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

[David H. Cook]

Rather than purchase the specific cigar-lighter-adapter power-adaptor
for (any)
specific device, I'd like to contruct my own. (e.g. by getting
necessary stuff
at Radio-Shack or where-ever)

So, let's say I've got an adapter plugged into the cigar-lighter socket
in a
car. I'd guess that this socket is protected by a fuse of 15 or 20
amps, right?
And, bare wires at the other end. And, let's say I've got the correct
geometry
of a 'tip' that will fit some product (say, a radio or scanner, or
whatever and
that it happens to want a voltage of 12-volts).

So, if my target-device wants/needs, say, 800 milli-amps, at 12-volts,
then
I want to leave the voltage alone, and just reduce the (possible)
current that
I allow to get to the device from the 15 or 20 amps down to (roughly)
1 amp, right?

So, what electrical components do I need buy and how do I connect them
(e.g. in series or in parallel) to safely reduce the possible current
to such a
target device?

(Or can I safely assume that the target device will have a FUSE and
protective
curcuitry of its own?)

TIA...

Dave

 

[[email protected]]

(Or can I safely assume that the target device will have a FUSE and
protective
curcuitry of its own?)

TIA...

Dave

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

 

[David H. Cook]

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

 

[Joel Kolstad]

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?

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.

 

[Tom Biasi]

Rather than purchase the specific cigar-lighter-adapter power-adaptor
for (any)
specific device, I'd like to contruct my own. (e.g. by getting
necessary stuff
at Radio-Shack or where-ever)

So, let's say I've got an adapter plugged into the cigar-lighter socket
in a
car. I'd guess that this socket is protected by a fuse of 15 or 20
amps, right?
And, bare wires at the other end. And, let's say I've got the correct
geometry
of a 'tip' that will fit some product (say, a radio or scanner, or
whatever and
that it happens to want a voltage of 12-volts).

So, if my target-device wants/needs, say, 800 milli-amps, at 12-volts,
then
I want to leave the voltage alone, and just reduce the (possible)
current that
I allow to get to the device from the 15 or 20 amps down to (roughly)
1 amp, right?

So, what electrical components do I need buy and how do I connect them
(e.g. in series or in parallel) to safely reduce the possible current
to such a
target device?

(Or can I safely assume that the target device will have a FUSE and
protective
curcuitry of its own?)

TIA...

Dave

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

 

[[email protected]]

Rather than purchase the specific cigar-lighter-adapter power-adaptor
for (any)
specific device, I'd like to contruct my own. (e.g. by getting
necessary stuff
at Radio-Shack or where-ever)

So, let's say I've got an adapter plugged into the cigar-lighter socket
in a
car. I'd guess that this socket is protected by a fuse of 15 or 20
amps, right?
And, bare wires at the other end. And, let's say I've got the correct
geometry
of a 'tip' that will fit some product (say, a radio or scanner, or
whatever and
that it happens to want a voltage of 12-volts).

So, if my target-device wants/needs, say, 800 milli-amps, at 12-volts,
then
I want to leave the voltage alone, and just reduce the (possible)
current that
I allow to get to the device from the 15 or 20 amps down to (roughly)
1 amp, right?

So, what electrical components do I need buy and how do I connect them
(e.g. in series or in parallel) to safely reduce the possible current
to such a
target device?

(Or can I safely assume that the target device will have a FUSE and
protective
curcuitry of its own?)

TIA...

Dave

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

 

[ehsjr]

Rather than purchase the specific cigar-lighter-adapter power-adaptor
for (any)
specific device, I'd like to contruct my own. (e.g. by getting
necessary stuff
at Radio-Shack or where-ever)

So, let's say I've got an adapter plugged into the cigar-lighter socket
in a
car. I'd guess that this socket is protected by a fuse of 15 or 20
amps, right?
And, bare wires at the other end. And, let's say I've got the correct
geometry
of a 'tip' that will fit some product (say, a radio or scanner, or
whatever and
that it happens to want a voltage of 12-volts).

So, if my target-device wants/needs, say, 800 milli-amps, at 12-volts,
then
I want to leave the voltage alone, and just reduce the (possible)
current that
I allow to get to the device from the 15 or 20 amps down to (roughly)
1 amp, right?

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

 

[Eeyore]

Rather than purchase the specific cigar-lighter-adapter power-adaptor
for (any) specific device, I'd like to contruct my own. (e.g. by getting
necessary stuff at Radio-Shack or where-ever)

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

 

[Eeyore]

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?

God yes.

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

Graham

 

[Guest]

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,,,

http://www.cvs1.uklinux.net/cgi-bin/calculators/ohms_law.cgi

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.

 

[Guest]

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.

 

[Eeyore]

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.

Sometimes.

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

Graham

 

[jasen]

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.

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

You can buy a regulator chip. A 3 pronged device that you attach to your
13.8 DC source.

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

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.

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....

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,

LS???

DYM LM7812C ???

http://www.cvs1.uklinux.net/cgi-bin/calculators/ohms_law.cgi

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

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.

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.

yeah, but those guys get the circuit right.

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.

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

Bye.
Jasen

 

[Guest]

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?

 

[Byron A Jeff]

Rather than purchase the specific cigar-lighter-adapter power-adaptor for
(any) specific device, I'd like to contruct my own. (e.g. by getting
necessary stuff at Radio-Shack or where-ever)

Sounds like interesting plan.

So, let's say I've got an adapter plugged into the cigar-lighter socket in a
car. I'd guess that this socket is protected by a fuse of 15 or 20 amps,
right?
Probably.

And, bare wires at the other end. And, let's say I've got the
correct geometry of a 'tip' that will fit some product (say, a radio or
scanner, or whatever and that it happens to want a voltage of 12-volts).

A typical scenario.

So, if my target-device wants/needs, say, 800 milli-amps, at 12-volts, then I
want to leave the voltage alone, and just reduce the (possible) current that
I allow to get to the device from the 15 or 20 amps down to (roughly) 1 amp,
right?

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.

So, what electrical components do I need buy and how do I connect them
(e.g. in series or in parallel) to safely reduce the possible current
to such a target device?

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

(Or can I safely assume that the target device will have a FUSE and
protective curcuitry of its own?)

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:

http://www.electronics-tutorials.com/basics/power-supply-regulated.htm

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