 Hello and welcome to this lecture on advanced electric drives. In the last lecture, we have just started the dynamic modeling of brushless DC motor drive. We will start from that. So, let us take a look at the dynamic modeling of brushless DC motor drive. For simplicity, we will consider a cylindrical rotor and the stators have three phase winding A, B and C. The rotor is a permanent magnet rotor, but having a cylindrical structure and hence the air gap is uniform. So, let us took a look at the dynamic modeling of brushless DC motor drive. So, we are talking about the dynamic modeling brushless DC motor drive. For simplicity, we will be taking a cylindrical structure in which the air gap is uniform. So, we have a motor in which the air gap is uniform. So, we will take uniform air gap which means cylindrical rotor. The stator has three phase winding which could be distributed and the coupling has the two phases will have coupling. Rotor is a cylindrical structure. So, we have a rotor and then we have the stator phase A, phase B and phase C and this is the rotor. So, the stator phases are A, B, C and this could be star connected. So, in that case we can connect them in star and this is a neutral point that is n. So, we have applied the voltage to the stator phase A, phase B and phase C and the stator currents are I A, I B and I C. Now, we will assume that each stator winding has a inductance of L, the self inductance is L and the mutual inductance between two phases will be m. So, the capital L is a self inductance and the mutual inductance between two phases would be m and this inductance is between say phase A and phase B and similarly between phase A and phase C is also m. Now, if we write down the dynamic equation of phase A, we will say the following that the voltage V A n is equal to each phase is having a resistance that is R S. So, we can say R S into I A, the self inductance is L d I A by d t and m is a mutual inductance we can say m d I B by d t that is a mutual between A and B, phase A and phase B and then we will have the mutual between phase A and phase C m d I C by d t and each phase will have a back m f because the rotor is a permanent magnet rotor. This could be a north pole here we have a south pole here and this is a permanent magnet structure. So, when the rotor is rotating the flux linkage will be this is the direction of the flux which is coming out of the rotor like this and this is linking the stator and as a result each phase will have a induced t m f and we know that induced t m f in a brush state DC motor is trapezoidal in nature. So, in fact the induced t m f nature we have seen in the last lecture is trapezoidal. So, it looks like this in every phase we have an induced t m f which is trapezoidal in nature. So, this could be the induced t m f is phase phase A. So, we will put here plus E A. Similarly, we can write down the equations for phase B and phase C for phase B we will have V B n the voltage of phase B with respect to the neutral point and the neutral here is n that is equal to R S into I B plus L D I B by D T. This is the self inductance the self inductance drop of phase B plus M D I C by D T plus M D I A by D T and this is phase B and in phase B also we have some induced t m f and induced t m f in phase B is E B plus E B. E B is also a trapezoidal wave form which is phase shifted from E A by 120. Similarly, we can have the voltage equation for phase C V C n is equal to R S I C plus L D I C by D T plus M D I A by D T plus M D I B by D T plus E C and E C is the induced t m f in phase C. Now, we have three equation the rotor is a permanent magnet the rotor does not have any winding. So, we do not have to write the equation for the rotor. So, these three equation can be represented in the form of a matrix equation. So, we can write down this in the form of a matrix and matrix equation. So, V A n V B n and V C n is a vector is a column vector that is equal to we have the resistance matrix that is R S 0 and 0 it is a diagonal matrix 0 R S and 0 0 0 and R S into I A I B and I C are the three currents of three phases plus we have the inductance drop and the inductance we have the self inductance of a phase and the mutual inductance between two phases. So, the inductance drop can be written again in the form of a matrix. So, we have the self inductance L and the mutual between A and B is m and the mutual between A and C is also m the inductance matrix and this is multiplied by the derivative operator is P and the currents are I A I B and I C. So, L m m is for the first row the second row we can write down the equation for phase B. So, this is the self inductance of phase B and then the mutual between B and A is m and the mutual between B and C is m. Similarly, we can write down the equation for the phase C the self inductance and the mutual inductance drops the self inductance is L. So, we can have here L P I C and the mutual between C and B is m and the mutual between C and A is m. In fact, if we see the three phases are symmetrical. So, the mutual between phase A and phase B is m phase B and phase C we can