 # Practice English Speaking&Listening with: Lecture - 27 Power Electronics

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In our last class we discussed the operation of buck-boost converter and cuk converter.

The transfer function of buck-boost converter is D divided by 1 minus D. The ratio of output

voltage, magnitude of output voltage to the input DC voltage is D divided by 1 minus D.

Another feature of buck-boost converter is that the output voltage is negative with respect

to the negative bus of the DC input. It is negative, output voltage is negative.

So therefore, ideal buck-boost converter, the output voltage tends to infinity as D

tends to 1, whereas, a non ideal buck-boost converter, output voltage tends to 0, similar

to a boost converter because the assumptions that we made are not valid for high values

of D. For D less than or equal to 0.5, the magnitude of output voltage is less than or

equal to the source voltage and for D greater than 0.5, the magnitude of output voltage

is higher than the input.

So, what is the relationship between the average source current and the average value of the

load current? I have derived this for buck as well as boost converters. The same procedure,

you equate the input power to output power assuming the converter is loss less. So, average

value of the input voltage is VDC. So, average value of the output voltage is D divided by

1 minus D. So therefore, the average value of the output current is the inverse of inverse

of the ratio of the voltages. It is 1 minus D divided by D.

So, input power is equal to output power. You just equate it. So, if voltages are related

by D divided by 1 minus D, currents are related by 1 minus D divided by D. What about the

cuk converter? Why why cuk converter is so popular? Though the transfer function of cuk

converter is same as that of a buck-boost converter, D divided by 1 minus D, current

and voltage relationship is the same. But then, why it is so popular? Why it became

instantly popular?

So, you see the slide here, this all we have discussed in the last class. Ratio is 1 at

D is equal to 0.5 tends to infinity, whereas, non ideal buck-boost reaches a peak, is again,

at depends it depends on the function of the internal resistance of the inductor and the

load resistance. Remember, Dmax for buck-boost is not the same as that of the boost and it

becomes 0 at D is equal to 1.

Now, coming to cuk converter, there is a capacitor connected in between this inductor and the

output stage or this is some sort of an intermediatory voltage source, an intermediatory voltage

source. So, when I close S, current is or energy stored in the inductor VC1 is applied

to the load because this point gets connected here. So, VC1 is applied to the load.

I said, there is an inductor here and a voltage source. So, I can represent this combination

by a current source. So, current varies smoothly unlike in buck and buck-boost. There are no

sudden changes in the source current. Source current jumps to a value which was flowing

through the diode just prior to closing the switch and becomes instantaneously 0 when

I open the switch, both in buck as well as buck-boost. So, that sort of a thing is absent

in in the cuk converter. Open S, stored energy in the inductor is transferred to the capacitor.

So, we had represented this case, this combination in a buck converter by current source. I told

you, there is an inductor is always present is always present across, always present in

the circuit. So, we can represent it by a current source. So, I can say that capacitor

C1 discharges discharges at a constant rate. Here, i2 does change over a very small band

...

So, I can assume that capacitor C1 discharges at a constant rate. Now, when I open S, stored

energy in the inductor is transferred to the capacitor VC1 and I told you that I can represent

this combination by a current source. So, capacitor charges here at a constant rate.

Just the opposite; when I close the switch, the intermediatory capacitor discharges and

when I open the switch, intermediatory capacitor charges. Both are at constant rate but then

the values of these 2 values, the load current and the source current are different.

So, if you see the equivalent circuits, so this is nothing but a boost converter where

VC1 can be represented by the load here. So, the relationship between VD and VC1 is is

given by 1 divided by 1 minus D or VC1 is equal to VDC divided by 1 minus D, whereas,

this is nothing but a buck converter with the input voltage or forcing function of VC1.

The relationship between VC1 and V0 is is proportional to D or VC2 or V0 is proportional

to D into VC1. Now, how about the current relationship?

Boost converter, voltages are related by 1 divided by 1 minus D. So therefore, currents

are proportional to 1 minus D, inverse of that. So therefore, average value of the source

current iS and the capacitor current, assume that capacitor current I0 is given by 1 minus

D. But then the same capacitor current is is is flowing through the load or the average

value of the capacitor current is nothing but average value of i2 itself. So, that is

we have a relationship between average value of the source current and average value of

the load current for a buck converter.

