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DC DC Converter

DC-DC converters are circuits that convert a DC voltage to another DC voltage level. They use switching elements like transistors and power switches to efficiently step up or step down voltage. The buck converter is a common DC-DC converter topology that can step down voltage. It uses a switch, inductor, diode, and capacitor. By periodically opening and closing the switch, the inductor filters the output to produce a lower average voltage. The output voltage of an ideal buck converter is equal to the input voltage multiplied by the duty cycle of the switch. Real converters have non-ideal components that cause additional voltage ripple. Proper component selection and design considerations are needed to minimize ripple.

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DC-DC Converters
DC-DC Converters DC-DC converters are power electronic circuits that convert a DC voltage to a different DC voltage level, often providing a regulated output. A BASIC SWITCHING CONVERTER An efficient alternative to the linear regulator Uses Power electronics switches like BJT,MOSFET IGBT… Also known as DC Chopper
DC-DC Converters Figure: (a) A basic DC-DC switching converter; (b) Switching equivalent; ( c) Output voltage.  Assuming the switch is ideal • The output is the same as the input when the switch ON • And the output is zero when the switch OFF  Periodic opening and closing of the switch gives the pulsed output waveform.  The average or DC component of the output voltage is 𝑉𝑜 = 1 𝑇 0 𝑇 𝑣 𝑜 𝑡 𝑑𝑡 = 1 𝑇 0 𝐷𝑇 𝑉𝑠 𝑑𝑡 = 𝑉𝑠 𝐷
DC-DC Converters 𝐷 ≡ 𝑡 𝑜𝑛 𝑡 𝑜𝑛 + 𝑡 𝑜𝑓𝑓 The DC component of the output voltage will be less than or equal to the input voltage for this circuit. ideal switch Zero loss Zero voltage across when ON Zero current through it when OFF But real switch has some power loss Considerable because it creates heat on the switch.
THE BUCK (STEP-DOWN) CONVERTERApplication Example: Controlling the speed of DC Motor Low-pass filter will be placed at the output • The diode provides a path for the inductor current when the switch is opened and is reverse-biased when the switch is closed.
THE BUCK (STEP-DOWN) CONVERTERVOLTAGE AND CURRENT RELATIONSHIPS: 𝑣 𝑥 in Fig. above, is 𝑉𝑠 when the switch is closed and is zero when the switch is open, provided that the inductor current remains positive, keeping the diode on. If the switch is closed periodically at a duty ratio D, the average voltage at the filter input is 𝑉𝑠 𝐷 An inductor current that remains positive throughout the switching period is known as continuous current. Conversely, discontinuous current is characterized by the inductor current’s returning to zero during each period.
THE BUCK (STEP-DOWN) CONVERTERBuck converters and DC-DC converters in general, have the following properties when operating in the steady state 1. The inductor current is periodic. 𝒊 𝑳 𝒕 + 𝑻 = 𝒊 𝑳 𝒕 2. The average inductor voltage is zero 𝑽 𝑳 = 𝟏 𝑻 𝒕 𝒕+𝑻 𝒗 𝑳 𝝀 𝒅𝝀 = 𝟎 3. The average capacitor current is zero 𝑰 𝑪 = 𝟏 𝑻 𝒕 𝒕+𝑻 𝒊 𝑪 𝝀 𝒅𝝀 = 𝟎 4. The power supplied by the source is the same as the power delivered to the load. For non-ideal components, the source also supplies the losses. 𝑷 𝒔 = 𝑷 𝒐 𝒊𝒅𝒆𝒂𝒍 𝑷 𝒔= 𝑷 𝒐 + 𝒍𝒐𝒔𝒔𝒆𝒔 𝒏𝒐𝒏𝒊𝒅𝒆𝒂𝒍
THE BUCK (STEP-DOWN) CONVERTERAssumptions for Analysis of Buck converter: • The circuit is operating in the steady state. • The inductor current is continuous (always positive) • The capacitor is very large, and the output voltage is held constant at voltage Vo. This restriction will be relaxed later to show the effects of finite capacitance. • The switching period is 𝑇; the switch is closed for time 𝐷𝑇 and open for time (1 − 𝐷)𝑇. • The components are ideal. The key to the analysis for determining the output 𝑉𝑜 is to examine the inductor current and inductor voltage first for the switch closed and then for the switch open.
THE BUCK (STEP-DOWN) CONVERTERANALYSIS FOR THE SWITCH CLOSED: • The voltage across the inductor is • 𝑣 𝐿 = 𝑉𝑆 − 𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 • Rearranging, • 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉 𝑆−𝑉𝑜 𝐿 𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑 • 𝑑𝑖 𝐿 𝑑𝑡 = ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 𝐷𝑇 = 𝑉 𝑆−𝑉𝑜 𝐿 • ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉 𝑆−𝑉𝑜 𝐿 𝐷𝑇
THE BUCK (STEP-DOWN) CONVERTERANALYSIS FOR THE SWITCH OPEN: • When the switch is open, the diode becomes forward-biased to carry the inductor current • The voltage across the inductor when the switch is open is 𝑣 𝐿 = −𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 • Rearranging, 𝑑𝑖 𝐿 𝑑𝑡 = −𝑉𝑜 𝐿 𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛 • The derivative of current in the inductor is a negative constant, and the current decreases linearly • The change in inductor current when the switch is open is ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 1−𝐷 𝑇 = −𝑉𝑜 𝐿 ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇
THE BUCK (STEP-DOWN) CONVERTERAt steady state ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 + ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = 0 𝑉 𝑆−𝑉𝑜 𝐿 𝐷𝑇 − 𝑉𝑜 𝐿 1 − 𝐷 = 0 • Solving for 𝑉𝑜, 𝑉𝑜 = 𝑉𝑠 𝐷 • Therefore, the buck converter produces an output voltage that is less than or equal to the input. • An alternative derivation of the output voltage is based on the inductor voltage. • Since the average inductor voltage is zero for periodic operation 𝑉𝐿 = 𝑉𝑆 − 𝑉𝑂 𝐷𝑇 + −𝑉𝑂 1 − 𝐷 𝑇 = 0 • Solving for 𝑉𝑂 𝑉𝑂= 𝑉𝑆 𝐷.
