How putting resistors operate when put in parallel with resistance and wattage.

Thread Starter

sirchuck

Joined Feb 14, 2016
131
From what I understand putting 4 1/8 watt 100ohm resistors in parallel will have the same effect as one 25ohm 1/2 watt resistor. Is that accurate?

Also, I tried putting a 39ohm 1/8watt resistor in a 5v circuit all by itself to increase amps being pulled from my power bank, but it got really hot really fast. So in order to solve the heat issue I want to use 4 100ohm 1/8 watt resistors in parallel so it doesn't get hot but still pulls about the same amount of amps.

Thanks. :)
 

Externet

Joined Nov 29, 2005
1,410
From what I understand putting 4 1/8 watt 100ohm resistors in parallel will have the same effect as one 25ohm 1/2 watt resistor. Is that accurate? ...
Yes, it is accurate. Same as four 1/8 Watt - 6.25 Ohms resistors in series.

Also, I tried putting a 39ohm 1/8watt resistor in a 5v circuit all by itself to increase amps being pulled from my power bank, but it got really hot really fast. So in order to solve the heat issue I want to use 4 100ohm 1/8 watt resistors in parallel so it doesn't get hot but still pulls about the same amount of amps.

Thanks.
The desire to increase Amperes makes not much sense, unless you want to discharge your power bank. Pulling the same amount of amperes trough resistance means heating the same amount. The amount of heat is the multiplication of the voltage times current.
What is solving the heat issue... Do you want more heat or less heat ? What do you want to do with your power bank ?
The 39 ohm 1/8 Watt resistor will produce less heat than the four 100 Ohm - 1/8 Watt resistors but all the heat will be concentrated to only that small resistor area and its temperature will be higher and perceived as more heat.

Even if you use 10 Watt resistors, the heat generated is the same, but will be distributed in a much larger area, perceived as less heat. But the reality is the heat would be the same; the temperature is the one that would then decrease.

Heat is NOT temperature. A match head burns at about the same temperature as a house on fire, but the amounts of heat are very different.
 
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OBW0549

Joined Mar 2, 2015
2,982
From what I understand putting 4 1/8 watt 100ohm resistors in parallel will have the same effect as one 25ohm 1/2 watt resistor. Is that accurate?
Yes, provided you don't bundle them all tightly together, which could decrease their power dissipation capability. Leave a bit of space between them for air flow.

Also, I tried putting a 39ohm 1/8watt resistor in a 5v circuit all by itself to increase amps being pulled from my power bank, but it got really hot really fast. So in order to solve the heat issue I want to use 4 100ohm 1/8 watt resistors in parallel so it doesn't get hot but still pulls about the same amount of amps.
5 volts across 39 ohms will draw 128 mA and dissipate 0.64 watts so yes, it will certainly get really hot really fast. 4 100 ohm resistors in parallel will give you 25 ohms and will draw 200 mA, not 128 mA.
 

Thread Starter

sirchuck

Joined Feb 14, 2016
131
Externet, putting resistors in series will also increase the amount of wattage they can handle? Or are you just saying the resistance would be the same as my example?

My power bank doesn't draw enough power to keep my project running, so I thought if I add a circuit that just has a resistor from positive to negative in it, I could allow enough current through to keep it active, but I don't want to burn the house down.

I figured if I put the resistors in parallel it would better dissipate the heat as if it were a resistor rated for 1/2 watt, therefor not burning up the resistor.

I think just putting the 1/8 watt 39ohm resistor would just burn up, and wanted a way to better dissipate the generated heat.

I'm ok with using 200ma instead of 168ma or whatever, I just need to know the principle, I can do the math to min/max the circuit later. Right now I'm not sure how much amps I need to keep it running.
 

mcgyvr

Joined Oct 15, 2009
5,394
My power bank doesn't draw enough power to keep my project running, so I thought if I add a circuit that just has a resistor from positive to negative in it, I could allow enough current through to keep it active, but I don't want to burn the house down.
Your logic is likely flawed.. A load (your project) typically "pulls" what it needs from a power source (power bank)..
The power bank doesn't "push" what it thinks you need..
Why don't you describe your "project" or post a schematic of it so better help can be given...

It could be that your power bank (whatever it is.. Give us specs on that too) may not be able to provide enough current for your project and so its dropping voltage or going into a shutdown or whatever..

Details.. Give us the details if you want proper help.. Don't assume you have it figured out..
Throwing resistors at something is likely not the solution..
 

Dodgydave

Joined Jun 22, 2012
8,383
Ohms Law mate!!

5V/ 39ohms = 128mA @ 640mW

5V / 25ohms = 200mA @ 1.W

Hence your 1/8w resistor burned up!
 
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Externet

Joined Nov 29, 2005
1,410
Externet, putting resistors in series will also increase the amount of wattage they can handle?
Yes.

My power bank doesn't draw enough power to keep my project running,...
Your project needs a more capable power bank than what you have, and adding resistors makes it much worse. Or your project is drawing too much current.
You cannot push a bulldozer with a bicycle.

I think just putting the 1/8 watt 39ohm resistor would just burn up, and wanted a way to better dissipate the generated heat.
What you should aim is the opposite. Avoid wasting power as resistive heat in a starving circuit. You are peeing out of the can. We still do not know what you are trying to do. Pulling more current from a power bank will NOT make your project run. The circuit itself will pull what needs to work IF the power bank is capable of such.
 

Thread Starter

sirchuck

Joined Feb 14, 2016
131
I just wanted to understand the principle of putting resistors in parallel and how it affects the wattage rating.

