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I am planning on wiring 4 pin momentary push switch button that has a LED as well. The switch itself is fine with 20V, but the only problem is that the power source input is 20V and the LED requires 12V.How should I lower the voltage?

I was thinking of a voltage regulator, but I thought it is easier to work with color coded resistors, since I will be working on tight space. Now I salvaged the resistors and found 47Ohm and 33.4Ohm resistors.

When I did the simple math

$$ V_{DIV} = 20V * {47 \over (47+33.4)} = 11.6915V $$

which is \$ 0.3V \$ lower than \$ 12V \$. Is this okay to be used?

Bear in mind I'm a newbie.

Schematic showing a voltage divider

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    \$\begingroup\$ This is actually a better duplicate: Using voltage divider in a circuit, but there are probably a dozen others. \$\endgroup\$
    – The Photon
    Aug 15 '17 at 4:14
  • \$\begingroup\$ In electronics, the words Voltage Divider and Power usually do not mix together. When you do the simple math, don't forgot the simple assumption: no current drawn from Vout! This means that there is no problem to use it to divide the voltage of a Signal, but you can't use it like that with a Power Supply, which is what you want to do here. \$\endgroup\$
    – Edesign
    Aug 15 '17 at 7:49
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You can't use a voltage divider to power a load. That will almost never work.

You have two options: - Step down 20V to 12V using a LDO or switching regulator - Get your 12V out of the resistor network in an op-amp buffered with a NPN transistor

This describes accurately the problem with voltage dividers: http://www.learningaboutelectronics.com/Articles/Voltage-follower

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