THE BUCK (STEP-DOWN) CONVERTERThe average inductor current must be the same as the average current in the load resistor, since the average capacitor current must be zero for steady-state operation. 𝐼𝐿 = 𝐼 𝑅 = 𝑉𝑂 𝑅 The maximum and minimum values of the inductor current are computed as ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉𝑆 − 𝑉𝑜 𝐿 𝐷𝑇 ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇  𝐼 𝑚𝑎𝑥= 𝐼𝐿 + ∆𝑖 𝐿 2 = 𝑉𝑂 + 1 𝑉𝑂 1 − 𝐷 𝑇 = 𝑉𝑂 1 + 1 − 𝐷
THE BUCK (STEP-DOWN) CONVERTER 𝐼 𝑚𝑖𝑛 = 𝐼𝐿 − ∆𝑖 𝐿 2 = 𝑉𝑂 𝑅 − 1 2 𝑉𝑂 𝐿 1 − 𝐷 𝑇 = 𝑉𝑂 1 𝑅 − 1 − 𝐷 2𝐿𝑓 where 𝑓 = 1/𝑇 is the switching frequency. • The above Eq. can be used to determine the combination of L and f that will result in continuous current. Since 𝐼 𝑚𝑖𝑛 = 0 is the boundary between continuous and discontinuous current, 𝐼 𝑚𝑖𝑛 = 0 = 𝑉𝑂 1 𝑅 − 1 − 𝐷 2𝐿𝑓 (𝐿𝑓) 𝑚𝑖𝑛= (1 − 𝐷)𝑅 2
THE BUCK (STEP-DOWN) CONVERTER• If the desired switching frequency is established, 𝐿 𝑚𝑖𝑛 = (1 − 𝐷)𝑅 2𝑓 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 where 𝐿 𝑚𝑖𝑛 is the minimum inductance required for continuous current. • In practice inductance is chosen greater than 𝐿 𝑚𝑖𝑛 • The peak-to-peak variation in the inductor current is often used as a design criterion in buck converter. • to determine the value of inductance for a specified peak-to-peak inductor current for continuous-current operation: ∆𝑖 𝐿 = 𝑉𝑠 − 𝑉𝑜 𝐿 𝐷𝑇 = 𝑉𝑠 − 𝑉𝑜 𝐿𝑓 𝐷 = 𝑉𝑂 1 − 𝐷 𝐿𝑓 𝑜𝑟 𝐿 = 𝑉𝑠 − 𝑉𝑜 ∆𝑖 𝐿 𝑓 𝐷 = 𝑉𝑂 1 − 𝐷 ∆𝑖 𝐿 𝑓
THE BUCK (STEP-DOWN) CONVERTER• Since the converter components are assumed to be ideal, the power supplied by the source must be the same as the power absorbed by the load resistor. 𝑃𝑆 = 𝑃𝑂 𝑉𝑠 𝐼𝑠 = 𝑉𝑜 𝐼 𝑜 𝑜𝑟 𝑉𝑜 𝑉𝑠 = 𝐼𝑠 𝐼 𝑜 • Basically from this expression we can say that DC-DC converter is DC transformer.
THE BUCK (STEP-DOWN) CONVERTER• OUTPUT VOLTAGE RIPPLE • In the preceding analysis, the capacitor was assumed to be very large to keep the output voltage constant. • The variation in output voltage is computed from the voltage- current relationship of the capacitor. • The current in the capacitor is 𝑖 𝐶 = 𝑖 𝐿 − 𝑖 𝑅 • While the capacitor current is positive, the capacitor is charging. From the definition of capacitance, 𝑄 = 𝐶𝑉𝑜 ∆𝑄 = 𝐶∆𝑉𝑜 ∆𝑉𝑜 = ∆𝑄 𝐶
THE BUCK (STEP-DOWN) CONVERTER• The change in charge ∆𝑄 is the area of the triangle above the time axis ∆𝑄 = 1 2 𝑇 2 ∆𝑖 𝐿 2 = 𝑇∆𝑖 𝐿 8 • resulting in ∆𝑉𝑜 = 𝑇∆𝑖 𝐿 8𝐶
THE BUCK (STEP-DOWN) CONVERTER • Using ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇 = ∆𝑖𝐿 ∆𝑉𝑜 = 𝑇𝑉𝑜 8𝐶𝐿 1 − 𝐷 𝑇 = 𝑉𝑜(1 − 𝐷) 8𝐿𝐶𝑓2 • It is also useful to express the ripple as a fraction of the output voltage ∆𝑉𝑜 𝑉𝑜 = 1 − 𝐷 8𝐿𝐶𝑓2 • 𝐶 = 1−𝐷 8𝐿(∆𝑉𝑜/𝑉𝑜)𝑓2
THE BUCK (STEP-DOWN) CONVERTER • Using ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇 for ∆𝑖𝐿 ∆𝑉𝑜 = 𝑇𝑉𝑜 8𝐶𝐿 1 − 𝐷 𝑇 = 𝑉𝑜(1 − 𝐷) 8𝐿𝐶𝑓2 • It is also useful to express the ripple as a fraction of the output voltage ∆𝑉𝑜 𝑉𝑜 = 1 − 𝐷 8𝐿𝐶𝑓2 • 𝐶 = 1−𝐷 8𝐿(∆𝑉𝑜/𝑉𝑜)𝑓2
CAPACITOR RESISTANCE—THE EFFECT ON RIPPLE VOLTAGE The ESR may have a significant effect on the output voltage rippleA real capacitor can be modelled as a capacitance with an equivalent series resistance (ESR) and an equivalent series inductance (ESL). ESR, often producing a ripple voltage greater than that of the ideal capacitance. The inductance in the capacitor is usually not a significant factor at typical switching frequencies. The ripple due to ESR can be approximated by first determining the ripple current assuming the capacitor ideal ∆𝑉𝑜,𝐸𝑆𝑅 = ∆𝑖 𝐶 𝑟𝐶 = ∆𝑖 𝐿 𝑟𝐶 ∆𝑉𝑜 < ∆𝑉𝑜,𝐶 + ∆𝑉𝑜,𝐸𝑆𝑅
CAPACITOR RESISTANCE—THE EFFECT ON RIPPLE VOLTAGE • The ripple voltage due to the ESR can be much larger than the ripple due to the pure capacitance. • In that case, the output capacitor is chosen on the basis of the equivalent series resistance rather than capacitance only. ∆𝑉𝑜 ≈ ∆𝑉𝑜,𝐸𝑆𝑅 = ∆𝑖 𝐶 𝑟𝐶
SYNCHRONOUS RECTIFICATION FOR THE BUCK CONVERTER Figure: A synchronous buck converter. The MOSFET S2 carries the inductor current when S1 is off to provide a lower voltage drop than a diode.
