ACC+ current?

[mrgenie]

Short version: I need to know, while the unit is connected to the permanent 12V power supply on the yellow wire, how much current (Ampere) the unit draws over the ACC? This is only a signal line right, not actually powering the device. So the Ampere could be as low as 0,05mA.

But how much is it really?


LONG VERSION (Only read if interested why I need this to be answered)

In my car when I start the car the Pumpkin HU goes into shutting down.
The Voltage on the yellow permanent power drops down to 11,4 Volts. I have a very strong battery compared to the engine.
To make sure it doesn't power off because of the car power supply on the yellow wire, I isolated the unit from the car.
powered it with a total isolated 12 Volt battery and put a 12 Volt on the ACC. Than I seperated the 12V from the ACC (keeping the 12V on the yellow)
and still got the shut down.

The ACC during starting the car engine drops down to 0 Volts.

So I'm 100% confident the car unit reboots simply because of the ACC drop.

I can even power the ACC with a 12 Volt battery I hold in my hands.. and the yellow with the car battery, then start the car.. and it doesn't shutdown because the 12Volt battery in my hand is isolated from the car.

So I'm very sure the shut down is simply because during the start of the car the voltage drops at 0Volts and even a small 12V battery can keep the unit running during start the car.

So to handle this, I want to use a capacitor with a resistor and a few diodes to make sure the current flows in the right direction, and the resistor in front of the capacitor of course to avoid 0 Ohm while starting.

This works in theory, but I also want to keep the unit running after I switch off the car. Let's say for another 30 seconds I need 12Volts on the ACC wire to make sure the unit remains on!!

Would a 22000uF capacitor be enough? At 12 volts minus 0,55 Volts on the Diode and assume the car radio switching off below 9,5 Volts I'd have 12 Volts - 9,5 minus 0,55 = roughly 2 Volts. 0,022F * 2 Volts = 0,044Coulomb = A * s..

so if A is 1mA it would be enough for 0,044/0,001 = 44 seconds but if it's 0,1mA on ACC it's enough for 440 seconds but if it's 10mA it would be only 4,4 seconds.

So how much is really drawn over the ACC if the unit is powered by the yellow line?

Basically I could do this math myself if I know how much the inner resistor value is on the Pumpkin HU on the ACC?

If it's 500 Ohms it would be 24mA but if the inner resistor on the red signal wire is 5kOhm we're talking 2,4mA or if you use a 50kOhm it would be 0,24mA.

[Jammy]

Hi,
you are clever and creative.
I am sorry that i don't know the current draws for ACC because we test the unit with the constant 12V battery as you.
Hope your kind understanding.
Regards

[stedon81]

Hi mrgenie,

first, if you want to know the current that is drawn, why don't you just measure it?
It might help to tell, which Pumpkin head unit you are using. I assume that there is a wiring harness included, which you can simply adapt to feed some multimeter into the circle.

Second: Are you sure, the voltage drop on the ACC wire is not intended? There are cars that switch off any unnecessary sink while you crank, just to have maximum power for starting the engine.

And lately, you can just start to experiment. Replacing the capacitor in your setup should not be the problem, and gold caps are usually quite cheap. I'm just not sure if you are really right that this head unit uses the ACC just as an indicator. Some units really draw their power from it, and then you're quickly somewhere at 50W, and for that you will need a big capacitor (around 1F or 2F), that also can deliver the required current (gold caps will not be sufficient, but rather a hifi powercap, which are expensive, and probably too big for your setup).

Best,
stedon81