physically smaller.

I noticed that when I looked at them.

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Man-wai Chang wrote:

It's only the resistance value of the resistor and the load that
determines the current drawn, not the wattage of the resistor.

Then why are resistors (in that shop at least) classified resistors by W?

To prevent overheating. If you take a resistor and operate it at its
rated power, it will easily reach 100 Celcius, and if it's rated for
at least about 1/2W it can also burn your fingers, melt wire
insulation, make circuit boards discolor or char, and even soften
solder. I've been told to use resistors rated for at least twice the
maximum power they'll have to dissipate (or 4x the max for hot
environments, like in a car's engine compartment), and usually when
I've found a cracked or burned resistor, it had been running at very
close to its power rating.

solder. I've been told to use resistors rated for at least twice the
maximum power they'll have to dissipate (or 4x the max for hot
environments, like in a car's engine compartment), and usually when

Thanks

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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.

"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*