19
\$\begingroup\$

For the sake providing as much current as possible to a small circuit, regulated down from ~9V to 5V (or 5V->1.5V), I have looked at some possible options. What I was originally going to do (maybe a regulator for a solar cell, or 9v battery) is the I assume standard used LM7805 (5v) IC. I have read that it does use up a small, but fair bit of current to do this especially when only 50-100mA peak current is available.

Would a Zener diode rated at ~5 volts be able to do this more efficiently, as it should keep the voltage at or very close to 5V for quite some time higher, "regulating" it?

Would a (MOS|J)FET/other transistor (if more efficient, ignoring the slightly weird use) or something of that sense be able to lower the voltage with a very simple energy conversion?

\$\endgroup\$

3 Answers 3

Reset to default
34
\$\begingroup\$

Linear regulators like the 7805 are inefficient, and more so when the input voltage is higher. It works as a variable resistor, which varies its value to keep the output voltage constant, here 5V. That means that the current consumed by your 5V circuit also flows through this variable resistor. If your circuit dissipates 1A then the power dissipation in the 7805 will be

\$ P = \Delta V \cdot I = (9V - 5V) \cdot 1A = 4W \$

4W in a single component is rather much, the 5W in your circuit will probably be distributed over several components. It means that the 7805 will need a heatsink, and that's most often a bad sign: too much power dissipation. This will be worse with higher input voltages, and the efficiency of the regulation can be calculated as

\$ \eta = \dfrac{P_{OUT}}{P_{IN}} = \dfrac{V_{OUT} \cdot I_{OUT}}{V_{IN} \cdot I_{IN}} = \dfrac{V_{OUT}}{V_{IN}}\$

since \$I_{OUT} = I_{IN}\$.
So in this case \$\eta = \dfrac{5V}{9V} = 0.56 \$ or 56%. With higher input voltages this efficiency will even get worse.

The solution is a switching regulator, or switcher for short. There are different types of switcher depending on the \$V_{IN}/V_{OUT}\$ ratio. If \$V_{OUT}\$ is less than \$V_{IN}\$ you use a buck converter.
While even an ideal linear regulator has a low efficiency, an ideal switcher has a 100% efficiency, and actual efficiency can be predicted by the properties of used components. For instance there's a voltage drop over the diode, and resistance of the coil. A well designed switcher may have an efficiency as high as 95%, like for the given 5V/9V ratio. Different voltage ratios may result in somewhat lower efficiencies. Anyway, 95% efficient means that the power dissipated in the regulator is

\$ P_{SWITCHER} = \left(\dfrac{1}{\eta} - 1\right) \cdot P_{OUT} = \left(\dfrac{1}{0.95} - 1\right) \cdot 5W = 0.26W \$

which is low enough not to need a heatsink. As a matter of fact the switching regulator itself may be in a SOT23 package, with the other components, like coil and diode SMDs as well.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ I appreciate the time you have taken to write out this answer, it is certainly one that has made be wonder - and I can always looking forward to improving efficiency down the road if need be! \$\endgroup\$
    – Hobbyist
    Oct 21, 2011 at 9:24
  • \$\begingroup\$ @John - You're welcome, glad to be of help. Just let us know if you have further questions on this. \$\endgroup\$
    – stevenvh
    Oct 21, 2011 at 9:32
  • \$\begingroup\$ @stevenh great answer! Most people probably don't care about this (?), but isn't it true that switchers typically (1) don't come in through hole packages, and (2) require more external circuitry and system design consideration (e.g. power ripple) than comparable linear regulators? \$\endgroup\$
    – vicatcu
    Oct 21, 2011 at 15:10
  • \$\begingroup\$ @vicatcu - indeed, they're mostly SMD. Design needs special attention if you want the highest efficiency and low radiation. May be the reasons why they're not PTH = bigger parts = longer loops. \$\endgroup\$
    – stevenvh
    Oct 26, 2011 at 20:21
  • \$\begingroup\$ Using a switcher however introduces noise unlike the linear regulators. In your case, dissipating 100mA in a linear regulator is actually not that bad (9-5)*0.1=400mW. I would use a linear regulator over a switcher in your situation. However, if you have circuit that takes 48V down to 5V and draining 200mA, you will need a switcher (8.6W !!!) \$\endgroup\$
    – lucas92
    May 24, 2016 at 20:22
8
\$\begingroup\$

The most efficient way would be to use a Buck Converter which is a type of Switching Regulator.

This type of regulator is far more efficient than a linear regulator, as it converts the power rather than intentionally dissipating the extra as heat.

Wasted power in your example if your circuit is drawing 100mA would be roughly:
(9V-5V) * 0.1A = 0.4W
Wasted power at 1A it would be around (9V-5V) * 1A = 4W.
This will be only about 55% efficient compared to maybe 80-95% efficiency for a switching regulator.

Here is an example part picked at random.
Some more examples here (select different check boxes to refine the search - I just selected 5V output)

\$\endgroup\$
2
  • \$\begingroup\$ Your fourth link returns invalid query, however I am taking a look now. Thank you for that insight, it appears some similar chips are available on eBay for not an awful cost (I just have to look, so many sections.) May wait awhile and test the two on different loads, may be only slightly beneficial. \$\endgroup\$
    – Hobbyist
    Oct 21, 2011 at 6:19
  • \$\begingroup\$ @JohnS. - fixed the link. eBay, Mouser, Digikey, etc - any decent electronics supplier should stock a suitable part for this. The operation/setup is a bit more complex than a linear regulator, but if you follow the datasheets advice carefully (inductor selection particularly important, usually there will be some recommended parts in the datasheet) you should be fine. You can also buy "ready rolled" modules if you don't mind spending a bit extra to make life simpler, eBay should have plenty of these too. \$\endgroup\$
    – Oli Glaser
    Oct 21, 2011 at 6:36
1
\$\begingroup\$

I understand that this is a pretty old thread, but I noticed that murata semiconductor makes a DC/DC converter that is a drop-in replacement for 78xx series ICs! "OKI-78SR Series" 85% efficiency and a 3-pin TO-220 style package. http://www.mouser.com/ds/2/281/oki-78sr-56393.pdf

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.