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Voltage regulator

  1. Feb 14, 2008 #1
    [SOLVED] Voltage regulator

    1. The problem statement, all variables and given/known data
    This isn't a homework or coursework question. It's just a general question. I have a circuit with a 15V power supply. The output has to be 12V. I've inserted a blocking diode to prevent current flowing in the opposite direction.

    The problem I'm having is how to attach a voltage regulator to the circuit. Also I don't know if the connection of the leads to the output is correct. My sketch of the circuit is attached.

    Sorry if I've posted it in the wrong forum.

    2. Relevant equations



    3. The attempt at a solution

    [​IMG]
    Sorry if the symbols are wrong
     
  2. jcsd
  3. Feb 14, 2008 #2
    This depends on what kind of voltage regulater you have. If you have one that gives an output of 12 V you have an IN pin that connects to 15V an out pin that connects to the output, and A GND pin that connects to the negative side of both the input and the output.
    You don't need R1 and R2 in this case. Indeed you don't want to use a voltage divider to for the output like this at all. The resistance of the output is unknown, so there's no way to tell what your output voltage is going to be.
    You can connect the GND pin to a voltage divider to adjust the output if you can't get a voltage regulator with the right voltager, or want one that is adjustable.
    I found a schematic for this here:

    http://ourworld.compuserve.com/homepages/Bill_Bowden/page12.htm
     
  4. Feb 14, 2008 #3
    So if the voltage regulator can output 12V then this would be the circuit

    [​IMG]
    If I use this voltage regulator, how do I adjust the Vout? I don't see a knob or anything. Forgive my ignorance.

    If the voltage regulator outputs 15V then this would be the circuit

    [​IMG]

    and if R1 = 100 ohms & R2 = 400 ohms the output should be 12V right? Thanks in advance.
     
  5. Feb 14, 2008 #4

    ranger

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    Gold Member

    Joe, the LM317 series of voltage regulators are very good. You can adjust Vout by using an potentiometer to the ADJ pin (1). The datasheet is an excellent source of information about the device.

    You can find tons of info here:
    http://www.national.com/mpf/LM/LM317.html

    A power supply using this voltage regulator:
    http://www.geocities.com/tomzi.geo/lm317/lm317.htm

    Keep in mind that there are other ways of building a power supply. You can use a simple zener diode as the regulator instead.

    EDIT: for problems of this nature. You're better off posting in the electrical engineering forum or the Science and Engineering homework forum.
     
    Last edited: Feb 14, 2008
  6. Feb 14, 2008 #5
    So basically when they say "Adjustable output down to 1.2V" it means the Vout can be adjusted using a potentiometer? If that's the case then I should use the second circuit with the LM317 right? Thanks again

    Sorry bout that. I'll keep that in mind. Is there any way to move the topic to the right forum?
     
  7. Feb 14, 2008 #6

    ranger

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    Gold Member

    Thats, right. It can work down to 1.2V with the range being adjusted by a pot to the appropriate pin. "Adjustable output down to 1.2V". The 1.2V would be the minimum to which the regulator can regulate efficiently and its called the dropout voltage. Hey, the output circuitry of a regulator needs power to work, right? This is where to 1.2V comes in. Lower dropouts are available by using a low dropout (LDO) voltage regulator.

    Your second circuit is not quite complete to support the LM317 just yet. A quick reference to the datasheet for the device will reveal that you have to add filter capacitors. Just use the links I gave you as a reference.

    Sure. Just use the Report button and report the initial post. A mentor would then review your request.
     
  8. Feb 15, 2008 #7
    Thanks a lot for your help. I just want to make sure I get it right. Is this circuit and the values for the capacitors and resistors correct?

    [​IMG]
     
  9. Feb 15, 2008 #8

    ranger

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    The caps look good. But the resistor values are off if you want 12V output. Vout is related to the resistors by the following:

    [tex]V_{out} = V_{ref} \cdot (1+\frac{R2}{R1}) + I_{adj} \cdot R2[/tex]

    Since [itex]I_{adj}[/tex] is very small (max about 100uA, typical 50uA), the error term [itex]I_{adj}R2[/itex] can be neglected. Vref is the internal reference voltage, which for the LM317 is 1.25V. Btw, there isn't any real need for the diode you've inserted.
     
    Last edited: Feb 15, 2008
  10. Feb 16, 2008 #9
    Brilliant thanks a lot mate.
     
  11. Feb 16, 2008 #10

    ranger

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    Gold Member

    Sure. You're welcome.
     
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