This is my power supply
https://www.amazon.com/gp/product/B01ARI17NI/ref=oh_aui_detailpage_o01_s00?ie=UTF8&psc=1

I'm using it to power 2 led's ( a very small circuit ). After a predetermined amount of time, the power bank turns off. I have seen circuits that will basically oscillate every few seconds to run through a small resistor for a short time to allow higher mA through to reset the power banks timer. I know those types of circuits work.

I know just putting 4 100ohm 1/8 watt resistors in parallel will cause the battery to discharge much faster, but I just wanted to know if it was possible to do without burning up any of the four 1/8 watt rated resistors. I think using one 25ohm 1/8 watt rated resistor would burn up. ( at least the 39ohm resistor got very hot )

So the only question I was trying to understand was, would using 4 1/8 watt rated resistors at 100ohms give me 25ohms resistance and have none of them burn up at 5v. :D I can test it, to be sure, but from what I understood from the initial responses, it looks like adding more resistors in parallel or even in series will share the heat load that I currently have passing through my single resistor.

So, really, there is no physical problem for you to solve. I just had a question on how to make my 1/8 watt resistors capable of handling 1/2 watt or 1 watt. It looks like just using more 1/8 watt resistors ( not clumped ) is the answer.
 

OBW0549

Joined Mar 2, 2015
2,982
My power bank doesn't draw enough power to keep my project running, so I thought if I add a circuit that just has a resistor from positive to negative in it, I could allow enough current through to keep it active, but I don't want to burn the house down.
If I'm reading you right, what you probably meant to say is that your project doesn't draw enough current from your power bank to keep it from going into automatic shutdown. Am I right?

I have a power bank (a 15,000 mAh Ravpower RP-PB19) that shuts down unless the load draws a certain minimum amount of current. I don't know what the threshold is, but most of my projects don't draw enough current to keep it alive. I doubt the keep-alive threshold is as high as 128 mA, though.

I gave up trying to use the Ravpower and switched to an el-cheapo 2200 mAh power bank that I got at the drugstore; being cheap, it doesn't have an automatic power-down circuit inside so it's always on. And it works great.
 

Thread Starter

sirchuck

Joined Feb 14, 2016
131
Yes OBW0549, exactly that. I have a cheap one too, works great.

I was just using the 100ohm as my example resistor, I didn't do any math yet to figure out what I really need. That part doesn't matter to me, just the part about how to keep the resistors from over heating because all my resistors are rated for 1/8 watt.
 

Externet

Joined Nov 29, 2005
1,410
So, really, there is no physical problem for you to solve.
The physical problem to solve is your understanding of the basics.

For your linked power bank, connect it this way :

(+)-----------------\/\/\/\/\-------------|>|----------------|>|-------------------(-)

Being (+) and (-) the output terminals of your 5V power bank.
\/\/\/\/\ being a 470 ohm resistor for the moment, to be tailored.
-|>|- being your LEDs
---------- is the interconnecting wires
(everything is in series !)
Note : If the LEDs are blue or white, it will not work ! Try only ONE LED first if that color !
 

Thread Starter

sirchuck

Joined Feb 14, 2016
131
(+)-----------------\/\/\/\/\-------------|>|----------------|>|-------------------(-)

Circuit runs for about 20 seconds then shuts off. I used two red led's, in the circuit you provided with a 470ohm resistor.
 

OBW0549

Joined Mar 2, 2015
2,982
I was just using the 100ohm as my example resistor, I didn't do any math yet to figure out what I really need. That part doesn't matter to me, just the part about how to keep the resistors from over heating because all my resistors are rated for 1/8 watt.
Just figure the power is equal to the square of the voltage divided by the resistance. So in this case,

P = 25/R.

Rearranging, and plugging in 1/8 watt for the power,

R = 25/0.125 = 200 ohms minimum to avoid overheating.
 

Externet

Joined Nov 29, 2005
1,410
OK. The circuit ran.
Then make sure the power bank is fully charged. If after a period of charging specified in the owner's instructions you still have a shut down; the power bank may be defective or was damaged by wrong use, or expects larger load, or data link on the remaining pair of the USB connector to stay on. Cannot guess which.

After shutting down in 20 seconds, unplugging and plugging again restores working ???
If yes, it is meant to shut down by your load condition.
If no, likely the cell inside is discharged.

Try about 270 ohms resistor instead after fully recharged and come back with results.
 

Thread Starter

sirchuck

Joined Feb 14, 2016
131
Title:
How do resistors operate when put in parallel with respect to overall resistance and wattage rating.

Just figure the power is equal to the square of the voltage divided by the resistance. So in this case,

P = 25/R.

Rearranging, and plugging in 1/8 watt for the power,

R = 25/0.125 = 200 ohms minimum to avoid overheating.
That's cool. I think you are saying if I used a single resistor (+) ----/\/\/\/\---- (-) at 5v, 200ohms & 1/8 Watts rating is the smallest I can use.
 

Thread Starter

sirchuck

Joined Feb 14, 2016
131
Externet,

This power bank is meant to be used to charge cell phones and has circuitry to shut off if it is not outputting (x) amount of mA.

It's fully charged up, and I could create one of the several circuits already on the web to get around the issue. I was really just trying to figure out the wattage ratings on resistors. At 5v on say a 39ohm resistor rated for 1/8 of a watt, so much current would pass that it would burn out the resistor - at least I assume. I mean, it got very hot.

So, what I wanted to know was how to lower the overall temperature being created by the current flowing through the resistor. One option I was trying to look into was putting four 100ohm resistors with 1/8 watt ratings in parallel.

The initial responses to this question answered my confusion. :) It seems that if you put enough resistors in parallel they will share the current so each resistor can bear the amount of current passing through. In other words, two 200 ohm 1/8 watt resistors in parallel would act like one 100 ohm 1/4 watt resistor, if I'm correct.
 
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