What is that? Average value of the source current is D times, D times the average value

of the load current. So, I will equate it here. You will get iS divided by 1 minus D

into D. So, this is nothing but a boost. You write a relationship between the currents,

capacitor current and iS. This is nothing but a buck converter. There is a relationship

between the source current and the load current, average values.

Now, substitute and you will get the relationship between iS and I0. So, this directly, so we

have a current source at the input and we have a current source at the output. Both

are the current source. So, how does the current, how does the various wave forms look like?

We will draw it for the continuous current because both, I said, input as well as output

is a current source. So, we can safely assume that assume that current is continuous. How

do they look like?

Close S, source current increases linearly. Same, even i2, the current that is flowing

through the inductor L2 also increases linearly because now capacitor VC1 is supplying power.

When I open the when I close the switch, input stage or input inductor is being charged by

the source voltage, whereas, at the load side, the power is being supplied by the intermediatory

capacitor VC1. So, there also i2 increase linearly. So, they look like, something like

this.

When I close S, current increases: when I open S, current decreases linearly. Similarly,

similarly at the at the load side, when I close this; this is the equivalent circuit,

VC1 supplies power, current increases linearly and when I open S, current freewheels through

D. So, current decreases linearly. By the way, the diode has to carry i2 as well as

iS, i2 as well as iS. If you see in this circuit, see, diode as to carry the current i2 as well

as iS, iS. The circuit as iS, when the switch is opened flows like this and i2 flows like

this.

So, how does IC1 look like? IC1 is or capacitor 1 is supplying power or i2 to the load. So,

if this is i2, iC is this. Same, the opposite direction, capacitor is discharging at a constant

rate or at a rate determined by i2. This is i2, capacitor current IC1. When I open S,

what happens to IC1? It is same as the source current now, it is charging. So, this is the

current that is that will flow through the capacitor C1 and I am assuming that the capacitor

current or the source current is continuous. So, when I open when I close the switch again,

IC1 instantaneously jumps to i2. It starts supplying i2. So, this is IC1, how does IC2

look like?

So, capacitor VC1 is discharging and it is charging here. How does IC2 look like? Now,

I need to apply KCL at this point, at this node. As long as i2 is higher than i0, the

difference in i2 minus i0 will flow through the capacitor. If i2 is greater than I0, VC2

will charge and when IC2 is less than I0, capacitor will discharge.

So see, this is the variation of i2. This is the average load current. So, in this duration,

capacitor will discharge. Mind you, capacitor current linearly changes and here also, beyond

this point, capacitor is discharging in this region. When i2 is higher than the average

load current, capacitor is charging. So, this is constantly increasing, this is constant.

So therefore, the capacitor current IC2 is also linearly changes. So, this sort of a

variation in voltage wave form, we have seen in the buck converter, buck converter.

So therefore, in a boost and a buck-boost converter, the output capacitor current changes

drastically, in the sense, the entire load current is being supplied by the output capacitor

when the switch is closed, both in boost as well as buck-boost converter, whereas here,

the capacitor current, even at the, which is connected at the load side gradually changes,

Now, what is the voltage that is coming across the diode as well as the switch? So, when

I close switch S, VC1 appears across the diode. If you see in this circuit, see, when I close

S, this point gets connected here or entire VC1 appears across D or in other words, D

should block VC1. Again, VC1 is a function of the duty cycle, is related to VDC by 1

divided by 1 minus D. So, maximum value of VC1 that the diode should block and what is

the voltage that is coming across S?

What happens when I open S? When I open S, diode starts conducting, diode starts conducting.

It carries both, the load current or i2 as well as iS. So this point gets connected here.

So, voltage across the switch is also the capacitor voltage VC1 because this point and

this point is the same. When the switch is opened, diode starts conducting. So, this

point gets connected here. So, voltage that is coming across S is VC1 itself. So, this

is the voltage across the diode VC1 and this is voltage across the switch. That is about

the cuk converter.

So, we have studied 4 DC to DC converters; buck, boost, buck-boost and cuk converters.

Now, let us solve few problems in these DC to DC converters.

The first problem is a buck converter feeding a DC machine. The problems says that R is

0, total inductance is 50 millihenries, switching frequency is 500 hertz, D is 0.5, average

current drawn by the motor is 10 amperes, Iaverage. Assume that iL is continuous or

assume that load current is continuous, assume that load current is continuous. Determine

the peak to peak ripple in the load current. We know that when I close S, load current

increases and when I open S, load current decreases. So, what is the peak to peak voltage

ripple?