THE BOOST CONVERTER Assumptions 1. Steady-state conditions exist. 2. The switching period is 𝑇, and the switch is closed for time 𝐷𝑇 and open for (1 − 𝐷)𝑇. 3. The inductor current is continuous (always positive). 4. The capacitor is very large, and the output voltage is held constant at voltage 𝑉𝑜. 5. The components are ideal.
THE BOOST CONVERTER • ANALYSIS FOR THE SWITCH CLOSED: 𝑣 𝐿 = 𝑉𝑆 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑜𝑟 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑆 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 𝐷𝑇 = 𝑉𝑆 𝐿 • Solving for ∆𝑖 𝐿 for the switch closed, ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉𝑆 𝐷𝑇 𝐿
THE BOOST CONVERTER • ANALYSIS FOR THE SWITCH OPEN: 𝑣 𝐿 = 𝑉𝑠 − 𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑠 − 𝑉𝑜 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 1 − 𝐷 𝑇 = 𝑉𝑠 − 𝑉𝑜 𝐿 • Solving for ∆𝑖 𝐿 (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= (𝑉𝑠 − 𝑉𝑜) 1 − 𝐷 𝑇 𝐿
THE BOOST CONVERTER For steady-state operation, the net change in inductor current must be zero. (∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛 = 0 𝑉𝑠 𝐷𝑇 𝐿 + (𝑉𝑠−𝑉𝑜)(1 − 𝐷)𝑇 𝐿 = 0 Solving for 𝑉𝑜, 𝑉𝑠 𝐷 + 1 − 𝐷 − 𝑉𝑜 1 − 𝐷 = 0 𝑉𝑜= 𝑉𝑠 1 − 𝐷 Expressing the average inductor voltage over one switching period, 𝑉𝐿 = 𝑉𝑠 𝐷 + 𝑉𝑠 − 𝑉𝑜 1 − 𝐷 = 0
THE BOOST CONVERTER Average inductor current can be obtained by assuming 𝑃𝑠 = 𝑃𝑜 Output power is 𝑃𝑜 = 𝑉𝑜 2 𝑅 = 𝑉𝑜 𝐼 𝑜 𝑉𝑆 𝐼𝑆= 𝑉𝑠 𝐼𝐿 Equating input and output powers 𝑉𝑠 𝐼𝐿 = 𝑉𝑜 2 𝑅 = [𝑉𝑠/(1 − 𝐷)]2 𝑅 = 𝑉𝑠 2 (1 − 𝐷)2 𝑅 𝐼𝐿 = 𝑉𝑠 (1−𝐷)2 𝑅 = 𝑉𝑜 2 𝑉𝑠 𝑅 = 𝑉𝑜 𝐼 𝑜 𝑉𝑠 • 𝐼 𝑚𝑎𝑥 = 𝐼𝐿 + ∆𝑖 𝐿 2 = 𝑉𝑠 (1−𝐷)2 𝑅 + 𝑉𝑠 𝐷𝑇 2𝐿 • 𝐼 𝑚𝑖𝑛 = 𝐼𝐿 − ∆𝑖 𝐿 2 = 𝑉𝑠 1−𝐷 2 𝑅 − 𝑉𝑠 𝐷𝑇 2𝐿
THE BOOST CONVERTER A condition necessary for continuous inductor current is for 𝐼 𝑚𝑖𝑛 to be positive. Therefore, the boundary between continuous and discontinuous inductor current is determined from 𝐼 𝑚𝑖𝑛 = 0 = 𝑉𝑠 1 − 𝐷 2 𝑅 − 𝑉𝑠 𝐷𝑇 2𝐿 𝑜𝑟 𝑉𝑠 1 − 𝐷 2 𝑅 = 𝑉𝑠 𝐷𝑇 2𝐿 = 𝑉𝑠 𝐷 2𝐿𝑓  𝐿𝑓 𝑚𝑖𝑛 = 𝐷 1−𝐷 2 𝑅 2 𝑜𝑟 𝐿 𝑚𝑖𝑛 = 𝐷 1−𝐷 2 𝑅 2𝑓 From a design perspective, it is useful to express 𝐿 in terms of a desired ∆𝑖 𝐿, 𝐿 = 𝑉𝑠 𝐷𝑇 ∆𝑖 𝐿 = 𝑉𝑠 𝐷 ∆𝑖 𝐿 𝑓
THE BOOST CONVERTER The peak-to-peak output voltage ripple can be calculated from the capacitor current waveform. The change in capacitor charge can be calculated from ∆𝑄 = 𝑉𝑜 𝑅 𝐷𝑇 = 𝐶∆𝑉𝑜 An expression for ripple voltage is then ∆𝑉𝑜= 𝑉𝑜 𝐷𝑇 𝑅𝐶 = 𝑉𝑜 𝐷 𝑅𝐶𝑓 𝑜𝑟 ∆𝑉𝑜 𝑉𝑜 = 𝐷 𝑅𝐶𝑓 where 𝑓 is the switching frequency. 𝐶 = 𝐷 𝑅(∆𝑉𝑜/𝑉𝑜)𝑓
THE BOOST CONVERTER As Buck Converter the voltage ripple due to the ESR is ∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶 Effects of Inductor Resistance • Inductors should be designed to have small resistance to minimize power loss and maximize efficiency. • Inductor resistance affects performance of the boost converter, especially at high duty ratios. • For the boost converter, recall that the output voltage for the ideal case is 𝑉𝑜 = 𝑉𝑠 1 − 𝐷 • The power supplied by the source must be the same as the power absorbed by the load and the inductor resistance, neglecting other losses. 𝑃𝑠= 𝑃𝑜 + 𝑃𝑟𝐿 𝑉𝑠 𝐼𝐿 = 𝑉𝑜 𝐼 𝐷 + 𝐼𝐿 2 𝑟𝐿 where 𝑟𝐿 is the series resistance of the inductor.
