24Volting a car...

[Stealer Of Souls]

So the ECU runs on 5V. And how persay would it get 5V? Methinks a 30c voltage regulator, which (normally) has a blocking voltage of around 35V... So, how would a 24V continuous fry something rated upto 35V? Slowly. That's how. Assuming it is done via regulator then the only issue you've got is thermal death. Which would be very slow given the operating voltages.

OMG, I wont bother going into it but if your got a reasonable understanding of electricity and its number of laws and principles you will understand why you wouldn't convert to 24v

You've lost me with that statement...

Ok basic rule of thumb p = v over i, so by doubling your voltage you will reduce the current. However because p = v over i just by doubling your v you won't increase your p (p is a rated constant after all) - just decrease your i.
The reason the escort worked better briefly on 24 volts to start is less current needed - ever looked on a battery it will give you cold cranking amps normally.
Also think about all your switch gear - ever seen how many relays, transistors, voltage sensors are in a car - good luck to you if you want to go through and change every single one of these.
Pertaining to the ecu it will be alright for spikes, and dips but try running it at a constant double voltage it won't last long at all - think of it as similar to running your pc at 400 volts instead of 230 - the transformer would simply be fried in a matter of milli seconds the bang would be worth watching though.
Thats enough from me now anyhow i'l sit back and watch what you make of it!

Yes but V=IxR. And the R of a starter is essentially fixed. So if you double V, you'll also double R. And the VxI product is therefore doubled too... Don't forget a starter is just a coil of wire with a current passing through it (on a real simple level).

[spencer]

So the ECU runs on 5V. And how persay would it get 5V? Methinks a 30c voltage regulator, which (normally) has a blocking voltage of around 35V... So, how would a 24V continuous fry something rated upto 35V? Slowly. That's how. Assuming it is done via regulator then the only issue you've got is thermal death. Which would be very slow given the operating voltages.

usually something like a LM2937 is used and most we use at uni (commonly used) are only rated to 26v.

[Stealer Of Souls]

The LM78XX series (available from Jaycar for about $1.5 retail) is rated to max 35V input, and can deliver upto 12V 1A (depending on version).
The LM29XX series is probably a lower output version (at a guess).

[spencer]

yea but the LM29 series is found in most mico-crontollers in cars why put a more expensive part that isn't needed? and what are you going to use to drive the injectors? you cant just use a voltage regulator

[NZ_AE86]

5v is only used for some sensors, your Injectors are still 12V as are Coils.

I think he should just go ahead and do it, then post up all his problems and costs and then a list of advantages his car has over another running 12V along with a list of his disadvantages?

[Perky]

I'm confused. Are some typos creeping into folks' calculations here?

Loudtoy wrote:
Ok basic rule of thumb p = v over i, so by doubling your voltage you will reduce the current. However because p = v over i just by doubling your v you won't increase your p (p is a rated constant after all) - just decrease your i.

I take it that this is just Ohm's law and that p is the resistance (R) of the starter motor, which is fixed for a given motor.

p=v/i, so if p is fixed then doubling v also doubles i. That is, current is increased.

Yes but V=IxR. And the R of a starter is essentially fixed. So if you double V, you'll also double R. And the VxI product is therefore doubled too

Ohm's law again. If R is fixed then doubling V does not double R (which, remember, is fixed) but it does double the IxR product. Since R is fixed this means that I (the current) must be doubled. Hence the VxI product (power) is also doubled.

Is that what you guys meant?

edit: As below -- power is quadrupled.

[Stealer Of Souls]

Mine has a terrible typo in it!!!
Doubling V, will double I assuming a constant R (which it should be). Therefore P quadruples. Since both V and I double.
Thanks Perky!!
e.g.
V = I x R
12(V) = 1(A) x 12(Ohm)
24 = 2 x 12

P = V x I
12(W) = 12(V) x 1(A)
48 = 24 x 2 ---> 4x POWER!

yea but the LM29 series is found in most mico-crontollers in cars why put a more expensive part that isn't needed? and what are you going to use to drive the injectors? you cant just use a voltage regulator

I could always just upgrade the regulator. It'd probably only be a 10 minute job. I'm very handy with a soldering iron.

5v is only used for some sensors, your Injectors are still 12V as are Coils.

Injectors. Another thing to add to the list of can't be replaced by a 24V unit...


What drives the injectors? An injector driver is PWM based isn't it? An injector is duty cycle driven. And therefore the injector driver would simply modulate the input to produced a pulse output which in turn delivers a voltage which appears to the injector as an average value. Which regulates the injector flow.
If this is correct. Then in theory putting 24V on the input to the driver (assuming 24V is within design spec) would simply force the micro's logic to reduce the duty cycles to half of what they previously were in order to achieve the same voltage.
So the BIG ASSUMPTION here is that all the components of the injector driver are capable of handling 24V.
???

An alternative is (assuming that there is less than 15A@12V total load) to fit a switchmode voltage reducer. Available from Jaycar for $120 (retail) and weighs approx 600gms. This way you would have a dual rail vehicle. Not the most elegant solution. But not impossible...


