# Convert 9V to 5V

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

1. ### Quillion Thread Starter Member

Jul 18, 2009
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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
Thank you very much

Jan 18, 2008
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3. ### ELECTRONERD Senior Member

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

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 Distinguished Member

Jan 11, 2009
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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
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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
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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
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THANK YOU VERY VERY MUCH it works awesomely and now I didn't even fry my microcontroller

8. ### Audioguru New Member

Dec 20, 2007
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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
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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
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I like zeners myself. Another approach for a low insertion loss regulator would look something like this.

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.

Apr 5, 2008
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12. ### Wendy Moderator

Mar 24, 2008
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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
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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
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Vectra Van Man,
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
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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?

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16. ### Audioguru New Member

Dec 20, 2007
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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
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translated:

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

yes it does, Thank You

18. ### vectravanman New Member

Mar 2, 2012
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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.