How to best reduce amperage from car-cigar-lighter adapter?

It will ar least be fused if you wire in series an inline fuse holder with a 1A fuse in it.

Ah, ok. Putting a fuse-holder and a 1-amp (fast-blow) fuse makes sense.

Are we saying that is the ONLY thing needed? Or should there be something MORE in the constructed circuit than just the fuse?

TIA again...

Dave

snipped-for-privacy@gmail.com wrote:

It depends on exactly how likely the item you're trying to protect is to short out (or otherwise fail in an "overcurrent" mode)... and the cost of replacing it!

A 1A fuse takes "almost forever" at 1A to actually blow; even at 2A it's often "many seconds" (I'm being vague here because the details depend on the fuse type -- fast, slow blow, etc.) -- even at 10A it's often in the "many milliseconds" range. In many cases, that lets enough energy through to damage other devices besides whatever it was that failed in the first place --> the purpose of fuses is more to protect *the car* than *the device*. For that purpose, any old 1A fuse is fine.

A somewhat more sophisticated approach is to build a current-limited power supply... you get ~12V up to 1A, and after that the output voltage is reduced such that no more than 1A is available. You can find plenty of schematics for this (it just takes a handful of components) if you Google something like "current limiting power supply schematic" or look in a standard text such as The Art of Electronics. A somewhat more aggressive approach -- that's almost as easy to implement -- is a "foldback current limiting" power supply that, once you hit 1A, will "trip" and requires the current draw to drop to much less (say, 250mA) before it'll "reset" and provide up to 1A again. (A small challenge with foldback power supplies, though, is that short, high-current spikes can trip it, even though this is probably OK -- you have to fiddle a little with foldbacks to make sure they don't respond *too* quickly.)

When large powers and expensive equipment is in question, the equipment's overall architecture starts to get involved in providing failure protection. For instance, if you have 2 1kW RF amplifiers, you could just put the two in parallel to obtain 2kW, but if one fails the other will most likely end up being shorted and be destroyed as well -- fancy hybrid coupling transformers are used to insure that this doesn't happen.

You just need to properly fuse your device. Adaptors for the vehicle power port (AKA cigarette lighter port) are available with a built in fuse. Ask the guy at Rat Shack for a fused cigarette lighter adaptor. (After he tries to sell you a cell phone.) Tom

Hi Dave

When powering something from a cigar lighter socket, I would always use a 'small' fuse in the cigar PLUG first. Dont put the fuse at the 'end' of your target device lead. I'll assume its thin and so could melt if a short occured. I'd also put a diode in series (ok for low current loads) with the power lead to protect your target device from an accidental reverse polarity. For me, I think thats all I'd bother with. Hope that helps

Steve Balstone

The device will take only the amperage it needs. You need nothing to protect the device, as you have described it.

The purpose of a fuse in the wire/adapter plug is to protect the *wire*. It has nothing to do with protecting the device. You want to protect the wire from overheating, in the event there is a short circuit in the device or connection to it. It is not that the wire is valuable - it is that a hot wire can burn other things/start a fire.

Ed

Little point. It'll cost you more to make your own and it won't save any amps unless you use switch-mode which is beyond most beginners' abilities. Quie a few adaptors are probably switch-mode already anyway.

Graham

God yes.

Frankly if you need to ask, you shouldn't be doing this.

Graham

So everyone says you need a properly rated fuse, and you do.

If you want to power an 800 mA device from a car battery power source, you must first realise that a car battery is NOT 12 volts. It will be 13.8 volts. So you will need to regulate the voltage to a proper 12 volts. You can buy a regulator chip. A 3 pronged device that you attach to your

