# Voltage regulation

• posted

Suppose you have a 12V supply and need to power a 5V device. If that device is a simple resistive load like say an incandescent light bulb, you can simply use the right size resistor to drop the voltage. If the device is more complex and does not draw a constant current, the resistor obviously won't work. Something like a 7805 would be the easiest solution. In both cases you are wasting a lot of power. If the first case over half the power is just generating heat in the resistor. I assume the same, or close to the same, amount of power is lost in the 7805 regulator.

If you needed to light two 5V light bulbs then you could hook them in series and waste a lot less power by using a smaller resistor.

Now for my question, can you do something similar by using 2 7805 regulators hooked up in series to power 2 separate loads? Alternatively and more what I want, can you also have the outputs hooked up in parallel to power one 5V device?

```--
Chris W
KE5GIX```
• posted

The use of a "BUCK" regulator works well in these cases to save power. Look at the below example.

"

• posted

As others have said, use of a buck switching regulator will save on the power waste.

Rather than connect two 5V regulators in series, it is possible (and simple) to make a variable supply from any regulator. See the datasheet for the LM317 regulator (essentially a 1.2V regulator) for how to do it.

And rather than connect two regulators in parallel, a greater current capacity can be realized by using a bypass transistor and a single regulator. You might find a schematic in a regulator datasheet, or perhaps through Google.

Regards,

Mark

• posted

You got some good answers, but I'm not sure if any explained why the series idea won't work. You cannot hook up 2 7805 regulators in series:

Input---7805---7805---output

The first one will produce 5V output. That output is too low to allow the second 7805 to work properly. The 7805 needs a minimum of about 8 volts input.

Ed

• posted

Perhaps he meant something more along these lines:

------

12V --o--------| 7805 |---- 10V out | | | | ------ | | | | | ------ | `--| 7805 |--' | | ------ | | ----- --- -

-- Mark

• posted

I think this might work too, and they're in series:

------

12V --| 7805 |---- 10V out | | ------ | | | ------ `--| 7805 |--o | | ------ | | ----- --- -

BUT here is what is actually suggested in datasheets. Choose R1 and R2 so that:

10V = 5V*( 1 + R2/R1) + Icom*R2

------

12V --| 7805 |--o-- 10V out | | | ------ \\ |com / | \\ R1 | / | \\ | | `------o | \\ / \\ R2 / \\ | | ----- --- -

-- Mark

• posted

You can do the first thing, connecting two of them (although ehsjr's point is a good one, and the 'drop out' of the 7805 regulators will cause a problem with 12V).

However, the second one, connecting them in parallel, does not work. The problem is that you need a ground reference, and the reference will be different for the two regulators. There is no good way to hook them up in parallel in the way you describe.

On the other hand, the industry has decided that the best way to do this kind of thing is with a 'switched mode power supply', or SMPS. That circuit uses devices that store energy (inductors) to offer power output at a lower voltage using a larger voltage. The power efficiency can be in the 90% range with good switchers. There are cheap chips that do most of the heavy lifting for you, and some companies specialize in building those chips, like