It is said that current is continuous. So, input voltage is given which is 200 volts,

D is given, RA is 0. So therefore, output voltage, average value of the output voltage

that is coming across the armature terminals is D into VDC. So, D is 0.5, VDC is 200 volts.

Therefore, applied voltage to the armature is 100 volts. It is said that RA is 0, armature

resistance is 0. Therefore, applied voltage to the armature is same as the induced emf

E, Eb or the back emf. Otherwise, it is Eb plus Ia Ra. So therefore, Eb also equal to

100 volts. Now, how do I determine the peak to peak ripple in the load current? So, what

happens when I close S?

Current starts from Imin, increases linearly and it attains a maximum value Imax. So, what

is the voltage equation? VDC is equal to L di by dt plus Eb. Resistance is 0 and when

I open S, current flowing through the load, it freewheels through the diode. So, KVL says

that L di by dt is minus Eb from DT to T. So, it looks like this; Son, Don, this is

the armature current, increases linearly, reaches the peak, comes down. So, this is

the average value of the armature current, 20 amperes. So, these are the 2 equations

that we wrote.

So, I di by dt, put a suffix here, that is increased and I put a decrease. Now, using

this equation subtract. What do you get? di by dt I minus L di by dt decrease is equal

to 200 divided by L, 200 divided by L. From these 2 equations, I will get this. So therefore,

L di by dt is 100, L di by dt is 100. How? See, it is very obvious here, 200 - average

value, Eb also is 100, so remaining L di by dt is 100 and minus L di by dt is also 100.

So, you substitute these values in this equation. You will get L di by dt is 100, minus L di

by dt is also 100. So, this is what we get here, if I substitute. So, di by dt is 2000

amperes per second. Value of L is known, value of L is known, it is how much? It is 50 millihenries,

L is 50 millihenries, it is given.

So, I know this slope, I know the average value, I know the time for which this switch

is closed. So, this point is DT by 2. This is linear, this is 20 amperes, I know the

slope, so I can calculate this as well as this. So, it is all you will get. Imin is

equal to 19 amperes and Imax is equal to 21 amperes. So, if I know the slope, if I know

the average value, I can always determine Imin and Imax because I know all these values.

So, they found to be 19 amperes and 21 amperes.

So, armature current is varies between 19 and 21, 19 and 21. So, if the armature current

varies from 19 and 21, therefore, torque also will pulsate, torque also will pulsate in

this. So, in order to reduce the torque pulsation, I need to reduce this ripple - I min difference

between Imin and Imax. So, that will call for a higher switching frequency or the different

value of L. So, we we did derive the expression for the current in the the the ripple in the

inductor current. So, use that expression and if you want to minimize the ripple, find

out the new value of L for the same input and output conditions.

A Second, very interesting problem, so what it says? Nothing but a boost convertor nothing

but a boost convertor, 100 volts, inductor of 100 micro henries, there is a switch connecting

to ground and through a diode it is connected to a 300 volts source.

So, how does this work? When I close S, diode is reverse biased because this point gets

connected here to ground. Cathode is connected to 300 volts. So, inductor charges at the

constant rate, L di by dt is 100 volts. Open S, the stored energy is being transferred

to the source. The essential condition for the boost convertor to work is output voltage

or V0 should be higher than VDC. So here, V0 is 300 volts, VDC is 100 volts.

Problem says, switching frequency is 20 kilo hertz, D is equal to 0.5. Calculate the power

transferred from 100 volts source to 300 volts source. Frequency is 20 kilo hertz, D is 0.5,

power transferred from 100 volts to 300 volts. Assume that circuit has attained a steady

state. Let us solve. It is not mentioned that whether the current is continuous or not.

Now, you need to tell me whether the current will be continuous or not, giving the circuit

equation.

Input is 100, output is 300, duty cycle is 0.5. In other words, time for which energy

stored is same as time for which energy is allowed to transfer to to the load. So, when

I close S, the forcing function is 100 volts. 100 volts is being applied to the inductor

for some time, current increases linearly and switch is opened for the same duration,

D is equal to 0.5. But then now, the voltage that is coming across the inductor is 200

volts. 100 at the input or source voltage, load voltage is 300 volts.

