AchillesGr

iTrader: (1)

i am instaling a pfc and i wonder why to use resistors with 5ohm injectors.
i see that the pfc is produced for a wide range of cars and most of them has low imbedanse resistors. what all these guys do? put extra resistors?

i am thinking that the pfc is capable of driving those low ohm injectors.
why the apexi would produse a product that will need extra resistors most of the time?
do anyone have any info aboyt that?
thanks

arghx

iTrader: (3)

the PFC manual gives an explanation of how the injector driver's loads can be calculated. I posted about it in some thread a while back, I'll just copy what I said here:

"
Some of the 1680's actually read more than the usual 2-3 ohms across the terminals, although they are not in the 12-15 ohm range of the stockers. I've seen varying numbers. But there are a couple variations/manufacturers of the 1680s and you would have to check resistance across your 1680s first then calculate how much wattage you would be up to. By these types of calculations, I would say you would want at least 6 or 7 ohm 1680's to run no resistors. I am unsure which particular 1680s have the higher resistance values.

I'll explain and go through a calculation of injector wattage with the stock injectors, and then stock primary's and 1680 secondarys rated at 7 ohms resistance. The original source of the information for these calculations is from page 80 of the Datalogit FC-Edit manual.

The max wattage allowed is 3 watts. First let's go through the easier calculation using the stock injectors first, which we are going to assume all have the same resistance. We have two formulas:

V=I*R (Ohm's law. in this case voltage equals the injector's current multiplied by resistance of the injector itself)
P=I^2*R (Joule's law. In this case, the injector's power/wattage equals the injector's current multiplied by itself, then multiplied by the resistance of the injector driver transistor)


V=I*R .

let's set V to 14V , probably the max you'd get to your PFC on crusty 15 year old wires.

Let's say that all the stock injectors would have 12 ohms of resistance each.

V=I*R ---> 14=I*12 . I=1.17A. so each stock injector needs about 1.17 amps.

Now we take the 1.17 amps and plug it into our power formula.

At 100% injector duty, the wattage is P=I^2*R . We plug in the 1.17amps we just calculated as being needed for each injector. Then for the resistance, we use .25 ohms across the transistor. That .25 ohms is a given value that comes from the Datalogit manual.

so P=I^2*R ---> P=(1.17)^2*.25 = .34 watts per injector.

So each of the stock injectors is going to use .34 watts . We multiply by 4 injectors, and we get:

.34 watts * 4 = 1.36 watts @ 100% duty cycle . Most people run up to about 85% duty cycle.

So 1.36 @ 100% duty cycle, multiplied by .85 , gets us the 85% duty cycle wattage of 1.156 watts. The max allowable is 3 watts according to the Datalogit document ^ .


Ok now let's do a calculation at 100% duty cycle for stock primary's and 1680 secondary's with 7 ohms of resistance. We just calculated that each stock injector is .34 watts . So multiply by 2 and we have .68 watts on the primary's.

Now we re-do the calculations for the secondary injectors:

V=I*R ---> 14=I*7 , I=2 . we are using 14 battery volts and 7 ohms across the 1680s . With a little algebra that gets us to 2 amps of current for each 1680.

P=I^2*R ---> P=(2)^2*.25 = 1 watt per 1680 injector @ 100% duty. We plugged in the 2 amps from the previous calculation, then multiplied by the .25 ohms that was given to us from the Datalogit document.

So stock primary's and 1680 secondarys (measuring 7 ohms resistance) at 100% duty gives us:

Front primary: .34 watts
Rear primary: .34 watts
Front secondary: 1 watt
Rear secondary: 1 watt

Total wattage at 100% duty: 2.68 watts . Total wattage at 85% duty: 2.68 * .85 = 2.278 watts . Max allowable is 3 watts.

So running a 550/1680 setup (secondary's measuring 7 ohms each) @ 85% duty would still put you significantly below the max power output of the injector driver circuit. Given the mathematical relationships though as resistance drops across those secondarys the current will shoot up quickly, so I wouldn't run less than 6 ohms. Someone else can post up more details about the different brands of 1680's and what resistance they measure."

AchillesGr

iTrader: (1)

hi,
that was a very interesting post. i read it before. my question us why you use R=0.25 ?
why not to use as R the injector resistanse? (=7ohms)
then the P=2^2*7=28w ?
the 0.25 where does come from?what i am doing wrong?
thanks

arghx

iTrader: (3)

Now I've got to go back and read through my own stuff to remember what I did, because I remember asking myself that very question when I first tried to do the calculations.



