Very simple question , how to design potential dividers ??

arunb · Joined: 08 Sep 2003 Posts: 492 Location: India

Hi,

I have a 24V source and I need to bring down this voltage to 5 V , I would like to do this using potential dividers , but I do not know how to calculate the resistor values

thanks
arun

Mark · Joined: 07 Sep 2003 Posts: 2838 Location: Atlanta, GA

If there is a resistor in the path, the more current draw by whatever is being powered by the 5V will cause more voltage to be dropped and the 5V will go down. Better stick to a regulator or you might get away with a resistor and a 5.1V zener.

Guest · Guest

ArunB -

Get a piece of paper and draw a voltage source of 24 Volts, connected to two resistors in series. Let one resistor have a value of 19 Kohms, and the other be 5 kohms. The current flow in this small loop circuit is the voltage divided by the total resistance, or 24V/(19K + 5K) = 24V/24Kohms = 1 mA.

Now, each resistor has 1 mA flowing through it. The voltage drop across the 19 Kohm resistor is V = 1 mA * 19 kohm = 19 volts. The drop across the 5 kohm resistor is V = 1 mA * 5kohm = 5 volts, which is what you wanted.

Take this simple example and work backwards to get simple algebraic expressions to compute the resistor value you need for various voltages, as well as current.

There may be fancier schemes, but I don't think it gets any simpler than this when using resistors.

HTH,
Bill

Sherpa Doug · Guest

Mark and Bill are both right depending on the usage. It all revolves around Ohms Law: E=IR, or Volts = Amps x Ohms. The catch is that the output of the divider has a considerable impedence of its own. As a rule of thumb for low accuracy applications the current through the bottom of the divider (the 5k in Bills post) should be ten times the current taken out the tap. If the tap is only driving an A/D converter which draws 100uA then the 1mA in Bills post is OK, for low accuracy. But if you are driving your whole PIC circuit from the tap, and your circuit draws 20mA to run an LED, then you need to run 10 * 20mA through the bottom resistor or 1/5 of an Amp! That is going to kill any batteries, generate a lot of heat, and still be a low accuracy 5V supply. In that case a regulator chip is definately the way to go.

Tell us more about the application.

Sherpa Doug