So, when I close the switch, voltage that is coming across the inductor is 100 volts.

When I open the switch, voltage that is coming across it is 200 volts. Duration for which

the switch is opened is same as the duration for which the switch is closed. Therefore,

inductor current has to be discontinuous.

See, for steady state, voltage across the inductor, average value should be 0. I can

have a positive voltage appearing across the inductor. That means current is building up

but then definitely, I cannot have a situation wherein, average value across the inductor

being negative. In this case, 100 volts is being applied for D into T across the across

the inductor. When the switch is closed, plus 100, when the switch is opened, voltage across

the inductor is difference of 2 voltages, 300 minus 100 that is 200 volts for 1 minus

D duration. So, D is 0.5, so 200 into 0.5 cannot be equal to 100 into 0.5.

What is the peak value of the inductor current? Peak value of inductor current is 100, the

voltage that is applied divided by L into time for which the switch is closed or 100

divided by L into D into T, D into T. So, D is 0.5. So, you will get 25 amperes.

So, at at 0.5 T, current is maximum. So, current, when I open S, current starts flowing through

the 300 volts source and current falls linearly. Voltage that is appearing across the inductor

is 200 volts, it follows linearly. So, slope of this line is 200 divided by L and let beta

be the instant where the current becomes 0. I know the slope, I know the peak current,

I know the peak current, I know the slope of this line. So, I can calculate this point

or T2.

So definitely, if this is 25, forcing function is 100. If this is 200, forcing function is

200. This should be half of this, 12.5 micro seconds 12.5 micro seconds. Previous one was

25 micro seconds, whereas, this is 12.5 micro seconds.

Now, what is the energy that is transferred to the 300 volts source? It is the area or

is proportional to the area under this curve, this. The average value of the output voltage

that is 300 volts into average value of the, this current or will give you will give the

output power. So, what is the average value that is or energy transferred? 300 into this,

half of ipeak into T, it is the area of this triangle divided by the whole time period

will give the average value of this current or the cycle. Is that okay? The area of this

triangle divided by the total time period T is the average value of the current that

is flowing into the source flowing into the source.

So, that is what I did, 300 half into ipeak into T2 divided by the total time period or

multiplied by the frequency 1 and the same, 20 kilo hertz or divided by the time period

1 over T. So, power that is received is 938 watts.

We will solve another problem, a problem on buck-boost. The input source voltage is 100

volts, output voltage is 500 volts, value of L is 100 micro henries, switching frequency

is 100 kilo hertz, the current is just continuous, current is just continuous. So, what is Ton

and the peak value of the inductor current? How do I solve?

Current is just continuous, so it starts from 0, reaches a peak just prior to opening the

switch. What is the voltage that is appearing across that inductor when the switch is closed?

It is the source voltage itself. So, in this case, source voltage is 100 volts, current

is just continuous, current is just continuous. So, it becomes 0 just prior to closing the

switch in the next cycle. But then the voltage that is appearing across the inductor when

the switch is open is the output voltage, is the output voltage.

So, in this circuit if you see, close S, diode cannot conduct, 100 volts appears across this

inductor. Open S, stored energy is transferred to this 500 volts source. So, voltage that

is appearing across this is 500 volts. Current is just continuous, so it starts from 0, reaches

a peak, touches 0 just prior to closing the switch again. So, I know the slope of this

line. What is the slope of this line? V divided by L. V is 100 volts, L is 100 micro henries.

So, peak value is D into T. So, this value is D into T. I know the slope of this line,

I know this peak value. So, equation for this line is same, 500 is the forcing function,

500 divided by L into T minus DT is the equation for this line. The slope of this is proportional

to output voltage 500 volts, whereas, this slope is the input voltage VDC, 100 volts.

So, peak value is 500 divided by 10, the value of inductor into T minus DT.

Now, both are the same both are the same. So, I will equate it, I will equate it and

you find that DT or Ton is 83.3 micro seconds. So therefore, the value of peak current, you

substitute here. You get as 83.3 amperes, 83.3 amperes, a very simple problem.

Again a buck convertor; input is 60 volts, output is 12 volts, inductor that is connected

is 20 micro henries, millihenries. 20 millihenries is the inductor that is connected in series.

