LECTURE 12 Lossy Converters (Static State Losses but Neglecting Switching Losses) HW #2 Erickson Chapter 3 Problems 6 & 7 A.

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1 ETUE 12 ossy onverters (Static State osses but Neglecting Switching osses) HW #2 Erickson hapter 3 Problems 6 & 7 A. General Effects Expected: (load) as (load) B. Switch and esistive osses hange M(D) Modified D oltage Transfer Functions M(D) with loss M(D) for no loss 1. General aspects of the origin of M(D) changes in static state loss terms:, d, d, and on a. max / min b. / Slopes 2. General solution for lossy Boost onverter a. only: M(D, ), η 1 b. Static State Device D losses: M(D, d, on, D) 3. Buck onverter a. Problem 3.3 of Erickson 4. General solution for lossy BuckBoost onverter a. M(D, d, on, ), Problem 3.4 of Erickson 5. ompare for a specific task: a. Buck vs. BuckBoost ircuits, Problem 3.5 of Erickson 1

2 2 A. GENEA EFFETS EXPETED ON M(D) ossy onverters: onsider the static state D loss only, no dynamic switch loss will be considered here Generally one expects, due to finite out, from any power supply for to fall as increases. n converters out is much more complex, especially with feedback. out effect on For fixed P out and in for a converter the input current increases proportional to the losses so that P in =P out losses. The losses we will consider will also include device losses but only in the static states due to finite on and on values. B. Effects on voltage D transfer function M(D) due to resistive and device voltage drops. Besides decreasing converter efficiency static state losses will also effect dc transfer functions. M(D) becomes a function of, on, on, etc. when losses are included as shown below. 1. max / min changes as seen by any converter inductor will cause di/dt slope changes. i = / (in units of A/sec) slopes change, affecting D and D themselves just by including resistance and device voltage drops. Hence, the value of M(D) changes. 2. ncluding inductor resistance, device on, as well as on changes M(D), from M(D) with these effects ignored. ets consider the inductor

3 3 resistance effects. a. llustrative (inductor series equivalent resistance) effects in boost topology M will change from M(D) to M(D, ) as shown below. is finite not only because of simple copper resistive losses. Wound inductors at high frequencies (0.1 1 MHz) have additional core and proximity effect losses. To a first approximation: ac/ dc =0.5 (wire dia. in cm)x f 1/2 x100 arger diameter wire has more effect than a small diameter Below we carefully distinguish between with the dominant for the load. Boost with during D Boost with during D i i i g v v g v i v v (t) g 1. sec balance across DT s D'T s t D[ g ] D [ g ] = 0 g D = 0 2. urrenttime balance across the capacitor i (t) DT s D'T s g / t /

4 4 D[/] D [ /] = 0 D / = 0 Usual Second term has all effects o Boost = of including finite 1 1. g D' lossless >D 2 big effect 1 term 2 Also big effects occur for D' D 0 or for D 1.0 Note some extreme converter operating conditions will amplify finite effects. Note well: (1) is the load resistor (2) = inductor losses For D 1.0 / has a big effect on M(D) As / 0 no effect on M(D) Note also as D fi 1.0, M(D) has big rollover if is even 1% of the load resistance. Note also in D gain at high D that for D< 0.2, effects are minimal. Summary The operating duty cycle, D value, amplifies equivalent series resistance,, effects. D > 0.5 and / even 1% you see big changes to M(D). D < 0.5 you see no effects.

5 5 g D' D' Equivalent ircuit with D Transformer : Use to get the operating efficiency η = P out / P in D':1 Equivalent circuit model of the boost converter, including a D :1 dc transformer and the inductor winding resistance. eferring to the secondary to get one loop equation. D' g /D' /D' 2 o learly want << D 2 Easy if D = 1.0, D = 0 Hard to achieve if D 0 D 1.0 deal o = g ase 1 1/D 2 D' 1

6 6 g deal (input) = D' ase D' 1 P η = out = o = P D ' 1 in g 2 D' 1 η(d, /) 1) As the on duty cycle D 0 one gets high η but efficiency is little effected by 2) D 1 and D 0 η 0 still we get much lower η for not negligible and much greater sensitivity to changes. ompare the above to the ossless Boost: 1 D':1 2 g o 2 = D 1 η = D o / g = P o /P in o = g /D = 1.0 (lossless) Next we include static device losses to the system loss. b. Static State Device oss Effects in the Boost Topology.

7 7 n our static states of the switching devices we only consider on from the transistor but both D and D for the diode when on, because on (T) 0. We expect to obtain M(D,, on, D, D ) and η(d,, on, D and D ). i 1 g 2 v i g v Fig Boost converter circuit, including inductor resistance. SW position 1: SW position 2: Fig Boost converter example. Transistor is on case: on () Diode on case: D and D Effects of on, D and D follow from revised circuits which change values of and hence circuit D and D periods as well. Period DT s ircuit onnections Period D T s ircuit onnections D i g on i v i g v D i v Next, we invoke both voltsec and currentsec balance on the inductor voltage,, and capacitor current,, respectively.

