Limiting DC Current for Car Battery Use

[John1397]

Is there any way to take 12 volts dc from a car battery and have it go thru some components so that what comes out amounts to still 12 volts dc, but is limited to only 20 ma output?

John

 

[Danger]

I'm no expert by any means, but my first thought is to put a 20mA fuse in the circuit. I've never had a problem with that sort of thing, though; nothing that I've ever hooked up to my car (and there have been some seriously weird ones) ever drew more current than it needed. The fuse is more to prevent fire in the event of something overloading the capacity of the wires and/or solder junctions.

 

[phyzguy]

Sure. The simplest way is with a fuse, but of course the fuse is destroyed if the 20 mA is exceeded. There are many current limiting circuits, try a Google search on current limiting, or try this wiki site.

Maybe it goes without saying, but you do realize that if the load attempts to exceed the 20mA limit, the circuit will need to drop the voltage below 12V in order to limit the current.

 

[Danger]

Thanks, Phyzguy. I was out nuking my supper with the intention of editing and telling John to wait for an expert to give him better advice. I'm glad that you showed up before he plugged anything in.

 

[vk6kro]

There are many circuits for current limiters.

Have a look here for some of them:
https://www.google.com.au/search?q=...po6IB7rmgJgE&ved=0CAoQ_AUoAQ&biw=1024&bih=566

Here is one circuit that works

You have to calculate a value for the sense resistor that will make it drop 0.6 volts at the required current.
In this case, it will be about 30 ohms.
Using Ohm's Law,
Resistance = Voltage / current = 0.6 volts / 0.02 amps = 30 ohms.

But, as Physguy mentioned, you can't have it both ways. If you supply 12 volts to a 100 ohm resistive load, it will draw 120 mA.
Using Ohm's Law,
current = voltage / resistance
current =12 volts / 100 ohms = 0.12 amps = 120 mA


To make it draw only 20 mA you must reduce the voltage to 2 volts.
Using Ohm's Law,
current = voltage / resistance
current = 2 volts / 100 ohms = 0.02 amps = 20 mA

Ideally, what you need is a circuit that will give you a steady 12 volts out except if the load resistance is low enough that it will draw more than 20 mA. In this case, the output voltage must reduce enough that the load cannot draw more than 20 mA

 

[John1397]

There are many circuits for current limiters.

Have a look here for some of them:
https://www.google.com.au/search?q=...po6IB7rmgJgE&ved=0CAoQ_AUoAQ&biw=1024&bih=566

Here is one circuit that works

You have to calculate a value for the sense resistor that will make it drop 0.6 volts at the required current.
In this case, it will be about 30 ohms.
Using Ohm's Law,
Resistance = Voltage / current = 0.6 volts / 0.02 amps = 30 ohms.

But, as Physguy mentioned, you can't have it both ways. If you supply 12 volts to a 100 ohm resistive load, it will draw 120 mA.
Using Ohm's Law,
current = voltage / resistance
current =12 volts / 100 ohms = 0.12 amps = 120 mA


To make it draw only 20 mA you must reduce the voltage to 2 volts.
Using Ohm's Law,
current = voltage / resistance
current = 2 volts / 100 ohms = 0.02 amps = 20 mA

Ideally, what you need is a circuit that will give you a steady 12 volts out except if the load resistance is low enough that it will draw more than 20 mA. In this case, the output voltage must reduce enough that the load cannot draw more than 20 mA

Looks like this would work would be nice if it came with resistor values diode type and transistor specifications as some of us folks are not that bright figuring this out ourselves.

 

[NascentOxygen]

The power ratings of components are also important in these power circuits.

 

[Runei]

What is it that you want 20 mA to go through?

 

[berkeman]

Looks like this would work would be nice if it came with resistor values diode type and transistor specifications as some of us folks are not that bright figuring this out ourselves.

That circuit will only give you about 10V output, though. You mentioned you still want 12V at the output, right? What are the specifications for the output? What tolerance can you tolerate in the output voltage.

If you want a true 12V at the output along with the 20mA current limiting, you will probably need to use a DC-DC converter, in a boost-buck or SEPIC topology. Also, 20mA is a pretty low current limit value -- what is the application?

 

[John1397]

From a 12 volt car battery you would have 300 amps all I want is 12 volts at 20 ma maximum the load will draw a maximum of 12 volts at 1 amp, but I do not want it drawing 1 amp only 20 ma and I do not care if voltage drops a lot this will not affect the circuit as the unit I am powering goes into shut down mode if the voltage drops below 11 volts, but still continues to work in sleep mode with only a small amount of voltage and amps.

 

[berkeman]

From a 12 volt car battery you would have 300 amps all I want is 12 volts at 20 ma maximum the load will draw a maximum of 12 volts at 1 amp, but I do not want it drawing 1 amp only 20 ma and I do not care if voltage drops a lot this will not affect the circuit as the unit I am powering goes into shut down mode if the voltage drops below 11 volts, but still continues to work in sleep mode with only a small amount of voltage and amps.

Well, I suppose you could just put a 25 Ohm resistor in series wiith the feed to your device. That will drop 0.5V at 20mA, to give you 11.5V at 20mA for your device. If your device tries to draw more, that will drop the voltage below 11V and your device will go into low-current mode.

0.5V = 20mA * 25 Ohms

Will that do what you want?

 

[Runei]

From a 12 volt car battery you would have 300 amps

Depends on what you connect? If you device wants to draw 20 mA, it will draw 20 mA. Just because the battery is capable of providing 300 amps, doesn't mean it will.

 

[jim hardy]

A little more information about your load would be helpful.
How close to 12 volts do you need voltage to be?
How close to 20 ma do you need the limit to be?

Your car battery is around 12-13 volts when parked and 14-15 while driving.

VK6 gave the basic approach in his post above - measure the current and control it with some active device. Often one sees a PNP transistor instead of NPN in his circuit; that allows lower voltage drop and is used in "Low Drop Out Voltage Regulators" - a search on that term will be instructive..

TI's LM2900 series is automotive LDO regulators with some nice features .
The electronics manufacturers learned about the need for these the hard way:
Reverse polarity protection: in case somebody hooks up the jumper cables backward it doesn't fry your regulator and its expensive load.
Overvoltage protection: When somebody leaves a battery terminal loose it will make intermittent connection. When the connection opens it's called a "load dump" and your alternator voltage spikes high before the car's voltage regulator can reduce excitation. That overvoltage wrecks non-automotive rated electronics in the car.
Temperature range to 125 degC. Under-hood is extreme environment, as is underdash near the heater .

Here's TI's LM2937 datasheet. It's available in 12 volt, or 10 volt if your load can work with that.
http://www.ti.com/lit/ds/symlink/lm2937.pdf
But it'll deliver around an amp into a short circuit until it gets hot and shuts itself down for thermal protection.
I often suggest an automotive regulator to hobbyists because of its accident-resisting features.
TI has a lot of new regulators since I last looked, so peruse their selection guides. "Zfoldback limiting" might be good for your application, sp read up on that.

If you need to get sophisticated with this project take a look at "High Side sensing"
http://www.maximintegrated.com/app-notes/index.mvp/id/746
http://www.analog.com/en/specialty-amplifiers/current-sense-amplifiers/adm4073/products/product.html

I envy you young fellows the sophisticated high-tech IC's you have available. But I am also aware there's so much of it as to be overwhelming.

Have fun, learn by doing.

old jim