rb26 injector resistor.

[getoffmyinternet]

2 banks? That's interesting, each bank splits off into three? Why did you decide to do it that way?

 

I'm sure everyone is talking about only putting resistors in parallel. Doesn't really matter how you do it, the the whole pack is going to be like 1 ohm and 60 watts. It could really be one fat resistor, but then each resistor has it's own wire anyway and radio shack doesn't sell 1 ohm 60 watt resistors... I would speculate too that the injectors aren't really pulling 4 amps because they are only actually on a fraction of the time, but I'm not sure how they figured that into the equation when they prescribed these resistance values so I figured when in rome, but I digress. We're all just too cheap to pay the 50 bucks for the actual resistor pack, heh.

 

Edit: I can't remember exactly how power is figured in a circuit, but if the whole circuit is 8ohm at 12v, then it's running 18w right? That doesn't mean the resistor is dissipating all of it though, 13.5w by my calculations, but again this would only be if the injector was always on.

[Zmanco]

2 banks? That's interesting, each bank splits off into three? Why did you decide to do it that way?

When I installed MS i wired the injectors into 2 banks of 3 each (each with its own set of wires back to MS) and configured MS so that each is fired separately. I added the resistors later and it was easier to add 2 packs of resistors (1 for each bank) than to add 1 pack for both.

 

You can use 1 pack of 2 ohms for both banks, but only if you wire it between the injectors and the +12V source on the relay board (or however you're supplying +12V to the injectors).

 

If you want to fire all 6 injectors as a single batch, then you would want the resistor pack to have a resistance of ~1 ohm. If you're installing from scratch, I'd advise against this approach as you're switching twice as much current and hence causing twice as much noise (actually it's 4 times as much energy since power in this case is a function of the square of current). I'm not saying you need to rewire if you've already done it this way and all is working well, just saying it's a better approach to handle them separately since MS already has that capability.)

 

Another potential issue with firing all 6 at one time is that it causes larger pressure pulses in the fuel supply, but I really don't know at what point this becomes significant. I'm running NA with the ZX turbo injectors (265 cc/min) and can't notice any difference firing all 6 in a single batch. Larger injectors would cause larger pulses and at some point could exceed the ability of the fuel supply (piping and pump behind it) to keep up and the pressure would drop.

 

It might be overkill, but I figured wiring 2 banks of 3 each minimized this issue so I wouldn't have to worry about it.

[zilvia_gt]

I found this ohms calculator, but I'm not sure how to plug in your numbers.

http://ourworld.compuserve.com/homepages/Bill_Bowden/ohmslaw.htm

 

The key is you want to get to ~2 ohms resistance with ~25W capacity.

[WizardBlack]

Well, as the formula states, V = IR. Since volts will always be 12 on a car, then you have I = 12/R. If you have 8 ohm resistors, then you are using 1.5 amps.

Power (watts) is simply current times voltage. P = IV. You can also use P = I^2 * R. So, power is 1.5*12 or 1.5^2 * 8. That's 18 watts.

If you run the resistors in series they just add up. If you route them in parallel, then it gets a bit more complicated. You have to square each resistor's value, then invert them, then add them, then take the square root.

[getoffmyinternet]

When all resistors in parallel are the same value, then you just divide one value by the total number.

[Zmanco]

Since volts will always be 12 on a car, then you have I = 12/R. If you have 8 ohm resistors, then you are using 1.5 amps.

Not exactly. In this case the output of the alternator (closer to 14V when the engine is running) is across the resistor and injectors combined because they are wired in series. Some of that 14V is dropped across the resistors, and the balance across the injectors.

 

The total resistance is the sum of the resistor pack plus the combined resistance of the injectors. If they are wired 3 per bank, then their combined resistance is 1/3 that of a single unit. If they're wired 6 per bank, then it's 1/6.

[getoffmyinternet]

You're gonna have to explain what you mean by banks I guess, I'm not totally getting it. Are you talking about having a series of groups? If you had 2 groups of 3 6ohm resistors and the banks were connected in series, the total resistance would be 4 ohms, for example.

[ktown z]

If I purchase a stock resistor box is it a plug and play thing? I see them on sale on ebay all the time for around 50.00. I haven't wired my rb yet (not that far into the swap). So there is a plug on the stock wiring harness for the resister box?