As for the coil. I have no solution as yet...

[Perky]

Ha! Serves me right for playing teacher.

Yes, power is quadrupled in this scenario.

[Loudtoy]

I'm confused. Are some typos creeping into folks' calculations here?

Loudtoy wrote:
Ok basic rule of thumb p = v over i, so by doubling your voltage you will reduce the current. However because p = v over i just by doubling your v you won't increase your p (p is a rated constant after all) - just decrease your i.

I take it that this is just Ohm's law and that p is the resistance (R) of the starter motor, which is fixed for a given motor.

p=v/i, so if p is fixed then doubling v also doubles i. That is, current is increased.

quote]

Yeah sorry my bad i was half pissed i meant p=vi which is p for power in watts v=volts obviously and i = current.

Mr Stealer your forgetting that coils aren't purely resistive circuits which you are implying and so a different formula altogether using parts of ohms, lenzs, inductance and other laws that i can't remeber from my apprenticeship papers apply.
I can't be bothered finding them as they are burried in the garage somewhere.

[Stealer Of Souls]

The majority of the resistance of a motor winding can be model as a plain resistance. The effects of inductance (and capacitance) play only a small part in the effective resistance. Also, during a start up the inductance and capacitance of a motor is negligible since the magnetic field hasn't fully stabilised.
The primary effects of inductance on a motor is normally only calculated during steady state running, where the inductance of the motor will influence power factor. (Does power factor have any real influence in DC???)

[JamesM]

hmm.. but its a race car lyndsay.. no heater.. ps.. etc.. u hardly need any wires..
i rewired my entire car and i think ive probably got max 2kgs of wiring now.. i think ur going to save more by doing that.. sort out what u dont need..
ive got a few wires for the lights,dash, rear lights.. and then the wires for the ecu..
24v or more maybe if u had the mega ice car... but u dont.. u have a pos 85

[Loudtoy]

The effects of inductance (and capacitance) play only a small part in the effective resistance. Also, during a start up the inductance and capacitance of a motor is negligible since the magnetic field hasn't fully stabilised.

So your saying inductance is caused by a steady magnetic flux then??
As far as i knew inductance is caused by magnetic fields forming and collapsing, Back emf on forming cause current to build slower, on collapsing causes it to reduce slower. Back emf basicly opposes current change and so will work in either direciton. Once the current is flowing and stabilised the inductance will be 0 in a dc circuit where it will be constantly changing in an ac circuit.
Cheapest motors are inductive motors using shaded pole and slip rings etc to keep the magnetic field circulating so the stator inside it will spin. I haven't taken a starter motor apart to check it but i would imagine that it works on some kind of inductance.

[Stealer Of Souls]

hmm.. but its a race car lyndsay.. no heater.. ps.. etc.. u hardly need any wires..
i rewired my entire car and i think ive probably got max 2kgs of wiring now.. i think ur going to save more by doing that.. sort out what u dont need..
ive got a few wires for the lights,dash, rear lights.. and then the wires for the ecu..
24v or more maybe if u had the mega ice car... but u dont.. u have a pos 85

You're always so negative about my poor car...
Don't forget I'm not running a full 100% race car. Mine is a daily driven streetable race car. So it still has a few things on it. And besides, a lot of modern race cars run a number of goodies, window heaters, PS etc...

So your saying inductance is caused by a steady magnetic flux then??
As far as i knew inductance is caused by magnetic fields forming and collapsing, Back emf on forming cause current to build slower, on collapsing causes it to reduce slower. Back emf basicly opposes current change and so will work in either direciton. Once the current is flowing and stabilised the inductance will be 0 in a dc circuit where it will be constantly changing in an ac circuit.
Cheapest motors are inductive motors using shaded pole and slip rings etc to keep the magnetic field circulating so the stator inside it will spin. I haven't taken a starter motor apart to check it but i would imagine that it works on some kind of inductance.

All true.
As a rule of thumb. A motor will take 6x normal running current to start. This starting current is usually calculated using just the resistance of the primary winding (no mag field, etc...). After initial transient phase (first .25-.5 seconds) motor will run at approximately normal running current. So if at 12V motor start takes 6A (2ohm), then at start up the same motor (2 Ohm) on 24V will draw 12A (remember we're modelling the start up using just the resistance since the actual difference when you include inductance is insignificant). Both will drop to approx 1/6 of start current.
The motor will most likely then catch fire if you keep it running since it's a 12V motor.

There is no argument with your theory. It's all very correct. But we're not cranking a start for more than a few seconds. During which if you model it as a plain resistance, you'll get a very very similar 'z' impedance to if you model it using the resistance and the 'j' impedance terms. More so with a DC motor. On an AC motor it's a little bit more important since you end up with lagging power factor which creates a greater power draw from the source. But if you compare the power draw (providing a power factor of not much more than 0.8 lagging) with a PF of 1.0 then it's still not that far off.

Disclaimer: It's been a couple of years since I last did electromagnetics - And motors weren't my strong point. I may have some details incorrect.