Voltage divider Circuit

Discussion in 'General Electronics Chat' started by jumbo, Dec 4, 2007.

  1. jumbo

    Thread Starter New Member

    Dec 4, 2007
    2
    0
    I have a DC adapter of output 7.6V, 600mA. Taking this I'm trying to get 3V 600mA. I thought just putting a voltage divider circuit at the output of the adapter would do the trick. I made the divider circuit by using 680 ohms and 1k Ohms, taking the voltage across the 680 as my new output. Well, I get the desired 3V, but I do not get the current. I would still be ok if the current output dropped to 300mA. My load consumes about 200mA. Could anyone please help me on this? I really appreciate any help.
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    When you hook up your voltage divider to a load, the load will become part of the voltage divider. This will alter you voltage divider values.

    I suggest you consider using a linear voltage regulator, such as the venerable LM317.
     
  3. Voltboy

    Active Member

    Jan 10, 2007
    197
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    They problem is that the 680 ohm resistor resist to current.
    Using different values you could get 3v at 600mA.
    Using this voltage divider calculator. the results are R1=7.66 ohm and R2 =5ohm
     
  4. jumbo

    Thread Starter New Member

    Dec 4, 2007
    2
    0
    Thanks guys. Appreciate your help. I'll try both the suggestions out.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    656
    But if you add a 200mA load to this, the voltage will drop to 2.4V. You will be wasting a lot of power trying to do it this way, and your output voltage will still vary as the load varies. Forget the voltage divider and use an LM317.
     
  6. Voltboy

    Active Member

    Jan 10, 2007
    197
    0
    Or maybe the LM350
    Btw, how you calculate that adding 200mA the voltage will drop to 2.4v?
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    The Thevenin equivalent resistance is 7.66 ohms in parallel with 5 ohms, which equals 3 ohms. In other words, we have a 3 volt source with 3 ohms series resistance. 200mA through 3 ohms will drop 0.6V.
    3V - 0.6V = 2.4V.
    An LM350 will work, but it's overkill for a 200mA load.
     
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