Trying to decipher resistor numbers?

[MMACH 5]

I have several resistors here in front of me...

This is what they say on the packages:
1: 1/2w 4.7 ohm 5%
2: 2w .47 ohm 5%
3: 5w 2.2 ohm 5%
4: 5w 2.7 ohm 5%
And both 5w resistors have a store label that reads "Power Wirewound Resistor. Voltage = 200."

I guess it just baffles me that a half watt resistor can have a higher resistance than the others. This thing is tiny, compared to the others.

What does the wattage on these mean?
What does the 5% refer to? -- Nevermind this question. I found a site telling me about tolerance.
Is the voltage the limit that these can pass without damage?
Wirewound? -- Nevermind this question, as well. I found a site about resistors' construction.

I'm so confused.

 

[Benson]

I guess it just baffles me that a half watt resistor can have a higher resistance than the others. This thing is tiny, compared to the others.

A 1mm air gap is effectively infinite resistance; maybe a teraohm or so. A piece of copper wire is practically 0. Size correlates with power, but has basically nothing to do with resistance.

What does the wattage on these mean?

Continuous power capability. If you dissipate more power, you risk burning it up.

What does the 5% refer to?

Permissible tolerance from the specified resistance.

Is the voltage the limit that these can pass without damage?

Yes, it's one of the limits. Trivially, you can calculate a voltage limit to avoid exceeding the power limit -- swrt(5W*2.7ohm) = 3.7 volts. But you can apply much higher voltage pulses with a low duty cycle without exceeding the power, so the 200V limit still applies. Above that, you risk breakdown.

Wirewound?

What it sounds like -- the resistor is made by winding a long wire. They can be made quite precise, and can handle relatively high currents, but a side effect of being a coiled wire is that makes a comparatively good inductor as well as a resistor, so you have to be careful if using these with AC.


PS. missed your edits. Hope this helps with the remaining two.

 

[Trekmeister]

The resistance has nothing to do with the size of the resistor. Regarding the wattage is how much power the resistor can dissipate without melting or burning up. It is quite easy to understand how a large resistor can handle more power, right?

Asking our good old friend mr Ohm he will tell you that the power (wattage) which the resistor has to get rid of depends on the voltage over it and the current through it. Or:

P = U * I

He also might point out one more thing about the voltage over something depends on the resistance and the current:

U = R * I

In other words running 1A through a 2.7 ohm resistor will drop 2.7V equalling to 2.7W in this case. Waaaay to much for a 0.5W resistor but ok for one of the bigger 5W ones.

Very basic stuff, there are probably tons of good tutorials all around the interweb you could go through.

 

[MMACH 5]

Thanks for the info. :thumbsup: It's all starting to make sense.

I purchased the resistors a few years ago for a project that never happened-life kept getting in the way. Anyway, here is what I'm trying to do with these:

I have several LED lights with circuit boards, designed to run off 2x AAA batteries. (Okay, they're bicycle taillights, but I figured this was the forum to find out more about resistors.)
Ever since I switched to rechargables, I've been disappointed with their brightness. Since they are supposed to run at 3v and the rechargables only kick out 2.4v.
I've had them wired to a central battery pack with 2x AA batteries. This extends the run-time and saves me from having to crack them open every week to change the batteries. However, they are still only getting 2.4v.

I want to replace the AAs with a single 18625 battery. The ones I have are 3.7v 2400mAh.
From what I've read about the length of wire from the battery pack to the lights, I'm already knocking down the volts to about 3.5-ish. I know the circuit boards on the lights will keep the LEDs from getting too much juice but I'm guessing that there's a risk of damaging the boards.
So, I'm left with an extra .5 volts to try and hold back.

Tell me if I calculated this correctly:

volts/amps = resistance
.5 volts is what I'm trying to dissipate
2400mAh = 2.4 amps
thus
.5/2.4 = .2083333

volts x amps = watts
.5 x 2.4 = 1.2

Meaning that all of the resistors I have on hand have too much resistance.

I need a resistor that says something along the lines of 2w .2 ohm, right?

Thanks again for the help.