Average current that is flowing is 5 amperes. The question is what is the peak to peak ripple

flowing through the load? Switching frequency is 1 kilo hertz, duty cycle is 0.2.

So, the problem, simple circuit buck convertor; 60 volts, battery is 12 volts, 20 millihenries

- the inductor that is connected, 5 amperes is the current that is flowing, switch is

controlled at or switched at 1 kilo hertz with D is equal to 0.2. What is the peak to

peak ripple? It is a straight 1 line problem. I know the input, I know the output, I know

the value of inductor, I know the time for which the switch is closed.

1 simple equation; L di by dt is equal to Vin minus Vout. So, whether the current is

starts from 0 or starts from any finite value, does not matter, it has to starts from Imin

and just prior to opening the switch, it has reached a peak in a buck convertor. When i

starts from the minimum value, it could be 0, does not matter. It reaches a peak just

prior to opening the switch. So, the difference between Imin to Imax is is peak to peak ripple

and this depends only on the voltage that is appearing across the inductor. Value of

the inductor and time for which it is closed, everything is known. L is known, time for

which it is closed, DT is also known. So, di is straight forward, 60 minus 12 is a voltage

appearing across inductor. This is this, L is 20 millihenries and D into T. D is 0.2

and this is T, is 0.48 amperes, this is the peak to peak ripple.

So, we have solved quite a few problems. Last and a very interesting problem in cuk convertor,

very interesting problem in Cuk convertor, input voltage in a Cuk converter is 50 volts,

output voltage V0 is 150 volts, peak to peak ripple in a current flowing through L1 and

L2 is 1 ampere. See, peak to peak ripple in both the inductors is 1 ampere. So, I can

assume it as if like, they are current sources and peak to peak ripple in the intermediatory

capacitor voltage is 10 volts or VC1, peak to peak ripple in VC1 is 10 volts and peak

to peak ripple in output voltage that is VC2 is 1 volt.

See, that is the reason I always said that output voltage in any converter can be assumed

to be constant and ripple free. It is always desirable or or it is expected that power

supply maintains a constant voltage across the load. So, VC2 is peak to peak ripple in

VC2 or V0 is 1 volt, intermediatory stage VC1, the ripple in VC1 is 10 volts.

The switch is switched at 25 kilo hertz, switching frequency is 25 kilo hertz and we have been

asked to neglect the internal resistance of L1 and L2. So, we will solve this problem.

The relationship between V0 and VDC is given by D divided by 1 minus D or V0 is equal to

VDC into D divided by 1 minus D. So, we know the source voltage 50 volts, output voltage

150 volts. So therefore, D is equal to 0.75. So, switching frequency is 25 kilo hertz,

D is 0.75. So, we know the time for which S is closed and opened.

So, what are they? The total time period is total time period is 40 micro seconds, Ton

is 30 micro seconds because D is 0.75 and Toff is 10 micro seconds. Now, how do I calculate

VC1 and the ripples in the inductor currents and the output voltage?

We know that average voltage across the inductor is 0 at steady state. So, what is the voltage

appearing across the inductor L1? I have to calculate VC1. I can calculate VC1 only from

the input and the time for which the switch is closed. For that I need to equate, I need

to equate the average voltage across the inductor to 0 or in other words, I can straight away

apply because I know the input voltage, I know D, I can calculate VC1, VC1.

So, voltage across the inductor when the switch is closed is VDC and when it is open, it is

VDC minus VC1. So, I will equate it, we find it to be 200 volts. This also should be equal

to, this also should be equal to VDC divided by 1 minus D. VC1 is nothing but VDC divided

by 1 minus D. This is the nothing but a boost converter. So, VC, supply voltage is 50 volts,

D is 0.75, so, 1 minus 0.75, 0.25. So, 50 divided by 0.25 is 200 volts.

How do I find out the average value of current that is flowing through the inductors? What

do I need to assume or what is the principle? Average current flowing through the capacitor

at steady state should be 0. VC1 or the capacitor C1 discharges at a constant rate. Current

that is flowing out of the capacitor C1 when the switch is off is the average load current

itself and capacitor charges at a constant rate and this current is proportional to the

source current. So, capacitor discharges at a constant rate and that current is average

value of this current is same as the load current and capacitor charges at a constant

rate and this current is proportional to the source current.

