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DC-DC PWM Converters Lecture Note 5 DC Choppers. The converters that achieve the voltage regulation by varying the on–off or time duty ratio of the switching element using a control technique called Pulse Width Modulation PWM. Objective – to efficiently reduce DC voltage ! The DC equivalent of an AC transformer Iin + Vin Iout DC−DC Converter − + Vout − Lossless objective: Pin = Pout, which means that VinIin = VoutIout and Vout I in Vin I out 3 Linear Conversion The load R1 + Vin + R2 − Vout − R2 Vout Vin R1 R2 Vout R2 R1 R2 Vin If Vin = 39V, and Vout = 13V, efficiency η is only 0.33 Unacceptable except in very low power applications 4 Linear Conversion From what explained above, it is clear that a DC conversion by a voltage divider presents some drawbacks: A DC voltage higher than the input voltage cannot be obtained; The output voltage depends on the load, in general; The efficiency is very poor. Linear Conversion the output voltage is given by: • only a step down conversion is possible • the efficiency remains low because all the power supplied by the source that it is not utilized by the load have to be dissipated by the power BJT. Switching Conversion 𝑉𝑜𝐴𝑉 𝑡𝑜𝑛 𝑡𝑜𝑛 = 𝑉𝑖𝑛 = 𝑉𝑖𝑛 = 𝑉𝑖𝑛 𝐷 𝑡𝑜𝑛 + 𝑡𝑜𝑓𝑓 𝑇 Duty Cycle Switching Conversion Transistor is operated in switched-mode: Switch closed: Fully on (saturated) Switch opened: Fully off (cut-off) When switch is open, no current flow in it When switch is closed no voltage drop across it. Since P=V.I, no losses occurs in the switch. Power is 100% transferred from source to load. Power loss is zero (for ideal switch): Switching regulator is the basis of all DC-DC converters Pulse Width Modulation The output DC voltage of DC chopper can be varied by controlling the width period (ton) with constant switching/chopping frequency fs. This method is called PWM method Pulse Width Modulation Choppers Types Two of the most popular categories of DC-DC converters are: Transformerless DC-DC Converters Insulated DC-DC Converters. Three basic types of non-isolated DC–DC converters are Step-down converter Step-up converter Step-up-down converter DC−DC Buck Converter Step-Down Converter Buck Chopper 12 On-State Off-State 13 Switch is turned on (closed) • Diode is reversed biased. • Switch conducts inductor current • This results in positive inductor voltage, i.e: • It causes linear increase in the inductor current 14 Switch turned off (opened) • Because of inductive energy storage, iL continues to flow. • Diode is forward biased • Current now flows (freewheeling) through the diode. • The inductor voltage can be derived as: 15 Analysis 16 Analysis 17 Steady-state Operation + L Realization using iL(t) power MOSFET and diode Vg +- + DTs Ts VL(t) D1 ic(t) R t Unstable Steady-state 18 Since the average voltage across L is zero VLavg D Vin Vout 1 D Vout 0 DVin D Vout Vout D Vout The input/output equation becomes Vout DVin From power balance, Vin I in Vout I out , so I in I out D Note – even though iin is not constant (i.e., iin has harmonics), the input power is still simply Vin • Iin because Vin has no harmonics 19 ! Output Voltage Ripple 20 Output Voltage Ripple 21 Examine the inductor current Examine the inductor current Switch closed, vL Vin Vout , diL Vout vL Vout , dt L Switch open, Vout A / sec L iL Imax Iavg = Iout Vin Vout A / sec L Imin DT diL Vin Vout dt L From geometry, Iavg = Iout is halfway between Imax and Imin ΔI Periodic – finishes a period where it started (1 − D)T T 23 Examine the inductor current Vout A / sec L iL Imax Vin Vout A / sec L Imin DT ΔI Periodic – finishes a period where it started (1 − D)T Taking the derivative of above equation with respect to D and setting it to zero shows that ΔI is maximum when D = 0.5 Examine the inductor current The boundary of continuous conduction is when ΔiLmin = 0, as shown below: vL Vout , diL Vout dt L The maximum required value of Lboundary occurs when D → 0. Therefore, the value of L will guarantee CCM for all D. Effect of raising and lowering L while holding Vin, Vout, Iout and f constant iL Lower L Raise L • Lowering L increases ΔI and moves the circuit toward discontinuous operation 26 Effect of raising and lowering f while holding Vin, Vout, Iout, and L constant iL Lower f Raise f • Slopes of iL are unchanged • Lowering f increases ΔI and moves the circuit toward discontinuous operation 27 the rms value inductor current: i (t ) Imax Imin the ripple i (t ) Imax 0 I avg = Imin + the minimum value I avg Imax Imin 2 Imin 0 the rms value inductor current: 2 I rms Avg i (t ) I min 2 2 2 I rms Avg i2 (t ) 2i (t ) I min I min 2 2 I rms Avg i2 (t ) 2 I min Avg i (t ) I min 2 I rms I max I min 2 2I 3 I max I min I 2 min min 2 Define I PP I max I min 2 I PP 2 2 I rms I min I PP I min 3 the rms value inductor current: I Recognize that I min I avg PP 2 2 I rms 2 I PP I I I avg PP I PP I avg PP 3 2 2 2 2 2 2 I PP I PP I PP 2 2 I rms I avg I PP I avg I avg I PP 3 2 2 2 I PP I PP 2 2 I rms I avg 3 i (t ) 4 2 I PP 2 2 I rms I avg 12 Or 4 I avg I avg I max I min 2 I PP I max I min Component