Hi everyone,
I am working on a project of my own and am completely new to circuitry.
My project is supplied by 3.7v battery, and 5v from outlet when not running on battery. I am using arduino powerboost 1000c, protrinket 5v, a stepper motor board, two infrared sensors and a push power led switch.

So my problem is that when I turn the device off and it is connected to the usb for charging the battery, the resistor gets hot. If its just connected to the battery alone, and the device is off, it still heats up.

I calculated the resistor I need through http://www.ohmslawcalculator.com/ohms-law-calculator. The voltage is 5v. The current and power I assumed to be 0.4 amp, and 2 watts. So it concluded 12.5 ohms resistor. As I have no background in this, most of my work and progress is through trial and error which got me to this point. I don't know what resistor to use for ohms and watts rating. I have spent lots of money buying various resistors, and driving 33 miles (66 miles round trip) to the closest radio shack to get them same day, while ordering others from online, but they don't seem to solve my problem. I then thought I needed a capacitor, but I tried and that didn't help either, unless I used it wrong, but doubt it.

In the pic the red wire to the right would connect to the end of the resistor, but I don't think this is a picture or schematic error.

Can someone please help me pick the right resistor

 

Welcome to AAC!

A block diagram or schematic showing connections would be much better than a photo of your breadboards.

The 2W resistor shouldn't be heating up because it isn't connected. I assume there's a current limiting resistor for the green LED, but dissipation in that resistor shouldn't be an issue.

EDIT: Just noticed that you said the resistor wasn't connected. What does the other end of the disconnected red wire connect to and how much current do you expect it to draw?

For your information, we usually derate components from their max power ratings. A very conservative design would derate a 2W resistor to 1W.

 

What is the purpose of the resistor?

 

Thank you for the welcome.

My apologies for the green led. That was for testing how to connect the capacitor if I needed it. It's not connected to anything. The resistor would heat up if I connect the wire, which I need connected at all times.
I have a schematic I made, but did not insert the resistor for the power yet. The 2 resistors you will see in the schematic are for the 2 or sensors.

Also I have a 68ohm 1/2 watt resistor that gets warm when device is off but not hot hot when connected to battery. But that resistor gets really hot when I connect to 5v and device is off.

 

The purpose of the resistor is to make the powerboost 1000c work correctly the way I need it to. It is wired with ENable pin to ground, so that when device is off, the powerboost does not turn on and waste charge from battery.

 

Also if I skip the resistor and just connect the 2 wires, my powerboost burns out. I have gone through about 6 of those. And they are pricey

 

Hello,

I do not see any resistor on the enable pin.
Also all other resistors in your drawing are 10K.
Please correct the schematic to the values and connections you are using.

Bertus

 

There shouldnt be a resistor there. I just tried that and almost burned out my powerboost again.
The 2x 10k are for the 2 infrared sensors.
I have not yet concluded what resistor to use between VS coming from the powerboost to the positive on the breadboard, where all positives form all components meet.

 

"For your information, we usually derate components from their max power ratings. A very conservative design would derate a 2W resistor to 1W.[/QUOTE]"

I am not sure what derate means or how to do this?

 

"For your information, we usually derate components from their max power ratings. A very conservative design would derate a 2W resistor to 1W.

"

I am not sure what derate means or how to do this?[/QUOTE]

 

How is this derating done?

I have tried 2 and even 3 resistors of same values in parallel niece and in series another time, but they either heat up the same in parallel, or they don't allow enough power to the device when in series.

 

I am not sure what derate means or how to do this?

In other words when your application calls for an X watt resistor, find one rated at about 2X.

Just to be clear, specifically what two points does the resistor connect between? And how exactly did you arrive at the 0.4A figure?

 

Also if I skip the resistor and just connect the 2 wires, my powerboost burns out. I have gone through about 6 of those. And they are pricey

Since you'd like to not keep burning things up, take a step back and figure out what you're doing wrong before throwing more parts at the problem.

What do you think this resistor that's getting hot is doing for you?

 

How is this derating done?

I have tried 2 and even 3 resistors of same values in parallel niece and in series another time, but they either heat up the same in parallel, or they don't allow enough power to the device when in series.

If your calculations call for a 2W resistor, you use a 5W (4W isn't a standard value). Or you use 2 2W resistor of twice the resistance you need in parallel.

 

Okay so doubling the power is derating. Thanks.
So I connected 2 resistors in parallel between the VS pin on the powerboost and the power wire to the rest of the components.
And the 0.4amp came from the ohmslawcalculator.com. I tried putting 68 ohm resistors in, and they were okay in providing adequate power to the device, and holding down heat when not on. So I added the watts of the 3x res

 

Right now I have 2x 12 ohm resistors of 2 watt ratings in series that work without burning anything, but they are too hot to touch. I know that can't be good.

 

Okay so doubling the power is derating.

That's one way, but not the usual way.

And the 0.4amp came from the ohmslawcalculator.com. I tried putting 68 ohm resistors in, and they were okay in providing adequate power to the device, and holding down heat when not on. So I added the watts of the 3x res

What do you expect the resistor to do?

I don't see a 12 ohm resistor in the schematic you attached.

 

Right now I have 2x 12 ohm resistors of 2 watt ratings in series that work without burning anything, but they are too hot to touch. I know that can't be good.

But that only gives you 6 ohms, not 12.

To do it with 12 ohm 2W resistors, you'd need 4 of them.

 

So adding resistors in series makes the ohms smaller? So if 2x 12 ohms in series become 6 ohms, then 4 resistors in series would bring it to 0 ohms? Is that what you recommend? Again I dont know how they work and have a hard time understanding, but do I need them to equal 0 ohms in the end for it to work correctly?

 

So adding resistors in series makes the ohms smaller? So if 2x 12 ohms in series become 6 ohms, then 4 resistors in series would bring it to 0 ohms? Is that what you recommend? Again I dont know how they work and have a hard time understanding, but do I need them to equal 0 ohms in the end for it to work correctly?

No! Resistors in series simply add together. In parallel, the total resistance is 1 / (1/R1 + 1/R2 + ...), so for example 4 resistors of 100 ohms in parallel gives you 1 / (1/100 + 1/100 + 1/100 + 1/100) = 25 ohms, and the total power dissipation of each resistor would be 1/4 of that used by a single 25 ohm resistor (since they're all equal values).