THE BOOST CONVERTER • The average diode current is 𝐼 𝐷 = 𝐼𝐿 1 − 𝐷 • Then 𝑉𝑠 𝐼𝐿 = 𝑉𝑜 𝐼𝐿 1 − 𝐷 + 𝐼𝐿 2 𝑟𝐿 • which becomes 𝑉𝑠 = 𝑉𝑜 1 − 𝐷 + 𝐼𝐿 𝑟𝐿 𝐼𝐿 = 𝐼 𝐷 1 − 𝐷 = 𝑉𝑜/𝑅 1 − 𝐷 • Substituting for 𝐼𝐿 𝑉𝑠 = 𝑉𝑜 𝑟𝐿 1 − 𝐷 𝑅 + 𝑉𝑜 1 − 𝐷 • Solving for 𝑉𝑜, 𝑉𝑜 = 𝑉𝑠 1 − 𝐷 1 1 + 𝑟𝐿 𝑅 1 − 𝐷 2
THE BOOST CONVERTER Figure: Boost converter for a nonideal inductor. (a) Output voltage; (b) Boost converter efficiency. 𝜂 = 𝑃𝑜 𝑃𝑜 + 𝑃𝑙𝑜𝑠𝑠 = 𝑉𝑜 2 /𝑅 𝑉𝑜 2 /𝑅 + 𝐼𝐿 2 𝑟𝐿 = 𝑉𝑜 2 /𝑅 𝑉𝑜 2 /𝑅 + [(𝑉𝑜 2 /𝑅)2/ 1 − 𝐷 2] 𝑟𝐿 = 1 1 + 𝑟𝐿[𝑅 1 − 𝐷 2]
BUCK-BOOST CONVERTER • The output voltage of the buck-boost converter can be either higher or lower than the input voltage. • Assumptions : 1. The circuit is operating in the steady state. 2. The inductor current is continuous. 3. The capacitor is large enough to assume a constant output voltage. 4. The switch is closed for time 𝐷𝑇 and open for (1 − 𝐷)𝑇. 5. The components are ideal.
BUCK-BOOST CONVERTER Figure: Buck-boost converter. (a) Circuit; (b) Equivalent circuit for the switch closed; (c) Equivalent circuit for the switch open.
BUCK-BOOST CONVERTER
BUCK-BOOST CONVERTER • ANALYSIS FOR THE SWITCH CLOSED: 𝑣 𝐿 = 𝑉𝑠 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑠 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 𝐷𝑇 = 𝑉𝑠 𝐿 (∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉𝑠 𝐷𝑇 𝐿 ANALYSIS FOR THE SWITCH OPEN: 𝑣 𝐿 = 𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑜 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 1 − 𝐷 𝑇 = 𝑉𝑜 𝐿 (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= 𝑉𝑜 1−𝐷 𝑇 𝐿
BUCK-BOOST CONVERTER • For steady-state operation, the net change in inductor current must be zero over one period. (∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= 0 𝑉𝑠 𝐷𝑇 𝐿 + 𝑉𝑜 1 − 𝐷 𝑇 𝐿 = 0 • Solving for 𝑉𝑜, 𝑉𝑜 = −𝑉𝑠 𝐷 1 − 𝐷 • The required duty ratio for specified input and output voltages can be expressed as 𝐷 = 𝑉𝑜 𝑉𝑠 + 𝑉𝑜
BUCK-BOOST CONVERTER • The average inductor voltage is zero for periodic operation, resulting in 𝑉𝐿 = 𝑉𝑠 𝐷 + 𝑉𝑜 1 − 𝐷 = 0 • Solving for 𝑉𝑜 yields 𝑉𝑜 = −𝑉𝑠 𝐷 1 − 𝐷 • The output voltage has opposite polarity from the source voltage. • If 𝐷 > 0.5, the output voltage is larger than the input; and if 𝐷 < 0.5, the output is smaller than the input.
BUCK-BOOST CONVERTER • Power absorbed by the load must be the same as that supplied by the source, where 𝑃0 = 𝑉𝑜 2 𝑅 𝑃𝑠 = 𝑉𝑠 𝐼𝑠 𝑉𝑜 2 𝑅 = 𝑉𝑠 𝐼𝑠 • Average source current is related to average inductor current by 𝐼𝑠 = 𝐼𝐿 𝐷 • resulting in 𝑉𝑜 2 𝑅 = 𝑉𝑠 𝐼𝐿 𝐷 • Substituting for 𝑉𝑜 and solving for 𝐼𝐿, we find 𝐼𝐿 = 𝑉𝑜 2 𝑉𝑠 𝑅𝐷 = 𝑃0 𝑉𝑠 𝐷 = 𝑉𝑠 𝐷 𝑅(1 − 𝐷)2
BUCK-BOOST CONVERTER 𝐼 𝑚𝑎𝑥 = 𝐼𝐿 + ∆𝑖 𝐿 2 = 𝑉𝑠 𝐷 𝑅(1 − 𝐷)2 + 𝑉𝑠 𝐷𝑇 2𝐿 𝐼 𝑚𝑖𝑛 = 𝐼𝐿 − ∆𝑖 𝐿 2 = 𝑉𝑠 𝐷 𝑅 1 − 𝐷 2 − 𝑉𝑠 𝐷𝑇 2𝐿 (𝐿𝑓) 𝑚𝑖𝑛= (1 − 𝐷)2 𝑅 2 𝐿 𝑚𝑖𝑛 = (1 − 𝐷)2 𝑅 2𝑓 where 𝑓 is the switching frequency.
BUCK-BOOST CONVERTER • OUTPUT VOLTAGE RIPPLE • The output voltage ripple for the buck-boost converter is computed from the capacitor current . ∆𝑄 = 𝑉𝑜 𝑅 𝐷𝑇 = 𝐶∆𝑉𝑜 • Solving for ∆𝑉𝑜, ∆𝑉𝑜 = 𝑉𝑜 𝐷𝑇 𝑅𝐶 = 𝑉𝑜 𝐷 𝑅𝐶𝑓 𝑜𝑟 ∆𝑉𝑜 𝑉𝑜 = 𝐷 𝑅𝐶𝑓
BUCK-BOOST CONVERTER • As is the case with other converters, the equivalent series resistance of the capacitor can contribute significantly to the output ripple voltage. The peak-to-peak variation in capacitor current is the same as the maximum inductor current. ∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
BUCK-BOOST CONVERTER •As is the case with other converters, the equivalent series resistance of the capacitor can contribute significantly to the output ripple voltage. The peak-to-peak variation in capacitor current is the same as the maximum inductor current. ∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
Other ConverterTopologies CÚk Converter Sepic Converter
Other ConverterTopologies Full-bridge DC-DC Converter
Other ConverterTopologies IsolatedFull-bridge DC-DC Converter
Other ConverterTopologies IsolatedHalf-bridge DC-DC Converter
DC DC Converter

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DC DC Converter

  • 1.
  • 2.
    DC-DC Converters DC-DC convertersare power electronic circuits that convert a DC voltage to a different DC voltage level, often providing a regulated output. A BASIC SWITCHING CONVERTER An efficient alternative to the linear regulator Uses Power electronics switches like BJT,MOSFET IGBT… Also known as DC Chopper
  • 3.
