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Author Topic: Distributor Vs. Distributorless (DIS) Ignition  (Read 1976 times)

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Offline Scientist - The Lab Racing

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Distributor Vs. Distributorless (DIS) Ignition
« on: October 12, 2004, 04:29:33 PM »
Many people have heard that a running a distributorless ignition setup is better than distributed and just blindly accept it. It is true but exactly why is it better? If you’re interested on why distributorless is better then this article will fill that void of knowledge. Before we get into the why one system is better we have to go over the basics of each system.

In a distributor system the igniter charges the coil and tells it when to release voltage in the form of a high tension current. This voltage is then sent through the coil wire to the distributor cap. From the distributor cap the voltage is transferred to the rotor. The rotor then distributes the high tension current bridges the gap between the rotor and the appropriate tower of the distributor cap. From that tower of the distributor the high tension current makes its way through the spark plug wire to the spark plug jumping one final gap across the spark plug gap, where it finally grounds itself to the block ending one ignition pulse. This sequence is repeated for all cylinders.

A Distributorless Ignition System (DIS) comes in two forms, Wastespark and Direct fire. A direct fire ignition has one coil and igniter per cylinder. In the direct fire system the car’s computer tells the igniter to charge the coil and fire it, sending the high tension voltage from the coil to the spark plug wire, from the spark plug wire the voltage travels through the spark plug jumping the spark plug gap and grounding itself to the block. A wastespark uses one igniter and one coil for 2 cylinders. The sequence of events is the same as Direct fire except it ignites not only the power stroke, but it ignites the exhaust stroke simultaneously.

DIS’s provide substantial reliability and performance gains over distributed systems. There’s no cap and rotor to wear out, a worn cap and rotor causes weak spark and leads to hesitation under acceleration. There is also one less gap for the charge to jump, between the cap and rotor, and less spark plug wire for the voltage to travel, this means less resistance, less resistance yields stronger spark at the plug. Some direct fire systems place the igniter and coil pack directly on the spark plug completely eliminating the spark plug wire. In some instances a DIS system provides enough additional power that the plug gap can be increased, creating a more exposed spark igniting more fuel that could have been left unconsumed before. DIS systems allow stronger spark and more precise ignition timing control which improves emissions and efficiency which will increase the overall power of the car.

Some cars can benefit from a DIS more than others. We are going to use a 4 cylinder turbo Honda with small displacement B16a, (1.6 liter DOHC VTEC), for an example. The reason for this is the motor makes power at high rpm, 6500-8200, Due to the small displacement, and will encounter high cylinder pressures, with dense Air Fuel mixtures due to the turbo. Quick fact, the higher the cylinder pressure makes it harder to ignite a spark strong enough to bridge the plug gap. This means the spark has to be extremely powerful in order to light such a dense air fuel mixture. That’s not too bad just use a stronger coil and igniter amplifier. But what happens when you have high cylinder pressure and high RPM, like a turbo B16a Honda engine? This situation provides inadequate coil charge time for reliable spark, which leads to hesitation. The remedy to this is a DIS. In a distributor system one coil has to x amount of time to charge between ignition pulses. In order to make more charge time, we just add another coil via DIS, now one coil will ignite cylinders 1 and 4 and the other coil will ignite cylinders 2 and 3. This is a waste spark system. We can make 4 times the charge time (not necessary) if we use 4 coils, direct fire, one per cylinder.

Offline Natzimus

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #1 on: October 12, 2004, 08:53:46 PM »
good info, i just started this topin in school also  [2thumbsup]

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #2 on: November 16, 2004, 11:46:58 PM »
Cool stuff, although i have some doubt about some stuff that was written.

"There is also one less gap for the charge to jump, between the cap and rotor, and less spark plug wire for the voltage to travel, this means less resistance, less resistance yields stronger spark at the plug."

I think that less resistance does NOT yield a stronger spark... According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.
This would mean that the higher the resistance, the higher the voltage output...for example, the secondary winding (high voltage circuit) in a ignition coil can produce 5000ohms of resistance (but check your specs of course)  and ignition wires are made to be highly resistant (about 50 000 ohms) to prevent radio frequency interference. ALTHOUGH excessive resistance such as an open circuit will cause a weak spark or a no spark situation. I guess what im trying to say is that less resistance does not cause a stronger spark but too much physical resistances, such as a worn cap and rotor, an open circuit or a wide spark plug gap which are excessive resistances can can cause a weak/no spark.