say that this mutual inductance is also m because of the symmetry they are phase shifted from each other by 120 degree and thus the mutual inductance between phase A and phase B phase B and phase C phase C and phase A are all equal and the mutual inductance here is m. So, this will be added with the back m f and the back m f of the respective phases we know that is E A for phase A, E B for the phase B, E C for the phase C. So, we have the voltage equation of a three phase machine where the rotor is a permanent magnet and the rotor rotation is reflected as a back m f in the three phases. Now, in fact, we can also say that the back m f E A, E B and E C are all proportional to omega r. So, we can say that E A is equal to there is a function which is k a, k a is not a constant which is a function and this is a function of theta r into omega r. Similarly, we can say that E B is again a function k B into omega r and E C is again a function that is k c into omega r. So, the induced m f here are the functions of the speed rotor speed and omega r in this case is electrical speed because we are assuming a two pole structure. In general, rotor could be a p pole structure. So, we have to appropriately take into account the pole pair. Now, these are the induced m f's. Now, can we simplify this set of equations? We have three equations, but there is a possibility of simplification and out of these three equations we need the currents. Voltages are not known. We have applied the voltage here. This voltage is V A n and this voltage is V B n phase B and the other voltage is V C n. So, the voltages are given. They are independent variables. We can apply the voltage to the machine by using a inverter. Now, when we apply the voltage, the currents will be flowing into the machine. Now, this we can simulate. So, the currents have to be calculated or have to be found out by numerical technique. So, we can simplify this set of equation in the following fashion. Now, we know that we have a star connected machine. The machine is star connected. So, we can say that I A plus I B plus I C is equal to 0. There is no neutral connection. The neutral, the fourth wire is not there. So, we have a three phase three wire system. So, we can say that I A plus I B plus I C is equal to 0 or we can say in this case I B plus I C is equal to minus I A. So, we take the first equation. The first equation is this voltage equation. If you take V A n is equal to R S I A. We will have here in this case plus L D I A by D T. We can have here plus M D by D T of I B plus I C plus E A. Now, here we have a term which is D by D T of I B plus I C and I B plus I C can be replaced by minus I A. So, we will have this and this we will be replacing by minus I A. So, if we do that and rewrite this equation, we will have the following equation. So, for the phase A we will have this following equation after simplification. V A n is equal to R S I A plus L D I A by D T minus M B I A by D T plus E A or we can also write down this equation in the following fashion R S I A plus L minus M B I A by D T plus E A. Now, this equation is much more simplified than the previous equation for phase A. In the phase A we had expression for I B and I C also, but this particular equation only involves I A. So, it is a simplified equation compared to the previous equation. The previous equation was this equation where we had both I A I B and I C. Now, we have simplified this equation and the equation that we have is only having D I A by D T and also I A. So, we can further rewrite this equation as R S I A plus L S D I A by D T plus E A. We can say L S is something like the inductance of phase A, because as it appears from this equation that phase A can be treated independently, because we have a star connected machine I A plus I B plus I C equal to 0. So, by virtue of that equation I A plus I B plus I C equal to 0, we can independently consider each phase. So, L S we can say the inductance of each phase. Similarly, for phase B we can have R S I B in a same way L S D I B by D T plus E B and also for phase C we can say V C N is equal to R S I C plus L S D I C by D T plus induced T A M that is E C. So, we have now three equations and three unknowns and unknowns are I A I B and I C. These three equation can also be written in the form of a matrix equation as follows. We can write down this as a matrix equation V A N V B N and V C N that is equal to R S 0 and 0 0 R S and 0 0 0 R S I A I B and I C. The resistance drop plus we have the inductance matrix is L S here L S in this case and we will see in this case D I A by D T D I B by D T D I C by D T. The inductance L S is the same in all three phases. So, we have the derivative of the current plus the back M F's E A E B and E C. Now, we need to find out the values of D I A by D T D I B by D T and D I C by D T. So, we can take this out in the left hand side we can write down D I A by D T D I B by D T D I C by D T. On the right hand side what we have here we have the voltages the voltages are V A N V B N and V C N and then minus the resistance drop R S 0 0 0 R S 0 0 0 and R S