Multiple voltage regulators?

Question

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.

Answer 1

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.

Answer 2

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.

Establish:

• 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.