A study on EV Drivetrain Components
Q1. Which types of power converter circuits are employed in electric and hybrid electric vehicles?
A converter is an electrical circuit that accepts a DC input and generates a DC output of a different voltage, usually achieved by high frequency switching action employing inductive and capacitive filter elements. Power conversion is converting electric energy from one form to another such as converting between AC and DC; or changing the voltage or frequency, or some combination of these.
A power converter is an electrical circuit that changes the electric energy from one form into the desired form optimized for the specific load. A converter may do one or more functions and give an output that differs from the input. It is used to increase or decrease the magnitude of the input voltage, invert polarity, or produce several output voltages of either the same polarity with the input, different polarity, or mixed polarities such as in the computer power supply unit.
The DC to DC converters is used in a wide range of applications including computer power supplies, board-level power conversion and regulation, dc motor control circuits, and much more.
The converter acts as the link or the transforming stage between the power source and the power supply output. There are several kings of converters based on the source input voltage and the output voltage and these fall into four categories namely the AC to DC converter known as the rectifier, the AC to AC cyclo-converter or frequency changer, the DC to DC voltage or current converter, and the DC to AC inverter
In electrical engineering, power engineering, and the electric power industry, power conversion is converting electric energy from one form to another such as converting between AC and DC; or changing the voltage or frequency; or some combination of these.
There are mainly three types of Power Converter circuit used in electric and hybrid electric vehicles are:-
1. Rectifier (AC to DC Converter)
A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction. The process is known as rectification since it "straightens" the direction of the current.
An on-board charger is used to charge the battery from an external AC Power Supply.
Simple Half Wave Rectifier Circuit
2. Inverter (DC to AC Converter)
An inverter is a power electronic device or circuitry that changes direct current (DC) to alternating current (AC). The input voltage, output voltage and frequency, and overall power handling depend on the design of the specific device or circuitry.
The traction inverter converts energy from the batteries to the traction motor, which propels the vehicle.
A simple inverter circuit
3. Chopper (DC to DC Converter)
A chopper circuit is used to refer to numerous types of electronic switching devices and circuits used in power control and signal applications. A chopper is a device that converts fixed DC input to a variable DC output voltage directly. Essentially, a chopper is an electronic switch that is used to interrupt one signal under the control of another circuit is used to refer to numerous types of electronic switching devices and circuits used in power control and signal applications. A chopper is a device that converts fixed DC input to a variable DC output voltage directly. Essentially, a chopper is an electronic switch that is used to interrupt one signal under the control of another.
A DC-to-DC converter steps down the power to lower voltages. This is used to control the doors, heater and windows.
Q2. An Electric Vehicle's powertrain with 72V battery pack in shown in the diagram below. The duty ratio for acceleration operation is 'd1' and for the braking operation the duty ratio is 'd2'.
Separately Excited DC Motor
separately excited dc motor are like other DC motors, these motors also have both stator and rotor. Stator refers to the static part of the motor, which consists of the field windings. And the rotor is the moving armature that contains armature windings or coils. Separately excited dc motor has field coils similar to that of shunt wound dc motor. The name suggests the construction of this type of motor. Usually, in other DC motors, the field coil and the armature coil both are energized from a single source. The field of them does not need any separate excitation. But, in a separately excited DC motor, a separate supply Provided for excitation of both field coil and armature coil.
Here, the field coil is energized from a separate DC voltage source and the armature coil is also energized from another source. The armature voltage source may be variable but, independent constant DC voltage is used for energizing the field coil. So, those coils are electrically isolated from each other, and this connection is the specialty of this type of DC motor.
Given Inputs
Motor and Controller Parameters
Rated Armature voltage= 72 V
Rated armature current= 400 A
Ra= 0.5Ω
KΦ= 0.7 Volt second
Chopper Switching frequency= 400 Hz
The vehicle speed-torque characteristics
`T_v=24.7+(0.0051)w^2` , consider this as equation-1 where
`T_v` =Torque in Nm
`w` =Angular velocity in rad/sec
Also, we know that Torque Speed Characteristics of a Motor can be given by the formula
`T=(K_mPhiE_s)/R_a-(K_mPhi)^2/R_aw` , considering this as Equation-2 where
`T` =Torque in Nm
`K_m` = Motor Constant in `(Nm)/sqrtW`
`Phi` =Magnetic Field Strength in Wb/m^2
`E_s` =Supply Volatge
`R_a` =Armature Resistance ohms
`w` = Angular Velocity in rad/sec
Given Duty-Cycle is 70%
`V_s=0.7Vol tag e=0.7*72`
`V_s=50.4V`
Substituting the given input values in Equation-2
`T=(0.7*50.4)/0.5-0.7^2/0.5w=70.56-0.98w`
Equating equation-1 and equation-2 to find a solution
`70.56-0.98w=24.7+0.0051w^2`
Transforming the equation we get a quadratic equation
`0.0051w^2+0.98w-45.86=0`
Solving the Quadratic Equation we get the values of `w` that satisfying the above equation.
