How to: DIY 3.3Ohm resistors

[jas0nn]

I noticed a few people were having problems on figuring out how to make these and alot of you guys have no soldering machine. Heres the ghetto way and it works for me.

How it works:
To get 3.3 Ohm, you use three 10Ohm resistors. Why 3? You use 3 because you divide the 10 by 3 to achieve the required 3.3. For example, if you want 2 ohms, you'll need five 10Ohm resistors to achieve the 2. 10 divided by 5 = 2.

What you need:
3 pieces of 1 watt 10Ohm resistors


Electrical tape
Scissor


1. Connect it all together without any tape.



2. Now start taping them all together.


3. Wrap it all up!


yay, now you made a 3.3Ohm resistor for all your ****!

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[Kiel]

I assume this is for the airbag sensor trick? You do know they actually make 3.3 ohm resistors, right?

[kimgt]

why would you not just solder them?

[Zand]

Originally Posted by kimgt

why would you not just solder them?

Originally Posted by jas0nn

alot of you guys have no soldering machine.

There

[jas0nn]

I know they make them but trying to look everywhere for it is a waste of time and effort. Plus, some of the companies I saw online are asking around $7 for shipping.

[cpunlamd]

I bought a bag of 100 high grade 3.3ohm resistors. For anyone that needs it, just send me a return envelope and stamp with your address already filled out + $1 and I'll send you a pair.

FYI: 10ohm x 3 = 3.0 ohms (That will throw an airbag code more often then not. Our ECUs are very sensitive. Therefore the 3.3ohm will work much better.)

[xxscaxx]

if your going to do it like that, at least twist them together so they dont come out of the tape. lol.

[NMZ_WRX]

MINE STILL DOES not work and i stil wonder how to do it. i got the ohms from a member on here and i just used ONE for the sensors. airbag light still on

[aboothby]

That's actually not the technically accepted method for calculating resistor values in parallel.

R=what you want

r1, r2, ... , rn = the series of 'n' resistors to be placed in parallel (of any value).

R = 1/(1/r1 + 1/r2 + ... + 1/rn)

The dividing method is only true if all of your parallel resistors are of the EXACT same size. Parallel impedence combinations are very sensitive and even small tolerances can make a fairly large impact on the resulting impedence. Especially with resistors from radio-shack which have a pretty high tolerance 5-10%. Which is fine for low sensitivity usage like lights and such.

When using resistors in a sensitive active device environment (anything with transistors) you should use low tolerance 1% resisitors at least.

www.mouser.com <- look under passive components and look for axial thick metal film resistors.

[div]

Sorry dude, that is right up there with one of the most ghetto things I have ever seen.

[Jack ffr1846]

This is NOT rocket science.



jack

[aboothby]

It's not "rocket science", but it IS science and there is alot of complex theory behind the simple little R= equations that you use every day. In circuit design it is important to use the correct theory in the projects. Even though everything in electronics is 'theory' it is the same theory used by all engineers who create these systems. It is important to stick to the same theory in order for the systems to operate as intended.

The everyday V=IR ("ohm's law") is actually just a simplification of the real ohms law: E + (v X B) = Jρ

In other words.. Use derivations of V=IR to calculate resistive loads if you want to be correct, its as easy as looking up the equations on google.

[Hines]

Originally Posted by aboothby

It's not "rocket science", but it IS science and there is alot of complex theory behind the simple little R= equations that you use every day. In circuit design it is important to use the correct theory in the projects. Even though everything in electronics is 'theory' it is the same theory used by all engineers who create these systems. It is important to stick to the same theory in order for the systems to operate as intended.

The everyday V=IR ("ohm's law") is actually just a simplification of the real ohms law: E + (v X B) = Jρ

In other words.. Use derivations of V=IR to calculate resistive loads if you want to be correct, its as easy as looking up the equations on google.

omg, ohms law, glad someone quoted it
on a second note, 1 watt resistors are not needed for this applicaton. using the 1% tolerance as aboothby recomended is the way to go. 1/4 watt is all that you would need, much smaller resistor, and if you dont have a soldering iron your best bet would be to stack the resistors on top of each other and twist them together, instead of the ladder you got going on. But reguardless i wouldnt do this without soldering anyways. but this is comming from an electronics engineer and i would probably be fired from my job if i did something like that :P

[jmikolich]

Originally Posted by Hines

omg, ohms law, glad someone quoted it
on a second note, 1 watt resistors are not needed for this applicaton. using the 1% tolerance as aboothby recomended is the way to go. 1/4 watt is all that you would need, much smaller resistor, and if you dont have a soldering iron your best bet would be to stack the resistors on top of each other and twist them together, instead of the ladder you got going on. But reguardless i wouldnt do this without soldering anyways. but this is comming from an electronics engineer and i would probably be fired from my job if i did something like that :P

Yes, Please as an Electrical Engineer too, I'd say, do this correct and don't have all that bare metal hanging out just waiting for something bad to happen...

for instance if your 12vdc line came across your 3.3ohm resistor, Ohm's law says 12/3.3 = 3.6amps is pulled. you can kiss those resistors goodbye and possible a fuse or two.

Seriously, do this right, some other sites would be.

DigiKey, Mouser, Arrow... just as a FYI

[aboothby]

Digikey has the best selection and prices, but mouser will ship out much faster.

[twizzstyle]

Originally Posted by div

Sorry dude, that is right up there with one of the most ghetto things I have ever seen.

What he said. It's worth the investment in spending $10 on a soldering iron, and the 20 seconds it would take to solder 3 resistors together.

[YourMother'sLover]

Holy Jesus....I bought a 10 pack of 3.3Ohm resistors at my local Radio Shack for like $1.26.


This is like going between your legs to pick your nose.....it'll get the job done, but REALLY!?!?!?!?!?!?!

[vision.dynamix]

Originally Posted by cpunlamd

I bought a bag of 100 high grade 3.3ohm resistors. For anyone that needs it, just send me a return envelope and stamp with your address already filled out + $1 and I'll send you a pair.

FYI: 10ohm x 3 = 3.0 ohms (That will throw an airbag code more often then not. Our ECUs are very sensitive. Therefore the 3.3ohm will work much better.)

Umm..10/3 is *NOT* 3. The difference between the 3x10 ohm resistor vs a 3.3 ohm resistor is 0.0333... ohms. Which, depending on where you get your resistors, is within tolerance anyway.

[twizzstyle]

Originally Posted by vision.dynamix

Umm..10/3 is *NOT* 3.

he just had his math wrong.

R = 1/((1/10)+(1/10)+(1/10)) = 3.3333

[kpluiten]

For what it's worth, an airbag igniter measures 2.0±0.2 ohms. This is the industry standard. 3.3 might work, but only 2.0 ± 0.2 is required.

[jason12085]

Just finished helping a buddy do a swap. We used the entire harness from the donor car 02 wrx, but we left out the airbag control module.

Do we need to put the resistors in every airbag plug?