They really don't explain what they're doing all that well as it was clearly written for someone who does circuit design on a regular basis. But as I pointed out, it is really a two-step calculation using two different resistance values:

"V=I*R (Ohm's law. in this case voltage equals the injector's current multiplied by resistance of the injector itself)
P=I^2*R (Joule's law. In this case, the injector's power/wattage equals the injector's current multiplied by itself, then multiplied by the resistance of the injector driver transistor)"

You have to rearrange the equations a bit. The goal is to figure out how much power the injector drivers are using, and to do that we need two resistance values: the resistance across an injectors and the resistance across an injector drivers. We need a two-step calculation. First you figure out how much current an injector uses, and that's where we used the measured resistance of the injectors themselves. We plug in an assumed battery voltage and the measured resistance across the injector, not across the injector drivers and solve for the current. The derived current per injector is used in the next step.

To find out how much power is being drawn per injector, you take the current per injector and you use the resistance across the injector driver itself. The FC-Edit manual lists the resistance across 1 (out of 4) injector driver as .25 ohm. We'll just have to trust that that is correct. So using the formula, you square the current you found in step 1, then multiply it by .25 ohms (the given resistance of the injector driver).

That process gives you the power draw for one injector. Since the primary and secondary injectors have different resistance values, you have to do that power calculation 4 times, one for each injector. Then you add the power consumption of each injector together and see if it is safely below 3 watts at a given duty cycle, which is what the FC-Edit manual says is the max acceptable wattage.

Now re-read the info from the FC Edit manual as well as my first explanation of the calculations and see if that helps clarify stuff. If anyone else can see any errors in my math please let me know.

AchillesGr

iTrader: (1)

i think you are wright about this . the watage in the pfc driver is the P1=2^2*0.25=1w
and the watage in the ingjector P2=2^2*7=28w
the combined circuit consumes P1+P2 but we are only interesting for the driver's watage P1=1w

the only question now is if this 1w is too much. if the pfc 3w is divided with 4 as the number of the drivers we take 0.75w per driver witch is < 1w . that means a problem? of course the injector does not draw 2amps all the time but that saves the case?

if we take the sum of all injectors we are still beneath the limit of the 3 watts but are we correct that way?are the circuits all independant? if they are i think we mistaken if we take as sum limit the 3w, we will need to take the 0.75w as a limit per injector.

arghx

iTrader: (3)

All I know is based on those instructions I posted there. Nobody else really knows how to calculate it.

If you read through the way they did it in the FC-Edit manual, as long as the combined wattage of all 4 injector drivers is less than 3 watts it is "safe," although I wouldn't run it at that limit. Since each factory injector is using between maybe .25 and .34 watts at the driver (depending on voltage, measured resistance, etc), multiply that by 4 and there is still a large safety factor in the PFC drivers. If we trust that it can handle the heat of 3 watts and it is only loaded with 1 watt on the factory injectors, I don't see any problems.

Remember that we have an exponent in Joule's law for power (P=I^2*R) . On a per-injector basis, the "R" is a fixed coefficient of .25 ohms so it doesn't matter that much. Because we have an I squared term, as the current across the injector itself increases (lower impedence/resistance), we rapidly increase power consumption of the injector driver.

So look at these numbers (max allowable power is 3 watts total):

9 ohm injectors (4): 14 volts / 9 ohm = 1.55 amps/injector
Power calculation: 1.55 amps * 1.55 amps * .25 ohms = .60 watts/injector driver
Total power @ 100% duty: .60 watts * 4 drivers = 2.4 watts

8 ohm injectors (4): 14 volts / 8 ohm = 1.75 amps/injector
Power calculation: 1.75 amps * 1.75 amps * .25ohms = .765 watts/driver
Total power @ 100% duty: .765 * 4 drivers = 3.06 watts

7 ohm injectors (4) 14 volts / 7 ohm = 2 amps/injector
Power calculation: 2 amps * 2 amps *.25 ohms = 1 watt/driver
Total power @ 100% duty: 1.0 * 4 drivers = 4.00 watts

Those are some simplified calculations to illustrate how the mathematical relationship in Joule's law affects things. In reality, we don't run sustained duty cycles of 100%, and once you upgrade injectors you probably have two injectors with one resistance value and two injectors of another.

cewrx7r1

iTrader: (2)

Look at page 46 of the latest notes version.