then we have the currents here I A I B I C minus the back M F E A E B and E C. Of course, we have an L S here L S can be taken to the right hand side we can multiply here 1 by L S. So, this equation we can say equation number 1. So, this equation gives us D I A by D T D I B by D T and D I C by D T. So, we have three simultaneous differential equations which can be solved by any numerical technique. For example, we can solve this equation using Runge-Cutte fourth order method. So, we can use Runge-Cutte fourth order method to solve this three equations and get the values of I A I B and I C. What about the speed? Unless we know the speed we cannot calculate the back M F because we have seen that E A here is a function k A into omega R. So, we need to know the value of the speed the speed can be obtained from torque equation. So, the torque can be obtained as follows. So, we can find out the value of the torque as follows. So, T E or the power in this case before that we can see the power expression the power in this case is P by 2 times we have mechanical power out. So, P A M here this is the mechanical power that is equal to we do not have any P by 2 term here because we are talking about the mechanical power that is E A I A plus E B I B plus E C I C this basically the respective back M F into the respective current. So, if we are the back M F and the current product in all three phases we get the total mechanical power and the torque can be found out by dividing this by the mechanical speed. So, the torque here P is equal to P M by the mechanical speed omega M that is equal to P by 2 into P M by omega R because we know that the mechanical speed is the electrical speed by pole pair. So, omega R by pi by 2 is omega R M. So, we have P by 2 coming here into this product we have the sum of the product of the back M F and the currents E A I A E B I B plus E C I C and that can be written as P by 2 this is what we have here that is equal to this divided by omega R and we have to multiply here P by 2 in this case. So, we can now replace E A by K A I A and E B by K B I B plus K C I C and omega R is in the numerator and omega R is also in the denominator they will get cancelled. So, finally, the expression for the torque is P by 2 K A I A plus K B I B plus K C I C. So, this is the expression for the torque of a brushless DC motor in which K A K B and K C are function of theta R. Now, how does this K A look like K A as we have seen it is basically a trapezoidal function and the K F of A S A if we see the function K A K A may look like this 120 here this is 60 degree in this case and this is 120 again and so on. So, this is of unity magnitude. So, this magnitude is 1. So, this is K A now when we multiply this by omega R we get the corresponding back K M F. So, this is the nature of K A this magnitude can be suitably scaled. So, for example, this depends upon the strength of the permanent magnet this could be in general this could be some K T. So, we can say that the magnitude of this is K T the torque constant K T is the peak magnitude. Similarly, in this case also this magnitude is minus K T and the constant for the phase B that is K B will be shifted from K A by 120. So, this is K B K B will have a pattern which is shifted from K A by 120. So, this is the nature of K B and so on. And similarly, we can have K C K C will be again shifted from K B by 120. So, we have the functions for phase A phase B and phase C when the respective functions are multiplied with the corresponding current we get the torque. And the torque in this case is K A I A plus K B I B plus K C I C into P by 2 and omega gate cancel. So, this is the expression for the torque. Now, when we get the torque we can get back the speed by having electro mechanical equation. So, the electro mechanical equation is the inertial torque D omega r by D T into J by P by 2 we neglect for simplicity the viscous friction, but the viscous friction will also come here B by P by 2 into omega r plus the load torque is equal to the motor torque the motor torque is T. So, from this we can get the expression for D omega r by D T. So, D omega r by D T can be obtained in the following fashion is equal to T minus T L minus B by P by 2 into omega r. This B is the coefficient of viscous friction any machine will have some viscous friction and that is represented as B. So, B into omega r m or omega r by P by 2 this is multiplied by P by 2 and divided by J. So, we have the expression for the speed and this speed equation can be solved again by a numerical integration technique to give us the value of speed. So, once we know the speed we can find out the back m f and the back m f are E A E B and E C the multiplication of corresponding constant k a k b and k c with the rotor electrical speed omega r. And then this back m f can be substituted in this equation to give us the derivative of the currents and the derivative of the currents can be solved using any numerical integration technique. Now, here we can understand that the electrical variables vary at a much faster rate speed is a mechanical variable. So, we