Things we learned on Dyno & 03 555 HO

[100 Proof]

Originally posted by DLeno

oh, nick: a zener diode and a series resistor is all you need ...

OK Doug, I'll bite since no one else chimed in. Where do I get the schematic for this gizmo and what are the values of the resistor and zener diode?



100 Proof

 

[DLeno]

well, first we have to find out what the maximum voltage is that corresponds to the ECM staying happy. then find a zener with that knee voltage, find out what reverse current is required to obtain that zener knee, and pick a resistor to provide that. Since I don't know what the right maxium voltage is, I'll pick one for fun and illustrate the design. the schemo would be:





860 ohms

MAP ----/\/\/\/\/-----------> to ECM MAP input and zener cathode







zener cathode

|

|

/\ 4. 7v zener diode, 5mA

|

|

gnd





sorry for the limits of acsii drawings on the web forum. the zener cathode is connected to the right (output) side of the resistor.



what happens in this circuit is that the ECM sees the output of the 860 ohm resistor, which tracks actual MAP voltage output until the zener value is reached. when the zener starts to conduct, it will clamp the voltage seen by the ECM to its zener value, in this case 4. 7 volts.



example:



SUPPOSE that 4. 7 volts was the right limiting voltage, and that you found a zener diode with a zener knee of 4. 7 volts at 5 mA and about 80 ohms series resistance (for sake of discussion). You have to put 5 mA through the zener diode when it has 4. 7 volts across it. so:



R = V/I = 4. 7V/5mA = 940 ohms. but if the zener itself presents 80 ohms, then you will need 940-80 = 860 ohms.



power considerations: 4. 7 volts across 80 ohms, means . 27 watts. so you better get at least a 1/2 watt zener. preferably a 1watt.



5 mA through 860 ohms is . 02 watts. so find a standard value 1/8 watt resistor close to 860 ohms.



It would be better to find a small signal zener with a smaller current. I just picked 5 mA for example's sake.





then experiment around to coax your boost fooler into higher maximum values without making the ECM unhappy. that will allow the most fueling.



note: the values here are only examples. I honestly DONT KNOW what the right max voltage is. so don't anyone go fabing up the above circuit without first finding out more information. also, zener knee voltages depend on current, so some experimentation will be in order to tweek things up right.

 

[100 Proof]

Originally posted by DLeno

... . so some experimentation will be in order to tweek things up right.

Volunteers? This would be a great contribution to "the cause" if anyone out there has the skill and inclination. Thanks Doug.



100 Proof

 

[bulabula]

Doug, I'm not following your rationale on needing to clamp a signal that is presumably clamped high already.



Looking at the connector pinout for the Manifold Absolute Pressure Sensor, there are 3 wires: sensor ground, MAP sensor signal, and 5V supply. 5V is probably max for the ECM.



I would think installing a resistor to simply reduce the raw voltage signal to the ECM is all that’s needed. It could be either in series with the supply voltage or the sensor output voltge. In this way, the ECM would never see a max MAP signal and not defuel.



For a starting point, and using Ohm’s law, pick a resistor value to reduce it lets say 25% to be safe. A variable resistor would work too.



With a 5V supply, its I'd say that 5V is enough to keep the ECM happy and avoid a smoke test; and I wouldn’t think the ECM would mind a 25%’ish cut on the low end, but maybe not.



It may be worth a try to find the appropriate connectors to splice in that resistor without cutting actual wires.

 

[RRoton]

Why not use two resistors to form a voltage divider.

Say for example a 100 ohm, and a 300 ohm.

That way you will ALWAYS be sending 3/4ths of the total voltage to the ECM.

I'm no EE but that sounds to me like it should work.

 

[DLeno]

guys, you're missing some things:



1. MAP sensor signal is an analog voltage that corresponds to measured Manifold Absolute Pressure. this is a 0-5v signal I believe. This signal is not clamped high, it is as continuously variable voltage. You want to pass this signal un-adulterated to the ECM, except that you want to impose a maximum ceiling.



