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Suppose we want to increase the voltage of our only battery.
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We have an inductor, a transistor, a diode, and a capacitor.
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We also have a device to which we have to deliver power,
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represented here by a light bulb.
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The connection to the gate of the transistor is not shown,
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but this is a voltage that we control.
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Suppose we control this gate voltage in a way such that
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the transistor behaves like a switch, which we can turn on or off.
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If we close the switch, we will cause a DC voltage to appear across the inductor.
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The current through the inductor will keep increasing,
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so long as the switch is closed.
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The current through an inductor can not change instantaneously.
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Therefore the moment we open the switch,
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the inductor will create a force causing the current to continue flowing.
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Suppose we keep turning the switch on and off.
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By continuously turning the switch on and off,
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we have created a DC output voltage at the light bulb
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that is higher than the voltage of the battery.
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We call this a “Boost Converter.”
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We can control the value of the DC output voltage at the light bulb
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by controlling the percentage of time that this switch is on.
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Suppose that we leave the switch off all the time.
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With the switch off 100% of the time,
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the output voltage will equal the voltage of the battery.
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What if we instead have the switch on most of the time?
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If the switch is almost always on,
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then assuming we have ideal components,
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the current will theoretically keep increasing to infinity.
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During the brief moment when we turn the switch off,
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an infinite current would charge the capacitor to an infinite voltage.
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By selecting the appropriate percentage of time that the switch is on,
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we can set the output voltage to any value that is larger than the battery voltage.
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But, we will need a different circuit to produce a
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steady state output voltage that is smaller than the battery voltage.
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At first, this may seem simple,
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as this can be done by using resistors to create a voltage divider.
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But, the problem is that we want to be efficient,
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and resistors dissipate energy as heat.
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Any energy that is lost as heat in a resistor
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is not delivered to the device which we wish to power.
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If we don't care about efficiency,
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but just want to be able to adjust the output voltage,
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we can do it through a circuit such as this one.
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The energy lost as heat in the transistor is the voltage across the transistor,
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multiplied by the current passing through it.
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Now, consider the following.
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If we operate and think of the transistor as an ideal switch,
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then when the switch is off, the current is zero…
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And when the switch is on, the voltage across it is zero.
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Therefore, in theory, if the transistor is either fully “on” or fully “off”,
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and never somewhere in between, the voltage multiplied by the current is always zero,
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and there is no energy dissipated as heat in the transistor.
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But, if we want to be 100% efficient, we also have to get rid of the resistor.
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Suppose that in this circuit, we always operate the transistor as an ideal switch.
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If we keep opening and closing the switch,
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the “average” voltage across the light bulb
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will be less than the voltage of the battery,
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but this is only because the voltage across the light bulb
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is zero when the switch is open.
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We can try to prevent the voltage from dropping to zero
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by adding a capacitor across the light bulb.
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But now, when we close the switch,
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we are trying to change the voltage of a capacitor instantaneously,
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which is impossible.
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As a result, an extremely large current will flow to charge the capacitor,
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and this can cause considerable damage.
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We can try to reduce this current by adding a resistor.
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But, as we said before, we do not want a resistor
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because resistors dissipate energy as heat.
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Instead of using a resistor, we can limit the current by using an inductor.
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An ideal inductor does not dissipate energy as heat.
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But, since the current through an inductor can't change instantaneously,
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the inductor will force the current to keep flowing through the switch
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even after we open it, which is an extremely dangerous phenomena.
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We can prevent the current from flowing through the open switch by adding a diode.
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The diode will give the current a different path to flow through.
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We call this circuit a “Buck converter.”
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We can control the value of the DC output voltage at the light bulb
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by controlling the percentage of time that this switch is on.
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This allows us to create any DC voltage at the light bulb
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that is lower than the voltage of the battery.
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Assuming that we had ideal components, this circuit would be 100% efficient.
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This is because ideal capacitors, ideal inductors, and ideal switches
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do not dissipate any energy as heat.
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We are thinking of the transistor as an ideal switch.
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An ideal diode can also be thought of as a switch.
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The diode is an “off switch” when it is blocking current
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from flowing in the reverse direction,
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and the diode is an “on switch” when current flows in the forward direction.
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The Buck converter uses the exact same components
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as the Boost converter, just arranged differently.
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In the real world, of course, these components are not ideal,
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and they do in fact dissipate power.
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Therefore, to maximize efficiency, actual power converts
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are typically more complicated than the circuits shown here.
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But, the basic principles are the same.
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These basic principles are that transistors are operated as switches.
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We never try to change the voltage across a capacitor instantaneously.
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And we never try to change the current through an inductor instantaneously.
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Much more information is available in the other videos on this channel,
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