It is well-know that switching regulators are more efficient then linear regulators. I also know that linear regulator have to dissipate the difference between the input voltage and output voltage times the current as heat.

But why does this not apply to switching regulators with the same conditions: same input voltage and output voltage and current?

I know switchers can get hot; I have one on a board that gets so hot you can barely touch it, but then again it's only 2 1/2 millimeters on each side and looks like an ant compared to a through-hole 7805 with its heat sink.


7 Answers 7


Linear regulators work by effectively putting a controlled variable resistor between the source and load. All of the current for the load flows through this resistive element. And the voltage across it is equal to the difference between source voltage and load voltage. So the power dissipated is

\$P_{lin} = I_{load}\times{}(V_{src}-V_{load})\$.

Switching regulators work by changing the duty cycle of the current flow over a switching cycle, then averaging out the output using a filter. During part of the cycle a high current flows with a low voltage drop. During the other part of the cycle almost no current flows with a high voltage drop. Neither of these conditions dissipates much power as heat. Ideally the power lost becomes

\$P_{sw}= \mathrm{DC}(I_{on})(0\ \mathrm{V}) + (1-\mathrm{DC})(0\ \mathrm{A})(V_{off})\$,

which is, of course, 0 W. Typically much of the inefficiency in the real world is due to power lost during the very short switching inteval between the "on" and "off" parts of the cycle.

  • 1
    \$\begingroup\$ +1. ...and a few more types of losses in SMPS: magnetic core losses, copper losses in magnetics (windings have parasitic resistance), losses caused by gate driving. \$\endgroup\$ Apr 7, 2014 at 21:49
  • \$\begingroup\$ I knew switchers chopped up the input, but didn't realize (duh) they varied the duty cycle to perform the regulation. \$\endgroup\$
    – tcrosley
    Apr 7, 2014 at 23:41
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    \$\begingroup\$ @tcrosley Even if you had already known the answer, this would have been an excellent question for educating future readers. \$\endgroup\$
    – The Photon
    Apr 8, 2014 at 0:15

Usually switching regulators are more efficient, but not always.

An ideal linear regulator has a voltage drop \$V_{IN} - V_{OUT}\$ and there is a linear pass element such as a transistor that acts like a resistor, so the power loss in the ideal case is P = \$I \cdot (V_{IN} - V_{OUT})\$, as you say. That's the ideal case, in reality the regulator needs a bit of current to work, and there may be a component that depends on the output current. Some LDO linear regulators that depend on lateral PNP pass elements can have very high consumption close to the dropout- perhaps 100mA wasted for 1A output current (because PNP transistors made with some IC processes tend to be have pretty crappy current gain).

An ideal switching (buck) regulator looks like this:


simulate this circuit – Schematic created using CircuitLab

Where the switch is a transistor, and D1 may be a diode or it may be another transistor. In the ideal case, there is no energy loss mechanism. The diode either blocks perfectly or conducts perfectly, the switch does the same, the inductor has no DC resistance, and the capacitor has no ESR. So the power in equals the power out. Of course reality can only approach that ideal. There will be losses that are 'overhead' and losses that increase with increasing current.

Note that the inductor is a critical part of this circuit- if you tried to omit it, the immovable (in the short term) voltage on C1 would come up against the immovable voltage on Vin and the current would become infinite. In a real circuit, SW1 would have some resistance and it would get as hot as the pass transistor in the linear regulator (except it would also be making tons of EMI).

  • \$\begingroup\$ Your last paragraph isn't exactly correct. A large capacitor with no inductor would still give you much better efficiency than a linear regulator. The drawback is a much higher amount of voltage ripple and more stress on the switch. \$\endgroup\$
    – horta
    Apr 7, 2014 at 18:49
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    \$\begingroup\$ @horta I disagree. Say the output current is 0.1A, duty cycle 0.1% (big capacitor). The switch current will be 100A for 0.1% of the time, and the voltage drop across the switch will be (VIN-VOUT) so the loss will be 0.1A * (VIN-VOUT) just as with a linear regulator. \$\endgroup\$ Apr 7, 2014 at 18:57
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    \$\begingroup\$ Ah, thanks for the example. That cleared it up. I definitely learn more when I'm wrong. \$\endgroup\$
    – horta
    Apr 7, 2014 at 19:07
  • \$\begingroup\$ I must have been thinking of a PWM without a capacitor or an inductor where it's just a chopper, but in that case, there's no voltage regulation at all. Since the full voltage then drops across the load, you still gain efficiency. \$\endgroup\$
    – horta
    Apr 7, 2014 at 19:15
  • 1
    \$\begingroup\$ @horta Yes, if you can PWM directly to a load (such as a heater or LED), you're way ahead of the game. \$\endgroup\$ Apr 7, 2014 at 19:39

It is well-know that switching regulators are more efficient then linear regulators.