    DC-DC Converters Figure: (a)A basic DC-DC switching converter; (b) Switching equivalent; ( c) Output voltage.  Assuming the switch is ideal • The output is the same as the input when the switch ON • And the output is zero when the switch OFF  Periodic opening and closing of the switch gives the pulsed output waveform.  The average or DC component of the output voltage is 𝑉𝑜 = 1 𝑇 0 𝑇 𝑣 𝑜 𝑡 𝑑𝑡 = 1 𝑇 0 𝐷𝑇 𝑉𝑠 𝑑𝑡 = 𝑉𝑠 𝐷
  • 4.
    DC-DC Converters 𝐷 ≡ 𝑡𝑜𝑛 𝑡 𝑜𝑛 + 𝑡 𝑜𝑓𝑓 The DC component of the output voltage will be less than or equal to the input voltage for this circuit. ideal switch Zero loss Zero voltage across when ON Zero current through it when OFF But real switch has some power loss Considerable because it creates heat on the switch.
  • 5.
    THE BUCK (STEP-DOWN) CONVERTERApplicationExample: Controlling the speed of DC Motor Low-pass filter will be placed at the output • The diode provides a path for the inductor current when the switch is opened and is reverse-biased when the switch is closed.
  • 6.
    THE BUCK (STEP-DOWN) CONVERTERVOLTAGEAND CURRENT RELATIONSHIPS: 𝑣 𝑥 in Fig. above, is 𝑉𝑠 when the switch is closed and is zero when the switch is open, provided that the inductor current remains positive, keeping the diode on. If the switch is closed periodically at a duty ratio D, the average voltage at the filter input is 𝑉𝑠 𝐷 An inductor current that remains positive throughout the switching period is known as continuous current. Conversely, discontinuous current is characterized by the inductor current’s returning to zero during each period.
  • 7.
    THE BUCK (STEP-DOWN) CONVERTERBuckconverters and DC-DC converters in general, have the following properties when operating in the steady state 1. The inductor current is periodic. 𝒊 𝑳 𝒕 + 𝑻 = 𝒊 𝑳 𝒕 2. The average inductor voltage is zero 𝑽 𝑳 = 𝟏 𝑻 𝒕 𝒕+𝑻 𝒗 𝑳 𝝀 𝒅𝝀 = 𝟎 3. The average capacitor current is zero 𝑰 𝑪 = 𝟏 𝑻 𝒕 𝒕+𝑻 𝒊 𝑪 𝝀 𝒅𝝀 = 𝟎 4. The power supplied by the source is the same as the power delivered to the load. For non-ideal components, the source also supplies the losses. 𝑷 𝒔 = 𝑷 𝒐 𝒊𝒅𝒆𝒂𝒍 𝑷 𝒔= 𝑷 𝒐 + 𝒍𝒐𝒔𝒔𝒆𝒔 𝒏𝒐𝒏𝒊𝒅𝒆𝒂𝒍
  • 8.
    THE BUCK (STEP-DOWN) CONVERTERAssumptionsfor Analysis of Buck converter: • The circuit is operating in the steady state. • The inductor current is continuous (always positive) • The capacitor is very large, and the output voltage is held constant at voltage Vo. This restriction will be relaxed later to show the effects of finite capacitance. • The switching period is 𝑇; the switch is closed for time 𝐷𝑇 and open for time (1 − 𝐷)𝑇. • The components are ideal. The key to the analysis for determining the output 𝑉𝑜 is to examine the inductor current and inductor voltage first for the switch closed and then for the switch open.
  • 9.
    THE BUCK (STEP-DOWN) CONVERTERANALYSISFOR THE SWITCH CLOSED: • The voltage across the inductor is • 𝑣 𝐿 = 𝑉𝑆 − 𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 • Rearranging, • 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉 𝑆−𝑉𝑜 𝐿 𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑑 • 𝑑𝑖 𝐿 𝑑𝑡 = ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 𝐷𝑇 = 𝑉 𝑆−𝑉𝑜 𝐿 • ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉 𝑆−𝑉𝑜 𝐿 𝐷𝑇
  • 10.
    THE BUCK (STEP-DOWN) CONVERTERANALYSISFOR THE SWITCH OPEN: • When the switch is open, the diode becomes forward-biased to carry the inductor current • The voltage across the inductor when the switch is open is 𝑣 𝐿 = −𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 • Rearranging, 𝑑𝑖 𝐿 𝑑𝑡 = −𝑉𝑜 𝐿 𝑠𝑤𝑖𝑡𝑐ℎ 𝑜𝑝𝑒𝑛 • The derivative of current in the inductor is a negative constant, and the current decreases linearly • The change in inductor current when the switch is open is ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 1−𝐷 𝑇 = −𝑉𝑜 𝐿 ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇
  • 11.
    THE BUCK (STEP-DOWN) CONVERTERAtsteady state ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 + ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = 0 𝑉 𝑆−𝑉𝑜 𝐿 𝐷𝑇 − 𝑉𝑜 𝐿 1 − 𝐷 = 0 • Solving for 𝑉𝑜, 𝑉𝑜 = 𝑉𝑠 𝐷 • Therefore, the buck converter produces an output voltage that is less than or equal to the input. • An alternative derivation of the output voltage is based on the inductor voltage. • Since the average inductor voltage is zero for periodic operation 𝑉𝐿 = 𝑉𝑆 − 𝑉𝑂 𝐷𝑇 + −𝑉𝑂 1 − 𝐷 𝑇 = 0 • Solving for 𝑉𝑂 𝑉𝑂= 𝑉𝑆 𝐷.
  • 12.
    THE BUCK (STEP-DOWN) CONVERTERTheaverage inductor current must be the same as the average current in the load resistor, since the average capacitor current must be zero for steady-state operation. 𝐼𝐿 = 𝐼 𝑅 = 𝑉𝑂 𝑅 The maximum and minimum values of the inductor current are computed as ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉𝑆 − 𝑉𝑜 𝐿 𝐷𝑇 ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇  𝐼 𝑚𝑎𝑥= 𝐼𝐿 + ∆𝑖 𝐿 2 = 𝑉𝑂 + 1 𝑉𝑂 1 − 𝐷 𝑇 = 𝑉𝑂 1 + 1 − 𝐷
  • 13.
