Convert 9V to 5V

Discussion in 'General Electronics Chat' started by Quillion, Jul 18, 2009.

  1. Quillion

    Thread Starter Member

    Jul 18, 2009
    10
    0
    Hello I am quiet nooby at the whole art of circuitry. All I had was a little robot to practise on, so he used a 9V battery and converted it to 5V. Now all of my equipment uses 5V and I recently brought 78M05 voltage regulator, so I have quiet a few parts so its not a problem, but can someone please show me how to build a circuit where I connect a 9V battery to the breadboard and by using my voltage regulator I change it to 5V
    Please help me
    Thank you very much
     
  2. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,681
    900
  3. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    You can simply use a voltage divider shown below:

    [​IMG]

    Now, you want a 5V output so we'll have that for Vout and Vin is 9V. We'll just pick a random resistor value for R2 because of the ratio (but we'll have it in the KΩ range)...so for R1 you'll need a 19KΩ and for R2 you'll need a 20KΩ resistor. This will give you 4.6V which is very close.
     
  4. russ_hensel

    Well-Known Member

    Jan 11, 2009
    818
    47
    Voltage divider is fine untill a load is applied. Thus the voltage regulator for varying loads ( and input voltages ).
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You must also use a capacitor on the output terminal to ground. 0.1uF to 10uF is good.
     
  6. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Thanks guys, after I put the post I thought that a voltage divider couldn't supply much current...especially if they are "resistors."

    I am learning day by day through trial and error and through everyone here. Thanks for making AAC such a success! I absolutely love learning about electronics and I hope that you will continue to support each person including me. I'm not the wizard here so may the more experienced tell me about the advantages and disadvantages of the things I suggest (like this voltage divider). Thanks again!
     
  7. Quillion

    Thread Starter Member

    Jul 18, 2009
    10
    0


    THANK YOU VERY VERY MUCH it works awesomely and now I didn't even fry my microcontroller
     
  8. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    A 9V alkaline battery quickly drops to 7.2V. But the 78M05 regulator performs poorly when its input is less than 8V and the datasheet does not have a minimum input voltage.

    Therefore you should use a "low dropout" 5V regulator that works fine until its input drops below 5.5V or less.
     
  9. darenw5

    Active Member

    Feb 2, 2008
    45
    0
    If you want a bad but cheap way to drop a 9V battery to 5V, a resistor and zener diode will do. I did exactly that in a high school project years ago; see http://www.darenscotwilson.com/spec/DigiDice/digidice.html This is like a voltage divider but handles a load (if not too great) just fine. Forget about getting the most out of battery life, though!
     
  10. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    I like zeners myself. Another approach for a low insertion loss regulator would look something like this.

    [​IMG]

    I got this from my LEDs, 555s, Flashers, and Light Chasers article. You would need to modify it (obviously), but the principle remains. You didn't say how much current you needed, so I can only speculate what the resistor would need to be, but the zener would need to be a 5.6V unit.
     
  11. bertus

    Administrator

    Apr 5, 2008
    15,647
    2,346
  12. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,536
    True enough.

    Voltage regulators do draw continuous current, and as audioguru pointed out, there are limits to what a 7805 can do with a low voltage.
     
  13. vectravanman

    New Member

    Mar 2, 2012
    12
    0
    Rather than create a new thread, i thought i would add to this thread as my problem regarding a similar issue. My task is to create a PCB to output data to an LCD display using a 9V battery and PIC. I am also new to designing schematics and working in the electronics field so any feedback would be helpful. If i use an LM78M05CDT regulator to convert 9V to 5V will i then be able to use a voltage divider circuit to convert 5V to 4V as i need 4V for my LCD but 5V most other devices in my circuit.

    Thanks
     
  14. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Vectra Van Man,
    Please read earlier in this thread:
    1) An old 78M05 does not work when a 9V battery is not brand new. Use a low dropout regulator instead.
    2) A voltage divider does not work when it has a load. Use a series silicon diode to reduce 5V to 4.3V.
     
  15. vectravanman

    New Member

    Mar 2, 2012
    12
    0
    Thanks for the reply Audioguru

    I have tried designing a circuit with two voltage regulators one with an adjustable output to give 4V. The circuit looks something like this.

    Is this a suitable circuit layout or do i need to add more to get the voltage im looking for?
     
  16. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Vectra Van Man,
    The LT1763 regulator is very small and has many pins.

    I guess you did not read the datasheet for the LT1763 series regulators:
    1) An LT1763 is the adjustable one. An LT1763-5 is the 5V one.
    You show an adjustable one with an unregulated output (its ADJ pin is not connected) and a second adjustable one with a 1.76V output because your arithmatic is wrong.
    2) Their MINIMUM output capacitor is 3.3uF but the datasheet shows that a 4.7uF cap is better and a 10uF cap is best.
    3) The datasheet says with a battery input then use a 1uF to 10uF input capacitor.

    You show the polarity of the battery backwards.
     
  17. hwy101

    Active Member

    May 23, 2009
    91
    28
    translated:

    "78m05 Usually have a circuit in the data table."

    yes it does, Thank You
     
  18. vectravanman

    New Member

    Mar 2, 2012
    12
    0
    Thanks audioguru

    Have actually decided to just use the 5V regulator and produce 4V through a resistor going to 2 series LED dies which are used on my LCD. This is the only part of my circuit that requires 4V. Its a pain but i think this will work.

    I am pretty new to Electronics Design so i appreciate this feedback as a lot of it is still over my head.
     
Loading...