To a point. Putting 3.5V into a LDO 3.3V linear regulator gives an efficiency of 94%. You'd be hard-pressed to find a switching regulator that can do that.

I also know that linear regulator have to dissipate the different between the input voltage and output voltage times the current as heat.

Yes, but linear regulators must draw as much or slightly more current for a given output current, whereas switching regulators trade the drop in output voltage for an decrease in input current, and therefore usually use less power than a similarly configured linear regulator overall.


Ideal switchers do not dissipate any power. They take a bit of power from the input side, store it and then release it on the output side.

Energy is stored either in a magnetic field inside an inductor or an electric field in a capacitor.

Due to non-idealities of real components, like ESR in inductors, they dissipate a bit of power. They also loose some power during transistor switching. Some energy is also lost in the controller.


But why does this not apply to switching regulators with the same conditions

For a series linear regulator, the source delivers power 100% of the time and some of this power must be wasted since (1) the source voltage (magnitude) is larger than the load and (2) the source current must be somewhat greater than the load current.

However, for a switching regulator, the source delivers power only over some fraction of a switching period. During this time, some of the power delivered by the source is delivered to the load and the rest is delivered to energy storage circuit elements - very little is wasted.

Then, during the off time, the energy storage circuit elements deliver power to the load.

This is the crucial difference - only enough power is drawn from the source during the on time to power the load continuously.

For example, if the load requires a continuous 5W, the source might deliver 10W 50% of the time and 0W the remaining 50% for an average power of 5W. The energy storage circuit elements 'smooth' the energy flow - absorbing excess power during the on time and then delivering it during the off time.


An ideal buck-boost switching regulator may be modeled as a pair of caps connected directly to the input and output, a coil, and some routing circuitry which can switch between three configurations (a buck-only, boost-only, or inverting circuit would only need two).

  1. The input connects to the output through the coil
  2. The coil is connected directly across the input
  3. The coil is connected directly across the output

Assume components behave in ideal fashion (no resistive or switching losses, etc.) the source caps sit at 10V, the output draws 1A, the switcher spends half its time in the first configuration, half in the third, and cycles fast enough that the cap voltages and coil current don't have a chance to change much during each cycle.

In "steady" state, subject to the above conditions, the coil will have one amp flowing through it all the time (since it will always be in series with a load drawing 1 amp). If the output cap is sitting at five volts, then half the time the coil will have +5V across it, and half the time it will have -5V, so on average its current will remain at 1 amp. Half the time the source cap will have one amp taken out of it (when it's connected to the coil), and half the time it will have none, so the source will see half an amp of current draw.

The simplest way to see how a switcher can draw less current from the source than the load draws from it is to look at where the electrons are flowing: half of the electrons which go through the load will come from the source, and half will be switched to bypass the source. Thus, the load will have twice as much current flowing through it as the source.


To bore everyone with the good old water-flow analogy I'll add this: assume we have three height levels H1, H½, H0; a supply of water comes from H1, then flows at H½ a bit to its destination, a mill or something, and then back all the way on H0. The regulator is at the transition from H1 to H½.

  • A linear regulator is a waterfall: the electrons just come thundering down and release their potential as thermal energy to the environment. The current on H½ will be the same as on H1.

  • A switcher doesn't just let the water flow down, but lowers it controlledly portionwise in buckets. Each bucket that comes down from H1 needs a counterweight, the natural thing to use is another bucket of water, from H0!

Depiction of water-flow analogy for switching power regulator

  • \$\begingroup\$ +1. nice illustration. Also, you might mention that a (switching) boost regulator can do something that no linear regulator can do, in the same way that a noria can do something that no waterfall can do. \$\endgroup\$
    – davidcary
    Apr 19, 2014 at 16:00

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