n00b capacitor question

[andrew_c]

I have a dash-cam in my car. It turns on and off with the ignition. Unfortunately there is a glitch when starting the engine, the cigarette lighter is disabled momentarily and the camera turns off - it does not turn back on when power returns.

I'm thinking of using a capacitor, either before or after the 5V transformer to stop the camera shutting down. I am making an assumption that the camera pulls no more than 500mAh @ 5V for the 2 seconds the cigarette lighter is disabled and will only operate so long as the voltage remains above 4V.

I have read a bunch of material around Coulombs and uF, just nothing to tell me what capacitance I need to keep my device powered. I read and tried following this calculation:

500mAh x 3.6C = 1800 Coulombs
1800 / 5V = 360F - this seems an insane capacitance, unless it was meant to be uF.

I'm also pretty certain that as soon as I pull any current from the capacitor, it's voltage is going to drop?

What am I doing wrong here?! Your help appreciated

Andy

[Zero999]

On what basis have you made that assumption?

I think you've confused mAh with mA.

In order to draw 1800 Coulombs in 2 seconds, the current would have to be 900A which is more than a car battery can provide.

If the camera draws 500mA for 2 seconds then that's only 1 Coulomb.

 Of course the voltage across the capacitor will drop. It's better to put the capacitor on the 12V side and let the DC-DC converter (transformers only work for AC) take care of the voltage drop. You should choose a capacitor which will not discharge below the minimum voltage the DC-DC can work from and still give 5V.

[T3sl4co1l]

Check your units!!!

Amperes are a unit of current flow.  Ah are charge, namely 1 amp for 1 hour (3600 s).

A device that uses 500mA will use 500mAh of capacity after one hour, and so on.

A coulomb is an amp-second.

So what you need is 500mA * 2s = 1 C, and 1C/(5V-4V) = 1F.

If it's getting 5V from a device that won't backflow, you can just tack on the capacitor, and it should be alright.  Well, the converter / charger might not be the happiest starting up into 1F... but that's a separate issue.

If it doesn't prevent backflow, then you'll need a diode, otherwise other loads on the accessory circuit will discharge it through the converter.  Which sucks, because if the load needs its 5V, you don't have many options there; you may need an "ideal diode" circuit.

On the upside, 10F+ caps aren't terrifically bulky or expensive, so you can spare plenty of extra.  And for those cold cranking mornings, if applicable.

Tim

[andrew_c]

Of course, you're both right and I've made some grave mistakes in my opening post.

The camera has a small internal 520mAh battery which is always topped up, I only need to keep the 5V input alive to keep the camera switched on - I chose 500mA as an example current to base some calculations off.

So I see, 500mA for 2 seconds is 1 amp-second (or 1C). I read it was then the Coulombs divided by the desired minimum Voltage? Is it in fact the difference between the min and max voltage as Tim has written?

So I'm assuming (again) the camera draws no more than 2.5W on the 5V side. I'd also like to make the assumption that it's a perfectly efficient DC-DC converter. We would be looking at around 210mA on the 12V side. Same calculation...

210mA for 2 seconds = 420mA-second / 0.42C - If I'm then dividing this by the difference in voltage potential I would have to guess a couple Volts.

0.42C / 2 = 0.21F - Actually looks like it's easier to store more "energy" at higher voltages - and cheaper?! The source of the 12V (yeah it's probably more than 14V) is the car battery so a diode would certainly be required. I would be tempted to use a 1n4002 I have on my desk, it's looks to be rated for the current and a manageable forward voltage drop.

Ignore the transformer on the right (DC-DC converter)


So a 0.22F 16V capacitor and a 1n4002 diode?

Andy

[max666]

Haven't checked your calculations, but since the camera has an internal battery, a different approach might be to look for a supply voltage sense pin and add a cap there. A cap there generally only needs to be very small compared to a cap bridging the whole device current. A picture of the PCB top and bottom would help to see if that can be done. The calculations on how biga cap you need would be the same.

[andrew_c]

These are not my photos, but it's a Mobius action camera. I would like to do this for a few of these cameras I have, so an external solution would be preferred.





Cheers, Andy

[max666]

Hmm ... yeah looks like an external solution will be easier :/

Yes, generally it's easier and cheaper to store more energy at higher voltages (SuperCaps might be exceptional):


Your calculation of 0.21F is correct, but that's still an awful lot of capacitance. The DC-DC converter might even work down to as low as 7V, but that would have to be tested. Also the 2s and the 210mA are only guesses so far.
The cigarette lighter is probably not switched separately, otherwise you would get away without the diode, but the 1n4002 should do fine (they are sturdy).
However a 16V rating for the cap seems dangerously low to me, I wouldn't go below 24V, but I will leave it to the automobile experts to decide what voltage rating would be required.

EDIT: Looking for a SuperCap, I found one with 1F @ 5.5V for 0,95€. Since it would be on the 5V side, it wouldn't need much higher voltage rating. Most certainly a lot cheaper than any +16V cap with 0.21F capacitance.
https://www.neuhold-elektronik.at/catshop/product_info.php?cPath=41_51_146&products_id=4558

[mikerj]

Rather than guessing, why not actually measure the power consumption of the camera, the lowest reliable operating voltage and the car voltage waveform during cranking?  You probably don't need 1/2 farad of capacitance on the 12v rail to keep it running.

Also consider the inrush current when such a large cap is connected to a high current supply in a discharged state, the poor little 1N4001 might not be happy.

[andrew_c]

In all fairness I probably am overcomplicating things. I would put £1 on a 25V 1000uF cap working - but you're right, I need to know for sure, and how/why it works.

I don't have a bench power supply to see the minimum voltage the DC converter needs - my last buck converter blew up a week ago. I should be able to pull some details from the board and grab a datasheet. I will measure the current draw and cranking voltage tonight and come back with the value.

I've had a think and it could actually be that the fuse I'm tapped into is part of the Aux circuit and is cut off when the motor is cranked. There might be a fuse which stays on permanently with enough voltage for the DC converter to stay alive.

Cheers,
Andy

[SeanB]

Lighter normally is a switched aux. Try the side lighting circuit ( easier to get to in the dashboard) as a supply. It does not get turned off with starting, or look for an IGN1 which powers the ECU and such which has to be there.

The diode and 1000uF 35V capacitor is a good idea, though I would also recommend a series 1R 1W resistor in series with the diode to tame the inrush current somewhat, so the spikes on the battery voltage will be clamped slightly.

[andrew_c]

Resolved! So the simplest solution was the winner. Unfortunately it didn't involve using a capacitor (but at least I learned something).

I was using F15 for the radio before, moved it one to the left which is the ECU and everything works like I want the voltage in the car does drop considerably when cranking, but not enough to drop the 5V to the camera. F17 might have been a better choice but it was easier to use F14 - I always carry spare fuses in the car just-in-case.

Fusebox with tap


Fusebox diagram


Thanks a lot for everyone's help!

Andy

[max666]

hehehe, many roads lead to Rome