[getoffmyinternet]

There should be a plug that the resistor pack goes right into, however it's quite possible that it wasn't included when you got the harness because there are a few sub engine harnesses aside from the main loom (mine came with neither) for things like hicas etc. However, it should be easy to make a new plug or just splice the wires directly together. All the injector resistor wires are on the same plug in the main engine harness (mixed with a few others), but that plug isn't the same shape as the resistor pack one. Look for a pair of rectangle plugs with 8 leads on one and 7 on the other. All six injector wires are on one side, I belive they are each yellow with a different stripe.

[z-ya]

You brought out my curiousity about the L28 FI resistor measurements. The L28 fuel injected resistor pack has two ceramic resistors. There is one white wire and two black wires. A resistance check on them measures out at 6 ohms.

 

Now here is the perfect junk yard fix as I see it. Take three of these packs, connect the white leads together and attach the remaining six black leads to the injector harness.

 

Actually the L28 resistor pack is made from a x4 and a x2 resistor pack bolted together (at least the ones I've used).

 

Pete

[WizardBlack]

Actually the L28 resistor pack is made from a x4 and a x2 resistor pack bolted together (at least the ones I've used).

 

Pete

 

Yup.

 

...

 

By banks, they are referring to groups of injectors that are fired together. Most people use the two channels that the MS has and fire in banks of three cylinders. A six cylinder car with sequential injection would have 6 "banks". Plain old batch fired would have one bank. Some people use the term "banks" to group things by cylinder head; left bank versus right bank, etc. Lots of ways to use the term.

 

If you add resistors in series, you add their ohm ratings.

[zilvia_gt]

Ok, so after much research I think I found the answer to this problem. I found some resistors that Jim Wolf sells with their ecu/injector upgrades. They're 6.8 ohm / 10 watts sold by VividRacing.

http://www.vividracing.com/catalog/jwt-dropping-resistor-nissan-240sx-p-12658.html

Upon reading numerous threads, this problem is actually a pretty universal thing with the import community. A lot of the DSM crowds, supra crowd, and even big turbo Honda crowd need to do this mod when updrading to bigger injectors. The consensus seems to be that the resistors need to be in the 6 ohm range and 10 watt range. So a 6 ohm / 6 watt resistor should work, but a 10 ohm / 10 watt resistor might be too much. I'm guessing the 6.8 ohm / 10 watt that Jim Wolf uses is the best bet.

 

Here a great in-depth article on resistors for those who are interested:

http://www.electronics-tutorials.ws/resistor/res_1.html

[Zmanco]

Did you see my post #39? You can buy the same resistors from mouser.com for $1.09 instead of $2.99.

[zilvia_gt]

Did you see my post #39? You can buy the same resistors from mouser.com for $1.09 instead of $2.99.

 

Yes I did, but you posted the the 6.2 ohms / 10 watts, part#280-CR10-6.2-RC. These are the 6.8 ohms / 10 watt, part#280-CR10-6.8-RC. The current price for both the 6.2 ohms and the 6.8 ohms / 10 watts from Mousers for a set of 6 is $3.30 (without shipping).

 

What still gets me (even with the multiple posts) is how to determine the resistor that will provide 2 ohms at the resistor. Since I will be wiring in parallel, one 12V power to the resistors, then from the resistors to the power side of the injectors, what still gets me is how to determine the resistor that will get 2 ohms (exact or as close to as possible) at the injector. Even with the formula and multiple posts, it still confuses me a bit. You mentioned that the ideal is to get around 2 ohms at the resistor and around 25 watts capacity. Ok, for the watt capacity we just add the individual capacity of the each resistors, so the ideal capacity would be a 5 or 6 watt resistor, but we opt to go 10 watts for safety and heating issues, that I get. So instead of the 30-36 watt total capacity of all six resistors, we now have a 60 watt total capacity. Wiring in parallel without banks like you did, would the 6.2 ohm or 6.8 ohm be more ideal?

[Zmanco]

In practical terms, it makes no difference if you use 6.2 or 6.8 ohm resistors. You will never be able to measure any meaningful difference in performance of the engine. You might be able to measure some very slight differences in opening times for the injector, but that time is programmable in Megasquirt so you can just adjust it if you have the tools to measure it that accurately.

 

But since you're curious about the theory, let's discuss it.

 

- Let's assume you're going to wire the 6 injectors into 2 banks of 3 each. This is the typical way to use megasquirt on an L6.

 

- Each bank will be wired independent of the other. That means that we want one resistor pack in series with 3 injectors that are wired in parallel with each other. See my very crude illustration attached.

