2
\$\begingroup\$

I've been reading for hours about how switching regulators work but I'm still unable to understand what it does to keep the power equity between the input and the output (ignoring internal losses).

Let's say you step down 5v to 2v with a buck converter. What happens with the 3v dropped? Are somehow "converted" to current?

Another simple example: Let's say I have a 3.7v battery and I want to feed an IC which has an input VDD range of 1.8 to 3v so I use a buck converter to step down the voltage. If I step down the voltage to 1.8v would the battery last longer than if I step down the voltage to 3v?

\$\endgroup\$
10
\$\begingroup\$

Yes, crudely put, it converts the voltage into current. Ignoring the internal losses, if you halve the voltage you double the (available) current.

So if you convert a 5V 1A input into a 2V output the current capacity becomes 2.5A minus internal losses.

Say you have a switching efficiency of 85%, then you can work it out as:

$$ \frac{5V}{2V} = 2.5 \times 0.85 = 2.125 $$ $$ 1A \times 2.125 = 2.125A $$

You can think of it as the DC equivalent of a transformer for AC.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ So what happens with the current excess if the load only sinks 1A of the 2.125A available? (This is probably related to the answer to the second example) \$\endgroup\$ – Mark Hardy Jan 25 '15 at 16:10
  • 1
    \$\begingroup\$ Ok, I think I got it. The buck converter will sink just enough current to satisfy the load demand, so following your example, if the load sinks 0.5A the buck converter will sink 0.5/2.125 = ~0.23A. I'm right? \$\endgroup\$ – Mark Hardy Jan 25 '15 at 16:27
  • 1
    \$\begingroup\$ Yep, you got it. Constant voltage, variable current. Unless of course you have a constant current source, when the voltage will change to satisfy P=VI. \$\endgroup\$ – Majenko Jan 25 '15 at 16:37
  • \$\begingroup\$ Thanks Majenko! Now it's crystal clear to me. Much appreciated. \$\endgroup\$ – Mark Hardy Jan 25 '15 at 16:42
5
\$\begingroup\$

In a switching regulator, energy is transferred indirectly using an inductor. In one part of the duty cycle, the source energizes the inductor, increasing its current. In the other part, the inductor drives the load. This is easiest to see in a buck-boost converter. Ideally, the switches and inductor/capacitor are lossless, so no power is lost. This is totally different from a linear regulator, which deliberately wastes power to control the voltage.

In a buck or boost converter, some energy is transferred directly from the source to the load as well, but the same principle applies.

You can also look at a buck converter as an L-C filter on a square wave from the source. Again, all components are lossless, so there's no waste. This is basically how a class D power amplifier works.

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

If an ideal continuous-mode buck regulator were e.g. turning 9 volts into 6 volts and the output were drawing 1 amp continuously (i.e. 6 watts), then the input would spend 2/3 of the time supplying one amp at nine volts (i.e. 9 watts), and the other 1/3 of the time supplying zero current.

Real regulators differ from ideal ones, of course, but the principle of operation would be the same. The biggest differences from an ideal regulator would be that the duty cycle would need to be slightly greater than 2/3, and that the output current wouldn't be an absolutely continuous one amp but would instead increase any time the device was drawing current from the source and decrease whenever it wasn't. A smoothing cap would generally be required on the output to prevent the voltage from rising and falling as a consequence of this.

Nonetheless, I think it's helpful to recognize the following principles:

  1. A continuous-mode buck regulator produces continuous current from intermittent current.

  2. A continuous-mode boost regulator produces intermittent current from continuous current.

  3. A flyback regulator, even a continuous-mode one, will draw current and supply current intermittently ("continuous mode" means that it's always either drawing or supply current, but unlike the other topologies a flyback-mode supply can never do both simultaneously).

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

Remember that voltage does not carry energy or power on its own, only in combination with current. Power is proportional to voltage times current, so the converter is allowed to use more voltage/less current on one side and less voltage/more current on the other side, with the power of each being equal (ideally, output actually has less power due to efficiency <100%).

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

In a linear regulator The current on the input == current on the output (if you ignore quiescent power of the linear reg for a moment).

The linear regulator drops the excess voltage via a pass transistor which dissipates this as heat.

In a switching regulator the current on the input != current on the output. This is done via switching an inductor ON and OFF to transfer energy from the input to the output. The tradeoff is voltage & current ripple.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

A "switching" regulator switches between two modes.

In the case of a buck converter with the switch closed the inductor is placed in series with the load. This stores energy in the inductor. The greater the voltage difference between input and output the more energy is stored in the inductor.

When the switch is opened. Current continues to flow through the inductor but the current now comes from the ground rail via the diode rather than from the input terminal. The voltage across the inductor is inverted and the energy stored in the inductor is discharged into the load.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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