Quick injector question

[Fusion Ed]

I've been looking into this in a bit more detail, and I think I can get away with paralleling the standard ones and still stay in spec ( can someone check my figures ? )

Can do:

Dropping resistor = 5 to 7 ohms
Injector = 1 to 3 Ohms
Assume V+ = 14 volts
Min possible resistance = 5 + 1 = 6 Ohms

Ok so far but the resistors won't have a tolerance of 300%!!!

Work out power required for each injector -

Lowest injector power required is when resistor = 7 ohm, injector = 1 ohm. This gives total resistance of 8 ohms and total current of V/R = 14/8 = 1.75 Amps.

No, lowest current condition would be 7+3 (10 Ω) not 7+1 (8Ω)

Minimum power required to fire injector is therefore = I2R = 1.75 * 1.75 * 1 = 3 Watts.

Min power total = P=V²/R ∴ P = 14²/10 = 19.6 W per injector. or 1.4 Amp.

Max power per channel is when resistor = 5 and injector =1 = 14*14/6 = 32.6 Watts

Min power total = P=V²/R ∴ P = 14²/6 = 32.6 W per injector. or 2.3 Amp, as you have.

Two injectors per channel at 3 Watts = 6 Watts.

Incorrect, Take worse case sinaro from above. Inj = 1Ω I=2.3 ∴ V=2.3 (V=IR) ∴ P=5.29 W (P=IV) per injector.

Dropping resistor has to dissipate 32.6 - 3 = 29.6 Watts.

No, its 32.6-5.29 = 27.31 Watts - for a single injectors using your worse case values.

If injector total resistance = 1 ohm and power dissipation is 6 watts, then the resistors must have 2.45 Amps running through them, 1.23 Amps each.

This falls over now as your values are not correct.

To effectivly open the injectors you need to keep the current flowing through them the same as it should be so, to take your worse case values. (its not the same in pratice)

You have worse case 0.5 Ω You need a current of 4.6amps (2.3x2) at a supply voltage of 14.

∴

V across injectors = 4.6x0.5=2.3 (we know that already from above) So:

14-2.3=11.7V Resistance = 11.7/4.6 = 2.5Ω (approx) which you should notice is half the current min of the worse case impedance you wrote above.

With 2.45 Amps flowing, and the dropping resistor must dissipate 29.6 Watts, then it must equal 4.93 Ohms.

So if I can keep the paralell injectors to more than 1 Ohm and the resistor to around 5 Ohms then I'm still within the spec of the ECU.

Technically no. You cannot add injectors, to a single driver output and expect the ecu to (on the face of it) cope. You need to maintain the current to open the injectors at a fast enough rate, which means you have to ask from more than the ECU was designed for origionally.

Better still would be to find some standard injectors that are just over 2 ohm to give a bit of margin.

Best of all is to get some high impedance injectors, of 10Ω + and then parallel two of them Or just use large injectors and ignore all this!

Ed

 

[AJ4]

Can do:

Quote:
Dropping resistor = 5 to 7 ohms
Injector = 1 to 3 Ohms
Assume V+ = 14 volts
Min possible resistance = 5 + 1 = 6 Ohms

Ok so far but the resistors won't have a tolerance of 300%!!!
300% ? A 6 Ohm nominal resistor with only 20% tolerance can be +/ 1 Ohm. According to the spec, anything within 5 to 7 ohms is ok ( as far as the Ecu spec is concerned )

Quote:
Work out power required for each injector -

Lowest injector power required is when resistor = 7 ohm, injector = 1 ohm. This gives total resistance of 8 ohms and total current of V/R = 14/8 = 1.75 Amps.

No, lowest current condition would be 7+3 (10 Ω) not 7+1 (8Ω)
I quoted injector power, not current -
10 Ohms = 14/10 = 1.4 amps across a 3 Ohm resistance is 5.88 Watts.
8 Ohms = 14/8 = 1.75 amps across a 1 Ohm resistance is 3.06 Watts

Quote:
Minimum power required to fire injector is therefore = I2R = 1.4 * 1.4 * 1 = 1.96 Watts.

