Voltage regulator to convert 12v/9v/5v to 3.7v

Discussion in 'The Projects Forum' started by JiffJaff, Sep 2, 2010.

  1. JiffJaff

    Thread Starter New Member

    Sep 2, 2010
    5
    0
    Hi,

    I studied electronics at school many years ago, but I'm a bit rusty to say the least. Hopefully you can excuse my basic questions and probably lack of clarity.

    I have two circuits that record video to SD card. Each is
    currently powered by a lithium ion battery that outputs at 3.7v, 2.0amps. The actual circuit consumes 130ma

    I want to remove the battery and instead power the circuit by taking power from mains powered equipment - one in my dj mixer that has 12v on it's main circuit.

    I'd like to power the other from a wall USB charger which is 5v. I know some computer usb ports are limited to 500ma which I suppose could be a problem if the circuit uses more than that?

    What would you suggest? I once read a tutorial that showed how to add usb power points to 12v circuits which used a 5v voltage regulator.

    Is that a good plan to follow?

    Can anyone suggest a voltage regulator that outputs at 3.7v that would be suitable? I found this, which appears to be a variable voltage regulator (not sure how it's adjusted), but has a low max output 0.1amp.
    http://uk.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=6616796

    I've also read about the importance of adding a capacitor to the voltage regulator to deal with any fluctuations. Can anyone explain how to decide what size/type of capacitor should be used.

    I've also heard that a heatsink might be required for the voltage regulator. Presumably going from 12v to 3.7 will generate more heat than going only to 5v?

    Am I on the right path here? Or should I be doing something else?

    Thanks!
     
    Last edited: Sep 2, 2010
  2. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    Adjustable voltage regulators like the LM317 are adjusted by choosing the values of two resistors connected to the output and ground terminals.

    This is a really old part and has a high dropout voltage. This is the minimum difference between the input voltage and the output voltage. A more modern part is called an LDO (Low Dropout) regulator. This is what you should be looking for.

    USB ports are limited to 100 mA unless you negotiate, using the USB protocol, for more current capability. Then you can get up to 500 mA
     
  3. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    A linear regulator (like LDO's and that LM317) dissipate (Vout - Vin)*Iload power, so yes going from 12V to 3.7V is less efficient than going down to 5V.

    The datasheet of the regulator you pick will recommend capacitor values and types usually.

    You'll need to calculate whether you need a heatsink, but using your numbers of 12V to 3.7V and 130mA gives about 1W. That actually exceeds the capability of the part you linked (LM317 in plastic packages), since it is limited to 100mA and limited from 2.4W to less than 1W dissipation depending on the package. Those are all small plastic packages which aren't meant for heatsinks by the way.

    If you still want to use such a part you can add a series resistor to reduce 12V down to something lower then follow it up with the regulator. This will make the resistor take up the bulk of the power dissipation rather than the regulator IC.
     
  4. JiffJaff

    Thread Starter New Member

    Sep 2, 2010
    5
    0
  5. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    The TO-92 package is unsuited for use with a heatsink. What you want is a TO-220 package. These can be bolted to a block of metal and be happy as a clam at Tj of 80-100 degrees C
     
  6. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    The actual circuit consumes 130ma when run at 3.7v

    Find an adjustable LDO in a TO-220 package and you should be able to take care of both circuits.

    Alternately you could run the 12V through a 64 ohm 2W resistor and a 3.7V zener diode provided that 130 mA is a fairly consistent current draw, if it varies a bit the resistor value could be reduced.

    You'd need to change the resistor to a 10 ohm 1/4W for the 5V supply.
     
  7. tom66

    Senior Member

    May 9, 2009
    2,613
    214
    This is technically true - but in reality, almost no USB ports implement this and allow you to draw up to 1 amp from them.
     
  8. JiffJaff

    Thread Starter New Member

    Sep 2, 2010
    5
    0
    Are these common compenents? If I went to my local store that sells components is it likely they'll have one of these I can buy in person.

    I found this online, is this the sort of thing I should get:
    http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=AP1186T5L-UDI-ND

    I want to get this circuit working with a 5v source this week. I thinking it might be easier to use the resister/zener diode combination. It seems easier to just solder those two components in series.

    Is there a downside to changing voltage with a resistor/diode as opposed to using a LDO?

    Thanks a lot!
     
  9. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Eats up a lot more power and usually requires large wattage components.
     
Loading...