# voltage drop help

Discussion in 'General Electronics Discussion' started by paul60, Apr 16, 2015.

1. ### paul60

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0
Dec 11, 2010
i have a 12v dc supply i need to drop to 4.5 volts what is the best way to do this
thanks paul

2. ### Gryd3

4,098
875
Jun 25, 2014
For a simple reference voltage. A voltage Divider.
A more ideal reference voltage. A Zener Diode.
To provide a little to moderate amount of current, an LM7805 voltage regulator. (Will give you 5V)
To provide a moderate to high amount of current, a switch mode converter. (Either pre-set at 4.5V or 'adjustable voltage'.) These can be built or bought.

Give us a little more detail on what you plan to use this for and we can elaborate further

3. ### paul60

22
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Dec 11, 2010
i have a small led light that uses 3 aaa batteries want to power it from a 12 volt dc supply looking for a simple circuit

4. ### Gryd3

4,098
875
Jun 25, 2014
Do you have any information on the light?
Using a 5V supply will most likely be fine. You can grab an LM 7805 for this.
If it is powered from 3 AAA it will most likely be well within the limits of that LM 7805.
You 'may' need a small heatsink on it though depending on the current draw from the LEDs.
How long do 3 AAAs last in the unit?

5. ### Laplace

1,252
184
Apr 4, 2010
Would a single resistor be a simple enough circuit? The resistance value would depend on the current needed to drive the LED.

hevans1944 likes this.
6. ### hevans1944Hop - AC8NS

4,543
2,116
Jun 21, 2012
Do the three AAA cells power anything else other than the red LED? Like, a circuit of some sort that might be sensitive to the voltage supply? If it is just the LED you want to power from 12 VDC, then @Laplace suggestion is the best way to do this.

7. ### mdjumethun

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May 27, 2013
@paul60 You can use voltage divider formula......
you say you have 12 volt dc. your need 12v-4.5v = 7.5 volt dc. please see blew .....

You can use ....
R1 = 1K ohms
R2 = 1.7K ohms
V1 = 12 V
V2 = 7.5 V

I hope this formula help for you
Thanks

8. ### Harald KappModeratorModerator

10,614
2,372
Nov 17, 2011
@mdjumethun The OP needs to drop the 12V down to 4.5V, not 4.5V from 12V. That would be the voltage acrossR1 in your circuit.
But: The OP wants to draw significant currrent (he hasn't stated how many mA) to light an LED light. The current to the LED light will significantly change your equations as they are valid for an unloaded resistive divider only.
In ascending order of circuit effort: possibly Laplace's idea is the simplest solution, followed by Gryd3's idea of using a zener diode (for varying load currents) followed by a full fledged voltage regulator.

9. ### paul60

22
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Dec 11, 2010
Ok guys thanks for all the help very much appreciated