    THE BUCK (STEP-DOWN) CONVERTER 𝐼𝑚𝑖𝑛 = 𝐼𝐿 − ∆𝑖 𝐿 2 = 𝑉𝑂 𝑅 − 1 2 𝑉𝑂 𝐿 1 − 𝐷 𝑇 = 𝑉𝑂 1 𝑅 − 1 − 𝐷 2𝐿𝑓 where 𝑓 = 1/𝑇 is the switching frequency. • The above Eq. can be used to determine the combination of L and f that will result in continuous current. Since 𝐼 𝑚𝑖𝑛 = 0 is the boundary between continuous and discontinuous current, 𝐼 𝑚𝑖𝑛 = 0 = 𝑉𝑂 1 𝑅 − 1 − 𝐷 2𝐿𝑓 (𝐿𝑓) 𝑚𝑖𝑛= (1 − 𝐷)𝑅 2
  • 14.
    THE BUCK (STEP-DOWN) CONVERTER•If the desired switching frequency is established, 𝐿 𝑚𝑖𝑛 = (1 − 𝐷)𝑅 2𝑓 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 where 𝐿 𝑚𝑖𝑛 is the minimum inductance required for continuous current. • In practice inductance is chosen greater than 𝐿 𝑚𝑖𝑛 • The peak-to-peak variation in the inductor current is often used as a design criterion in buck converter. • to determine the value of inductance for a specified peak-to-peak inductor current for continuous-current operation: ∆𝑖 𝐿 = 𝑉𝑠 − 𝑉𝑜 𝐿 𝐷𝑇 = 𝑉𝑠 − 𝑉𝑜 𝐿𝑓 𝐷 = 𝑉𝑂 1 − 𝐷 𝐿𝑓 𝑜𝑟 𝐿 = 𝑉𝑠 − 𝑉𝑜 ∆𝑖 𝐿 𝑓 𝐷 = 𝑉𝑂 1 − 𝐷 ∆𝑖 𝐿 𝑓
  • 15.
    THE BUCK (STEP-DOWN) CONVERTER•Since the converter components are assumed to be ideal, the power supplied by the source must be the same as the power absorbed by the load resistor. 𝑃𝑆 = 𝑃𝑂 𝑉𝑠 𝐼𝑠 = 𝑉𝑜 𝐼 𝑜 𝑜𝑟 𝑉𝑜 𝑉𝑠 = 𝐼𝑠 𝐼 𝑜 • Basically from this expression we can say that DC-DC converter is DC transformer.
  • 16.
    THE BUCK (STEP-DOWN) CONVERTER•OUTPUT VOLTAGE RIPPLE • In the preceding analysis, the capacitor was assumed to be very large to keep the output voltage constant. • The variation in output voltage is computed from the voltage- current relationship of the capacitor. • The current in the capacitor is 𝑖 𝐶 = 𝑖 𝐿 − 𝑖 𝑅 • While the capacitor current is positive, the capacitor is charging. From the definition of capacitance, 𝑄 = 𝐶𝑉𝑜 ∆𝑄 = 𝐶∆𝑉𝑜 ∆𝑉𝑜 = ∆𝑄 𝐶
  • 17.
    THE BUCK (STEP-DOWN) CONVERTER•The change in charge ∆𝑄 is the area of the triangle above the time axis ∆𝑄 = 1 2 𝑇 2 ∆𝑖 𝐿 2 = 𝑇∆𝑖 𝐿 8 • resulting in ∆𝑉𝑜 = 𝑇∆𝑖 𝐿 8𝐶
  • 18.
    THE BUCK (STEP-DOWN) CONVERTER •Using ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇 = ∆𝑖𝐿 ∆𝑉𝑜 = 𝑇𝑉𝑜 8𝐶𝐿 1 − 𝐷 𝑇 = 𝑉𝑜(1 − 𝐷) 8𝐿𝐶𝑓2 • It is also useful to express the ripple as a fraction of the output voltage ∆𝑉𝑜 𝑉𝑜 = 1 − 𝐷 8𝐿𝐶𝑓2 • 𝐶 = 1−𝐷 8𝐿(∆𝑉𝑜/𝑉𝑜)𝑓2
  • 19.
    THE BUCK (STEP-DOWN) CONVERTER •Using ∆𝑖 𝐿 𝑜𝑝𝑒𝑛 = − 𝑉𝑜 𝐿 (1 − 𝐷)𝑇 for ∆𝑖𝐿 ∆𝑉𝑜 = 𝑇𝑉𝑜 8𝐶𝐿 1 − 𝐷 𝑇 = 𝑉𝑜(1 − 𝐷) 8𝐿𝐶𝑓2 • It is also useful to express the ripple as a fraction of the output voltage ∆𝑉𝑜 𝑉𝑜 = 1 − 𝐷 8𝐿𝐶𝑓2 • 𝐶 = 1−𝐷 8𝐿(∆𝑉𝑜/𝑉𝑜)𝑓2
  • 20.
    CAPACITOR RESISTANCE—THE EFFECT ONRIPPLE VOLTAGE The ESR may have a significant effect on the output voltage rippleA real capacitor can be modelled as a capacitance with an equivalent series resistance (ESR) and an equivalent series inductance (ESL). ESR, often producing a ripple voltage greater than that of the ideal capacitance. The inductance in the capacitor is usually not a significant factor at typical switching frequencies. The ripple due to ESR can be approximated by first determining the ripple current assuming the capacitor ideal ∆𝑉𝑜,𝐸𝑆𝑅 = ∆𝑖 𝐶 𝑟𝐶 = ∆𝑖 𝐿 𝑟𝐶 ∆𝑉𝑜 < ∆𝑉𝑜,𝐶 + ∆𝑉𝑜,𝐸𝑆𝑅
  • 21.
    CAPACITOR RESISTANCE—THE EFFECT ONRIPPLE VOLTAGE • The ripple voltage due to the ESR can be much larger than the ripple due to the pure capacitance. • In that case, the output capacitor is chosen on the basis of the equivalent series resistance rather than capacitance only. ∆𝑉𝑜 ≈ ∆𝑉𝑜,𝐸𝑆𝑅 = ∆𝑖 𝐶 𝑟𝐶
  • 22.
    SYNCHRONOUS RECTIFICATION FOR THEBUCK CONVERTER Figure: A synchronous buck converter. The MOSFET S2 carries the inductor current when S1 is off to provide a lower voltage drop than a diode.
  • 23.
    THE BOOST CONVERTER Assumptions 1.Steady-state conditions exist. 2. The switching period is 𝑇, and the switch is closed for time 𝐷𝑇 and open for (1 − 𝐷)𝑇. 3. The inductor current is continuous (always positive). 4. The capacitor is very large, and the output voltage is held constant at voltage 𝑉𝑜. 5. The components are ideal.
  • 24.
