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DC-DC PWM Converters Lecture Note 5
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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.
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Objective – to efficiently reduce DC voltage
! Objective – to efficiently reduce DC voltage The DC equivalent of an AC transformer Iin Iout DC−DC Converter + Vin − + Vout − Lossless objective: Pin = Pout, which means that VinIin = VoutIout and
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Linear Conversion + Vin − Vout R1 R2 The load
If Vin = 39V, and Vout = 13V, efficiency η is only 0.33 Unacceptable except in very low power applications
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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.
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Linear Conversion only a step down conversion is possible
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.
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Switching Conversion 𝑉 𝑜𝐴𝑉 = 𝑉 𝑖𝑛 𝑡 𝑜𝑛 𝑡 𝑜𝑛 + 𝑡 𝑜𝑓𝑓 = 𝑉 𝑖𝑛 𝑡 𝑜𝑛 𝑇 = 𝑉 𝑖𝑛 𝐷 Duty Cycle
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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
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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
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Pulse Width Modulation
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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
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DC−DC Buck Converter Step-Down Converter Buck Chopper
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On-State Off-State
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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
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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:
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Analysis
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Analysis
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Steady-state Operation
Unstable Steady-state
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Since the average voltage across L is zero
! Since the average voltage across L is zero The input/output equation becomes From power balance, , so 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
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Output Voltage Ripple
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Output Voltage Ripple
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Examine the inductor current
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Examine the inductor current
Switch closed, Switch open, From geometry, Iavg = Iout is halfway between Imax and Imin iL Imax Iavg = Iout ΔI Periodic – finishes a period where it started Imin DT (1 − D)T T
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Examine the inductor current
iL Imax ΔI Periodic – finishes a period where it started Imin DT (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
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Examine the inductor current
The boundary of continuous conduction is when ΔiLmin = 0, as shown below: The maximum required value of Lboundary occurs when D → 0. Therefore, the value of L will guarantee CCM for all D.
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Effect of raising and lowering L while holding Vin, Vout, Iout and f constant
Lower L Raise L Lowering L increases ΔI and moves the circuit toward discontinuous operation
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Effect of raising and lowering f while holding Vin, Vout, Iout, and L constant
Lower f Raise f Slopes of iL are unchanged Lowering f increases ΔI and moves the circuit toward discontinuous operation
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the rms value inductor current:
the ripple + = the minimum value
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the rms value inductor current:
Define
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the rms value inductor current:
Recognize that Or
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Component Ratings Inductor current rating Capacitor current rating MOSFET and diode currents ratings Voltage ratings
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Inductor current rating
Max impact of ΔI on the rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout iL 2Iout Iavg = Iout ΔI Use max
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Capacitor current rating
L out L C (iL – Iout) iC = (iL – Iout) Note – raising f or L, which lowers ΔI, reduces the capacitor current Iout ΔI −Iout Max rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout Use max
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MOSFET and diode currents and current ratings
L out in L C (iL – Iout) 2Iout Iout 2Iout Iout Use max Take worst case D for each
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Voltage ratings i i I L + V V C i – i I L + V V C i – C sees Vout
Switch Closed L + V V in C out i C – Diode sees Vin MOSFET sees Vin i I L out Switch Open L + V V in C out i C – Diode and MOSFET, use 2Vin Capacitor, use 1.5Vout
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There is a 3rd state – discontinuous
! There is a 3rd state – discontinuous I MOSFET out + L V V in C out DIODE I out – 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 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 vL = (Vin – Vout) Switch closed vL = –Vout Switch open 650kHz. With L = 100µH, this corresponds to net parasitic C = 0.6nF
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! Impedance matching Iout = Iin / D Iin DC−DC Buck Converter + +
Vin − + Vout = DVin − Source Iin + Vin − Equivalent from source perspective So, the buck converter makes the load resistance look larger to the source
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Example 1: Step-Down DC-DC Converter supplied by 230V DC voltage
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: Average and RMS values of output voltage Power delivered to the load and Chopper efficiency.
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DC−DC Boost Converter Step-Down Converter Boost Chopper
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Buck converter Boost converter i + v – I i + V V C i – i + v – I i + V
L + v – I i out Buck converter in + V V in C out i C – i L + v – I out Boost converter i in + V V in C out i C –
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Boost (step-up) converter
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Boost Analysis: Switch Closed
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Boost Analysis: Switch Opened
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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 toff ton
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Output Characteristics
Vo Infinity Vs D 1
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Output Characteristics
Io Is D 1
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As 1 → D , the width of the ΔQ area increases to fill almost the entire cycle, and the maximum peak-to-peak ripple becomes
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Examine the inductor current
Switch closed, Switch open, Iavg = Iin is half way between Imax and Imin iL Imax Iavg = Iin ΔI Imin DT (1 − D)T T
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Continuous current in L
iL 2Iin Iavg = Iin (1 − D)T Then, considering the worst case (i.e., D → 1), use max guarantees continuous conduction use min
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Inductor current rating
Max impact of ΔI on the rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iin iL 2Iin Iavg = Iin ΔI Use max
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MOSFET and diode currents ratings
+ v L – i i I i L D out in L + V V in C out i C – 2Iin 2Iin Use max Take worst case D for each
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Capacitor current and current rating
L out i D in L + V V in C out i C – iC = (iD – Iout) 2Iin −Iout −Iout Max rms current occurs at the boundary of continuous/discontinuous conduction, where ΔI =2Iout Use max
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Voltage ratings i I i L + V V C – i I i L + V V C – Diode sees Vout
C sees Vout L + V V in C out – i I out i L in L + V V in C out – MOSFET sees Vout Diode and MOSFET, use 2Vout Capacitor, use 1.5Vout
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Impedance matching Iin DC−DC Boost Converter + + Vin − − Source Iin +
Equivalent from source perspective
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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.
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