To step-down AC voltage we use a transformer,
but, how do you step down DC voltage? We use a buck converter. What is a buck converter? In this video, we will be exploring the design
process and working of a Buck converter which is used to lower the DC voltage efficiently. If you have a source, let say DC, and a switch,
that is turned on and off periodically, you get a PWM signal. The amount of time a digital signal is in
the active state relative to the period of the signal is called its duty cycle. If the switch is on for a long duration, the
duty cycle increases, and if it's on for a short duration,
the duty cycle decreases. Now, if you calculate the average of a cycle,
with duty cycle 50 percent, it's half of the input voltage. That is, we have reduced the voltage from
12 volts to 6 volts with just a switch. For efficiency reasons, we will replace it
with an electrical switch, or a MOSFET. This MOSFET is controlled by a PWM signal. How it is generated and controlled will be
explained later in the video. But, This is just the amplified PWM signal
and it has high voltage peaks. To smooth this we add an inductor in series
with the load. This is an inductor. Inductor wants to keep the current constant
through itself and for that it will instantly change the voltage across itself. As the switch closes, the current starts to
flow. To resist this flow, the inductor drops the
voltage at another end to zero, creating equal and opposite voltage to the battery. This is possible due to the magnetic fields
which are generated in the inductor. But it can't resist it for a long time, thus
the current starts flowing and voltage at the other end starts rising. Also, the inductor starts storing the energy
in its magnetic fields. After some time the magnetic field stabilizes
and the inductor will act as a closed switch, allowing maximum current to flow. Now, If you open the switch, then there is
no source to supply current and thus current starts falling. But as you know current through the inductor
cannot change instantly. Hence, now the inductor act as a battery supplying
current but it slowly runs out of energy. As this end is open, the electrons accumulate
here creating a high negative voltage. This may damage the components. Thus we add a low voltage drop, Schottky diode
to create a path for electrons. But the voltage at the load still has high
voltage spikes. Hence, we increase the frequency of the PWM
signal, such that the voltage and current by the inductor remains somewhat stable. This is also the reason why the switch mode
power supply uses high frequencies. To further smooth-out them we add a capacitor
in parallel. This is a capacitor. Capacitor wants to keep the potential difference
constant across itself and for that it will change the current through it. As the switch closes, the voltage increases
to 5 volts. To resist this, the capacitor flows current
through itself, raising the other terminal to 5 volts. This is possible due to the electric fields
which are generated in the capacitor. But it can't flow current for a long time
as the plates get charged, thus the current reduces, and voltage at the other end starts
dropping to the ground. Also, the capacitor starts storing the energy
in its plates. After some time the plates get fully charged,
thus no current can flow, and the capacitor act as an open switch. Now, If you open the switch, then there is
no source to supply voltage and thus voltage starts falling. But as you know voltage across the capacitor
cannot change instantly. Hence, now the capacitor act as a battery
supplying current. But it slowly runs out of energy and thus
the flow of current reduces and after some time it stops, thus the voltage drops. And we have created is the buck converter. But, there are some issues, as the load changes
the voltage across the load also changes, so we need to create feedback to change the
duty cycle of the PWM signal with respect to the load, also, how is this PWM signal
getting generated? This is the complete circuit. The voltage at the output is reduced by a
voltage divider and it is fed into an op-amp which acts as an error amplifier. This component is known as an Operational
amplifier, or an Op-amp in short. It has two inputs and one output, the other
two terminals are for the supply voltage. It is used to amplify the difference between
the inputs. It compares both the input. If the voltage at the non-inverting input
or the Positive terminal is greater than the inverting one, then the output is the positive
supply voltage. Or, if the voltage at the inverting input
or the Negative terminal is greater than the non-inverting one, then the output is the
negative supply voltage. In this configuration, The op-amp wants to
keep both of its inputs at the same voltage and for that, it will change its output. As the feedback is to negative input, the
output varies according to the voltage at the negative input. This can be seen as a see-saw. When the input voltage falls, the output rises. And when the input rises, the output falls. This is how the differential amplifier works. This is a Triangle wave generator. The output of the error amplifier and wave generator
is fed to another Op-Amp. Which acts as a comparator and outputs
the PWM signal which controls the MOSFET. Here the negative input is a constant voltage
from the differential amplifier, and the positive input is the wave form
triangle wave generator. Hence, when the triangle wave is higher than
the constant voltage, The output is the positive supply voltage. Here it is plus five volts. And, when the constant voltage is higher than
the triangle wave, the output is the negative supply voltage. Here it is ground or zero volts. This is how the comparator works. If the constant voltage increases, then the
width of On-time decreases, and if the voltage decreases, the duty cycle of PWM also decreases. This is a P-channel MOSFET. An N-channel MOSFET will turn on when the
gate voltage is above the source voltage, and a P-channel MOSFET will turn on when the
gate voltage is below the source voltage. So when the PWM is high the MOSFET is off
and when the PWM is low MOSFET is on. Hence, the duty cycle of PWM and MOSFET is
the opposite. As one increases other decreases. But you may think, why not use an N channel
MOSFET, because, it is on when the gate is high, and off when the gate is low? Because the voltage at the gate required to
turn on the n-MOSFET should be higher than the source voltage, And during operation,
the drain and source voltage will be almost the same. Thus we need higher than Vcc at the gate to
turn on the MOSFET. Hence, we use a P-MOSFET. Also, we add a pull-up resistor between source
and gate. This is the positive supply from the battery. These voltages are created with the help of
voltage regulators. As these voltages are used only for references,
thus there is no current drawn from them and no power is lost. In real cases, due to non-ideal components,
some power is lost as they draw a small amount of current. Now, Let's look at the working of the complete circuit. We will have an input voltage of 12 Volts,
and we want the output of 5 Volts irrespective of the load. Now, If we increase the load, the current
increases, but the PWM is the same as before, hence the voltage falls. This causes the voltage at the reference to
fall and thus increasing the difference between this and reference at the differential amplifier. This increases the voltage at the output and
then the duty cycle of PWM decreases from the comparator thus increasing the on-time
of MOSFET and increasing the voltage at the load. Also, we can change the voltage by changing
the value of the potentiometer. This change in resistance will affect the
feedback voltage, thus it will change the output of the error amplifier and then change
the PWM signal. Now, while designing this some care should
be taken. First, the voltage at the error amplifier
should always be below this reference voltage. Hence, we select the feedback voltage divider
such that the voltage does not exceed 2 volts. Thus the resistor value turns out to be 100
Kilo ohm and 20 k trimmer. But, this will reduce the voltage to a minimum
of 6 volts. Hence, we use a 100-kilo ohm trimmer. For this arrangement of the op-amp and the
value of the resistor, the minimum output voltage is nearly 4.5 volts. For these values, If the load is reduced by
a large amount, the reference voltage may rise above 2.5 volts as the energy storage
components release their energy. This isn't an issue with this analog circuit
as the error amplifier will output zero volts and this will turn off the MOSFET. But, for a digital PWM controller, this may
cause some problems or damage them. The second is the inputs of the comparator. In this arrangement, we get a range from 12
to 5 volts for this feedback resistor values. And, If the inputs of the comparator are reversed
then the range becomes 8 to almost zero volts. Also, the high voltage output is towards the
high end of the trimmer. This is how the closed-loop buck converter works. The voltage and current from the source are
converted to lower voltage and higher current at the output.