can afford to solve the electrical variable with a better integration method. So, these three equations can be solved by Runge-Coutet fourth order technique. If we are trying to solve these three equations we can use Runge-Coutet fourth order integration technique to solve these three equations. And then the speed equation after we have obtained I A I B and I C this I A I B and I C can be substituted here to get the value of t and t is known here t l is a load torque omega r is the previous value. So, this equation can be solved by a first order integration method first order integration to give us the value of omega r. In fact, when we are simulating the first set of equation that is equation number 1 this is equation number 2 and this is equation number 3. So, when you are solving the first equation we are assuming that E A E B and E C are constant for the time delta t delta t is a simulation step. So, we are we are simulating this in step of delta t this is the simulation step and for that small delta t we can assume the back m f to be constant. And that is only true with the speed is constant. So, while solving equation 1 we are assuming that the speed is constant we after solving this equation 1 we get I A I B I C and these three currents can be used to get the speed omega r. So, this is the dynamic modeling of a brush state DC motor. Now, as you have already seen that we can have a trapezoidal excited permanent magnet motor also we can have a sinusoidal excited permanent magnet motor. And the sinusoidal excited permanent magnet motor are called permanent magnet synchronous motor. So, we will be now discussing about the sinusoidal excited permanent magnet motor or permanent magnet synchronous motor. So, our next discussion is permanent magnet synchronous motor. And we call this to be PMSM and these are sinusoidally excited which means the back m f in this case is a sinusoid the back m f E A E B and E C are not trapezoidal they are sinusoidal. So, we will say that here in this case the back m f are sinusoidal. Now, we will be right now we will be only confine into the steady state analysis of this motor then this will be followed by the dynamical analysis. In steady state we assuming that the speed is constant and we can represent the motor by its equivalent circuit. So, we have the synchronous motor where the field is a permanent magnet we can replace this by a single phase steady state equivalent circuit. So, we can have an equivalent circuit like this here we have let us say we excite this current source and then we have the synchronous reactant that is J X S and the field induced E M F is E F its A C quantity. And we excite this from let us say a current source we have a current source inverter or an inverter which can be current controlled. So, we are injecting the current into the stator winding. So, in that case we can assume that the current in the stator is constant its A C current, but it is having fixed amplitude and fixed frequency let us say in the steady state condition. So, in the steady state condition we can say it is I S. So, this is the steady state single phase equivalent circuit. So, we are discussing about the steady state operation steady state single phase equivalent circuit and in this case since we have a sinusoidal excited motor the voltage which is applied and the currents which are applied will also be sinusoidal. So, in fact in this case I S I S will be sinusoidal. So, we have I S in phase A phase B and phase C. So, they all will be sinusoidal. So, this is a phasor and we represent the phasor by a bar on the top I S bar and this voltage is the voltage here that is V is the terminal voltage and E F is the field induced T M F. Now, we can rewrite this or read this equivalent circuit in a different way what we will do here we will be replacing this part by its Norton's equivalent circuit. So, what we will do we will take this branch and we will be replacing that branch by again a current source this is I F this is also a phasor here and then we have a parallel branch in this case and the source here is I S this is a voltage this is a current source. So, what we have here is a current I S and this reactance is X S the synchronous reactance. Now, in this case we have two currents one is coming this way other is coming from the rotor and this current which is flowing through this branch is called the magnetization current. So, we have I M. So, we can have the expression for I F and I M also. So, what is I F here I F is the Norton's equivalent current that is equal to E F by J X S and this I M is the sum of the other two currents the straighter current is I S and this is the field current which is coming from the field side this is I F. So, we can say I M is the sum of I S and I F. So, we can now draw the equivalent circuit after drawing the equivalent circuit we can draw the phasor diagram. In the phasor diagram what we have we have the field here this is the field is in the D axis and this is the current I F and the induced T M F will be right angle to the D axis this is E F and the straighter current