Power Supply Question

Is there any way I just use a resistor to drop the voltage?

Sure, but you'll waste a ton of power by the time you get it regulated. And, it'll never be as stiff as if you just use a regulator like previously suggested.

You'll still waste the same ton of power using a linear regulator (P=IV, no getting away from it).

I agree a regulator's usually a better approach (depending on the possible load variations). Resistor's a damn sight cheaper & easier if we're talking constant milliamps.

I see your point. I was going by the old rule of using 1/10 of the current for the load and wasting the rest in a parallel leg in order to get some stiffness. If you have a constant load, you wouldn't have to do that.

It seriously depends on the type of current draw your device has. If it fluctuates at all the resistor approach is pretty much out.

I agree with the LM7805 recommendation above. It is about the simplest circuit you can build. 1 LM7805 and 1 capacitor should do it. You could even skip the cap if you don't care about the quality of the power. Worst case you can just solder them together and wrap them in electrical tape. I don't know where you live, but in the states you can get all the parts at your local Radio Shack for a few bucks. The package from Radio Shack even shows you how to do the circuit.

Good luck.

-Doug

Thanks for all the reply.

MY load is basically constant. How will I compute the resistor that I need to use then?

thanks

1. Measure your load current I_L, at 5V.

2. You need to "lose" 5V from your 10V supply, at this current. The voltage across a resistor is given by V = I x R, so we have 5 = I_L x R, where R is your unknown series resistor. Re-arranging, R = 5 / I_L. Remember that I_L is in Amps (with V in Volts and R in Ohms).

Examples:

I_L = 1mA. R = 5 / 0.001 = 5000Ω. 4700Ω is a commonly-available value (often written 4k7), which would almost certainly do the job.

I_L = 500mA. R = 5 / 0.5 = 10Ω (often written 10R).

You must also consider the power to be "lost" in (or dissipated by) the resistor (to make sure it doesn't overheat).

The power (in Watts) is given by P = I x V, so in example (1), P = 5mW (no problem for pretty well any resistor), while in example (2), P = 2.5W, so you'd need a power resistor of at least this rating, and you'd have to allow for the fact that it'll get pretty hot.

Hope this helps.

A single resistor and a 5 volt zenor diode should work just fine to drop the voltage from 10 to 5. Calculate the size of the resistor from R = (Vin–Vz)/I. Hook the resistor to the 10 volt supply, and connect the zenor to the other end of the resistor and ground. Tap off between the resistor and zenor for 5 volts.

Don't just connect a single resistor in order to have a 5V voltage drop on it and, thus, to use the remained 5V to supply your circuit... Unless the "load current" (the operation current of your circuit) is extremely constant... But, almost always, the current consumption is not constant at all... During the operation of your circuit, it's current varies... And even if you have calculate your series resistor for a 5V voltage drop on it, concerning the typical (average) current, the unavoidable current variations will make this 5V voltage drop to vary too... Thus the (remaining) 5V voltage (which is used to supply your circuit) will vary too...

Skeeter said about a zener with a series resistor... That's a solution but there is a disadvantage: you'll have a greater current consumption (given by the 10V power supply) due to the (pretty much) current which is consumed by the zener in order to do its job (e.g. regulation)... Thus a greater total power consumption...

Jman said about the 7805 regulator: maybe the better solution as it gives a better regulation (and lower total power consumption than the zener)... Again it depends of the current demands of your circuit: it will not be sufficient if the current is really large (more than the 7805 (or any other single chip regulator) can support)... If so the things becomes more complicated as you need to add a power transistor (usually a MOSFET) in order to pass the load current through it... and, anyway, you need to design a "power linear regulator" circuit... For very large load currents, though, you need to design a "power switching regulator" circuit (or use a "stand alone" module) as it is (by far) more efficient than a linear regulator circuit...

Is this a "one off" problem? Do you need the 10 V for other devices in the unit? Where are you? When you say "device": do you mean a component or a consumer product?

You should be able to get a 5V supply for about $15 like this one:-

http://cpc.farnell.com/PW01911/batteries-power-supplies/product.us0?sku=UNBRANDED-DSA-0051-05UK

You have lots of good answers here. As a designer of fairly complex power supplies I would go with the simple 7805 or maybe an LM317 with a couple small set point resistors if the power is low enough. The resistor approach varies linearly with load current. Lots of simple options, if your power is higher or you need some efficiency it will get more complicated. For more detailed answers you need to provide a more detailed question.