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• Suppose we want to increase the voltage of our only battery.

• 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 fullyonor fullyoff”,

• 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,

• theaveragevoltage 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 anoff switchwhen it is blocking current

• from flowing in the reverse direction,

• and the diode is anon switchwhen 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.

• Much more information is available in the other videos on this channel,

Suppose we want to increase the voltage of our only battery.

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B1 US voltage switch transistor current ideal light bulb

# Boost Converters and Buck Converters - Power Electronics

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Amy.Lin posted on 2021/01/06
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