 

[Trekmeister]

Sorry but you are mixing up current and capacity here. 2400mAh is 2400 milli-ampere-hours, so it is a measure of capacity. In theory you would be able to draw 2.4A for one hour, or 1A for 2.4 hours etc.


What you need to do first is to find out the current usage of your bicycle light. Without knowing anything about your light it is very hard to make any good guesses. However a small red LED might be using around 30mA at 2V, a rough estimate.

Provided the light is from the most basic possible construction with only a series resistor and the LEDs we can calculate a suitable resistor.

If there are 2 LEDs in parallel in the light it should draw 60mA. You want to drop at least 1V over your resistor (assume at least 4V from the battery, closer to 4.2V fully charged). Do not worry about voltage drop in the cables. Unless you have many many meters worth of very thin wires this will not be a problem when running this kind of low current devices.

R = U/I => R = 1/60 = 0.017 kOhm = 17 Ohm
P = U*I => P = 1*60 = 60mW = 0.06W (very little, any resistor will be able to handle this)

Since resistors don't come in just any value but are organised in series, you will probably not be able to find an exact match. However an 18 Ohm resistor should be easy to find and will work just fine. However this all depends on your light. The best thing is if you can take your multimeter and measure just how much current the light draws and use those values instead.

 

[MMACH 5]

Yikes, now I'm confused again!

Seriously, thanks so much for the help.

Two of the blinkies have 9 LEDs on their circuit boards and one has 3 LEDs.
But all of them take 2x AAA batteries. Does this mean their circuit boards are designed to dissipate any extra current coming at them?

The only meter I have is an analog multimeter. Do I just take measurements from the back of the circuit board, where the LEDs are soldered on and then multiply by 9 (or by 3 on the other light)?

I might have to change my screen name to NNOOB 5.

Thanks again.

 

[MMACH 5]

Okay, I found a thread here on CPF that explains measuring draw. I'll do this when I get home tonight.

 

[MMACH 5]

Even though I found how to TAKE measurements for draw, I don't know how to INTERPRET the readings.

I had the selector set at DCA-250m (the other two DCA settings are 500u and 10m, but on both of these, the needle would peg out to max).

Here are the readings from my measurements:

Thanks again.

 

[jason 77]

Even though I found how to TAKE measurements for draw, I don't know how to INTERPRET the readings.

I had the selector set at DCA-250m (the other two DCA settings are 500u and 10m, but on both of these, the needle would peg out to max).

Here are the readings from my measurements:

Thanks again.

Well your meter is saying that the 3 led light is drawing 85mA the 9 led light is drawing 120mA and when all the lights are on the total current draw is 195mA.

 

[MMACH 5]

Whoops. The 18650 battery that I was testing with, last night was rather drained. I realized this when I was fiddling with the setup tonight and had a different battery. When I took a current reading it pegged out the meter.

So after some checking and measuring, I realized the one from last night was almost dead.

Since the resistors are about $.40 apiece, I decided to just grab a variety of them and do some testing.

These lights are right off the shelf and the only modification is holes for wiring. The reading I did not measure last night was the current of 2x AAAs running all three lights. I measured this tonight and got 220mA. Since this is what these lights were designed to run off of, my goal is to bring the current from the 18650 down to 220mA or at least very close.

I began clipping in different rated resistors until I found the right one. These were all 1/4 & 1/2 watt resistors. 2.2ohm didn't pull it down enough and 5.6ohm pulled it down too far. In the end, one of the resistors I had on hand already brought the current to 220mA. I'll be using the 1/2w 4.7ohm resistor.

So tomorrow, I'll get it all put together, mounted on the bike and wired in.

Thanks again for all the help.

 

[MMACH 5]

I'm sure y'all are all tired of hearing about my stumbling through this, (nobody likes to watch sausage being made) but I have another question.

Even though the 4.7ohm resistor brings the current usage down to 220mA, when I take a reading from the 18650 battery with the resistor in place, it still measures 4v. So the voltage is the same whether the resistor is wired in or not. Does the resistor not have any effect on voltage, only current?

Thanks again.