13.8 DC source. Make a little circuit board using that strip pre drilled stuff. The live / positive lead will go from the car positive, to the front left prong. Heat sink at back, writing on front. The regulated 12 volts will now be restricted to 1 Amp, and it comes out at the right prong, you will attach the positive out to the right prong, but then you need to restrict the current to 800 mA. You must attach the negative wire to the centre /=/ negative, If you are happy with 12 volts at 1 Amp you just connect the negative wire to the centre prong and there you go. For lower current you will need ( a 'PROPERLY CALCULATED' current limiting resistor.) The centre prong is common/ to us/ this means the negative. A resistor attached to the 12 volt 1 Amp out 'right prong' is to be fixed /shorted the centre prong. You still take your power out from the left prong, its just shorted to negative / centre prong via a resistor. The current limiting resistor value is calculated using 'Ohms law'. (Don't panic) So now you are using the regulator you will soon have a nice steady 12 volts. you will have bought a 1 Amp positive chip, with a number something like LS7812C? and be getting 12 volts / 1 Amp. So now using Ohms law you need to work out the resistor value. The on line Ohms law calculator is here, so its not going to be a nightmare,,,

I did your suggested math. Your Answers: voltage 12 V current 800 mA resistance 15 Ohms wattage 9.6 W

So you need a resistor of 15 Ohms to reduce you 1 Amp current to 800mA. also give some scope on the Wattage of the resistor. Use at least a 12 Watt or bigger resistor. I will repeat myself here. This resistor is taken from the positive out / left lead and fixed to the centre 'negative lead. This shorting out is what causes current limiting. Not the most power saving way of doing it bit it will work and probably cost you less than £5 to knock up. Similar circuitry is used for building posh battery chargers. Best of luck hope this helped, with the ohms law calculator you will be able to work out the correct limiting resistor value for any appliance.

SORRY chum, did get my rights 'n' lefts mixed up. Should read You still take your power out from the RIGHT prong, its just shorted to negative / centre prong via a resistor.

You still take your power out from the RIGHT prong, its just shorted to negative / centre prong via a resistor.

Sometimes.

Also it will be 9 volts and 40-50 volts.

Graham

it's not always 13.8 either, that'll fluctuate.

for a 13.8V source to 12V it'll need to be a low drop-out type

with a 7812 that's not going to work. (they want atleast 14v on the input) and perform better with some headroom. so really 15V would be where to start.

with a low drop-out regulator that's still not going to work without a heatsink and capacitors....

LS???

DYM LM7812C ???

and that's wrong too... if the 7812 performed an exact 1A current limit (it doesn't) installing that "room heater" resistor would use up 80% of that capacity and leave them only 200ma to power their device with.

or 1.8 (ish) volts at 800 mA (and more volts at lower currents) if the common terminal is disconnected from the -ve terminal of the battery and output is taken from it.

yeah, but those guys get the circuit right.

Ohms law is easy enough to figure with a cheap pocket calculator, (or without when the need arises)

Bye. Jasen

Its a funny old world, no one wants to tell the bloke anything that will help him, then when an enthusiastic amateur suggest a way the criticism starts. If your all so clever to know how now brown cow... why not just tell the chap in the first place?

Sounds like interesting plan.

Probably.

A typical scenario.

Wrong on two counts. First is that you are almost guaranteed not to get 12V from that socket. Car voltages are all over the place, though it's usually somewhere around 14V.

Second is that in general you don't need to worry about reducing the amount of available current. Quite the opposite. You need to make sure that your source can supply the required current. So instead of thinking:

"I need to make sure that the current does not exceed 800 ma."

You need to think:

"I need to make sure that I can supply at least 800ma at 12V."

Since you've properly surmised that you have 15-20A of current available you should have no problems.

Now the only overcurrent issue you need to worry about is if somehow you get a short in your circuitry. Then the full weight of 15-20A of current becomes available. And that amount of current can fry a lot of stuff. So you may want to put a 1-2A inline fuse with your circuit so that if you get a short, that fuse will blow protecting your circuit in some fashion.

You don't need to reduce the current. You need to regulate the voltage.

I wouldn't. I'd put in at least a fuse.

Now back to voltage regulation. The simplest way to do it is to use a zener diode regulator. However I think 800mA is pushing even a high powered zener. So a zener driving a pass transistor is probably the ticket. You can find such a circuit here:

Note that all you need is the circuitry starting from C1 and working your way to the right. You want to get a 12.6V zener. A 12V zener will regulate at 11.4V and a 13V zener will regulate at 12.4V.

Put your inline fuse between your cable and the C1 cap and you'll be good to go.

Hope this helps,

BAJ