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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,

  • and please subscribe for notifications when new videos are ready.

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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