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Capacitor to power a device during vehicle cranking.

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czwalga00gt

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I have a device that needs to hold 9.5 volts during cranking or it will shut off. The catch is that the device has to work in order for the car to start.

When the battery has a low charge (still enough to turn the car over), the car still will not start because it's not getting a crank sensor voltage coming from the conversion module i'm using.


So, I want someone to double check my thinking here.

I'm going to put a diode running to the module, then a capacitor in parrallel with the module, then a resistor between the cap and module.


Just trying to figure out the values needed to power the module for 1 second.
The modules draws, 500mA at 12v. P= 6 watts.

P=v^2/r
r=24ohms

t=F*r
1=F*24
C=.05~ F
 
The cap needs to be larger than that.

C=q/V. For 500mA for 1 second, q = 0.5 x 1 = 0.5 coulombs. For a 3V capacitor discharge drop during that 1 second, the required capacitor size is 0.5/3 = 0.166F, which is a very large capacitor. You would likely use some type of super-capacitor for that.

You don't need a resistor between the cap and module. That will just reduce the available energy from the cap.
 
You need a large super cap to get that current, as they like to be discharged slowly.

I have a 50F 2.3v super cap that will power a red led for around 45 minutes.
 
What about using a voltage boosting/regulating circuit?
It would be much cheaper than a big cap!
 
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What about a seperate small chargeable 12v battery with a big diode.


I pretty much want the battery to be able to be charged, but when the voltage drops, the 2nd little battery still power the module?
 
What about a seperate small chargeable 12v battery with a big diode.


I pretty much want the battery to be able to be charged, but when the voltage drops, the 2nd little battery still power the module?

That is possible too.
A few considerations:

1. You'll eventually need to replace the battery.
2. You'll need to regulate the current to it to keep it from getting overcharged.
 
^ Looks like it should. I'll have to give it a shot, or I've found a few voltage boosting circuits like you said. Eitherway neither seem very complicated. Thanks!
 
The battery and diode should work. It could be a small lead-acid, maintenance-free type battery.

You could also conceivably use a NiMH battery (for example 10 AA cells in series would give 12V) but you would need to be careful about not over-charging them. One way would be to use two diodes, one from the car battery and one from the NiMh battery to the module. That way there is no direct path from the car battery to the NiMH battery so it couldn't overcharge. To charge it, connect a resistor from the car battery to the NiMh battery. The resistor would be 33Ω for about 75mA charge current when the generator is putting out 14.5V (suggested trickle charge is .03C to 0.05C where C is the battery mAH capacity).

Since the battery will never be significantly discharged, it won't require a lot of charging current to maintain its charge. When the generator is not charging, the two battery voltages will be close to the same value so very little current will flow through the resistor.
 
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I'm curious what this "module" is and why it needs to see crank sensor voltage.



I have an rx7 with a corvette engine. Originally it was a 2006 engine, then i turbo'd the car. I bought a built shortblock for the turbo that was out of a 2007. Between 2006 and 07 they switched the crank reluctor wheel. So to use the computer out of the 05 I needed this converter module for the crank sensor and cam sensor.
 
On bad days cranking voltage can fall as low as 6 volts. That's what makes starter relays special.
If OP has the schematic of his module maybe an automotive voltage regulator can be substituted for what he has now.
 
If OP has the schematic of his module maybe an automotive voltage regulator can be substituted for what he has now.
How would an automotive voltage regulator help the low-voltage-during-starting problem?
 
How would an automotive voltage regulator help the low-voltage-during-starting problem?
If you want 5 V available during cranking, an automotive voltage regulator will be low dropout and contain reverse polarity protection.

A standard voltage regulator, with about 2.5 V drop, plus a series diode, and you need around 8.5 V to work, which can be difficult during starting.
 
Another reason that the battery voltage may be dropping low is that the thing draws 500mA. If it is wired to the battery and not to a switched ACC source, it could be draining the battery. If he doesn't drive the car for a few days, that might be enough to drop the voltage.
 
If you want 5 V available during cranking, an automotive voltage regulator will be low dropout and contain reverse polarity protection.
Where does a 5V requirement come from? He stated in the first post that his module needed 9.5V.
 
Another reason that the battery voltage may be dropping low is that the thing draws 500mA. If it is wired to the battery and not to a switched ACC source, it could be draining the battery. If he doesn't drive the car for a few days, that might be enough to drop the voltage.


I actually have the module on a switch that I turn off along with the ignition, but you are right... if it forget to turn the switch off and leave it sit for a few days, it will drain the battery.



To the above comments, i've tested it 9.5 is about the threshold for the module to function. 12v is the standard operating voltage.
 
500ma to massage a cps and a cmp is crazy. The current requirement for an ECU sensor input is negligeable.
 
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