    THE BOOST CONVERTER •ANALYSIS FOR THE SWITCH CLOSED: 𝑣 𝐿 = 𝑉𝑆 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑜𝑟 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑆 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 𝐷𝑇 = 𝑉𝑆 𝐿 • Solving for ∆𝑖 𝐿 for the switch closed, ∆𝑖 𝐿 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉𝑆 𝐷𝑇 𝐿
  • 25.
    THE BOOST CONVERTER •ANALYSIS FOR THE SWITCH OPEN: 𝑣 𝐿 = 𝑉𝑠 − 𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑠 − 𝑉𝑜 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 1 − 𝐷 𝑇 = 𝑉𝑠 − 𝑉𝑜 𝐿 • Solving for ∆𝑖 𝐿 (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= (𝑉𝑠 − 𝑉𝑜) 1 − 𝐷 𝑇 𝐿
  • 26.
    THE BOOST CONVERTER Forsteady-state operation, the net change in inductor current must be zero. (∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛 = 0 𝑉𝑠 𝐷𝑇 𝐿 + (𝑉𝑠−𝑉𝑜)(1 − 𝐷)𝑇 𝐿 = 0 Solving for 𝑉𝑜, 𝑉𝑠 𝐷 + 1 − 𝐷 − 𝑉𝑜 1 − 𝐷 = 0 𝑉𝑜= 𝑉𝑠 1 − 𝐷 Expressing the average inductor voltage over one switching period, 𝑉𝐿 = 𝑉𝑠 𝐷 + 𝑉𝑠 − 𝑉𝑜 1 − 𝐷 = 0
  • 27.
    THE BOOST CONVERTER Averageinductor current can be obtained by assuming 𝑃𝑠 = 𝑃𝑜 Output power is 𝑃𝑜 = 𝑉𝑜 2 𝑅 = 𝑉𝑜 𝐼 𝑜 𝑉𝑆 𝐼𝑆= 𝑉𝑠 𝐼𝐿 Equating input and output powers 𝑉𝑠 𝐼𝐿 = 𝑉𝑜 2 𝑅 = [𝑉𝑠/(1 − 𝐷)]2 𝑅 = 𝑉𝑠 2 (1 − 𝐷)2 𝑅 𝐼𝐿 = 𝑉𝑠 (1−𝐷)2 𝑅 = 𝑉𝑜 2 𝑉𝑠 𝑅 = 𝑉𝑜 𝐼 𝑜 𝑉𝑠 • 𝐼 𝑚𝑎𝑥 = 𝐼𝐿 + ∆𝑖 𝐿 2 = 𝑉𝑠 (1−𝐷)2 𝑅 + 𝑉𝑠 𝐷𝑇 2𝐿 • 𝐼 𝑚𝑖𝑛 = 𝐼𝐿 − ∆𝑖 𝐿 2 = 𝑉𝑠 1−𝐷 2 𝑅 − 𝑉𝑠 𝐷𝑇 2𝐿
  • 28.
    THE BOOST CONVERTER Acondition necessary for continuous inductor current is for 𝐼 𝑚𝑖𝑛 to be positive. Therefore, the boundary between continuous and discontinuous inductor current is determined from 𝐼 𝑚𝑖𝑛 = 0 = 𝑉𝑠 1 − 𝐷 2 𝑅 − 𝑉𝑠 𝐷𝑇 2𝐿 𝑜𝑟 𝑉𝑠 1 − 𝐷 2 𝑅 = 𝑉𝑠 𝐷𝑇 2𝐿 = 𝑉𝑠 𝐷 2𝐿𝑓  𝐿𝑓 𝑚𝑖𝑛 = 𝐷 1−𝐷 2 𝑅 2 𝑜𝑟 𝐿 𝑚𝑖𝑛 = 𝐷 1−𝐷 2 𝑅 2𝑓 From a design perspective, it is useful to express 𝐿 in terms of a desired ∆𝑖 𝐿, 𝐿 = 𝑉𝑠 𝐷𝑇 ∆𝑖 𝐿 = 𝑉𝑠 𝐷 ∆𝑖 𝐿 𝑓
  • 29.
    THE BOOST CONVERTER Thepeak-to-peak output voltage ripple can be calculated from the capacitor current waveform. The change in capacitor charge can be calculated from ∆𝑄 = 𝑉𝑜 𝑅 𝐷𝑇 = 𝐶∆𝑉𝑜 An expression for ripple voltage is then ∆𝑉𝑜= 𝑉𝑜 𝐷𝑇 𝑅𝐶 = 𝑉𝑜 𝐷 𝑅𝐶𝑓 𝑜𝑟 ∆𝑉𝑜 𝑉𝑜 = 𝐷 𝑅𝐶𝑓 where 𝑓 is the switching frequency. 𝐶 = 𝐷 𝑅(∆𝑉𝑜/𝑉𝑜)𝑓
  • 30.
    THE BOOST CONVERTER AsBuck Converter the voltage ripple due to the ESR is ∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶 Effects of Inductor Resistance • Inductors should be designed to have small resistance to minimize power loss and maximize efficiency. • Inductor resistance affects performance of the boost converter, especially at high duty ratios. • For the boost converter, recall that the output voltage for the ideal case is 𝑉𝑜 = 𝑉𝑠 1 − 𝐷 • The power supplied by the source must be the same as the power absorbed by the load and the inductor resistance, neglecting other losses. 𝑃𝑠= 𝑃𝑜 + 𝑃𝑟𝐿 𝑉𝑠 𝐼𝐿 = 𝑉𝑜 𝐼 𝐷 + 𝐼𝐿 2 𝑟𝐿 where 𝑟𝐿 is the series resistance of the inductor.
  • 31.
    THE BOOST CONVERTER •The average diode current is 𝐼 𝐷 = 𝐼𝐿 1 − 𝐷 • Then 𝑉𝑠 𝐼𝐿 = 𝑉𝑜 𝐼𝐿 1 − 𝐷 + 𝐼𝐿 2 𝑟𝐿 • which becomes 𝑉𝑠 = 𝑉𝑜 1 − 𝐷 + 𝐼𝐿 𝑟𝐿 𝐼𝐿 = 𝐼 𝐷 1 − 𝐷 = 𝑉𝑜/𝑅 1 − 𝐷 • Substituting for 𝐼𝐿 𝑉𝑠 = 𝑉𝑜 𝑟𝐿 1 − 𝐷 𝑅 + 𝑉𝑜 1 − 𝐷 • Solving for 𝑉𝑜, 𝑉𝑜 = 𝑉𝑠 1 − 𝐷 1 1 + 𝑟𝐿 𝑅 1 − 𝐷 2
  • 32.
