Using a Capacitor to Prevent Voltage Drop for 2A Load During Car Engine Cranking

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#1 21667199 20 Mar 2013 11:17

Nick Durkin Nick Durkin

Anonymous

Post #1
21667199 20 Mar 2013 11:17

I have a couple small things running off of the car battery in my car. When I start the car, the battery voltage drops during the cranking period. This sometimes causes the things I have running off of the battery to stop working for a fraction of a second. This is an issue for me. I think I have a solution which would be to somehow use a capacitor to make up for that loss of power while the car is cranking. I don't know anything about capacitors or how I would use it so if you guys could give me some guidance that would be great. I guess I just need to know where to place it and what the rating should be. In case you need to know, the things I have running off of the battery are not pulling any more than two amps of current.

Thanks.
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#2 21667200 02 Aug 2013 14:36

stephen Van Buskirk stephen Van Buskirk

Anonymous

Post #2
21667200 02 Aug 2013 14:36

Adding just a capacitor won't have much effect as the load that's draining the battery would just drain the capacitor. If you could tolerate a diode between the battery and cap that would work, so long as the cap has sufficient capacity. The diode passes power from the battery and alternator to the cap and devices, but won't allow the lower starting voltage to drain the battery.

The diode needs to be sized so it will pass the current you need plus the capacitor charging current, probably about 3A.

Most diodes drop about 0.6V. This shouldn't be a problem. The battery puts out about 12.6V, unless it's really loaded, and the alternator puts out about 14.5V. Devices in a car need to tolerate voltages in a wider range and should work below 12v.
#3 21667201 02 Aug 2013 17:27

Frank Bushnell Frank Bushnell

Anonymous

Post #3
21667201 02 Aug 2013 17:27

I agree with Stephen in principle, but winter starting can take few seconds, and to supply two amps for a few seconds, a very big capacitor would be needed.

So I suggest a secondary battery rather then a capacitor. A very small motorcycle battery, or lead acid battery as supplied for alarm systems would do the job quite adequately.
It will be charged from the car's main system, and the diode will prevent the secondary battery voltage dipping due to the car starter current.

Circuit attached to show the principle.
#4 21667202 03 Aug 2013 08:38

Rohit Dubla Rohit Dubla

Anonymous

Post #4
21667202 03 Aug 2013 08:38

1N4001 might not be suited to handle the current - maybe parallel two or three or use bigger diodes (1N4007 upwards, or the IN5406/5407 types).
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#5 21667203 03 Aug 2013 10:52

stephen Van Buskirk stephen Van Buskirk

Anonymous

Post #5
21667203 03 Aug 2013 10:52

It occurs to me that if you used a second small 12v battery the diode drop might interfere with charging. There is a simple circuit that reverse biases the gate to an mosfet and creates a diode with virtually no drop at high current. (Mosfets pass current in either polarity if they are biased correctly.) If there's interest I can draw it. I used it for reverse voltage protection for a motor driver that drew 20A.
#6 21667204 03 Aug 2013 14:02

Frank Bushnell Frank Bushnell

Anonymous

Post #6
21667204 03 Aug 2013 14:02

Well spotted, Rohiit, my mistake. A 10 Amp diode should be sufficient, perhaps with heatsink.

Diodes do not generally work well in parallel, as manufacturing differences result in slightly different volt drops, so current does not divide evenly.

Stephen, I would have thought that as as the car charging system usually provides about 14V, the secondary battery would receive quite sufficient charge to sustain it for it's purpose. However, for a lower volt drop, a Schottky diode would be better.

However, I would be interested to see your Mosfet solution.
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#7 21667205 03 Aug 2013 16:59

stephen Van Buskirk stephen Van Buskirk

Anonymous

Post #7
21667205 03 Aug 2013 16:59

1N4007 and 1N4001 have the same current rating, just different maximum reverse voltages. Note than a 1A 1N400* drops 1.1V @ 1A according to the data sheet. As with the 1N400* the last digit of the 1N540* gives the voltage. It drops 1.2V at 3A. At 2A it drops about 0.9V.

The output voltage for alternators is generally 14.5 to 14.8V. A common 12V battery actually has about 13.2V output with no load. The voltage varies with the cell type, charge and load. The charge rate varies with the voltage and the charge. Assuming 14.5V and a diode drop of 0.9V the differential charge voltage is 13.6, not all that good for a standard battery and well below the recommended charge voltage. Remember the diode passes 2A to drive the load and whatever current the battery draws when it's charging so the diode will always drop at least 0.9V, assuming the load is always on.

Would a very large capacitor work better?

As for my suggestion about a mosfet, I had my head in a bucket - forget I said anything. The design is really good about reversed voltage but has a threshold that makes it unworkable for this.
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#8 21667206 03 Aug 2013 18:45

Frank Bushnell Frank Bushnell

Anonymous

Post #8
21667206 03 Aug 2013 18:45

Stephen, as far as I know, a car alternator is designed to provide up to about 100A to simultaneously run systems and charge the battery.

