ohms law help

Discussion in 'Homework Help' started by Gregfox, Jul 28, 2013.

  1. Gregfox

    Thread Starter Member

    Jul 23, 2011
    14
    1
    Hello,
    I'm trying to determine what resistor I would have to use to get 500mV from a 3 Volt
    Supply.
    E=IR requires a current, so I don’t know how to calculate for the resistance. I would like to know the methodology used.
    Thanks!
     
  2. odinhg

    Active Member

    Jul 22, 2009
    65
    15
  3. wayneh

    Expert

    Sep 9, 2010
    12,148
    3,058
    You DO know the current, since it is the 3V voltage drop divided by the total resistance. Well OK, you will know it once you work out the algebra.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,763
    4,800
    Pick a current that is reasonable, such 0.1mA or 1mA, and find the total resistance you need. The figure out how much of it needs to be in the pot to drop 1V across it. That will put 500mV as the midpoint of the pot's range.
     
  5. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    For a voltage divider to feed an op-amp input something like 1 ma current is typically OK. (This can vary depending on the op-amp used).

    EDIT: I should probably look before I Post eh... :rolleyes:
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,763
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    You were probably already typing your response when I posted mine. Happens pretty frequently. I've had times when I've taken ten or twenty minutes to put a response together and after posting it find that there are four or five posts and mine is now just a repeat of stuff that wasn't there when I started responding. Oh well. If nothing else it reinforces the point being made and serves as an indicator that there is at least some agreement between responders.
     
  7. Gregfox

    Thread Starter Member

    Jul 23, 2011
    14
    1
    Thank you all. From what was posted I guess I have to guess the current, then work out the R value. I was hoping there was a more scientific way of doing that.
    I guess I could measure the input current and go from there.
    Regards.
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,763
    4,800
    It's not that you are "guessing" anything. You are CHOOSING a current level through the voltage divider.

    It's like if I come up to you and tell you to find two numbers, x and y, that add to 100. Well, you can pick 50 and 50, 0 and 100, 10 and 90, -1000 and 1100, 33.33 and 66.67. All of those are equally correct. There is no "science" to it. You are given one constraint and two unknowns, so you get to pick one of the unknowns as you see fit.

    In the case of a voltage divider, the desired voltage division ratio sets the ratio of the resistors, but you can multiply both resistor values by whatever constant you want as long as you use the same constant for both. That constant will dictate the total current flowing in the divider chain. Under ideal conditions, the choice is completely arbitrary and you could choose 100A or you could choose 100pA. But as you start allowing for non-ideal realities, both of those bounds get moved in. The current-delivering capability of the supply and the power dissipation considerations bring down the upper limit and the noise considerations and op-amp input bias current requirements bring up the lower end. But you will still be left with a range of two or three orders of magnitude where you can operate and be fine. You generally go for a current that is in the lower portion of that range.

    With experience you will get a feel for what reasonable values to use are when you have arbitrary choices that have to be made.
     
  9. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,363
    There is an infinite number of solutions. The values of R9 and R10 are interdependent. We say the answer is ratiometric.

    The ratio of R10/(R9 + R10) must be the same ratio as 0.5/3, i.e. 1/6.

    Choose a value for R10 and R9 must be 5 times R10.
     
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