Ratings Inductor current rating Capacitor current rating MOSFET and diode currents ratings Voltage ratings Inductor current rating 2 2 I Lrms I avg 1 2 1 2 I pp I out I 2 12 12 Max impact of ΔI on the rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout 2Iout iL Iavg = Iout ΔI 0 2 2 I Lrms I out 1 2 2I out 2 4 I out 12 3 2 I Lrms I out 3 Use max 32 Capacitor current rating iL Iout L C Iout iC = (iL – Iout) 0 −Iout (iL – Iout) Note – raising f or L, which lowers ΔI, reduces the capacitor current ΔI Max rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout Use max 2 2 I Crms I avg 1 2 2 I out 2 02 1 I out 12 3 I I Crms out 3 33 MOSFET and diode currents and current ratings iL iin Iout L C (iL – Iout) 2Iout Iout 0 2Iout Iout 0 Use max Take worst case D for each I rms 2 I out 3 34 Voltage ratings iL iin Iout C sees Vout Switch Closed L Vin C iC + Vout – Diode sees Vin MOSFET sees Vin iL Switch Open Iout L Vin C iC + Vout – • Diode and MOSFET, use 2Vin • Capacitor, use 1.5Vout 35 ! There is a 3rd state – discontinuous Iout MOSFET L Vin DIODE 36 C Iout + Vout – • Occurs for light loads, or low operating frequencies, where the inductor current eventually hits zero during the switch-open state • The diode opens to prevent backward current flow • The small capacitances of the MOSFET and diode, acting in parallel with each other as a net parasitic capacitance, interact with L to produce an oscillation vL = (Vin – Vout) Switch closed vL = –Vout Switch open • The output C is in series with the net parasitic capacitance, but C is so large that it can be ignored in the oscillation phenomenon 650kHz. With L = 100µH, this corresponds to net parasitic C = 0.6nF Impedance matching Iout = Iin / D Iin + + Source ! DC−DC Buck Converter Vin Vout = DVin − − V Rload out I out Iin + Vin Equivalent from source perspective Requiv − Vout Vin Vout Rload D Requiv 2 I in I out D I out D D2 So, the buck converter makes the load resistance look 37larger to the source Example 1: Step-Down DC-DC Converter supplied by 230V DC voltage. The load resistance equal to10Ω. Voltage drop across the chopper when it is ON equal to 2V. For a duty cycle of 0.4, calculate: a) Average and RMS values of output voltage b) Power delivered to the load and c) Chopper efficiency. DC−DC Boost Converter Step-Down Converter Boost Chopper 39 Buck converter + vL – iL iin Iout L Vin Boost converter C iin + vL – iL iC Iout L Vin + Vout – C iC + Vout – Boost (step-up) converter 41 Boost Analysis: Switch Closed 42 Boost Analysis: Switch Opened 43 Average Output voltage Expression The net energy in the inductor is should be equal to zero over T period Average voltage across inductor is 0 VLavg ton Vs toff Vs Vout 0 Vout ton DT Vout ton toff Vs Vo Vout toff Vs toff Vs ton T Vs toff Vs T Vs T ton Vout T Vs T DT Vout Vs 1 D Output Characteristics Vout Vo Vs 1 D Infinity Vs D 0 1 Output Characteristics Io Pout Ps Is Vout I out Vs I s Vs I out Vs I s (1 D) I out (1 D) I s D 0 1 As 1 → D , the width of the ΔQ area increases to fill almost the entire cycle, and the maximum peak-to-peak ripple becomes V Q I out T I out C C Cf 47 Examine the inductor current Switch closed, diL Vin vL Vin , dt L Switch open, diL Vin Vout vL Vin Vout , dt L Vin Vout A / sec L iL Imax Iavg = Iin Vin A / sec L Imin DT Iavg = Iin is half way between Imax and Imin ΔI (1 − D)T T 48 Continuous current in L Vin Vout A / sec L iL 2Iin Iavg = Iin 0 (1 − D)T 1 Vin Vin 11 D Vin V V 1 D 2 I in out in 1 D T 1 D 1 D T Lboundary Lboundary Lboundary f 2 I in Vin D Lboundary f , V D Lboundary in 2 I in f Then, considering the worst case (i.e., D → 1), V L in 2 I in f use max guarantees continuous conduction 49 use min Inductor current rating 2 2 I Lrms I avg 1 2 1 2 I pp I in I 2 12 12 Max impact of ΔI on the rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iin 2Iin iL Iavg = Iin ΔI 0 2 2 I Lrms I in I Lrms 1 2Iin 2 4 Iin2 12 3 2 I in 3 Use max 50 MOSFET and diode currents ratings iin + vL – iL iD Iout L Vin C iC + Vout – 2Iin 0 2Iin 0 Use max Take worst case D for each I rms 2 I in 3 51 Capacitor current and current rating iin iL iD Iout L Vin C iC + Vout – iC = (iD – Iout) 2Iin −Iout 0 −Iout Max rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout Use max I Crms I out 52 Voltage ratings Diode sees Vout iin iL Iout C sees Vout + Vout – L Vin C iin iL Iout L Vin C + Vout – MOSFET sees Vout • Diode and MOSFET, use 2Vout • Capacitor, use 1.5Vout 53 Impedance matching I out 1 DIin Iin + + Source DC−DC Boost Converter Vin − Vin 1 D − Vout V Rload out I out Iin + Vin Equivalent from source perspective Requiv − 1 D Vout 1 D 2 Vout 1 D 2 R V Requiv in load I out I in I out 54 1 D Example 2: A boost chopper has input voltage of 20 V with switching frequency equal to 1 kHz. Calculate: • The required duty cycle that can be applied to the switch to boost the input voltage to 60V. • The ON and OFF period for the constant switching frequency operation. • Output current if the resistance load equal to 10 Ω. • Average input inductor current. • The maximum and minimum currents via the input inductor if the inductance is 10mH.