Offline Darkeye

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #3 on: December 2, 2004, 09:36:27 AM »


I think that less resistance does NOT yield a stronger spark... According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.
This would mean that the higher the resistance, the higher the voltage output

or the higher the Resistance  the lower the Amperage, leaving the voltage constant [thumbsup]

bossman if u increase  the resistanace u go have less flow which mean amperage is going to decrease which mean voltage go stay near the same, now if there is less resistance u get higher Amperage which mean a stronger punch with the same voltage [boxing]. i see how u could get mixed up  ::)
« Last Edit: December 2, 2004, 09:41:15 AM by darkeye »
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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #4 on: December 2, 2004, 09:58:56 AM »
as dark eye said..the voltage will remain the same if we lived in a perfect world.  but if i can still remember my physics from 3 years ago...Take for example....High tension wires...over a long distance of wires, there is more resistance than with shorter wires.  So when electric companies wire electricity to distance  areas they 1.  Step up the voltage to travel, and 2. because there is a voltage drop due to resistance and heat generated from resistance (some voltage is lost through heat energy) they use a step down transformer on the other end rated at the new voltage.  its also like batteries...you will never get the rated voltage of a battery due to internal resistance and the more wiring used u get the idea.

Offline lightning - r

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #5 on: December 2, 2004, 04:20:54 PM »
Cool stuff, although i have some doubt about some stuff that was written.

"There is also one less gap for the charge to jump, between the cap and rotor, and less spark plug wire for the voltage to travel, this means less resistance, less resistance yields stronger spark at the plug."

I think that less resistance does NOT yield a stronger spark... According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.
This would mean that the higher the resistance, the higher the voltage output...for example, the secondary winding (high voltage circuit) in a ignition coil can produce 5000ohms of resistance (but check your specs of course)  and ignition wires are made to be highly resistant (about 50 000 ohms) to prevent radio frequency interference. ALTHOUGH excessive resistance such as an open circuit will cause a weak spark or a no spark situation. I guess what im trying to say is that less resistance does not cause a stronger spark but too much physical resistances, such as a worn cap and rotor, an open circuit or a wide spark plug gap which are excessive resistances can can cause a weak/no spark.






Where did you get all this rubbish from. [laugh] [laugh] [laugh]

ohms law:   I=V/R , therefore same voltage with less resistance = more current and with more current you get stronger sparks.

Offline lightning - r

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #6 on: December 2, 2004, 04:28:28 PM »
 eg1

 V = 12volts
 R = 6ohms
 I = 12/6
ans = 2 amps

eg 2

 V =12volts
 R = 2ohms
 I =12/2
ans = 6 amps

Offline Natzimus

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #7 on: December 2, 2004, 04:30:18 PM »
lightning - r

 what to say about u man.. u r came from a donkey, i know it for a fact.
 

hear what if u dont know what the hell u talking about shut the **censored** up, Dis ute is in an auto tech program wid me, what do u do for school ?
 man, u r so stupid dawg...  obv ohms law is:
  voltage = amp x resistance.
 to find amperage = volt/resistance
 to find resistance = volt/amp
 to find voltage     = amp x resistance


 stupid stupid lightn -r i pity ur family.

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #8 on: December 2, 2004, 04:32:28 PM »
eg1

 V = 12volts
 R = 6ohms
 I = 12/6
ans = 2 amps

eg 2

 V =12volts
 R = 2ohms
 I =12/2
ans = 6 amps


 u r a Phuking poket, the bro stated OHms law, not how to find aperage.
 in these examples above u r finding amperage,
so ur formula is right, but its not ohms law.

Offline Darkeye

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #9 on: December 2, 2004, 08:24:23 PM »

 According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.


So, Natzimus u saying u bredrin wrong in the above quote. :-\

me knoe evryting I said is true until u can prove me wrong  [nod]
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Offline Natzimus

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #10 on: December 2, 2004, 09:11:15 PM »

 According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.


So, Natzimus u saying u bredrin wrong in the above quote. :-\

me knoe evryting I said is true until u can prove me wrong  [nod]

 bro he is right, check ur info, that is legit

Offline lightning - r

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #11 on: December 2, 2004, 10:33:57 PM »
lightning - r

 what to say about u man.. u r came from a donkey, i know it for a fact.
 

hear what if u dont know what the hell u talking about shut the f**k up, Dis ute is in an auto tech program wid me, what do u do for school ?
 man, u r so stupid dawg...  obv ohms law is:
  voltage = amp x resistance.
 to find amperage = volt/resistance
 to find resistance = volt/amp
 to find voltage     = amp x resistance


 stupid stupid lightn -r i pity ur family.


OHMS LAW state that the current flowing in a ckt. is directly proportional to the voltage and inversely proportional to the resistance.

I = V/R
R = V/I
V = IR

Before u make stupid statement you could have simply do a search on the net  for ohms law. Why the personnal attack, you must have spend a lot of time sleeping with  donkeys to now that  I came from one.

Offline Natzimus

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #12 on: December 2, 2004, 11:13:38 PM »
 :) :) :) :) [thumbsup] i know, i saw u tippin 1 , thats why, anyways,


dude, what the hell, i wrote the same thing already

to find amperage = volt/resistance
to find resistance = volt/amp
to find voltage    = amp x resistance


 i dont need to do research.. i got this straight from my head, i am worried about u..
geez, i stated what i had already stated.. anyways,, i cant argue with u, as u are MR KNOW IT ALLL OHMS LAW.

Offline Darkeye

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #13 on: December 3, 2004, 12:16:04 AM »

 According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.


So, Natzimus u saying u bredrin wrong in the above quote. :-\

me knoe evryting I said is true until u can prove me wrong  [nod]

 bro he is right, check ur info, that is legit

1. lightning - r and Impact Blue stated the same ohm's law but u say lightning - r is wrong and Impact Blue is right

2.  If u increase the resistance of the wire ur voltage will not increase, good example can be found in scientist (the wj member) example above. this is in the ideal situation and is true

i was just thinking it over and i think wat happen was we were talking about the ideal situation and u guys were taking the actual situation *still want to know more details from u guys as education is key

so we r both right to some degree increasing resistance will increase voltage and decrease amerage, but i still dont think that lenghening the distance is a good way to increase resistance due to heat loss and amperage drop which is wat packs the punch thus less effecient
[thumbsup]
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Offline sweetestaboo

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #14 on: December 3, 2004, 09:43:48 AM »
Guys tone down a bit. im trying to learn here.  ::)
unuh confusing me  :'(
te reto a que me quites la ropa

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #15 on: December 3, 2004, 09:49:22 AM »

 According to Ohm's law, 1volt of electrical pressure will cause 1 amp of current flow through 1 ohm of resistance...or Voltage = Amperage x Resistance.


So, Natzimus u saying u bredrin wrong in the above quote. :-\

me knoe evryting I said is true until u can prove me wrong  [nod]

 bro he is right, check ur info, that is legit

1. lightning - r and Impact Blue stated the same ohm's law but u say lightning - r is wrong and Impact Blue is right

2.  If u increase the resistance of the wire ur voltage will not increase, good example can be found in scientist (the wj member) example above. this is in the ideal situation and is true

i was just thinking it over and i think wat happen was we were talking about the ideal situation and u guys were taking the actual situation *still want to know more details from u guys as education is key

so we r both right to some degree increasing resistance will increase voltage and decrease amerage, but i still dont think that lenghening the distance is a good way to increase resistance due to heat loss and amperage drop which is wat packs the punch thus less effecient
[thumbsup]

 U r both wrong
 V = IR  is an equation ,therefore both side of the equation must be balance. IF V is known then IR must be equal to V ,in this case V is known (12volts or whatever voltage your coil is pushing) so if you increase resistance then the current must decrease for you to get V. likewise if you decrease resistance then the current must increase.
 So people sell low resistance plug leads,non - resistor racing plugs etc . to get more spark.

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #16 on: December 3, 2004, 01:12:52 PM »
Again that is in the ideal situation. V is subject to change along the length of a the wire due to heat (which is not in ur equation because  the equation is for ideal situations) and resistance. weather V falls or rises is the quetion here. V will fall over distances in a wire but if  a resistor is in the path of the current, directly after the resistor there will be a jump in votage and the voltage again will continue to fall over the distance of the wire

If V is know then IR must be equal to V . is for an ideal situation not reality. Heat and other factors are not present in the equation
« Last Edit: December 3, 2004, 01:13:25 PM by darkeye »
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Offline lightning - r

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #17 on: December 3, 2004, 03:30:29 PM »
Did u read the man's post. The man said that less resistance will NOT give a stronger spark which is what I'm saying is WRONG. I am not talking about voltage drop accross a peice of wire.

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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #18 on: December 3, 2004, 05:03:01 PM »
Did u read the man's post ? ???.

Cool stuff, although i have some doubt about some stuff that was written.

"There is also one less gap for the charge to jump, between the cap and rotor, and less spark plug wire for the voltage to travel, this means less resistance, less resistance yields stronger spark at the plug."

