Bridging the gap between Ignition Keyswitch off and Timed Relay power

Thread Starter

Cormac Champion

Joined Jul 28, 2016
24
Hi all,

I've a CarPC installed in my car and it's powered from the cars' 12v so it's a simple setup. However, after turning off the ignition, I want to power the PC for a further 5 minutes for the purpose of being able to make quick stops (like a petrol fillup) without having to shut the PC down. I have installed a Turbo Timer which looks after the countdown but I have a problem with a momentary loss of power between the ignition Off and the Timer power On.

I think my best bet is a capacitor to just bridge the potential 0.5 second (or less) gap - but what size of capacitor do I need ? The voltage is obviously about 12v (since the engine and alternator will be off at this point). The PC (a thin client model) could be pulling up to 3A. I bought a 16v 4700uf cap but once I connected it to the feed supply 12v wire, at the minus leg of the cap, it only measured just over 4v.
So do I need three of them ? In series or in parallel ? Or something else ?
 

#12

Joined Nov 30, 2010
18,190
at the minus leg of the cap, it only measured just over 4v.
The minus leg of the cap should be at ground. 12V---diode pointing to the right-----capacitor to ground----input to computer.
From dVout = dVapplied times e to the (-t/RC) I get 81,000 uf to carry 3 amps for half a second and only lose 3 volts.
That will take you from the battery charging voltage (about 14V after the diode) to 11 volts still on the computer input.
Fortunately, the car stereo industry has made available huge capacitors.
 

#12

Joined Nov 30, 2010
18,190
Auto electric systems are horribly noisy. When you turn the headlights off, the alternator typically over charges to about 60 volts for a millisecond.
Considering the resistance of the wire and this massive capacitor, you can probably get away with less than a 60 volt capacitor.
You're in the range of $10 for 81,000 uf at 25 volts and $30 for a 60 volt capacitor.
If you want a better nail in this, somebody else is going to have to help because I have never done a car stereo.
 

Thread Starter

Cormac Champion

Joined Jul 28, 2016
24
I have found ads for 2.7v SuperCapacitors.

I assume I need to get about 6 x 2.7v and connect them in Series - but what size of 2.7v SuperCaps do I need to have a final total of around 100,000uf ?
 

ion54

Joined Apr 9, 2013
4
Super caps are rated 2.7V only for temperatures not higher than 65C. If the location is such as the temperature can go above that value you shall derate them. Taking in consideration you charge them when engine is running you shall estimate a maximum voltage of 16V when alternator is going full blast, This is not taking in consideration possible failures of the car's electrical system and possible jump start conditions.
You should install the capacitors in series and use balancing resistors in parallel with them (1K should do the job). To make sure I would use 7 capacitors. The final value of your equivalent capacitor is determined by the minimum voltage your computer can operate at.
You need energy from the caps at a rate of 3A for 0,5 sec. that is a charge of 1.5C.
Assuming you switch from engine run to engine off your voltage at capacitor terminals goes from 13.5V to let's say 10V (minimum functional voltage of your PC). That is a delta V of 3.5V. Capacitance value you need is calculated by dividing the charge to delta V: C = 1.5/3.5 = 0.43F
Because you have a series of 7 capacitors the value of each capacitor shall be 7 times the required value, or 3F.
Wouldn't be easier to power the PC from the battery regardless of engine status? You will only need to disconnect when parking the car for longer periods.
 

crutschow

Joined Mar 14, 2008
24,362
For a maximum of 3V drop in capacitor voltage during the switch-over you need a minimum capacitance of (3A*0.5s) / 3V = 500,000μF or 1/2F (where C = Q/V and Q = I*t).
 

#12

Joined Nov 30, 2010
18,190
For a maximum of 3V drop in capacitor voltage during the switch-over you need a minimum capacitance of (3A*0.5s) / 3V = 500,000μF or 1/2F (where C = Q/V and Q = I*t).
Did I slip a digit in post #2?

Edit: I calculated for an RC decay, not a constant current load. But I calculated an R value for the computer...which should be pretty much the same...right?
 
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crutschow

Joined Mar 14, 2008
24,362
Did I slip a digit in post #2?

Edit: I calculated for an RC decay, not a constant current load. But I calculated an R value for the computer...which should be pretty much the same...right?
Right.
But you apparently solved for ΔV, not V/Vo.
V/Vo = e^-t/RC where Vo is the voltage at time zero, V is the voltage at the end of the time period, and 4Ω is the load, giving
ln(V/Vo) = -t/RC or
C=-t/(ln(V/Vo)*R) or
C = -0.5/( (ln(11/14) *4) = -0.5/(-.241*4) =518mF = 0.518F
rather close to the 0.5F calculated using a constant-current load.

A simulation is shown below.

upload_2016-8-12_19-50-16.png
 
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#12

Joined Nov 30, 2010
18,190
I have messed up the use of that formula so many times that I began to hate it a long time ago.:mad: Every time I think I understand it, I stop thinking and do it wrong again.:(
 
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MaxHeadRoom

Joined Jul 18, 2013
20,023
I've a CarPC installed in my car and it's powered from the cars' 12v so it's a simple setup. However, after turning off the ignition, I want to power the PC for a further 5 minutes for the purpose of being able to make quick stops (like a petrol fillup) without having to shut the PC down.
Why not take it off of a spare fuse that is Hot at all times, rather than just Hot in Run.
Max.
 
Or... Slow down the power down of the PC, by isolating it with a Schottky Diode (Low voltage drop) diode and big cap.

it looks like the glitch your trying to overcome is going from ON to ACC

This http://www.digikey.com/en/product-highlight/l/linear-tech/ltc4353-dual-low-volt-ideal-diode-controller is a replacement for a high power schotkey diode. It may likely work. The likely advantage would be turn on would be faster and the large cap won't be needed (Maybe). It's a suggestion that needs more work
 

catfibres

Joined Oct 17, 2012
7
That LTC4353 looks like a spiffy part - one I would have loved to have 13 years ago while designing a redundant CPCI 350W converter! AS to the issue at hand, I sort of agree with MAXHR - wouldn't an Always Hot connection, a simple SPST switch and a diode or two (Schottky would be nice) take care of the whole situation? Half a Farad of Super Cap is not cheap.
 

#12

Joined Nov 30, 2010
18,190
It's fairly common around here that people ask for what they imagine they want and don't ask us what they might need. When they find out the price of seven supercaps and then find there are no auto stereo stores in the neighborhood, they might consider an SPST switch on a wire from the battery.
 

Thread Starter

Cormac Champion

Joined Jul 28, 2016
24
Why not take it off of a spare fuse that is Hot at all times, rather than just Hot in Run.
Max.
If the PC fails to shut down properly or somehow restarts (it's set to restart automatically at power on), then it could (and did !!) drain the battery. In addition, I need the power off to be able to turn it on automatically with power resume, then next time I get into the car

This http://www.digikey.com/en/product-highlight/l/linear-tech/ltc4353-dual-low-volt-ideal-diode-controller is a replacement for a high power schotkey diode. It may likely work. The likely advantage would be turn on would be faster and the large cap won't be needed (Maybe). It's a suggestion that needs more work
Hmmmm, how many connections would I need to make on this ? And onto which pins ? I'm assuming two - 1 In and 1 Out.
 
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