Min power total = P=V²/R ∴ P = 14²/10 = 19.6 W per injector. or 1.4 Amp.
Agree with 1.4 amp as the lowest current, only I quoted power per injector, not total ( which includes power dissipation in the resistor and we're only interested in how much power to fire the injector ).

Quote:
Max power per channel is when resistor = 5 and injector =1 = 14*14/6 = 32.6 Watts

The rest of the figures wont add up either because we differ on the minimum injector power ( I say when its 7/1 and you say 5/1 )

Cheers for having a look through, much appreciated 8)

 

[sypher]

can I ask what is the preffered applied voltage to those Bosch injectors, as i thought about keeping one dropping resistor @ 6 Ohm and two injectors @ 2 ohm in parrallel, giving 2 volt applied to injectors thus the the ECU would need to sink 2 A instead of 1.75 A...

given 14V altenator output.
stock;.....................................3.5 V applied to 2 Ohm giving 1.75 A
One dropping R' & 2 injector; 2.0 V applied to 1 Ohm giving 2.0 A
Two dropping R' & 2 injector;3.5 V appiled to 1 Ohm giving 3.5 A ( or a 3 Ohm custom resistor pack).

From this you can then work out wattage.

 

[Fusion Ed]

hmm, Your missing the point. Wattage has nothing to do with it. Unless your concerned by power dissipation, but other problems will occour before this.

The ECU has pull to ground bipolar transistors as its output devices. (old technology) Now I dont have a GTIR ecu so I cannot tell you the specifications, so to reliably establish WHAT can actually be driven by the ECU requires an understanding at component level how this section of the ECU works. A typical example of what some Nissan ECCS ECUs use is a device called a PU4424 darlington driver. Each output has a peak collector current of 8A and a rating of 4A. The device dissipation is 15w max.
Now if the GTIR ecu has the same (which it may do)a pratical safe limit is its rated 4A per channel.
This is the only thing that concerns you other than not burning out the injector coils (which is very unlikely since they are cooled by fuel), so choose a resistor value that allows a decent amount of current to them they will open quicker and beable to work with stronger fuel rail pressures should you ever want to.

If you really wanted to not have to worry about all this you should make a FET driver and bypass the internal old one then you could pratically have as many injectors if you wanted.

 

[AJ4]

hmm, Your missing the point. Wattage has nothing to do with it

lol, all these years in electronics and I thought it was power and heat that destroyed devices, not current :lol:

 

[Fusion Ed]

Sorry, but if your such an expert then why ask here? Like yourself I have my own fair share of electronics experience. I was an RF engineer with SMD high power MOSFET devices and PIC control and PLL design for 5 years. With RF, dissipation and efficiency is everything. I now do part time sub-contract system control work for people like Wizards of Nos etc..

You are right that power dissipation is that which kills components, but if you keep under their max current ratings here as I said then that will not be an issue. Have you yet opened the PFC to see what output drivers it uses? The E-manage for example uses discrete bi-polar drivers and will happily fire low impedance injectors directly (even if greddy say otherwise).

Still knowing full well that 700+cc will work perfectly on the pulsar, why do you not just fit a set of those and rescale the injector size? You say they don't atomise as well, do you have proof of this? i.e. HC emissions at idle? It certaintly will not be anywhere near as bad if at all as you claim, many stock injectors will have less than ideal spray pattens anyway considering their age, rather than a new set of 700s or what ever.

 

[AJ4]

well it did kinda start as a nice tech discussion because a few other people were interested in doing the same thing and we were bouncing a few ideas around ( before it turned into a childish pissing contest about 'who knows best' :lol: )

Personally, I dont want to use big injectors if I dont have to, and who said I needed only 700cc anyway ? :lol: Even if it was physically possible to have a 700cc injector that atomised as well as 400cc I couldn't care less For my application, I want to use smaller more efficient injectors.

Or maybe as a 'tuner' your softening me up to try and sell me a set of 700cc's :lol::lol: