Ignition loads current verses input voltage
05-05-2010, 06:42 PM
#1
Ignition loads current verses input voltage
Ignition loads current verses input voltage.
It was stated in the Evolution 58cc dying thread by a member that higher battery voltage results in lower current draw. Quote “Check out ohms law!!! The higher the voltage the lower the current draw, FACT!!! “
This is backwards. A higher source voltage with a constant load always means a higher current draw.
First you need to know what the load resistance is. To do that ohms law says R=E/I . Lets use 4.8 volts for the voltage. For the current I’ll use what bcchi stated (since he builds what most consider the finest ignition systems on the market).
“The RCEXL for example a clone of mine. On average will draw 400MA at 8000 RPM,4.8 volts ,500 MA at 8000RPM 6 volts,600MA 8000RPM.7.4 volts.Go figure.I just went and checked 10 new CHRCEXL ignitions..This was average on all of them,not all exactly the same but close.”
So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms.
If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw.
Higher voltage results in higher current. Sorry guys but this was really bugging me.
Later,
John
It was stated in the Evolution 58cc dying thread by a member that higher battery voltage results in lower current draw. Quote “Check out ohms law!!! The higher the voltage the lower the current draw, FACT!!! “
This is backwards. A higher source voltage with a constant load always means a higher current draw.
First you need to know what the load resistance is. To do that ohms law says R=E/I . Lets use 4.8 volts for the voltage. For the current I’ll use what bcchi stated (since he builds what most consider the finest ignition systems on the market).
“The RCEXL for example a clone of mine. On average will draw 400MA at 8000 RPM,4.8 volts ,500 MA at 8000RPM 6 volts,600MA 8000RPM.7.4 volts.Go figure.I just went and checked 10 new CHRCEXL ignitions..This was average on all of them,not all exactly the same but close.”
So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms.
If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw.
Higher voltage results in higher current. Sorry guys but this was really bugging me.
Later,
John
05-05-2010, 06:50 PM
#2
Senior Member
My Feedback: (1)
RE: Ignition loads current verses input voltage
Good explanation.
Others have been trying to tell people for years it's similar to plumbing to make mental imagery easier. Bigger pipe, more flow. They don't like to listen and instead figure more is always better. Ain't so.
Others have been trying to tell people for years it's similar to plumbing to make mental imagery easier. Bigger pipe, more flow. They don't like to listen and instead figure more is always better. Ain't so.
05-05-2010, 06:59 PM
#3
RE: Ignition loads current verses input voltage
I had to say something. Sometimes it easy to get a wrong conclusion and then state it as fact.
I simply wanted to show how to arrive at the correct conclusion. Something I should have done in my first reply on the 58 cc thread.
John
I simply wanted to show how to arrive at the correct conclusion. Something I should have done in my first reply on the 58 cc thread.
John
05-05-2010, 07:58 PM
#6
Senior Member
RE: Ignition loads current verses input voltage
ORIGINAL: JNorton
Ignition loads current verses input voltage.
It was stated in the Evolution 58cc dying thread by a member that higher battery voltage results in lower current draw. Quote “Check out ohms law!!! The higher the voltage the lower the current draw, FACT!!! “
This is backwards. A higher source voltage with a constant load always means a higher current draw.
First you need to know what the load resistance is. To do that ohms law says R=E/I . Lets use 4.8 volts for the voltage. For the current I’ll use what bcchi stated (since he builds what most consider the finest ignition systems on the market).
“The RCEXL for example a clone of mine. On average will draw 400MA at 8000 RPM,4.8 volts ,500 MA at 8000RPM 6 volts,600MA 8000RPM.7.4 volts.Go figure.I just went and checked 10 new CHRCEXL ignitions..This was average on all of them,not all exactly the same but close.”
So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms.
If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw.
Higher voltage results in higher current. Sorry guys but this was really bugging me.
Later,
John
Ignition loads current verses input voltage.
It was stated in the Evolution 58cc dying thread by a member that higher battery voltage results in lower current draw. Quote “Check out ohms law!!! The higher the voltage the lower the current draw, FACT!!! “
This is backwards. A higher source voltage with a constant load always means a higher current draw.
First you need to know what the load resistance is. To do that ohms law says R=E/I . Lets use 4.8 volts for the voltage. For the current I’ll use what bcchi stated (since he builds what most consider the finest ignition systems on the market).
“The RCEXL for example a clone of mine. On average will draw 400MA at 8000 RPM,4.8 volts ,500 MA at 8000RPM 6 volts,600MA 8000RPM.7.4 volts.Go figure.I just went and checked 10 new CHRCEXL ignitions..This was average on all of them,not all exactly the same but close.”
So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms.
If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw.
Higher voltage results in higher current. Sorry guys but this was really bugging me.
Later,
John
BCCHI
05-05-2010, 08:04 PM
#7
Senior Member
RE: Ignition loads current verses input voltage
ORIGINAL: av8tor1977
Yep, good answer and good explanation, though I can't believe you went to all the trouble to test 10 ignitions!! Ohm's law is Ohm's law! But maybe it takes 10 times to convince some people huh?
AV8TOR
Yep, good answer and good explanation, though I can't believe you went to all the trouble to test 10 ignitions!! Ohm's law is Ohm's law! But maybe it takes 10 times to convince some people huh?
AV8TOR
BCCHI
05-05-2010, 11:24 PM
#9
Senior Member
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RE: Ignition loads current verses input voltage
Many folks sort of know that for a given wattage, higher voltage means lower current. Additionally, lots of folks who are not electrical professionals have also heard about I square R losses and realize that running a higher voltage for a given wattage, with its reduced current, is more efficient than obtaining the same wattage with lower voltage and higher current.
Ed Cregger
Ed Cregger
05-06-2010, 03:30 AM
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RE: Ignition loads current verses input voltage
Ed,
What makes you think the Wattage remains the same if you increase voltage? That would depend on the appliance design wouldn't it?
What makes you think the Wattage remains the same if you increase voltage? That would depend on the appliance design wouldn't it?
ORIGINAL: NM2K
Many folks sort of know that for a given wattage, higher voltage means lower current. Additionally, lots of folks who are not electrical professionals have also heard about I square R losses and realize that running a higher voltage for a given wattage, with its reduced current, is more efficient than obtaining the same wattage with lower voltage and higher current.
Ed Cregger
Many folks sort of know that for a given wattage, higher voltage means lower current. Additionally, lots of folks who are not electrical professionals have also heard about I square R losses and realize that running a higher voltage for a given wattage, with its reduced current, is more efficient than obtaining the same wattage with lower voltage and higher current.
Ed Cregger
05-06-2010, 05:53 AM
#12
RE: Ignition loads current verses input voltage
Shezzee Ed thanks for muddying the waters! You are talking about transformers or the transmission line itself. Wattage with the same load will also increase when you raise the voltage for a fixed load. Lets take the same example.
So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms. P=IXE .5 A X 4.8 volts = 2.4 watts
If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw. .625 A X 6 volts = 3 watts
IF THE LOAD REMAINS THE SAME - AND YOU INCREASE THE VOLTAGE THE CURRENT WILL ALSO INCREASE. PERIOD.
Kurt it does not prove your point. Look at the math. Please look at the math.
John
So lets use 500 mA or .5 amps. The load resistance is therefore 4.8/.5 or 9.6 ohms. P=IXE .5 A X 4.8 volts = 2.4 watts
If we use a 6 volt battery pack the current draw is I=E/R or 6/9.6= .625 or 625 mA current draw. .625 A X 6 volts = 3 watts
IF THE LOAD REMAINS THE SAME - AND YOU INCREASE THE VOLTAGE THE CURRENT WILL ALSO INCREASE. PERIOD.
Kurt it does not prove your point. Look at the math. Please look at the math.
John
05-06-2010, 09:28 AM
#14
RE: Ignition loads current verses input voltage
ORIGINAL: Tired Old Man
Good explanation.
Others have been trying to tell people for years it's similar to plumbing to make mental imagery easier. Bigger pipe, more flow. They don't like to listen and instead figure more is always better. Ain't so.
Good explanation.
Others have been trying to tell people for years it's similar to plumbing to make mental imagery easier. Bigger pipe, more flow. They don't like to listen and instead figure more is always better. Ain't so.
For a simple explanation of amperage, the water pipe/plumbing description works as advertised but in this discussion guys are talking about voltage which would equate to pressure in a pipe/plumbing description. That's a whole 'nuther ball game as has been pointed out above. Increasing the water pressure (voltage) increases friction loss (resistance) in a given size hose/pipe.
Where the newbie 'lectric guys really get into trouble with these ideas/computations, is when the load is variable like most 'lectric powered models as that changes things again vs. the example JNorton gave which is a fixed load.
Being a retired firefighter, I know all 'bout friction loss for computing fire flow. Being an A&P required me to learn more about electricity than I wanted to and there is still lots more I don't know, but don't 'wanna learn any more than I have to now to to keep my models in the air, like when it took Dick Hanson 6 months to convince me to give A123's a try.
The plumbing description compared to electricity stuff does work for complete neophytes or those who think they can't grasp some of the basic concepts...
05-06-2010, 10:17 AM
#15
RE: Ignition loads current verses input voltage
Even in a variable load for any giving point in that load if you increase the voltage for that given point the current will rise to. A variable motor loads gets tricky determining battery life and ESC capacity.
I just want to nail this concept down - if you increase voltage to a load the current will increase. It doesn't matter if you use power or current as everything is related in the formulas given. They are not separate stand alone entities. Change one value the other will change. You cannot have current flow without having power being dissapated through a resistance. Power is related to the voltage times current drawn. In addition every component also has capacity and inductive values present.
Look at the chart.
John
I just want to nail this concept down - if you increase voltage to a load the current will increase. It doesn't matter if you use power or current as everything is related in the formulas given. They are not separate stand alone entities. Change one value the other will change. You cannot have current flow without having power being dissapated through a resistance. Power is related to the voltage times current drawn. In addition every component also has capacity and inductive values present.
Look at the chart.
John
05-06-2010, 10:38 AM
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RE: Ignition loads current verses input voltage
norton, That also means the output will increase also, in a motor that means higher rpm and in an ignition coil that means a hotter spark, if you take the regulator or heat sink out of the equation.
05-06-2010, 10:48 AM
#17
RE: Ignition loads current verses input voltage
Higher voltage will give you higher power or output to the load.
I don't have a disagreement with a brushed motor, it will speed up. A 3 phase DC motor may or may not depending upon the ECS - usually it will also speed up.
However an ignition circuit depending upon how its designed it may or may not provide a hotter spark. If the ignition uses a switching designed boost source for the coil the output will tend to be self stabilized. Ask Bcchi what his does.
John
I don't have a disagreement with a brushed motor, it will speed up. A 3 phase DC motor may or may not depending upon the ECS - usually it will also speed up.
However an ignition circuit depending upon how its designed it may or may not provide a hotter spark. If the ignition uses a switching designed boost source for the coil the output will tend to be self stabilized. Ask Bcchi what his does.
John
05-06-2010, 11:02 AM
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RE: Ignition loads current verses input voltage
The way I look at it is this way, if you had an ignition coil wrapped or wound for 4.8v input and 10,000v output and ran with a 4.8v battery, verses an ignition coil wrapped or wound for 6v input and 10,000v output ran on a 6v battery you should have 20% less current draw on the 6v coil. That takes the power loss(heat generated) from the regulator out, when 6v is dropped to 4.8v by the internal regulator! Am I right?
05-06-2010, 11:03 AM
#19
RE: Ignition loads current verses input voltage
ORIGINAL: kurt2022
norton, That also means the output will increase also, in a motor that means higher rpm and in an ignition coil that means a hotter spark, if you take the regulator or heat sink out of the equation.
norton, That also means the output will increase also, in a motor that means higher rpm and in an ignition coil that means a hotter spark, if you take the regulator or heat sink out of the equation.
if this held true you would have less power and the motor would run slower with higher voltage applied.
John
05-06-2010, 11:10 AM
#23
RE: Ignition loads current verses input voltage
The coil will only discharge enough power to hit ground, the higher the resistance the more power used up to its full capacity. Higher input voltage will provide a higher available secondary voltage but it will not be used if the resistance is low, excess is disipated as heat and not good for the ignition over time.
05-06-2010, 11:15 AM
#24
RE: Ignition loads current verses input voltage
ORIGINAL: kurt2022
It would run faster until it smoked, if it was not rated for the higher voltage
It would run faster until it smoked, if it was not rated for the higher voltage
An observed fact but why does it smoke? Because the current increases so much that the motor windings can no longer handle it - they glow red and insulation is burned off. Smoking the motor.
Do you now agree with higher voltage comes higher current?
John
05-06-2010, 11:18 AM
#25
RE: Ignition loads current verses input voltage
ORIGINAL: jedijody
The coil will only discharge enough power to hit ground, the higher the resistance the more power used up to its full capacity. Higher input voltage will provide a higher available secondary voltage but it will not be used if the resistance is low, excess is disipated as heat and not good for the ignition over time.
The coil will only discharge enough power to hit ground, the higher the resistance the more power used up to its full capacity. Higher input voltage will provide a higher available secondary voltage but it will not be used if the resistance is low, excess is disipated as heat and not good for the ignition over time.
I think this topic has been pretty well beaten to death.
Later,
John