I need to make an 11 volt voltage regulator. I would really like this to be step-down, not a linear voltage regulator because I am going to be drawing between 1-2 amps at 11 volts, and if I was using a linear voltage regulator it would get very warm. The input voltage that I am using is going to be about 12.8 volts.

I know a linear regulator under these circumstances would have 86% efficiency, but they are not reliable when working with voltages close together.

The last option I can think of (which is still linear...) would be to use resistors to lower the voltage, but I don't think that that would be any cheaper than using a step down converter. Because the resistors would have to "handle" 3.6watts.

There are regulators like this on eBay (which I am bidding on, but only up to $2), should I just buy something like that? Or would it be cheaper to make it myself?

Even if it is cheaper, I don't know how, and that's why I'm here :)

  • 4
    \$\begingroup\$ 22.6 Watt ?? 2 Amp at 1.8 Volt difference is almost 4 Watt. Are you sure the input voltage will always be 12.8 Volts? A linear regulator will be 86% efficient, and probably much easier. \$\endgroup\$ Jun 29, 2012 at 18:20
  • 2
    \$\begingroup\$ You sure you need to step down the voltage? what are you powering? cannot it not handle 12.8 V input? \$\endgroup\$ Jun 29, 2012 at 22:35

6 Answers 6


That eBay thing is a switching regulator, aka a "switcher", aka SMPS (Switch-Mode Power Supply). These things can indeed reach efficiencies of 95 %, exceptionally 96 %. Lot of it depends on input and output voltage, and the highest efficiency is with parts that are designed for a specific input and output voltage. So the eBay thing won't always be as efficient, especially not at low output voltage or high input/output voltage ratio.

You can make them yourself; as you can see they only require a few parts, but designing a high efficiency switcher requires some experience to choose the right parts and make the right PCB layout. So I would suggest you buy one. I guess a component cost for the eBay module will be around 6 or 7 dollar, so 2 dollar would really be a bargain.

You can't use the series resistor to regulate to 11 V. At 2 A and 12.8 V in you would need a 0.9 Ω resistor, but when the current drops to 1 A that resistor's voltage drop will be reduced to 0.9 V, and the output will rise to 11.9 V.


Your numbers don't make sense.

With 12.8 V in and 11 V out, the efficiency of a linear regulator would be 11V/12.8V = 86%. That's in the range of ordinary switchers, so unless you can get or make a really good one you might as well use a linear regulator. Just as with your 92% number, if have no idea where you got the idea that linear regulators can't be reliable with these voltages. That just plain nonsense.

With a linear regulator in the worst case it will dissipate (12.8V - 11V) * 2A = 3.6 W. If you can arrange to live with that, it is probably your simplest option. As Steven said, switching regulators, which would be buck converters in this case, can do better, but that will not be a trivial design to get the efficiency you want. You will probably need synchronous rectification, good low Rdson pass elements, etc. It is doable, but is not a novice project.

Perhaps there are switcher chips out there that provide the drive for the high side switch and the low side syncronous rectifier all built in, and you add the FETs and inductor. In that case it could be worth a try, but you have to read the datasheet carefully and look at the app notes and understand why things were done as they were. We can help here if you run into specific questions.

  • \$\begingroup\$ Yeah. I have no idea where I got 92%... Thanks. \$\endgroup\$
    – Sponge Bob
    Jun 29, 2012 at 19:30
  • \$\begingroup\$ spot on - 86 % is pretty good and you are unlikely to get a switcher to be more efficient than that. Even if precisely designed for that specific input output ratio it can be difficult \$\endgroup\$ Jun 29, 2012 at 22:34
  • 1
    \$\begingroup\$ Synchronous rectification is always better, but in this case a diode won't do much harm: the output voltage is relatively high, and the diode's duty cycle is low due to the low in-out voltage difference. \$\endgroup\$
    – stevenvh
    Jun 30, 2012 at 5:27
  • \$\begingroup\$ @stevenvh: Yeah, good point. The most important thing in this case is a very efficient switch, like a very low Rdson FET, or possible multiple in parallel. However as you parallel them you get more gate drive losses. Finding the sweet spot would be part of the detailed design task. \$\endgroup\$ Jun 30, 2012 at 14:14

Let's do some math and figure out if we can design this ourselves.

For simplicity, let's assume that your specified efficiency \$ \eta = 0.95 \$ applies to full power.

\$ 11.6V \times 2A = 23.2W\$ maximum output power.

\$ \dfrac{23.2W}{0.95} = 24.421W \$ maximum input power.

\$ 24.421W - 23.2W = 1.221W\$ which is your total power loss budget. This represents the total losses available for the entire converter - conduction losses, switching losses, gate drive losses, the losses of the control circuitry, etc.

The best chance you have would be with a buck converter using synchronous rectification, and the lowest \$ R_{ds(on)}\$ and \$Q_g\$ MOSFETs you can get to minimize conduction and gate drive losses.

There are tons of application notes and design aids out there that explain how to design a synchronous buck.

Remember: the most efficient regulator is no regulator at all. Input = output hence 100% efficiency. : )


One consideration that has not been addressed is just how good a supply is your 12.8 volts. If it can get higher, then the efficiency of the linear regulators discussed will decrease. If it gets too close to your output requirement of 11 volts, then the regulator may lose regulation unless it is carefully designed to handle a small output/input differential. How sensitive is your load to the actual voltage of the power supply? A circuit designed for 11 volts may very well work fine at 12.8 volts and thus there is no need for a regulator.


I think that you should use a power rated diode to drop the voltage. These won't have any effect other than dropping the voltage and wont generate the heat associated with resistors in series. Its fairly easy to get 12.8v down to 11v. Diodes present a constant voltage drop and lower power dissipation than rsistors without causing voltage fluctiation on the current draw. If you've got a steady 12.8 volts and need 11, use a large barrier diode with a voltage drop of 1.8 volts...though 0.7 to 0.9 volts is more common. You may have to use two.

  • \$\begingroup\$ This also has a lower overall efficiency the same as a linear regulator as mentioned in another answer. \$\endgroup\$
    – PeterJ
    Jul 19, 2014 at 4:11

Some very advanced mechanical motor-generator pair on rubys in vacuum in weightless conditions can possibly reach 95% efficiency. But cost will certainly exceed the $2 budget.

  • \$\begingroup\$ It's not too hard these days to get 95% efficiency in a buck converter when Vout is close to Vin. \$\endgroup\$
    – John D
    Mar 10, 2014 at 18:28

Your Answer

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

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