`w_1=(-b+sqrt(b^2-4ac))/(2a)=(-0.98+sqrt(0.98^2-4*0.0051*(-45.86)))/(2*0.0051)=38.9150`
`w_2=(-b-sqrt(b^2-4ac))/(2a)=(-0.98-sqrt(0.98^2-4*0.0051*(-45.86)))/(2*0.0051)=-231.0718`
Matlab Script for solving the Quadratic Equation and solving the equation by finding the intersection of the two curves
Let Equation-1, `T_v=24.7+(0.0051)w^2` represent a Quadratic function and Equation-2, `T=(K_mPhiE_s)/R_a-(K_mPhi)^2/R_aw` represent a linear function.
The point of intersection of the two functions gives us the required solution for the Steady State, satisfying both functions.
Matlab Script
%Program to find Vehicle Steady State
%Let the rpm ranges from 0 to 15k
rpm=linspace(0,3000,20000);
%Converting rpm to angular velocity
w=2*pi.*rpm/60;
%The Given Formula for Vehicle Torque is
T_given=24.7+0.0051.*(w.^2);
%given data
v=72; %rated volatge in volt
a=400; %rated armature current in ampere
ra=0.5; %Ohm
km_phi=0.7; %Volt second
%supply volatge when duty cycle=0.7
vs=0.7*v;
%the torque can be given by the formula
t=(km_phi*vs/ra)-(km_phi^2/ra).*w;
%plotting the two curve
plot(w,T_given)
hold on
plot(w,t)
title('Torque vs Angular Velocity');
xlabel('Angular Velocity');
ylabel('Torque');
%finding the intersection point for two curves
dfr=abs(T_given-t);
index=find(dfr<0.01);
w_sol=w(index);
str={'Steady State at w= ',num2str(w_sol)};
plot(w(index),t(index),'r*');
text(w(index),t(index),str);
legend('Quadratic Torque Function','Linear Torque Function','Steady State Solution');
grid on
%finding the roots for the quadratic equation
a=0.0051;
b=0.98;
c=-45.86;
d=sqrt(b^2-4*a*c);
w1=(-b+d)/(2*a);
w2=(-b-d)/(2*a);Result
The solution obtained for the steady-state is at w=38.91rad/sec, both the methods gave the same output as explained above.
Q3. Explain in brief the author’s perspective on Induction Versus DC Brushless Motors.
The author makes a detailed comparison between Induction and DC Brushless motor in order to decide which one is better. The author compares each aspect of the two machines from construction, operation, efficiency, and control. The
Differences Between DC and Induction Motor
- DC Brushless Motors uses permanent magnets as the rotor whereas the induction motor uses a distributed windings.
- DC motor is more expensive than induction motors as the cost of the permanent magnet is comparatively higher than that of copper windings.
- Permanent magnet (PM) rotors are also difficult to handle due to very large forces that come into play when anything ferromagnetic gets close to them.
- Induction motors have a cost advantage over PM machines. The development costs of an Induction motor is high, but likely little or no recurring costs.
- Induction machines are more difficult to control. The control laws are more complex and difficult to understand. Achieving stability over the entire torque-speed range and over temperature is more difficult with induction than with DC brushless.
- Changing magnetic field strength (B) in a permanent magnet (PM) motor is difficult compared to the induction motor. For example, when maximum torque is required, especially at low speeds, the magnetic field strength (B) should be maximum – so that inverter and motor currents are maintained at their lowest possible values.
- DC brushless, as machine size grows, the magnetic losses increase proportionately and part load efficiency drops. With induction, as machine size grows, losses do not necessarily grow.
- Induction motor drives are favored where high-performance is desired, peak efficiency will be a little less than with DC brushless, but average efficiency may actually be better.
- Rotor heat generated in a DC brushless drive is less compared to an induction motor and cooling is easier is also easier.
In the end, the author concludes that DC brushless drives will likely continue to dominate in the hybrid and coming plug-in hybrid markets, and that induction drives will likely maintain dominance for the high-performance pure electrics.
References
1. https://semiengineering.com/electric-cars-gain-traction-but-challenges-remain/
2. https://en.wikipedia.org/wiki/Power_inverter
3. https://www.sunpower-uk.com/glossary/what-is-a-power-converter/