Need to reduce voltage using as few and as small components as possible

[TerryMathews]

I'm going from 12V (cigarette lighter power) to 1.5V. Original application was AAA battery, so I don't need to limit current. I need a way to step it down using as few components and as few wires as possible. Ideally a discrete component with an in and an out, but I know that's not too likely.

 

[FrankSchwab]

How much current do you need? Assuming that it's fairly low, you might want to try an
LM317

It's a three-terminal device with a maximum input voltage around 30 V, and a minimum output voltage of 1.2V. You can certainly buy one at Radio Shack.

The datasheet shows a typical hookup diagram. Leave out R1, R2, C2, and connect Vout directly to Adj. It should give you a steady, regulated 1.2V. Assuming your device pulls 100 ma, the LM317 will be dissipating (13.8V - 1.5V) * 0.1 = 1.23 watts. It's gonna get hot - so buy a heat sink for it.

Parts:
LM317 - $2.50
0.1 uf Capacitor - $1.30
Heat Sink - $1.70

Good luck,

/frank

 

[AnthraX101]

It depends on how regulated you need the voltage to be, along with how regular your load is going to be.

You could do it with something as simple as a voltage divider circuit. For "ideal" 12v, you need something close to a 2:7 relationship of the two drop down resistors, with an identical load to your resistor. Generaly you should consider a cars power to be closer to 13.5v at idle. That means you want a 1:4 ratio.

This isn't going to provide clean power, nor is it going to be ideal, however that's probably the least amount of components (2 resistors) that you're going to be able to do it with.

AnthraX101

 

[Mark R]

LM317 is definitely the way to go.

You get nice clean 1.2V power with just a chip and capacitor. If you add 2 resistors you could tweak the voltage as required.

If the power consumption of your device is low (if it runs on a single AAA, it probably is) then you don't really need a heatsink but it's a good idea - it doesn't need to be big. The LM317 also has thermal protection and will shut down if it overheats.

This is undoubtedly the easiest and most reliable way to do this voltage conversion.

 

[FrankSchwab]

Actually, I can do better with two components....

Oh, darn, can't find a 1.5V zener. Let's make it three passive components.

+12
----Dropping resistor--------------1.2V
|
Diode
|
Diode
|
------------------------------------------
Gnd

(amn, can't format that right. Move the diodes to the right of the dropping resistor)

Use a high-power resistor (power = R * I^2). Calculate it's value so that some amount of current (say, 2 to 3 times the amount drawn by your device) is continuously running through the diodes (this, of course, only works for very low powered devices). Attach your device across the diodes, and you'll have a pretty well regulated 1.2V source. If the battery voltage goes up (very common in a car electrical system to see microsecond long spikes above 24V), the diodes keep the voltage around 1.2V as opposed to the resistor-divider network. If the battery voltage goes down, less current flows through the diodes and 1.2V is still available to your device.

As with the LM317 suggestion, I'd certainly test this with a Voltmeter and a resistor as the load, rather than your fancy schmancy electronic device.

/frank

 

[KalTorak]

Be careful doing it this way... what everyone's describing here is a linear regulator from 12V to 1.5V, and you're gonna be burning 7 watts as heat in your pass element (the resistor or LM317, depending on what ya pick) for every watt you deliver to your device.

I realize that's probably the right thing for your application, but it makes me feel all icky to see a 15% efficient implementation....

 

[TerryMathews]

Originally posted by: KalTorak
Be careful doing it this way... what everyone's describing here is a linear regulator from 12V to 1.5V, and you're gonna be burning 7 watts as heat in your pass element (the resistor or LM317, depending on what ya pick) for every watt you deliver to your device.

I realize that's probably the right thing for your application, but it makes me feel all icky to see a 15% efficient implementation....

I see what you're saying, but I don't know of anyplace convient to pull less than 12V from either my car (91 Corvette) or my iPod's car adapter.

 

[Mark R]

If you tell us what it is that you want to power, we could estimate its power consumption.

If you only need 0.1 W of power, then your regulator would only need to dissipate about 1W - that's not a problem. You don't have a small battery that needs conserving, and 1W of heat can easily be shifted even without a heatsink.

If your device needs 1W, then you'll need to pay attention to the temperature of the circuit (careful choice of components and heatsinking) but power consumption from the car is unlikely to be a problem.

 

[TerryMathews]

Originally posted by: Mark R
If you tell us what it is that you want to power, we could estimate its power consumption.

If you only need 0.1 W of power, then your regulator would only need to dissipate about 1W - that's not a problem. You don't have a small battery that needs conserving, and 1W of heat can easily be shifted even without a heatsink.

If your device needs 1W, then you'll need to pay attention to the temperature of the circuit (careful choice of components and heatsinking) but power consumption from the car is unlikely to be a problem.

It's an FM transmitter - Digiana AudioX.

 

[Torched]

You could just pass the positive leg through a TVS diode. Weve done that at our work with a fan circuit
The output of a bridgerectifier @ 24v AC is roughly 33v DC. We stepped the 33 down to roughly 24v with a 7.5v TVS diode only to power a 2 watt fan though. I imagine if your power req are larger you would need the 317 circuit.

The LM317 is pretty cool because it comes with overload protection. You can short the output directly to ground and the chip will shut down!

The diodes come in different values. I imagine you would need about 10v of stepdown... oh and dont get the bidirectional kind...
take a look here: http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=19866&Row=399083&Site=US

 

[blahblah99]

I would definately also go with the linear regulator, but tie the tab of the device to something that can dissapate heat.

Otherwise, I'd go with a switcher w/ built in transistor. That becomes a 1 IC + 5 passive component solution.

 

[PowerEngineer]

I should start with a disclaimer here...but I have wired up some simple devices (like garage door openers) to the 12V car system with just a in-line resistor.

Assuming that the device load is fairly consistent and that it'll easily tolerate a little voltage variation from 1.5 volts, then all you need is a resistor that will drop ~10.5 volts when the device load current is passing through it. The best way to determine this to start with a 3.0 volt source (like a two-cell flashlight) and connect it through a 100 Mohm variable resistor (set at maximum ohms) to the device. Check the voltage across the device (hopefully starting at less than 1.5 volts). Slowly reduce the resistance until you measure 1.5 volts. Measure the resistance, and then multiply by 7 (10.5/1.5). That resistance value should drop the 12 volts to 1.5 volts.

As already pointed out, you'll be dissapating a lot more power in this dropping resistor (7 times as much) than in the device itself. You want to be sure that the resistor you install is rated for at least twice that amount.

 

[TerryMathews]

I went the LM317 + heatsink route. Thanks for all your help guys.