I think that less resistance does NOT yield a stronger spark


The man is saying that less resistance in the form of  SHORT WIRES and no Distributor will NOT give a stronger spark . Im also saying he is wrong. and im not just telling him that him wrong (like u r) im trying to explain y i think he is wrong thats were the whole long wire thing come in. y u cant just throw in ohm's law and calculate crap cause there are other factors [nod]
« Last Edit: December 3, 2004, 05:04:43 PM by darkeye »
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Re:Distributor Vs. Distributorless (DIS) Ignition
« Reply #19 on: December 21, 2004, 11:16:30 PM »
whoaa...i see theres been some friction going on here...lol

well first of all, the secondary circuit in the ignition system is also known as a high voltage circuit (which i did mention)...which also means that as secondary voltage rises, current decreases. To prove that, say 2 amps of current flowing to the primary circuit at 12V would equal 24 Watts of electrical power (wattage= Voltage x Amperage) Sorry for introducing a new equation....but what im trying to say is....secondary current will equal total primary wattage divided by secondary voltage...which is a very low amperage...

I think that more amperage will cause a more powerful spark when accompanied by high resistance... perhaps thats what i should have said...but again, there isnt much amperage in the secondary circuit anyway, also, thats why touching a firing spark plus is also fun. lol (that was unecessary)

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #21 on: August 15, 2005, 11:26:35 AM »
i hate my distributorless set up, cant wait to change it

Offline G.R

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #22 on: August 15, 2005, 09:34:26 PM »
you people amuse me. first off all changing the resistance cannot change the voltage!!! or raise it.
all the voltage is supplied by the coil. this is the MAX VOLTAGE.

1. WE can only ALTER the RESISTANCE and the CURRENT in the wires. got that? we can't get more voltage from no where!!

2. Since we are in the real world there is heat loss due to resistance in the wires.
    so power is lost through the wires as heat.

3. by using thicker cables we reduce the heat loss. and keep a high current as stated above.

4. Now At the gap of the spark plug WE WANT POWER LOSS! but in the form of a spark!


 ... power used for the spark= I x R2
    where I is the current and R the resistance of the spark plug gap.
  now R is constant... but now the current is higher due to more conductive wires...

 let R= 2 ohms  current before = 1A  current after = 2A
spark power before = 1A x 2= 2W          Power after = 2A x 2 = 4W

So thicker wires give more power at the saprk plugs...

5. when them say the distributor add an extra gap for the current to go cross, they are saying that in a sense you have 2 spark plug gaps!

now you have 2 resistors in series on the spark plug cable!

DIS...

coil----------------->------------spark plug gap--------->-------engine ground
EMF                                       R2

Distributor

coil ---------->-------------- distributor gap------------------>-------------- sparkplug gap------------->-------engine ground
EMF(max voltage                 R1                                                 R2

V=I x R,  total voltage EMF= V1 + V2 = I R1 + IR2

Since the emf is constant the voltage has to be shared between the two resistors, R1 and R2 for the distributor's case. But For DIS, R2(spark plug) Gets all the voltage to itself!!
Simple as that!





                             
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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #23 on: August 16, 2005, 02:01:16 AM »
i hate my distributorless set up, cant wait to change it

DIS is more precise cause if you look at a distributor its design is flawed..... you just want something that  you can put a spanner on so that you cant adjust ur timing better yuh seh dat instead.

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #24 on: August 16, 2005, 07:22:51 AM »
thats exactly why i hate it, i dont feel in control

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #25 on: August 21, 2005, 04:33:36 PM »
DIS should allow you more precise control of ignition.

by design, wouldn't a distributer system be one static adjustment?

In DIS the Cars ECU would control the ignition from the crank angle sensor right? and chooses when to fire?

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #26 on: August 21, 2005, 05:57:14 PM »
yeah, but you dont control that, you feel powerless

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #27 on: September 2, 2005, 10:44:54 AM »
wow
so much info. At least my eyes are wide open now to all this stuff,  distributor vs dis setup.

im moving from a distributor setup to a dis setupe now (changing engine) but let me ask something n i dont wanna sound weird.

can one who as a distributor setup, convert it to a DIS setup ?
if yes how ? ::) :o :o

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Re: Distributor Vs. Distributorless (DIS) Ignition
« Reply #28 on: September 5, 2005, 11:52:24 AM »
I have never actually seen it done but it should be possible, but damn costly. I guess you would have to get the hole that the distributor was in pluged so how. Maybe a ECU flash or Rom chip bought or just new ECU altogeather. Coil packs in 1 per every two cylinders or individual coil packs for each cylinder. I prefer individual but they are also the most expensive. Sensors are the thing that would make or break this system, CAS - Crank Angle Sensor or CPS - Crank Position Sensor (same thing different names by manufactures) CPS - Cam Position Sensor the proper awg wire to splice it inot your wire loom I am not sure if I am missing anything. Just know that a decent amount of wiring will be needed and a reputable automotive electrician.