2. series resistance won't work for two reasons: (1) The Cummins design is for a low output impedance device (the MAP sensor) to drive a high impedance device (the ECM). by the time you find the input impedance of the ECM (very small) in order to pick the right series resistance value, you'll end up with a very large resistor. Thus you will blow the original design to pieces, creating electronic instability and susceptability to noise. (2) If you put in a series resistance, all you'e done is voltage divided the MAP output (your additional resistor plus the input impedance of the ECM which appears as the other resistor in the voltage divider). If you do this, you will guarentee that the ECM never sees the correct boost level. thats bad.



3. a voltage divider is a bad idea for two reasons: (1) it guarentees that the ECM will always see boost levels below what they really are. This will have the effect of ham-stringing the ECM so that the engine will be in virtual limp mode (ECM won't fuel at high enough levels because you're telling it that boost is always low). You want to give the ECM unprocessed signal (below the maximum) so that it fuels correctly. (2) a voltage divider is a really poor voltage limiter because it imposes the same scale factor across the entire signal range. For example, if you can acheive 35 lbs of boost and you want the ECM to read only 25, then you must put in a . 7 to 1 voltage divider. If you do that, when boost is really 22 lbs (perfectly normal) you will tell the ECM that boost is really 15. 4 lbs. and it will fuel accordingly.



4. putting a resistor in series with the supply voltage will cause everything dependant on that 5V supply to not work properly.



5. if your're commited to two components already (two resistors), make one of them a zener diode and all will be happy. the ECM will read actual boost levels (and fuel properly) until boost gets too high, and then the ECM will see only 25 lbs (or whatever the max is). don't be afraid of the lowly zener. it is your freind

 

[RRoton]

I see your point. My thinking was, if you reduced the boost voltage the ECM saw throughout the range, you would increase fueling throughout the range. You are saying the opposite would happen. If that is the case, a clamping zener would be the way to go. I suppose that would be safe as long as the waste gate never failed.

 

[100 Proof]

Originally posted by DLeno

guys, you're missing some things:



1. MAP sensor signal is an analog voltage that corresponds to measured Manifold Absolute Pressure. this is a 0-5v signal I believe. This signal is not clamped high, it is as continuously variable voltage. You want to pass this signal un-adulterated to the ECM, except that you want to impose a maximum ceiling.



2. series resistance won't work for two reasons: (1) The Cummins design is for a low output impedance device (the MAP sensor) to drive a high impedance device (the ECM). by the time you find the input impedance of the ECM (very small) in order to pick the right series resistance value, you'll end up with a very large resistor. Thus you will blow the original design to pieces, creating electronic instability and susceptability to noise. (2) If you put in a series resistance, all you'e done is voltage divided the MAP output (your additional resistor plus the input impedance of the ECM which appears as the other resistor in the voltage divider). If you do this, you will guarentee that the ECM never sees the correct boost level. thats bad.



3. a voltage divider is a bad idea for two reasons: (1) it guarentees that the ECM will always see boost levels below what they really are. This will have the effect of ham-stringing the ECM so that the engine will be in virtual limp mode (ECM won't fuel at high enough levels because you're telling it that boost is always low). You want to give the ECM unprocessed signal (below the maximum) so that it fuels correctly. (2) a voltage divider is a really poor voltage limiter because it imposes the same scale factor across the entire signal range. For example, if you can acheive 35 lbs of boost and you want the ECM to read only 25, then you must put in a . 7 to 1 voltage divider. If you do that, when boost is really 22 lbs (perfectly normal) you will tell the ECM that boost is really 15. 4 lbs. and it will fuel accordingly.



4. putting a resistor in series with the supply voltage will cause everything dependant on that 5V supply to not work properly.



5. if your're commited to two components already (two resistors), make one of them a zener diode and all will be happy. the ECM will read actual boost levels (and fuel properly) until boost gets too high, and then the ECM will see only 25 lbs (or whatever the max is). don't be afraid of the lowly zener. it is your freind

And now you know why I'm not volunteering to take this on. I'm a mechanical kind of guy.



100 Proof

 

[DHerzog]

This is my first diesel and I was seriously thinking about a HP muffler to lower EGT... . glad I read this thread, but now I am confused.



As a newby, what is DEFUELING... . what gets hurt if it does exist? Detrimental to the fuel system, the engine, or???



