I have a 12V DC voltage and need to draw around 1A at 1.2V. A quick calculation gives me the following results:

\$P_{dissipated} = (12V-1.2V) \times 1A = 10.8W\$

That's huge!

Can I use multiple regulators like 12 to 9, 9 to 7, 7 to 5, 5 to 3, and 3 to 1.2?

That would give me the following power dissipations in separate regulators: 3W, 2W, 2W, 2W, 1.8W.

Is that ok? And the two voltage stages are higher than the drop out voltage of the regulators.

  • 2
    \$\begingroup\$ en.wikipedia.org/wiki/Rube_Goldberg_machine \$\endgroup\$ Dec 16, 2013 at 3:08
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    \$\begingroup\$ Can you use a switch-mode step down regulator (buck converter)? They are usually 90% efficient, so the 1.2W output would cause a 0.12W loss. \$\endgroup\$ Dec 16, 2013 at 3:10
  • \$\begingroup\$ A far more sane approach would be to use an inexpensive buck converter such as advertised here (for just $5.09!): ebay.com/itm/… \$\endgroup\$ Dec 16, 2013 at 3:22
  • \$\begingroup\$ @AlfredCentauri : Haha, really funny :D \$\endgroup\$ Dec 16, 2013 at 3:22
  • \$\begingroup\$ @NickAlexeev : Thank you, wasn't aware of them ! I will take a look but shouldn't be a problem from what I've read for the moment. \$\endgroup\$ Dec 16, 2013 at 3:23

2 Answers 2


There's nothing conceptually wrong with using a string of voltage regulators to drop voltages as long as you take the drop-out voltage between each stage into account. It's quite a common approach when multiple voltages are needed by different parts of a circuit.

If you only require the lowest voltage and want to stick with a linear regulator there normally isn't a lot of point. Your total power dissipated has just been shared across devices and a more direct approach would be to use something like a LM317K in a TO-3 package with a suitable heatsink.

But usually when dropping that amount of voltage at any significant current a switching step-down converter also known as a buck converter is more efficient and ends up being cheaper. Many pre-built adjustable output voltage modules are available on e-bay.

If you prefer something you can include in your design and solder to your PCB there are also many easy to use devices around such as the TI Simple Switcher range that include a web-based design tool. While I'm not aware of any with a 1.2V output there is also the Murata 78xxSR Series and similar devices that are switching regulators with a LM78xx pinout.

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    \$\begingroup\$ even if you can't easily get a switcher to output the 1.2v you can have it output something close and get a linear regulator to trim it down from there, that will reduce the power dissipation on the linear and result in less ripple. \$\endgroup\$
    – Gorloth
    Dec 16, 2013 at 6:14
  • \$\begingroup\$ A quick search at LT gave 63 hits for buck converters matching the spec. \$\endgroup\$
    – Lundin
    Dec 17, 2013 at 12:15

A buck converter is the 'best' solution. But if you want to use a linear regulator you can arrange to drop most of the voltage and power in a resistor.


  • Say, \$V_{in}min = 12V\$
  • Regulator dropout voltage. Typically \$V_{dropout} = V_{out} + 2V\$, but varies greatly with regulator.
    Use say \$V_{dropout} = 2V + 1.2V = 3.2V\$.
  • Say, \$I_{max} = 1A \$.

The maximum allowed resistor size is given by this formula:

$$Rs=\frac{V}{I}=\frac{$V_{in}min-V_{dropout}}{1A}=\frac{12-3.2}{1} = 8.8 \Omega$$

Use a slightly smaller value to give slightly more "headroom" rather than too little. Say 8R2 or 6R8. Wattage of resistor needs to be chosen appropriately.

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    \$\begingroup\$ If energy is to be dissipated, a dedicated power resistor is a much better idea than giant heatsinks on active elements. \$\endgroup\$
    – oakad
    Dec 18, 2013 at 1:28
  • \$\begingroup\$ @oakad - Yes. As above. BUT the resistor needs to be designed for heat-sink-less dissipation. Some power resistors need to bolt to a suitable heatsink to dissipate more than a small % of their rated power. You can buy power resistors that will free-air dissipate hundreds of Watts without a heatsink, but cost is usually high. Nichrome wire can be used to make your own free-air dissipating power resistors. \$\endgroup\$
    – Russell McMahon
    Dec 18, 2013 at 9:55

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