So, capacitor is being charged for 10 micro seconds, duration for which the switch is

opened and capacitor C1 is being discharged for 30 microseconds or the duration for which

the switch is closed. So, I will neglect the ripple in iL1 as well as iL2, I will neglect

the ripple in iL1 as well as iL2. It is said that ripple in current is 1 ampere, so I will

neglect it. So, iL2 is the current that is flowing in the inductor 2, L2. So, iL2 into

30 micro seconds for which the device is closed or this is the period the capacitor is discharging

and 10 micro second is the period for which the capacitor is charging at iL1.

It is mentioned that average load current is 10 amperes. We know that average value

of the capacitor current is 0. So, average value of the load current should be equal

to average value of this inductor current I2. So, iL1 comes out to be 30 amperes, iL1

comes out to be 30 amperes.

What is next? How do I determine L1? How do I determine L1? Ripple is given, ripple is

given, ripple is 1 ampere, time for which switch is closed is also given. So, what is

the circuit equation? Circuit equation is Ldi by dt is equal to VDC, Ldi by dt is equal

to VDC. So, 50 volts is the input voltage L1, ripple in the L1 is 1 ampere or di is

1 ampere in 30 micro seconds in 30 micro seconds. 50 is the input voltage, L1 is the inductor

value, 1 is the ripple in the current, 30 micro second is the time for which the device

is closed. So, you will get L1 to be 1.5 micro or 1.5 millihenries. I do not know, may be,

micro, milli. Find out, it is this. Looks like, 1.5 millihenries.

Similarly, ripple in the inductor 2, it is 1 ampere and what is this ripple proportional

to? See, when I when I see when I close this switch, voltage that is appearing across L2

is VC1 minus VC2. So, that is the voltage that is coming across the L2. I know VC1 which

is 200 volts, I know V0 150 volts, time for which this circuit is known or is same as

the time for which the switch is closed.

So, I can calculate the value of L2. VC1 minus V0 divided by L2 is the rate of change of

current. Rate of increase in current, di by dt. Switch is closed for 30 micro seconds.

This is 1 ampere, 1 ampere. VC1 is 200 volts, V0 is 150. So, you can calculate L2. Now,

what is the value of C 1 and C2? How do I calculate C1 and C2?

Now, to determine C2, I need to know the time for which C2 is charging and C2 is discharging.

Similar to buck converter, C2 charges when the inductor current I2 is less than the load

current sorry the capacitor C2 charges when the inductor current is higher than the load

So, that can happen that happens from DT by 2 to 1 plus D divided by 2 into T. We are

assuming the current increases. See here, current increases and decreases. iC2 is charging

in this period and during this period iL2 is higher than higher than the load current.

So, this is DT by 2, this is 1 plus DT by 2, 1 plus DT by 2.

So, the charge that is lost is given by this 0.5 is the, see, this is peak to peak ripple

is peak to peak ripple is 1 ampere. So, this is definitely 0.5 amperes. Peak to peak ripple

in both of them is 1 ampere, so this is 0.5. So, this is known, I need to find out the

area. So, it is this, this is the charge, this is the charge micro coulomb, micro coulomb.

So, C2 is given by 5 microfarad. So, delta V is known is 1 volt, 1 volt, 1 volt, 1 volt.

Ripple in delta V0 is 1 volt. So, capacitor C2 is 5 microfarads, delta q is Similarly,

C1 is discharged by an average current of 10 amperes. This is the average of load current,

average value of i2 is same as average value of current that is flowing through C1 that

is capacitor is being, this current is supplied by a capacitor, entire i0 and this is the

charging.

So, this is the discharge period, I know the current, I know the time for which this occurs.

So, I can calculate I can calculate the value of C. So, this is the charge, 10 ampere into

30, charge that is lost divided by the voltage. Ripple in the voltage is 10 volts, ripple

in the voltage is 10 volts this is 30 micro farads.

So, that is about it. A very interesting and very educative problem, we solved almost all

the aspects in DC to DC conversion. First you find out the maximum and minimum ripple

in the load current. Then, a very good problem in cuk converter, then a good problem in boost

converter. It was a very educative problem, in the sense, D was it was just mentioned

D is equal to 0.5, it was not mentioned that whether the current is continuous or not.

We had to deduce. So, more about it we will see in the next class.

Thank you.

The Description of Lecture - 27 Power Electronics

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