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.
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#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).
#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.
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#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.
#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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FAQ

TL;DR: During cranking, a ~12.6V battery can sag while alternators target ~14.5V; “Adding just a capacitor won't have much effect.” Use a diode-isolated aux battery or a large supercap to hold a 2A load. [Elektroda, stephen Van Buskirk, post #21667200]

Why it matters: This FAQ helps car DIYers keep small 12V devices alive during engine start without resets or brownouts.

Typical alternator output is ~14–14.8V; resting battery ~12.5–13.2V. Expect voltage dips during cranking. [Elektroda, stephen Van Buskirk, post #21667205]
Supplying 2A for “a few seconds” needs a very large capacitor; a small auxiliary battery is more practical. [Elektroda, Frank Bushnell, post #21667201]
Standard silicon diodes drop about 0.9–1.2V at 2–3A; size for load plus charging current. [Elektroda, stephen Van Buskirk, post #21667205]
Schottky diodes reduce forward drop versus standard diodes, easing heat and voltage loss. [Elektroda, Steve Lawson, post #21667207]
Paralleling diodes without ballast can cause thermal runaway; prefer one higher‑current device. [Elektroda, Frank Bushnell, post #21667204]

Quick Facts

Typical alternator output is ~14–14.8V; resting battery ~12.5–13.2V. Expect voltage dips during cranking. [Elektroda, stephen Van Buskirk, post #21667205]
Supplying 2A for “a few seconds” needs a very large capacitor; a small auxiliary battery is more practical. [Elektroda, Frank Bushnell, post #21667201]
Standard silicon diodes drop about 0.9–1.2V at 2–3A; size for load plus charging current. [Elektroda, stephen Van Buskirk, post #21667205]
Schottky diodes reduce forward drop versus standard diodes, easing heat and voltage loss. [Elektroda, Steve Lawson, post #21667207]
Paralleling diodes without ballast can cause thermal runaway; prefer one higher‑current device. [Elektroda, Frank Bushnell, post #21667204]

Can a single capacitor keep my 2A device alive during cranking?

Not reliably. The starter can drag voltage for several seconds. Delivering 2A that long requires a very large capacitor. A small secondary battery or supercap bank behind a diode is more effective for ride‑through. “Adding just a capacitor won’t have much effect.” [Elektroda, stephen Van Buskirk, post #21667200]

How large would the capacitor need to be for a few seconds at 2A?

Very large—on the order suggested as impractical in the thread. That is why contributors recommend a small auxiliary 12V battery instead, which charges from the vehicle and holds up the load during cranking through a diode. [Elektroda, Frank Bushnell, post #21667201]

Is a secondary battery better than a capacitor here?

Yes. A small motorcycle or alarm lead‑acid battery can supply 2A for several seconds easily. Isolate it with a diode so starter current cannot pull it down. It recharges from the car once the engine runs. [Elektroda, Frank Bushnell, post #21667201]

What diode size and placement should I use?

Place the diode between the main battery/alternator and the aux capacitor or secondary battery plus your devices. Size it to pass your 2A device current and the capacitor charging current—about 3A total in this case. [Elektroda, stephen Van Buskirk, post #21667200]

Which diode type minimizes voltage loss and heat?

Use a Schottky diode with adequate current rating. It has a lower forward drop than standard silicon diodes, helping your devices see higher voltage and reducing heating. “Consider using a Schottky low forward drop high current diode.” [Elektroda, Steve Lawson, post #21667207]

Does a 0.6V diode drop matter for 12V electronics?

Usually not. Automotive devices are designed for a wide range and often run below 12V. With charging at ~14.5V and typical device tolerance, the small drop still leaves enough headroom in normal operation. [Elektroda, stephen Van Buskirk, post #21667200]

Can I parallel small rectifier diodes to handle more current?

Avoid it. Manufacturing differences make current share uneven. One diode can hog current, heat up, drop voltage further, and fail. Use a single higher‑current diode instead, with a heatsink if needed. [Elektroda, Frank Bushnell, post #21667204]

If I must parallel diodes, how do I balance them?

Add small series resistors as ballast. Forward voltage falls about 2.1 mV/°C, so temperature can cause runaway without resistors. Example math in the thread suggests ~42 mΩ per leg for three diodes at 3A. [Elektroda, Steve Lawson, post #21667208]

Will the diode drop prevent charging a small secondary battery?

It reduces charge voltage across the aux battery, slowing charge. With ~14.5V system and ~0.9V drop at 2A, the aux may see ~13.6V, which is sub‑optimal for standard lead‑acid. Expect slower charging performance. [Elektroda, stephen Van Buskirk, post #21667205]

Is a MOSFET “ideal diode” a good idea here?

Not for this case per the thread author who proposed and then withdrew it. The scheme that works for reverse‑polarity protection had a threshold making it unsuitable for maintaining the aux supply here. [Elektroda, stephen Van Buskirk, post #21667205]

What about using a 5F “car audio” capacitor and an SCR?

One suggestion is a 5F or larger audio‑grade capacitor and an SCR that disconnects during cranking, preserving charge for your devices. It reduces diode heating and keeps the cap energized until the starter event ends. [Elektroda, Mark Harrington, post #21667210]

How much current can a typical alternator deliver?

The discussion cites alternators designed for up to about 100A to simultaneously run vehicle systems and recharge the battery. This capacity explains why loads bounce back quickly once the engine is running. [Elektroda, Frank Bushnell, post #21667206]

What small battery type works as the auxiliary source?

Use a compact 12V lead‑acid, such as a motorcycle or alarm system battery. It easily handles a 2A accessory load during cranking and stays charged via the vehicle system through an isolating diode. [Elektroda, Frank Bushnell, post #21667201]

How do I build a simple diode‑isolated aux battery setup?

Wire a small 12V aux lead‑acid battery in parallel with your devices.
Insert a diode from the car battery/alternator to the aux battery/device node, oriented to allow charging.
Secure wiring; add a heatsink if the diode runs warm. [Elektroda, Frank Bushnell, post #21667201]

Is a 1N4001 rectifier okay for this?

No. A 1N4001 is undersized for the charging and load currents here. Use higher‑current parts like the 1N540x series or an appropriately rated Schottky diode. [Elektroda, Rohit Dubla, post #21667202]

What forward‑drop should I expect at 2–3A through common diodes?

Standard silicon rectifiers often drop about 0.9–1.2V at 2–3A. This affects both device voltage and heat in the diode, so choose ratings and cooling accordingly. [Elektroda, stephen Van Buskirk, post #21667205]

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