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I am a high school student. Please help me, i want shift a 24 volts at the input to 6v on the output!! i need a dc level shifter but i dont know its contents!!

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    \$\begingroup\$ Is this a signal or a power supply? \$\endgroup\$
    – endolith
    Commented May 30, 2010 at 11:50
  • \$\begingroup\$ this is a technicality, but what I think you're actually asking for here is a DC-DC converter. A level-shifter doesn't "scale" a voltage level, it merely "offsets" it. \$\endgroup\$
    – vicatcu
    Commented Jun 1, 2010 at 16:26

4 Answers 4

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just put in a 7806 and you're done

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This really depends on what the 6v output is going to be used for. If it is purely an act of showing you can step 24v down to 6v and show on a volt meter that it is actually 6v, you will probably want a simple voltage divider. If this is the case, Google will be your friend.

If you are needing to power something off of the 6v output, you will need a voltage regulator. As Erik said, a 7806 would do the job. There are many other things that will work also, this is another case that Google will do wonders. Just make sure to use the term "voltage regulator" and not DC level shifter. The advantage of a voltage regulator over a voltage divider is that you can have your load (what your powering) change over time and still have the same voltage output. In the case of a voltage divider, you have to treat your load as a resistor, capacitor, and inductor. All of these will effect the voltage output of a voltage divider.

If you have some sort of signal at 24v that you want to step down to 6v, then there are a lot more aspects that we would need to know in order to help you out.

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  • \$\begingroup\$ Note that regulating 24v down to 6v with a linear regulator could have serious heat dissipation issues. \$\endgroup\$
    – Mark
    Commented Jul 13, 2011 at 20:00
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There is another aspect to consider, and that is how much current you need at 6V.

As already correctly answered, to demonstrate 6 V, a voltage divider will work fine, for small amounts of current a linear regulator will be ok. For larger currents a switching regulator will be more efficient and not require a heatsink.

For example using a linear regulator: 100 mA * 18 V (24 - 6) = 1.8 W of thermal energy that you need to dissipate. Using a 10 K/W heatsink will result in a temperature raise of 18 K over ambient. That will be ok for room temperature through to hot areas, circa 80 deg C. Above this temperature you will need to check the thermal resistance of the regulator and add that to the equation to find the junction temperature and keep it below the maximum, normally 125 deg C.

Alternative using a switching regulator will increase the efficiency and not require a heatsink. An example of one is here: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=PT78ST106V-ND

This one is expensive, but there will be similar products out there that are cheaper.

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refer to lm3524 of national semiconductor data sheet to see a simple sample circuit. better refer to a datasheet book 1 and 2 there are examples along with hint on the circuit calculations.

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  • \$\begingroup\$ use pwm to convert voltage level with low dissipation. \$\endgroup\$
    – kavous
    Commented Mar 16, 2013 at 16:12

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