Constructing a voltage divider that takes into account internal resistance

Discussion in 'Homework Help' started by ztjust, Jul 26, 2012.

  1. ztjust

    Thread Starter New Member

    Apr 1, 2012
    Hello. So I'm trying to create a small resistor network (voltage divider) that halves the voltage of a measuring device. I have a measuring device that can only go up to 20 Volts, and the battery I need to measure can go up to 21 volts. So essentially I want to create a voltage divider to halve the voltage, and then just multiply by two when I analyze the data.

    The measuring device has an input impedance of 122k. If I want to halve the 20 volts to 10 volts, wouldn't I just need another resistor of 122k in series?

    If not, how would I go about compensating for internal resistance?

    Also, how would I create the actual circuit? I don't have much experience with soldering or building physical circuits.

    I know this is a lot, and I thank you for any time you have given me today.
  2. mlog


    Feb 11, 2012
    You don't say anything about the impedance of the device you are measuring. It could matter. Let's look at an example.

    Let's use your idea of a resistor divider by simpling adding a 122k resistor in series. What if the device under measurement had an source impedance of 122k Ω? So now you have 3 resistances in series. You have the 122k source impedance from your device, the 122k you added in series, and the 122k from your measuring device. What's the voltage across your measuring device? Hint: It's not half.

    You really need to know the source impedance of the device that is to be measured, and then you can begin to design something.

    By the way, I would consider an active buffer amplifier, but if I was limited to passive devices, then I would use a voltage divider that did not directly include the input resistance of the measuring device. Think Thevenin's Theorem. I would add 2 resistors rather than just one. Go ahead and experiment with some values.
  3. ramancini8

    Active Member

    Jul 18, 2012
    The way I read the post the device being measured has an impedance of 122K, so the series resistance of 122k is correct if the source impedance is negliable.