The new 12V battery in my car shows about 12.5 volts at rest when I switch on (ACCS only, so almost no load), so anything above that voltage will charge it. Soon after starting the engine, dependent on load, it rises to and remains between 13.8 and 14 volts.

So I am very certain that the lower charging voltage for the secondary battery in my circuit will merely result in a lower charge rate, but as it is a very much smaller battery for a very much smaller load, I think this will cause no problem.
#9 21667207 04 Aug 2013 11:51

Steve Lawson Steve Lawson

Anonymous

Post #9
21667207 04 Aug 2013 11:51

0.6V is merely an approximation and is the typical voltage for very low current. At higher currents, the voltage is higher, more like 1 to 1.5 Volts at 2 to 3 Amps. Consider using a Schottky low forward drop high current diode.
#10 21667208 04 Aug 2013 12:08

Steve Lawson Steve Lawson

Anonymous

Post #10
21667208 04 Aug 2013 12:08

If you parallel diodes it's wise to include a small series resistance to balance out the current. Because diodes do not have a linear transfer curve and because the forward voltage changes with temperature (the voltage drops as the temperature goes up), thus, one of the diodes will start to conduct before the others and will heat up, causing it's forward voltage to drop which will channel even more current into that diode causing it to heat up more -- thus thermal runaway and silicon smoke!

Chose a resister that will offset the potential voltage difference between the paralleled diodes. The forward voltage of a Si diode will drop by about 2.1 mV/°C so say the anticipated temp increase is around 20°C -- 2.1 x 20 = 42mv. R = V/I, so for 3 Amps and three 1N4001 (I would use 1N4004's or above), that would be R = 42mv/(3Amps/3) = 0.042Ω (i.e. one 42mΩ resistor in series with each diode -- which could be achieved with a length of smaller gauge wire -- just make sure the gauge is rated to handle that amount of current).
#11 21667209 04 Aug 2013 13:52

Frank Bushnell Frank Bushnell

Anonymous

Post #11
21667209 04 Aug 2013 13:52

Steve, I agree with your basic analysis of parallel diodes (up in smoke).

But isn't the easiest solution to just use one big diode, with heatsink if required ?
#12 21667210 04 Aug 2013 15:52

Mark Harrington Mark Harrington

Anonymous

Post #12
21667210 04 Aug 2013 15:52

Use a 5 (or Higher value ) farad Capacitor such as they use in automobile top end hi fi installations

http://compare.ebay.co.uk/like/251149965080?v...&ff3=1&ff11=ICEP3.0.0&ff12=67&ff13=80&ff14=65

That’s the easy answer with Rohits recommendation on the diode size and type Or ( Far better Idea ) use a SCR Higher current handling capabilities with minimum holding current such that when the ignition is turned it fires the SCR by applying voltage to the gate When the current drops on the battery side(Ignition Start)

The minimum holding current comes into play switching off the SCR leaving the cap charged thus ensuring you still have supply left on the cap to keep other electronics live and youve solved the heat dissipation problem at the same time

See below for typical SCR I would use in this application

http://www.crydom.com/en/Products/Catalog/Adv...%20Series%2050-100%20Amp%20SCR/Diode%20Module

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#13 21667211 04 Aug 2013 22:24

Steve Lawson Steve Lawson

Anonymous

Post #13
21667211 04 Aug 2013 22:24

Hi Frank: I was commenting on Rohit's suggestion to parallel 1N4001 rectifiers (i.e. the reason why I placed my comment beneath his comment).
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Topic summary

The discussion addresses preventing voltage drop during car engine cranking for a 2A load powered by the car battery. Using a capacitor alone is insufficient because the load drains the capacitor quickly. A diode placed between the battery and capacitor can isolate the capacitor, allowing it to supply power during cranking without discharging back into the battery. The diode must handle at least 3A and typically causes a voltage drop of 0.6 to 1.5V depending on current, which may affect charging voltage. Schottky diodes are recommended for their lower forward voltage drop. Paralleling diodes is generally discouraged due to uneven current sharing and thermal runaway risks, but if done, small series resistors can help balance current. An alternative solution is using a secondary small 12V battery (e.g., motorcycle or alarm system battery) with a diode to maintain charge and supply power during cranking. A large capacitor (5 farads or higher), commonly used in automotive audio systems, can also help but may require additional circuitry such as an SCR to manage charging and discharging efficiently. MOSFET-based diode replacements with near-zero voltage drop were mentioned but later reconsidered. Overall, the best practical approach involves a diode-isolated secondary battery or a large capacitor with appropriate current handling and isolation components to maintain stable voltage during engine start.
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Using a Capacitor to Prevent Voltage Drop for 2A Load During Car Engine Cranking

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