Power Electronics and Drives Assignment Table of Contents 1 Assignment Questions 1 1.1 AC to DC Uncontrolled Rectification 2 1.2 AC to DC Controlled Rectification 3 1.3 AC to AC Voltage Controllers 4 1.4 DC to DC Converters 5 1.5 DC to AC Inverters 6 1.6 DC Drives 7 1.7 AC Drives 8 2 Formula Sheet 9 2.1 Fourier Series 9 2.2 Trigonometric Rules 9 2.3 Differentiation and Integration Rules 9 2.4 DC and RMS 9 2.5 Performance Parameters 10 2.6 DC Motor 11 2.7 AC Induction Motor 12 Assignment Questions The assignment questions are based on a past examination paper and are typical of what you should expect in your exam. If you are sufficiently prepared, then you should be able to complete this assignment in less than 4 hours. All assignments shall be submitted individually. Group assignments are not allowed. A formula sheet is provided in section 2.   AC to DC Uncontrolled Rectification Consider the following Single Phase Full Wave Bridge uncontrolled rectifier. The following measurements were taken: Parameter Value RMS Output Current 0.686 amps DC Output Current 0.680 amps RMS Output Voltage 27.5 volts DC Output Voltage 24.5 volts Ignore the losses in the diodes. Provide approximate sketches of the following: Supply voltage Output voltage Output current Current through diode D1 Supply Current Calculate the following: AC Output Voltage AC Output Current Voltage Ripple Factor Current Ripple Factor DC current through diode D1 RMS current through diode D1 RMS supply current DC output power Total output power Transformer Utilisation Factor Power Factor AC to DC Controlled Rectification Consider the following Single Phase Full Wave Bridge controlled rectifier. The supply voltage is V_s (θ)=V_m sin⁡θ, where θ=ωt. Thyristors TH1 and TH2 are fired at θ=α and the current continues to flow until θ=β (the extinction angle). V_dc=V_m/π(cos⁡〖α-cos⁡〖β)〗 〗. Derive the expression for the DC output voltage. V_rms=V_m √(1/2π (β-α+((sin⁡2α-sin⁡2β ))/2) ). Derive the expression for the RMS output voltage. Assume that the firing angle α=60°, the load is purely resistive with R = 10 ohms (L = 0), and Vs = 30 volts rms. What is the value of the extinction angle (β) for a purely resistive load and why is this so? Sketch the shape of the output current. Calculate: Peak supply voltage (V_m ), Output DC current, Output RMS current, RMS Supply current, Peak Supply current, Transformer Utilisation Factor, Power Factor, and Crest Factor.   AC to AC Voltage Controllers The figure below shows the simulation circuit for a single phase full wave AC to AC voltage controller. The supply voltage is 240 volts RMS, 50 Hz and the load consists of a 100 ohm resistor in series with a 0.3 H inductor. The firing angle is 60 degrees. The RMS output current is 1.41 amps. The following figure shows the output voltage (Vo = red) and the current (blue). Ignore any losses in the thyristors. Calculate the input power factor Explain the observed waveforms for the output voltage and output current Derive the general expression for the output current for α≤θ=ωt≤β as a function