    THE BOOST CONVERTER Figure:Boost converter for a nonideal inductor. (a) Output voltage; (b) Boost converter efficiency. 𝜂 = 𝑃𝑜 𝑃𝑜 + 𝑃𝑙𝑜𝑠𝑠 = 𝑉𝑜 2 /𝑅 𝑉𝑜 2 /𝑅 + 𝐼𝐿 2 𝑟𝐿 = 𝑉𝑜 2 /𝑅 𝑉𝑜 2 /𝑅 + [(𝑉𝑜 2 /𝑅)2/ 1 − 𝐷 2] 𝑟𝐿 = 1 1 + 𝑟𝐿[𝑅 1 − 𝐷 2]
  • 33.
    BUCK-BOOST CONVERTER • Theoutput voltage of the buck-boost converter can be either higher or lower than the input voltage. • Assumptions : 1. The circuit is operating in the steady state. 2. The inductor current is continuous. 3. The capacitor is large enough to assume a constant output voltage. 4. The switch is closed for time 𝐷𝑇 and open for (1 − 𝐷)𝑇. 5. The components are ideal.
  • 34.
    BUCK-BOOST CONVERTER Figure: Buck-boostconverter. (a) Circuit; (b) Equivalent circuit for the switch closed; (c) Equivalent circuit for the switch open.
  • 35.
  • 36.
    BUCK-BOOST CONVERTER • ANALYSISFOR THE SWITCH CLOSED: 𝑣 𝐿 = 𝑉𝑠 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑠 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 𝐷𝑇 = 𝑉𝑠 𝐿 (∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 = 𝑉𝑠 𝐷𝑇 𝐿 ANALYSIS FOR THE SWITCH OPEN: 𝑣 𝐿 = 𝑉𝑜 = 𝐿 𝑑𝑖 𝐿 𝑑𝑡 𝑑𝑖 𝐿 𝑑𝑡 = 𝑉𝑜 𝐿 ∆𝑖 𝐿 ∆𝑡 = ∆𝑖 𝐿 1 − 𝐷 𝑇 = 𝑉𝑜 𝐿 (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= 𝑉𝑜 1−𝐷 𝑇 𝐿
  • 37.
    BUCK-BOOST CONVERTER • Forsteady-state operation, the net change in inductor current must be zero over one period. (∆𝑖 𝐿) 𝑐𝑙𝑜𝑠𝑒𝑑 + (∆𝑖 𝐿) 𝑜𝑝𝑒𝑛= 0 𝑉𝑠 𝐷𝑇 𝐿 + 𝑉𝑜 1 − 𝐷 𝑇 𝐿 = 0 • Solving for 𝑉𝑜, 𝑉𝑜 = −𝑉𝑠 𝐷 1 − 𝐷 • The required duty ratio for specified input and output voltages can be expressed as 𝐷 = 𝑉𝑜 𝑉𝑠 + 𝑉𝑜
  • 38.
    BUCK-BOOST CONVERTER • Theaverage inductor voltage is zero for periodic operation, resulting in 𝑉𝐿 = 𝑉𝑠 𝐷 + 𝑉𝑜 1 − 𝐷 = 0 • Solving for 𝑉𝑜 yields 𝑉𝑜 = −𝑉𝑠 𝐷 1 − 𝐷 • The output voltage has opposite polarity from the source voltage. • If 𝐷 > 0.5, the output voltage is larger than the input; and if 𝐷 < 0.5, the output is smaller than the input.
  • 39.
    BUCK-BOOST CONVERTER • Powerabsorbed by the load must be the same as that supplied by the source, where 𝑃0 = 𝑉𝑜 2 𝑅 𝑃𝑠 = 𝑉𝑠 𝐼𝑠 𝑉𝑜 2 𝑅 = 𝑉𝑠 𝐼𝑠 • Average source current is related to average inductor current by 𝐼𝑠 = 𝐼𝐿 𝐷 • resulting in 𝑉𝑜 2 𝑅 = 𝑉𝑠 𝐼𝐿 𝐷 • Substituting for 𝑉𝑜 and solving for 𝐼𝐿, we find 𝐼𝐿 = 𝑉𝑜 2 𝑉𝑠 𝑅𝐷 = 𝑃0 𝑉𝑠 𝐷 = 𝑉𝑠 𝐷 𝑅(1 − 𝐷)2
  • 40.
    BUCK-BOOST CONVERTER 𝐼 𝑚𝑎𝑥= 𝐼𝐿 + ∆𝑖 𝐿 2 = 𝑉𝑠 𝐷 𝑅(1 − 𝐷)2 + 𝑉𝑠 𝐷𝑇 2𝐿 𝐼 𝑚𝑖𝑛 = 𝐼𝐿 − ∆𝑖 𝐿 2 = 𝑉𝑠 𝐷 𝑅 1 − 𝐷 2 − 𝑉𝑠 𝐷𝑇 2𝐿 (𝐿𝑓) 𝑚𝑖𝑛= (1 − 𝐷)2 𝑅 2 𝐿 𝑚𝑖𝑛 = (1 − 𝐷)2 𝑅 2𝑓 where 𝑓 is the switching frequency.
  • 41.
    BUCK-BOOST CONVERTER • OUTPUTVOLTAGE RIPPLE • The output voltage ripple for the buck-boost converter is computed from the capacitor current . ∆𝑄 = 𝑉𝑜 𝑅 𝐷𝑇 = 𝐶∆𝑉𝑜 • Solving for ∆𝑉𝑜, ∆𝑉𝑜 = 𝑉𝑜 𝐷𝑇 𝑅𝐶 = 𝑉𝑜 𝐷 𝑅𝐶𝑓 𝑜𝑟 ∆𝑉𝑜 𝑉𝑜 = 𝐷 𝑅𝐶𝑓
  • 42.
    BUCK-BOOST CONVERTER • Asis the case with other converters, the equivalent series resistance of the capacitor can contribute significantly to the output ripple voltage. The peak-to-peak variation in capacitor current is the same as the maximum inductor current. ∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
  • 43.
    BUCK-BOOST CONVERTER •As isthe case with other converters, the equivalent series resistance of the capacitor can contribute significantly to the output ripple voltage. The peak-to-peak variation in capacitor current is the same as the maximum inductor current. ∆𝑉𝑜,𝐸𝑆𝑅= ∆𝑖 𝐶 𝑟𝐶 = 𝐼𝐿,𝑚𝑎𝑥 𝑟𝐶
  • 44.
  • 45.
  • 46.
  • 47.