voltage divider with a twist

Discussion in 'Homework Help' started by n8tron, Mar 7, 2010.

  1. n8tron

    Thread Starter New Member

    Jan 18, 2010
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    0
    I need everyone's help. I have been working on a problem for the last three weeks and can't figure it out. I need to build a voltage divider for the following values. 12V 50ma, 9V 50 ma, 3.3V 20 ma, and 3.3V 20 ma. I have calculated and built the simple voltage divider, but I need the circuit to be able to compensate for changes in the currents. I have been told to use the 10 to 1 rule, but that's where I get lost. If anybody can help me I would appreciate it. The basic voltage divider is easy. It's adding the resistors in parallel that throws me off. Thanks in advance and sorry if this is an easy answer that I'm just not seeing.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi n8tron,

    Your question isn't all that clear regarding the design requirements.

    Are we to understand that the voltage divider chain is concurrently supplying the given voltages at the nominated load currents. Or is it a case of any one (but only one) of the outputs at a time?

    Were you assigned a primary DC source voltage or is the question completely open ended with respect to component & source selection.

    Your additional point about being able to compensate for changes in load currents (presumably meaning no change in voltages) would seem beyond the scope of a simple resistive divider network design.

    Regards,

    t_n_k
     
  3. n8tron

    Thread Starter New Member

    Jan 18, 2010
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    0
    Sorry for not including that information. Yes, the loads are all being supplied at the same time (all four devices drawing current simultaneously). And no power supply has been selected for us. We are supposed to decide which one would work best. I usually picked an oversized supply when I was testing circuits with multi sim.
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    I saw another voltage divider problem in another forum. I'm guessing that when your instructor uses the word "compensate", he means just that the voltage divider outputs shouldn't vary much for a moderate change in load.

    The 10 to 1 rule simply means that you pass a current through the divider chain 10 times greater than the currents the loads will take.

    In the limit, you could make a divider chain have very good output voltage stability by just passing 100 times, or 1000 times as much current through the divider chain as the loads will take. But, of course, this is an inefficient way of doing things.

    I think what you need to do is design a divider chain of resistors, supplied from a stable voltage source at the top, with the current in the chain set at 10 times the load current at each node.

    The highest output voltage required is 12 volts, so why not use a 12 volt supply at the top of the chain, and supply the 12 volt load from there. That way, the 12 volt load will receive a perfectly regulated 12 volts, independent of load. Then the rest of the loads will be supplied by a divider chain.

    The current in the divider chain isn't constant from top to bottom since the various loads siphon off some current as you move from top to bottom of the chain, but you could make the current in the bottom resistor 10 times the load current of the bottom load as a start.

    You say that "I need to build a voltage divider for the following values. 12V 50ma, 9V 50 ma, 3.3V 20 ma, and 3.3V 20 ma." Do you really have two loads of 20 mA @ 3.3 volts? If so, that would the same as 1 load of 40 mA @ 3.3 volts.

    Assuming the lowest voltage load is as I described, select your bottom resistor to carry 10 times the load current being supplied at that node.

    This will give 3.3 volts/400 mA = 8.25Ω for the bottom resistor of the divider chain.

    The next higher voltage required is 9 volts. The difference between the 9 volt tap and the 3.3 volt tap is 5.7 volts. The next resistor up the chain must carry the 400 mA in the bottom resistor plus the 40 mA supplied to the 3.3 volt load. So, we need a resistor that will drop 5.7 volts while carrying 440 mA. That resistor will have a value of 5.7/.440 = 12.95 ohms.

    You can continue this process all the way up the chain and get your resistor values for the divider chain.

    You could choose larger currents in the chain, and change the numbers, but the process of starting at the bottom and working your way up is the same.
     
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