of time for a single half cycle DC to DC Converters A step up converter is shown in the following figure. The step up converter is being used to charge a 1 ampere-hour, 30 Volt battery from a 10 Volt DC source. The GTO is being switched at 1 kHz with a 50% duty cycle. The source inductance is 0.01H. The average voltage during mode 2 is V_(d(ave))=V_s/(1-D), where D is the duty cycle. During Mode 1, the GTO is on. Calculate the peak ripple current and Sketch the GTO current for a single cycle. After the GTO is turned off, What is the initial value of Vd? How long will it take for the current to reach zero? Sketch the shape of the battery current over a single cycle. What is the average current flowing into the battery? How long will it take to charge the battery? Sketch the shape of Vd over a single cycle. What is the maximum duty cycle for stable operation?   DC to AC Inverters A single phase, 50 Hz, half bridge DC/AC inverter has a resistive load with R = 10 ohms. The DC input voltage is Vs = +/-100 V. Sketch the output voltage waveform (Vo). Then determine The total RMS output voltage – Vo(rms) The power (Po) dissipated in load resistor R The Fourier Series for the output voltage (Vo(t)) The RMS output voltage at the fundamental frequency – Vo1(rms) The average current through each transistor The RMS current through each transistor The Total Harmonic Distortion (THD) Note: THD=1/V_o1 √(∑_(n=2)^∞▒V_on^2 ), where V_on= the RMS amplitude of n_th harmonic. Hint: V_(0(rms))=√(∑_(n=1)^∞▒V_on^2 ).   DC Drives A full wave, 3 phase bridge controlled rectifier is supplying the armature of a separately excited DC motor. The rectifier is being fed by a 3phase supply with a line to line voltage of 415 volts rms. The DC output voltage of the rectifier is V_dc=(3√3 V_m)/π cos⁡α and the firing angle is 60°. The separately excited dc motor has the following characteristics: Field Resistance Rf = 150 ohms Armature resistance Ra = 0.02 ohms Voltage constant Kv = 0.5 V/A rad /sec No load losses are negligible The rectifier is delivering 50 amps ripple free dc current to the armature winding and the motor is running at 3000 revs per minute. Calculate: The armature voltage (Va) produced by the rectifier, The back emf (Ea) of the DC motor, The field current (If), The required field voltage, Torque delivered to the load, The power delivered to the load. The efficiency of the motor, including field losses.   AC Drives A 3 phase DC/AC inverter is supplying a 3 phase, 50 Hz square wave to an induction motor. The 3 phase voltage waveforms each have a peak value of 300 volts as shown in the following figure. The 3 phase, 50 Hz, 6-pole Y connected induction motor has the following equivalent circuit parameters: Rs = 0.5 ohms, R'r = 0.5 ohms, Xs = X'r = 1.0 ohms. Ignore Rm and Xm. The rotor speed is 950 rpm. The no load losses (Pnl) are negligible. Use the approximate equivalent circuit shown below: Assume that only the fundamental voltage generated by the inverter is significant and ignore the higher harmonics. Determine: The rms phase to neutral voltage generated by the inverter for the 1st harmonic( ie the fundamental), Slip, s, Magnitude of the input impedance, The rotor current (I_r^' ) The developed power (P_d), and The developed torque (T_d ). Formula Sheet Fourier Series a_0=2/T ∫_0^T▒v(t)dt=1/π ∫_0^2π▒v(θ)dθ a_n=2/T ∫_0^T▒〖v(t) cos⁡nωt dt〗=1/π ∫_0^2π▒〖v(θ) cos⁡nθ dθ〗, n=1..