Thanks for all the great info!!! Oo.

 

[Ted Jannetty]

This is my first diesel and I was seriously thinking about a HP muffler to lower EGT... . glad I read this thread, but now I am confused.



As a newby, what is DEFUELING... . what gets hurt if it does exist? Detrimental to the fuel system, the engine, or???



Thanks for all the great info!!! Oo.

When the computer sees more boost than is programmed in the software it pulls fuel away to control turbo speed or boost.



Nothing is hurt by it, it is a safety built in to the software incase something fails like a wastegate to prevent overboost.

 

[DHerzog]

Thanks for the quick reply, Ted...



I don't know a lot yet about diesels, and I was concerned that it might cause a run "LEAN" condition, which, on a gasoline engine, can burn the valves and cause other havoc... .



So, I guess I can safely add the flow through muffler without any mechanical consequences??? Thanks again

 

[rbattelle]

I don't know a lot yet about diesels, and I was concerned that it might cause a run "LEAN" condition, which, on a gasoline engine, can burn the valves and cause other havoc... .

Lean conditions are of little or no concern on a diesel, since they will make heat directly in proportion to the amount of fuel injected. Little fuel = little heat. Whereas a gas engine is run at constant fuel/air ratio, diesels run a variable fuel/air ratio. At idle they run extrememly lean.



-Ryan

 

[Yo Hoot]

Why the hell didnt cummins just use a bigger turbo... less heat... make it so computer doesnt send the message to de-fuel. .



Nick

Emissions... faster spool cleans up the exhaust.



Anybody know if there is any low end torque management in first gear with the 48RE?

 

[DLeno]

so the answer to your question, DHERZOG, is that when you reduce exhaust restriction by changing mufflers, increasing pipe diameters, etc. there are two consequences for the Cummins turbo Diesel:



1. its true that small improvements in running exhaust gas temperatures can be obtained, but these improvements are not significant at stock fueling levels. The stock exhaust system is fine indeed.



2. The turbocharger will probably spool up quicker by some small amount. while this appears to be good (and is, for the largest portion of the operating range), a potential consequence is a very small loss in peak horspower at full WOT. this is because the ECM de-fueling behavior will back off in response to the new spool behavior. see the most recent issue of the TDR "technical topics)

 

[KKlepfer]

DLeno,

So does a fueling box that does boost fooling solve the de-fueling behavior?

 

[JCleary]

DLeno,

So does a fueling box that does boost fooling solve the de-fueling behavior?

My quad box gave me a 1 hp gain on the dyno in the "off" position as opposed to no box at all. In the off position it does boost fooling only.



This was with the AFE and Banks 4" turbo-back exhaust, which eliminates the 4" to 3. 5" elbow that connects to the stock downpipe.

 

[TCluff]

Ya, everyone kept saying that a bigger exhaust gave you so much more power on top end that it made the bottom end feel weeker and that it was all in our heads. Well i staright piped my truck (the factory 3. 5 exhaust is more than adequate, just the muffler is the big restriction) and I noticed a definite drop in power pulling from low rpms. And now finally we have the answer why

why would you go with a larger ex with a basically stock motor, cummins designed the exhaust for that stock motor to run the way it was designed, so in other words if i put 3. 5 inch ex on my truck i will gain power, probably not, it would spit the head gasket out from extremely high drive pressure, and the pyro would be outa numbers before you had it on the floor... . bigger ex is a must for trucks with extra fuel and boost, not a stock truck. . :-{}

 

[Yo Hoot]

why would you go with a larger ex with a basically stock motor, cummins designed the exhaust for that stock motor to run the way it was designed, so in other words if i put 3. 5 inch ex on my truck i will gain power, probably not, it would spit the head gasket out from extremely high drive pressure, and the pyro would be outa numbers before you had it on the floor... . bigger ex is a must for trucks with extra fuel and boost, not a stock truck. . :-{}

Yea but dam does a big pipe look good. :-laf



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[nickleinonen]

Yea but dam does a big pipe look good. :-laf

ain't that the truth

 

[gilliland]

Just put a 6 inch tip on the stock exhaust, looks big, runs stock!