We have an inductor, a transistor, a diode, and a capacitor.

We also have a device to which we have to deliver power,

represented here by a light bulb.

The connection to the gate of the transistor is not shown,

but this is a voltage that we control.

Suppose we control this gate voltage in a way such that

the transistor behaves like a switch, which we can turn on or off.

If we close the switch, we will cause a DC voltage to appear across the inductor.

The current through the inductor will keep increasing,

so long as the switch is closed.

The current through an inductor can not change instantaneously.

Therefore the moment we open the switch,

the inductor will create a force causing the current to continue flowing.

Suppose we keep turning the switch on and off.

By continuously turning the switch on and off,

we have created a DC output voltage at the light bulb

that is higher than the voltage of the battery.

We call this a “Boost Converter.”

We can control the value of the DC output voltage at the light bulb

by controlling the percentage of time that this switch is on.

Suppose that we leave the switch off all the time.

With the switch off 100% of the time,

the output voltage will equal the voltage of the battery.

What if we instead have the switch on most of the time?

If the switch is almost always on,

then assuming we have ideal components,

the current will theoretically keep increasing to infinity.

During the brief moment when we turn the switch off,

an infinite current would charge the capacitor to an infinite voltage.

By selecting the appropriate percentage of time that the switch is on,

we can set the output voltage to any value that is larger than the battery voltage.

But, we will need a different circuit to produce a

steady state output voltage that is smaller than the battery voltage.

At first, this may seem simple,

as this can be done by using resistors to create a voltage divider.

But, the problem is that we want to be efficient,

and resistors dissipate energy as heat.

Any energy that is lost as heat in a resistor

is not delivered to the device which we wish to power.

If we don't care about efficiency,

but just want to be able to adjust the output voltage,

we can do it through a circuit such as this one.

The energy lost as heat in the transistor is the voltage across the transistor,

multiplied by the current passing through it.

Now, consider the following.

If we operate and think of the transistor as an ideal switch,

then when the switch is off, the current is zero…

And when the switch is on, the voltage across it is zero.

Therefore, in theory, if the transistor is either fully “on” or fully “off”,

and never somewhere in between, the voltage multiplied by the current is always zero,

and there is no energy dissipated as heat in the transistor.

But, if we want to be 100% efficient, we also have to get rid of the resistor.

Suppose that in this circuit, we always operate the transistor as an ideal switch.

If we keep opening and closing the switch,

the “average” voltage across the light bulb

will be less than the voltage of the battery,

but this is only because the voltage across the light bulb

is zero when the switch is open.

We can try to prevent the voltage from dropping to zero

by adding a capacitor across the light bulb.

But now, when we close the switch,

we are trying to change the voltage of a capacitor instantaneously,

which is impossible.

As a result, an extremely large current will flow to charge the capacitor,

and this can cause considerable damage.

We can try to reduce this current by adding a resistor.

But, as we said before, we do not want a resistor

because resistors dissipate energy as heat.

Instead of using a resistor, we can limit the current by using an inductor.

An ideal inductor does not dissipate energy as heat.

But, since the current through an inductor can't change instantaneously,

the inductor will force the current to keep flowing through the switch

even after we open it, which is an extremely dangerous phenomena.

We can prevent the current from flowing through the open switch by adding a diode.

The diode will give the current a different path to flow through.

We call this circuit a “Buck converter.”

We can control the value of the DC output voltage at the light bulb

by controlling the percentage of time that this switch is on.

This allows us to create any DC voltage at the light bulb

that is lower than the voltage of the battery.

Assuming that we had ideal components, this circuit would be 100% efficient.

This is because ideal capacitors, ideal inductors, and ideal switches

do not dissipate any energy as heat.

We are thinking of the transistor as an ideal switch.

An ideal diode can also be thought of as a switch.

The diode is an “off switch” when it is blocking current

from flowing in the reverse direction,

and the diode is an “on switch” when current flows in the forward direction.

The Buck converter uses the exact same components

as the Boost converter, just arranged differently.

In the real world, of course, these components are not ideal,

and they do in fact dissipate power.

Therefore, to maximize efficiency, actual power converts

are typically more complicated than the circuits shown here.

But, the basic principles are the same.

These basic principles are that transistors are operated as switches.

We never try to change the voltage across a capacitor instantaneously.

And we never try to change the current through an inductor instantaneously.

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