is somewhere along this axis I S they are all phasors and the resultant of this two we are we are having a vector addition in this case. So, I F plus I S is I M and I M is given as I M phasor. So, I S plus I F is equal to I M and I M is the magnetization current. Now, we are discussing the phasor diagram of a permanent magnet synchronous motor we have the field current that is I F I F is the field current it is along the field axis. We have shown that by an pole structure here this is the field and then the field induced T M F is right angle to the field. So, this is the field induced T M F that is E F and the straighter current I S will be along a direction which is shown here that is I S I S bar and the summation of I S and I F will give us the magnetizing current that is I M. What about the voltage the voltage which is available here this voltage V is because of I M. So, in this case the relationship between I M and the voltage V will be 90 degree. So, what we do here is that we draw here a line which is again pi by 2 with respect to I M and this voltage this is voltage V in this case. So, this is the voltage V here and V can also be evaluated in the following fashion. In this equivalent circuit we can say that V is equal to this is I S J I S X S plus E F. So, in fact if we add the synchronous reactance drop that is I S into X S we get V. So, we know the I S here. So, this is J I S X S. So, if we are this J I S X S with E F we get the terminal voltage that is V. So, we have we have a vector triangle here in one side we have E F other side we have J I S X S and the third side we have the voltage that is V applied voltage. We can either apply a voltage or apply a current does not matter. Once we apply a voltage we can find out the current and even if we are applying a current to the machine depending upon the control we can either apply a voltage or a current, but however the operation of the machine will not change. So, in this case this angle that we usually know the angle between E F and V is called the torque angle or delta and we also have an angle between I F and I S. This angle between I S and I F this angle is called delta prime we can call this to be delta prime. So, we can write down the power equation here. Now, we have ignored the resistance of the machine usually the resistance is a small quantity. So, we can say that the resistance is very small which can be neglected. Now, under this situation we can write down the power which is coming out of the machine and the power which is going into the machine. So, in this case we will say that the power in is equal to power out we are assuming that the loss is not there. So, what is P in here? P in the input power here is the power which is absorbed by this back M F. This back M F is observing power if it is a motor we will have a mechanical output and the mechanical output will be coming out of this. So, this is P M or P out and P out is basically the power which is absorbed in the field induced E M F and we can write down this from this phasor diagram. So, we have E F and I S we can say for is a single phase circuit for three phase will multiply with three E F amplitude and I S amplitude and what is the angle between them? The angle between this is the angle and this angle is if we see this angle in this case is delta prime minus pi by 2. So, we have this is the delta prime here and this angle is pi by 2. So, the angle which is shown here this angle is delta prime minus pi by 2. So, we can write down this equation. This equation is E F I S for three phase we can multiply with three cos of delta prime minus of pi by 2 and that we can write as 3 E F I S sin of delta prime. Now, we know what is E F E F is X S into I F. So, we can say here that is equal to 3 X S into I F this is the output power which is the mechanical power we have ignored the losses in this case. What about the torque? The torque is the power by the mechanical speed. So, we can find out the torque here the torque in this case is given by p out by omega r m that is equal to we have 3 into X S into I F into I S into sin delta prime divided by omega r m. And we can say that is the constant K here into I F I S into sin delta prime. Now, in the steady state condition we know that X S is equal to the synchronous speed that is omega s into L S. And what is omega r m? Omega r means the rotor speed that is equal to omega r by p by 2. Mechanical speed is the electrical speed by p by 2 and the electrical speed is nothing but omega s by p by 2. So, in the numerator we have omega s in the denominator also we have omega s they will get cancelled. So, hence we will get a constant here K into I F into I S into sin delta prime. Now, torque looks like the equation of a decimation the torque equation looks like the equation of a decimation torque equation. So, here we can say that p is equal to K K F I S sin delta prime because the rotor is a permanent magnet rotor. So, we can club this I S with K that is a constant quantity because the flux of a permanent magnet does not change you cannot control the flux of a permanent