How Do I get 3V From 30V?

[bhjazz]

So I want to add a custom touch to my next amplifier. This custom touch...some lighting (as always with me, geez) does ship with it's own power supply: a box which holds two AA batteries and supplies 3V. I'd rather use the power already inside and dispense with the battery holder. The problem is, I have 30 volts.

How do I calculate the correct resistor to use to drop the voltage down so that I can wire the power unit directly in? I have access to 1/2 and 1W resistors, if necessary...!

Thanks!

 

[grawk]

For that big a voltage drop, use a transformer.

 

[tomb]

Grawk is essentially correct, but it depends on the amperage. For instance, we can use an LED in our amp circuits when the LED only uses 1.4 - 1.7 volts. We can get away with that kind of huge voltage drop with a 1/4, 1/2, or 1W resistor - because the LED may only draw 20ma or less.

So, you'd have to know the current draw of your device to size the resistor properly - assuming you can do it.

V = I * R, but power is the limiting factor.

Also, it's standard practice to size the resistors at close to 1/2 of their rating. You can fudge at this, but you may cook a resistor or two - your calculations and the amp draw may not be that certain in all situations.

27 Volt Drop:
1W supports 37ma
1/2W supports 16.5ma
1/4W supports ~9ma,
these are absolute ratings - as mentioned, you might want to select the resistor at twice that power rating. However, it gives you the amperage rating that might be supportable with a simple resistor performing the voltage drop - more than that and you're talking a big power resistor, maybe with a heat sink or worse - and your power supply would have to supply all of the excess current getting burned through the resistor, too.

The value of the resistor in ohms will be deterimined by the specific current draw.

For instance, if the device drew 20ma, then R = V/I, R = 27/0.02 = 1.35Kohms. The power at that current and resistance would be (I^2)*R, or (0.02^2)*1.35K, or 0.54W. As mentioned, that should probably be a 1W resistor to be really safe.

 

[bhjazz]

Thank you, tomb. I am familiar with V = I * R, but had no idea where to start! Once this item arrives in the mail I'll get a better look at the specs and test results. Thanks!

 

[amb]

If you use a "superbright" LED, all you need is a couple of mAs and they will be very bright. Dropping 30V down to 3V (typical for a blue or white LED) with a 10K ohm resistor will flow 27V / 10K = 2.7mA. The power dissipation on the resistor will only be 27V * 2.7mA = 73mW. You can safely use a 1/8W resistor.

 

[Alcaudon]

You can use a LM317 regulator and a couple of resistors too, but you'll need to mount the regulator on a heatsink (as big as you can). I'm using this in some hardware of my final project (converting from 30V to 5V) and it has been working (almost non stop) for months (of course well heatsinked).

Hope it helps

 

[fordgtlover]

http://www.hebeiltd.com.cn/?p=zz.led...tor.calculator

This links to a LED resistor calculator.

 

[error401]

Quote:

Originally Posted by bhjazz /img/forum/go_quote.gif Thank you, tomb. I am familiar with V = I * R, but had no idea where to start! Once this item arrives in the mail I'll get a better look at the specs and test results. Thanks!

What you need to know is the current draw of this device. Current through the circuit will be the same at all points along the path, but voltage will be different across each component. If you know the current, then you can choose the correct resistor to drop the right amount of voltage for the part.

The LED case is a bit different because its voltage is constant and it is basically a short circuit. The resistor is used to regulate current, not voltage. As a side effect of the LED having a (relatively) constant voltage across it, the resistor drops the remaining voltage.

This all falls apart if the part you want to use draws variable current, since the resistor's voltage drop directly depends on it. If current draw changes too much, the voltage will too.

As was mentioned, if the current draw isn't too large, you can use a linear regulator like the LM317, but be careful of the temperature of the device. Otherwise you'll probably have to use a switcher.

 

[bhjazz]

First: Thanks for the great answers, all. I really appreciate them.

Second: It wont be an LED. It will be another type of lighting. My initial LED plan for this amp had a major setback so I am researching a few other ideas for lighting. I'm bummed because the LED configuration I had in mind would have rocked all but the most stuffy of DIY aficionados. Alas, the parts just didn't fit together correctly and quite frankly would have looked like dreck from all but the most perfect of angles.

Thanks again. This place rocks!