# How do I provide 7 volts from a 12 volt source?

I want to power a GHI Panda II single board computer that has a GHI FEZ Connect shield attached for Ethernet access. GHI Documentation specifies 6.5 V to 7.5 V input for the Panda and shield that can use a total of 238mA. I have a nice regulated power supply providing 12 Volts. What is a decent way to change that 12V to 7V (give or take 0.5V)? Is a simple resistive voltage divider adequate? If so, what size of resistors? It seems to me that the resistors need to be in a ratio of 5:7, but not sure what to use to provide the required current. Or maybe i should be using something totally different. Thanks.

• The term you should be searching for is regulator. Either a buck converter, or a LDO with a heatsink. – Matt Young Oct 16 '13 at 22:57
• So is there a difference between a linear voltage regulator and a voltage divider? – Matt Oct 17 '13 at 3:36
• A voltage divider is NOT a regulator. The voltage between the two resistors will vary with load current. – Matt Young Oct 17 '13 at 4:27

You could use a resistor divider but that doesn't regulate the output voltage at all - the more current you take, the lower the output voltage is.

If maximum current is 238mA you could use a resistor from the 12V feeding into a zener diode connected to ground. 1N5342 is a 6.8V zener diode rated at 5W. If this is connected to 12V via an 18 ohm resistor you should get a reasonable 6.8V output across the range of currents. In fact, at the point where 6.8V is on the verge of dropping, the current output will be about 288mA. If you put a schottky diode in series with the zener you could lift the voltage up to about 7.2v. This is an optional idea should tolerances on the 6.8V zener work against you and it ends up more like 6.46V (down 5%).

Power ratings are 5W for the specified zener although a 3W device would be OK. The resistor power rating needs to be greater than 2W to avoid it frying too much.

You could use a linear voltage regulator like the 7805 - these can easily be adjusted to produce 7V with a small pot in series with the ground pin. How much power will this self-generate - it's dropping 5V whilst supplying up to 238mA so it'll dissipate over 1.2W so a heat sink is going to be needed - just a small one.

Or maybe even consider a power efficient buck regulator like the LM2575 (simple switcher). These are probably about 85% power efficient so it'll dissipate about 0.3W and not need a heat-sink.

Personally I'd go for a switcher unless your application was particularly prone to high-frequency ripple noise on the regulated output.

• Great info, thanks for taking the time. i understand about the variable voltage supplied by the volt divider and how the zener would buffer that. I, too, am leaning to the LM2575, but it seems to me that i'd need to supply an inductor, a diode (schottky?) and a capacitor(s). Also, Digikey supplies a variety of LM2575s. Is there any product that supplies the LM2575 with components or some spec somewhere for applicable LM2575 and components? – Matt Oct 17 '13 at 17:59
• @Matt the data sheet will tell you these things if you wish to learn. It's very simple once you spend a little while dipping your toe in. – Andy aka Oct 17 '13 at 18:16
• Absolutely. I had glanced at a data sheet and saw a diagram with no specs, but, wow, lots of info. Only concern is line noise as plan is to put it inside an anti-EMI enclosure with the Panda II. Not sure if the LM2575 would create noise in the 20+ GPIO and analog signal lines inside the aluminum box. – Matt Oct 18 '13 at 13:12
• BTW, here's a switching reg product w/ integrated components including a small potentiometer. robotshop.com/productinfo.aspx?pc=RB-Dim-14&lang=en-US – Matt Oct 18 '13 at 15:47