magnet. So, that is K F I S is the stator current and delta prime is the angle between the field axis and the stator M M F that is I S. So, in this case if we want to have maximum torque delta prime should be 90. So, the maximum torque is obtained maximum torque for ampere is obtained for delta prime is either plus or minus because for motoring we will have plus 90 for braking we will have minus 90 degree plus minus pi by 2. So, in this case so T is equal to plus minus K F into I S. So, let us see the phasor diagram when delta prime is 90 delta prime is 90 is a preferred condition where we get maximum torque per ampere. So, we maximize the torque. So, torque can be maximize if delta prime is 90. So, we will have the phasor diagram here this is the field axis and this is the field current we have I F and this is the field induced M F that is E F and what we say in this case is that the stator current should be in phase with the field induced M F. This is I S these are all phasor quantities we have the current I F is also a phasor and then what about V? V can be found out by adding the synchronous reactance drop that is J I S X S to E F. So, if we complete this triangle the vector triangle we get the applied voltage that is V. So, this is the condition for maximum torque per ampere and this is the motoring region. Now, we see in this case the internal power factor is unity E F is in phase with I S and hence the internal power factor of the motor is unity. However, the external power factor which is the angle between the voltage and the current V and I S that is non-zero and in fact we have a lagging power factor here. So, this is the external power factor angle this is theta and I S is lagging behind V although we have a internal unity power factor the external power factor is lagging. So, we can say here this results in unity internal power factor. However, the external power factor is non-unity that is lagging. Now, this is for the motoring operation what about breaking? Breaking means the torque has to be negative. So, for negative torque we have to make delta prime equal to minus pi by 2 because we have a simple equation here. So, if delta prime is pi by 2 sin pi by 2 is plus 1 if delta prime is minus pi by 2 sin minus pi by 2 is minus 1 for breaking we have negative torque. So, we can see what is the breaking situation phase the diagram for breaking operation. So, we have the field here this is the field current and this is the induced T m f that is E f these are all phasor quality I f and E f and the stator current is at this direction 180 degree with respect to E f. So, what we have here is the following that we have I S in this case and what about the phasor here. So, if we have I S and I f we can find out the vector sum of this 2 this is I m because I f plus I s is equal to I m and this is the applied voltage that is V and here we have the synchronous reactance drop in this case J I s X s. So, when we add J I s X s to E f we get the terminal voltage that is equal to V. Now, this is for the breaking operation and in breaking the induced T m f and the currents are in phase opposition this angle is pi. So, when this angle is pi cos of pi is minus 1 and hence the power is fed back to the electric source. We have a motor motor is under breaking condition the kinetic energy of the motor is fed back to the electrical source and here the torque is negative it means the forward motion of the motor is arrested and the energy in the mechanical parts of the motor is fed back to the source. Now, we have a condition in which we can also go for field weakening which is usually done for higher speed. So, in field weakening we go for higher value of delta prime. So, we can briefly discuss about field weakening for field weakening the flux is reduced or the magnetization current is reduced I m is reduced. So, how to reduce I m as we know that I m is a vector sum of I f and I f. So, we have this vector equation I m is equal to I s plus I f. Now, to reduce I m we have to reduce I f. So, in a wound full wound field synchronous machine the magnetizing current can be reduced by reducing the field current and that is done for higher speed, but in the case of a permanent magnet synchronous motor the field is a permanent magnet. So, I f cannot be reduced. So, we can have a equivalent reduced I f by making a large delta prime. So, we have the field here I f. So, what we do in this case is that we make delta prime large. So, as a result this is I s in this case and this is the field induced T m f that is E f and I s will have two parts one is quadrature to the field that is this part and other part is in phase opposition to the field. So, I s is now having two parts one is right angle to the field winding that is this part is 90 degree. So, we can call this is I s x let us say and we can have another part here I s y this I s y is demaintaging the field winding and hence we have a reduced magnetization current. So, in the next lecture we will try to discuss in detail the field weakening operation and also the dynamic control of permanent magnet synchronous motor.