∞ b_n=2/T ∫_0^T▒〖v(t) sin⁡nωt dt〗=1/π ∫_0^2π▒〖v(θ) sin⁡nθ dθ〗, n=1..∞ v(t)=a_0/2+∑_(n=1)^∞▒(a_n cos⁡nωt+b_n sin⁡nωt ) V_rms=√((a_0/2)^2+∑_(n=1)^∞▒〖(a_n^2+b_n^2 )/2〗) For a square wave of magnitude +/- V, v(θ)=V∑_(n=1,3,5…)^∞▒〖4/nπ sin⁡nθ 〗 Trigonometric Rules cos⁡〖(A-B)=cos⁡A cos⁡B+sin⁡A sin⁡B 〗 cos⁡〖(A+B)=cos⁡A cos⁡B-sin⁡A sin⁡B 〗 sin⁡(A+B)=sin⁡A cos⁡B+cos⁡A sin⁡B sin⁡(A-B)=sin⁡A cos⁡B-cos⁡A sin⁡B sin⁡2A=2 sin⁡A cos⁡A cos⁡2A=(cos⁡A )^2-(sin⁡A )^2=2(cos⁡A )^2-1=1-2(sin⁡A )^2 (sin⁡A )^2=1/2 (1-cos⁡2A ) (cos⁡A )^2=1/2 (1+cos⁡2A ) sin⁡A+sin⁡B=2 sin⁡〖(A+B)/2〗 cos⁡〖(A-B)/2〗 sin⁡A-sin⁡B=2 cos⁡〖(A+B)/2〗 sin⁡〖(A-B)/2〗 cos⁡A+cos⁡B=2 cos⁡〖(A+B)/2〗 cos⁡〖(A-B)/2〗 cos⁡A-cos⁡B=2 sin⁡〖(A+B)/2〗 sin⁡〖(B-A)/2〗 sin⁡A sin⁡B=1/2 ⌊cos⁡〖(A-B)-cos⁡(A+B) 〗 ⌋ cos⁡A cos⁡B=1/2 [cos⁡〖(A-B)+cos⁡(A+B) 〗 ] sin⁡A cos⁡B=1/2 [sin⁡(A-B)+sin⁡(A+B) ] Differentiation and Integration Rules d/dθ sin⁡nθ=n cos⁡nθ d/dθ cos⁡nθ=-n sin⁡nθ ∫▒sin⁡nθ dθ=-cos⁡nθ/n ∫▒〖cos⁡nθ dθ〗=sin⁡nθ/n DC and RMS V_av=V_dc=1/T ∫_0^T▒v(t)dt=1/2π ∫_0^2π▒v(θ)dθ V_rms=√(1/T ∫_0^T▒〖v^2 (t)dt〗)=√(1/2π ∫_0^2π▒〖v^2 (θ)dθ〗) I_ac=√(I_rms^2-I_dc^2 ) Performance Parameters Displacement factor DF=cos⁡ϕ Power Factor PF=I_rms1/I_rms cos⁡ϕ=P_in/(V_(s(rms)) I_rms ) (single phase) Power Factor PF=I_rms1/I_rms cos⁡ϕ=P_in/(3V_(s(rms)) I_rms ) (3 phase) Harmonic Factor HF=√((I_rms/I_rms1 )^2-1) Crest Factor CF=I_peak/I_rms Efficiency η=P_dc/P_in Form Factor FF=V_rms/V_dc ≥1 (ouput voltages) Ripple Factor = λ_v=V_(ac(RMS))/V_dc =√(〖FF〗^2-1) For single phase TUF=P_dc/(V_s I_s ) For 3 phase TUF=P_dc/(√3 V_(s(l-l)) I_s ) DC Motor T_d=K_t i_f i_a E_a=K_v i_f ω_m V_f=R_f i_f+L_f (di_f)/dt V_a=R_a i_a+L_a (di_a)/dt+E_a T_d=J (dω_m)/dt+Bω_m+T_L Where T_d= Developed torque (N-m) T_L= Load torque (N-m) E_a= Back EMF or speed voltage V_f= Field voltage V_a= Armature voltage Note: K_t = torque constant K_v = voltage constant (Volts/A/rad/sec) However K_t=K_v  AC Induction Motor RF speed is f_s=f/(p/2) (revs/sec), where f= supply frequency p= no of poles f_s is referred to as the synchronous speed Slip s=(ω_s-ω_m)/ω_s , where ω_s= the synchronous speed (radians/sec) ω_m= rotor mechanical rotational speed (radians/sec) Thus ω_m=(1-s)ω_s (radians/sec) Stator Cu loss =P_su=3I_s^2 R_s Rotor Cu loss =P_ru=3(I_r^' )^2 R_r^' Core loss =P_c=(3V_m^2)/R_m ≈(3V_s^2)/R_m Gap power =P_g=3(I_r^' )^2 (R_r^')/s= the power passing from the stator to the rotor through the air gap Developed power =P_d=3(I_r^' )^2 (R_r^' (1-s))/s Developed torque =T_d=P_d/ω_m =(P_g (1-s))/(ω_s (1-s) )=P_g/ω_s =(3(I_r^' )^2 R_r^')/(sω_s ) Input power =P_i=3V_s I_s cos⁡〖θ_m 〗=P_c+P_su+P_g where θ_m= the angle between V_s and I_s Output power =P_o=P_d-P_no load Efficiency η=P_o/P_i =(P_d-P_no load)/(P_c+P_su+P_g )