voltage divider

Discussion in 'Homework Help' started by Joe Treadweorth, Feb 19, 2010.

  1. Joe Treadweorth

    Thread Starter New Member

    Feb 19, 2010
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    Having a little problem getting started with this and could use some help doing so.

    I need to design a voltage divider.

    40V dc source, outputs to 20v with 6.8K load, 15V with 4.7K load, 10v with 3.3K load and 5v with a 2.2K load. Max source current to be 24ma.

    I have went over my notes, searched the book but cannot seem to get started. We were told to start at the bottom and work our way from the last resistor to the first, but I cant seem to get started. Am I missing information?

    I need to determine the resistor values for the circuits. all I need is help getting started.

    Any help would be greatly apprecaited.
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    63
    Are your loads just resistors?
     
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    Be sure to provide us with your efforts. I few attached sketches showing the circuit or circuits would help avoid misinterpretation of the actual circuitry involved.

    hgmjr
     
  4. Joe Treadweorth

    Thread Starter New Member

    Feb 19, 2010
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    0
    Yes the loads are just resistors. Here is a sketch of what I believe it will look like.

    I need to calculate the values for R1, R2, R3, and R4.

    Again thanks for any help.
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You're going to need another resistor between your 40V supply and the top of the divider chain, otherwise the 6.8k will be at 40 volts, and there won't be any way to make it 20 volts.

    I'll give you a hint. Since you know the value of each resistive load at each output voltage, you can calculate the current each of the resistive loads will draw. That gives you a big step toward a solution. Don't forget the 24 mA max source load.
     
  6. Joe Treadweorth

    Thread Starter New Member

    Feb 19, 2010
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    I think we were told to start at R4 and work backwards is this correct?

    I thought I needed to at least know the bleeder current to start though.
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    This is one way to do it, but it's crucial that you understand that:

    "You're going to need another resistor between your 40V supply and the top of the divider chain, otherwise the 6.8k will be at 40 volts, and there won't be any way to make it 20 volts."

    If you don't do that, it won't be possible to make it work. Do you understand why?
     
  8. Joe Treadweorth

    Thread Starter New Member

    Feb 19, 2010
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    yes i understand that. I need a resistor to drop the voltage down.

    I greatly appreciate the help. We were only given one example of the loaded voltage divider and the loads were unknown and we only solved for current which we were given the bleeder current to start with.
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    This problem doesn't have just one solution. You have 5 resistors to choose, but there are only 4 node voltages. This means that you have to choose one of the resistors arbitrarily. The fact that the problem mentions a maximum source current gives you a clue that there must be more than one solution.

    The problem can be solved from the bottom up, or from the top down.

    I'll get you started with the bottom up method.

    A moments thought will persuade you that you don't even need R4; we arbitrarily select its value to be ∞, in other words. You could choose some other value, but it can't be too small or you will end up with more than 24 mA of source current.

    Without R4, how can we get 5 volts across R8? By passing a current of 5/2700 amps through it. So, if we have 5/2700 amps in R8, then we must also have 5/2700 amps in R3. We need 5 volts across R3, so what is its resistance? 5/(5/2700) = 2700 ohms.

    Next, calculate how much current R7 is taking; this is possible because we know its resistance and how much voltage is across it. Then the current coming down through R2 must be the sum of the current in R7 and R3, and we know how much voltage must be across R2, so with this information we can calculate its value.

    You should now see how to proceed a step at a time like I've shown.
     
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