MikeInCape

Thanks!

I actually went to radioshack before posting looking for 10ohms... Might have been a posting of yours that I found that on if you have that somewhere on the internet. I couldn't find anyone else who has done this. Seems like it would be more common to me?

They didn't have any 10ohm so I blindly went with the 8ohms. Well I plugged those in to try it, and it threw a code immediately pretty much, versus the normal couple of runs to throw it (with nothing hooked up).

Ok... So my I did not have enough resistance. Understood.

And I did expect the heat. This is simulating a heater. I will mount on solid surface.

Can I put a 8ohm+2ohm together to get a 10ohm resistor? Or just look somewhere else to find the 10s?

Also... Am I correct in understanding that I can wire these to any 12V ON source, and I do not have to get the specific wire coming from the ECU for 12+ to that o2 sensor? (in looking the cars schematic the whole sensor system's 12V+ looks tied together on the 4.6s.)

cliffyk

The PCM monitors the heater current through its grounding of the heater circuits, you can supply the 12V+ from anywhere but the PCM must be aware of the current being drawn (by its supplying the ground path) to prevent the heater related DTCs.

Radio Shack doesn't stock anything that can do it (from a resistance and power capacity perspective), in any reasonable combination.

The problem is that the PCM expects a 3.3 Ohm (60W) heater, and frankly the easiest (and probaly cheapest in the long run) way to make this happen is to screw a functional O2 sensor into the bung and wire it up--and then use a MIL eliminator (or an O2 simulator if one insists) to trick the PCM.

60W is quite a bit of power, if you doubt this find a 60W incandescent bulb in your house and grab it with your hand.

In the long run it's a whole lot easier, and not that at much more money once all is said and done, to just buy a tuner and shut off the rear sensors (and heaters)...

MikeInCape

Most definitely... Just a fun learning project for me... Temporary...

I am planning on getting Sniper Tuning Software soon, so it's not a huge deal. I figured my Hypertech MaxPower would do some things such as that and speedometer correction. But nope, on a 98 it just loads a "PowerProgram". 99+ They say it allows changes.

I am just curious by nature, and this deep in, I figure I will finish it.

I enjoy tinkering, that is why cars are so much fun.

I will look around to find the correct resistors. I've been thinking, and I might mount those under the hood.

Thanks for the expert knowledge! Seriously.

lol... "or use an O2 simulator if one insist"


I was with you with 10Ohm x ~14V =~60W. (see how exact I am) I do not know the electrical math, or do not have direct knowledge. Just a tinkerer in a garage.

cliffyk

eBay is a good source for power resistors, in fact here's a Buy-it-Now for a 3.3Ω 60W package--$5.99 the seller has 9 left.

I like it when a plan comes together!

MikeInCape

Sweet... Thanks Cliffyk!

MikeInCape

Hey Cliff... Since I'm brushing up on my electronics here.

So 6x 10Watt 20Ohm in parrallel.

So our wattage is multiplied, because it is parrallel. And our Ohmage is divided.


Correct?

Out of curiosity what happens in series?

So... I know you've given me the solution for real cheap up there... But once again tinkerer...

I was thinking... And if my math is wrong let me know...

I can get 10watt 10 ohms from Radioshack

If I put 3 parralell then I would have 30 watts x 3.33

Now If I put 3 more just like that in a series with the other 3... Would that give me 60watts x 3.33ohms?

And still only 6 components? Does this sound right? Or am I off my rocker?


Sad thing is if I build the ciruit instead I will spend $4 more total, plus time.

cliffyk

For resistors in parallel the formula is:

Rrslt = 1 / (1/R1 + 1/R2 + 1/R3 + 1/R...)

I.e. the resulting value is the reciprocal of the sum of the reciprocals of each value...

So for six 20Ω resistors in parallel it's:

Rrslt = 1 / (1/20 + 1/20 + 1/20 + 1/20 + 1/20 + 1/20)

Rrslt = 1 / 0.3 = 3.3Ω

But since all the resistors are the same it can be simplified to the value of one resistor divided by the number of resistors.

In series the resulting resistance is just the sum of all the individual values.

Wattage is somewhat counter-intuitive, because in both series and parallel resistor packs the total power capacity is the sum of the individual resistors' capacity.

So--You are correct that three 10Ω 10W resistors in parallel will result in 3.3Ω with a 30W capacity--however placing 2 such packs in series would result in a 6.6Ω 60W capacity pack. You could get back to 3.3Ω by placing two of the parallel/series packs in parallel, and get 120W capacity to boot.

This would of course require twelve 10Ω 10W resistors, or twelve 10Ω 5W resistors to get 3.3Ω with a 60W capacity ($24 of resistors from the Shack)

To simplify things you could just make six pairs of 10Ω resistors in series, and then connect all six pairs in parallel.



One last comment, the power resistors sold by Radio Shack are ceramic, sand filled wirewound units. These are not especially vibration resistant and will often crack when subjected to prolonged periods of even microscopic levels of vibration.

MikeInCape

Thanks... Exactly what I was looking for. I will buy the ones on ebay, I appreciate the knowledge for the future.

dsbsurfn

Check out National Semiconductor. Not sure if they have resistors, but if you go to Order, then click on samples, you may be able to find some for cheap. I used them for all of my EE projects in college, mostly specialty ICs and Op-Amps.

ct_white_stang

Does anyone else here understand what is going on? LOL