 

For 3 typical low impedance injectors wired in parallel, the target value of the Resistor Pack is 2 ohms. Ideally we'd like to just buy a 2 ohm resistor with a rating of 25W. In fact these aren't readily available so we'll make a pack ourselves by using several resistors wired in parallel.

 

Note: from a purely electrical point of view, you could also wire a single 6 ohm resistor in series with each injector to achieve the same thing. But since Megasquirt is structured to support only 2 injector circuits (banks), we'll continue to model 2 Resistor Packs.

 

The math is R(total) = 1/(1/R1 + 1/R2 + ...).

 

10W is a typical power rating for inexpensive wire wound resistors, so if we use 3 in parallel, we have 30W of dissipation capacity.

 

Let's make the math easier by assuming that all the resistors have the same resistance. R(total) = 1/(1/R1 + 1/R1 + 1/R1).

 

Simplifying yields R(total) = (1/3)R1

 

So using 6.2 ohm resistors produces a pack with resistance = 2.1 ohms

6.8 ohm resistors produce a pack with 2.3 ohms.

 

As I said up front, the difference between 2.1 and 2.3 ohms in this application is insignificant to the operation of the injectors.

 

Does that help?

 

Note: I watched a PBS Nova special on Fractals last night so I'm feeling rather academic today Hope this was at least somewhat useful.

[chrism]

zmanco,

You beat me to the punch on this one. I would add that using 1% resistors would be better.

[getoffmyinternet]

For what it's worth, I finally found the page where I saw the injector check a long time ago. I don't know where 6 ohms came from...

 

Side note, the injectors obviously aren't that picky, since the tolerance is 1ohm all around. The dropping resistor for each however, definitely 5 ohms. I suppose you could play it safe and put in a 100w pack of them so long as they are 5 ohms each, or .8 ohms all together. If you go cheap and get bulk silver band resistors one could be off by a whole .5 ohms, however. Other than that I doubt pack arrangement does much of anything.

 

 

[Zmanco]

Let's go back to basics for a minute: Why are the resistors needed?

 

Answer: the injectors are basically a big solenoid. One of the characteristics of a solenoid is that when power is first applied, it has a very low resistance and then as the current ramps up, the effective resistance rises which in turn reduces the current. In other words, there's an initial surge that levels off.

 

The resistors are required for low impedance injectors to limit the initial surge to protect the transistors that do the switching in the FI controller (such as megasquirt). There is no magic exact value that is best. There is a tradeoff between limiting the current and slowing down the time it takes for the injector to open. This really should be measured for a given injector because beyond a given level, allowing a larger initial current (smaller resistor) won't cause the injector to open any faster but will put a larger load on the switching transistors.

 

Read this thread if you're interested in how this can be measured:

http://forums.hybridz.org/showthread.php?t=121843

 

The point of all of this discussion is that as long as you're in the ballpark with resistor values there's very little to be gained worrying about tweaking them further.

[zilvia_gt]

Thanks everyone for putting up with my curiosity.

 

Zmanco, by breaking it up like that it makes more sense. However, in your case you're using MS that fires 2 banks of 3 injectors at a time. So the question now is on a stock rb26dett using the stock ecu, is it sequential fire, batch fire, or 2 banks like your MS? Following your math, I calculated that sequential fire (6 banks or 1 injector at once) will need 12 ohms resistors to get 2 ohms at the injector and a batch fire (1 bank or all 6 injectors at once) will need a 2 ohm resistor to get 2 ohms at the injector. I'm I confusing myself again?

 

GetoffMyInternet, good find!

 

On a side-note: If I did find a 2 ohm 25 watt resistor such as below:

I could just use this one resistor, connect one end to a 12V and connect all the + end of the injectors to the other?

[Zmanco]

I'm afraid I have no experience with RB motors so can't help you with how the injectors are set up. But for the two examples you list (and assuming the injectors are typical low impedance), for all 6 fired simultaneously using a single resistor, it should be 1 ohm. Note, this resistor would need to be wired between the +12V source and the injectors and not between the injectors and the FI controller. I say this because it's unlikely that all 6 will be switched by a single circuit.

 

For the opposite with each injector fired independently, each should have a 6 ohm resistor.

 

FWIW, I doubt all 6 fire as a single batch. Unless the injectors are rather small, there's enough time in a cycle, even at 7k rpm, to fire 2 banks of 3 each, and electrically (from a load and noise point of view), that is more desirable as well.

 

BTW, nice find of the 2 ohm 25w resistors. How much do those cost? They look easier to mount.