8 8 v (t) (1) oltsec balance on vs. time DT s D'T s g on D[ g on ] D [ g D D o ] t g D D nput onditions = 0 yields: g D on D D D D D o = 0 This is the input circuit loop for the above equation: D on D' D D' D on D' D g D' o i (t) (2) urrentsec balance on c vs. time / DT s D'T s t D[/] [ /] = 0 / Output conditions yield the simple equivalent output circuit result: D' D / = 0

9 9 ombining nput/output with a D transformer: D on D' D D' D D':1 g 1 1 D' o D 1 = =M(D, D' D D' g g ON, on, D, D ) D 1 2 D' lossless D esistive Effects ideal factor Only for do we get full output For the extreme D = 0 D on = 0, D D = D and resistive factor is slightly simplified P η = o D' = = D', but / g differs with losses Pin g g D' η = 1 D 1 Don D' D = η(d,, on, D, D ) 2 g D' Diode on esistive oltage Effects (3) The following is problem 3.3 of hapter 3 Erickson Buck onverter with: nput 1 1 filter 1 has 1 winding resistance due to copper wire

10 10 Transistor with on Diode with D and D ossless Buck: 1 1:D 2 g o o = D g 1 = 2 /D η = o /(D g ) = P o /P in = 1.0 (lossless) g i 1 v1 i 1 1 i 2 D1 2 i 2 a) Q(on) 1 on g i 1 1 i = g = 2 on = = 2 / Next we will consider the switch case when the transistor is off and the diode is on.

11 11 Q(off) g i 1 i 1 1 D D 2 i 2 1 = g = D 2 D = 1 2 = 2 / Do oltsecond balance on s / harge balance on s 1. 1 : D( g ) D ( g ) = 0 g = : D( 2 on ) D ( D 2 D 2 2 ) = 0 D 2 on D 1 D D D 2 D 2 2 = : 1 D 2 = 0 1 = D : 2 / = 0 2 = / 1 nput ircuit From 1, 3, and 4 1 = g 1 1 = g D 2 1 = g D 1 1 = g D 1 Output ircuit: Then: (2) D on D g D 2 1 D D D D 2 = 0 D 2

12 12 D on D 2 1 D' D D' D 2 Dg 2 Fig. 1 D Trf Model: g 1 : D D on D 2 1 D' D D' D 2 2 Fig. 2 b) From Fig. 1 = D on D g D 2 1 D D D D 2 c) Using the oltage Divider: = (D 2 g D D ) Don D 1 D'D 2 Filtered Buck = (D D D ) g 2 D D D' g on 1 D 2 η = = (1 D' D ) D g D 2 D D D' g on 1 D 2

13 13 Next we solve the general equations for the BuckBoost 4. BuckBoost solved in general g D 1:D D':1 o (a) Equivalent Series esistance,, losses only Assume lossless except for eflect all to load side. For D short and open. D' For D conditions: g (D/D') /D'2 (b) Device losses only BuckBoost with device loss ( 0) Transistor: on Diode: D o o = D' 2 D g D' D 1:D D' D D on D':1 g o

14 14 D D' g (D/D') D on /D' 2 o D D o = g D ' Don D' 2 (c) BuckBoost onverter including all static losses. This is problem 3.4 from Erickson 1 A g = µh f s = 40 khz load 5 out required is 5 and g = 1.5 Diode: D = 0.5 Schottky D = 0 (approx.) Transistor: on = 35 mω deal BuckBoost: o D D = = 1 D D' g eal BuckBoost: we will find in the next few pages D D o = g D ' other ( on, ) Diode esistive Effect Effect

15 Below we work out Erickson problem 3.4 for illustration of how to do the HW for hapter 3 HW # 2 will be Erickson hapter 3 Pbms.6 and 7. This will be due next week along with any questions asked in the lectures 15

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17 17 (5) ompare Buck vs. BuckBoost Topology for D onditions Only. This is problem 3.5 of Erickson. 500 D Bus DD onverter 10A 400 Transistor: on = 0.5 Ω D 0 Diode: D 0 nductor: 0

18 18 Find D and η for placing different topologies inside the D D converter. Then compare why one topology might be favored over the other for the specified conditions. Prob. 3.5 Only loss is Mosfet with on = 0.5Ω Buck i g 500v D 1 Mosfet on and Diode off. g i on v v oad=10a i oad=10a i 400v i g 500v BuckBoost v Mosfet on and Diode off. g i on i v D1 i oad=10a i 400v oad=10a i v (t)= g i on g on v (t)= g i on g on i c (t) = i i c (t) = 10 Mosfet off and Diode on. Mosfet off and Diode on. i v i 10A v (t)= v (t)= i c (t) = i i c (t) = i Thus averaging we get: Thus averaging we get: <v > = (g on )D ()D = 0 <v > = (g on)d ()D = 0 <i c > = D( 10) D ( 10) = 0 <i c > = D(10) D ( 10) = 0 So Dg Don = 0 So Dg on D D = 0 10 = 0 = 10 D 10 = 0 = 10/D g v i i 10A

19 19 Don Don Dg Dg D' g. 1:D. Don g g. 1:D. Don. D':1. First we need D: Dg 10onD = 0 Dg 10onD/D D = 0 Dg gd 2 10onD 2D D 2 = 0 D = /(g on) D 2 (g ) D(2 g 10on) = 0 Using the quadratic equation for D: D = 0.81; D = 0.19 D = 0.45; D = 0.55 Solving for efficiency: g = D Solving for efficiency: g = D η = P out /P in = load /gd η = P out /P in = load /gd = (4000)/[Dg(10)] =(4000)/[gD(10/D )] η = 0.99 = 99% η = = 97.8% P loss = P in P out P loss = P in P out P loss = = 50 Watts P loss = = 90 Watts The BUK is better for this application! Another way to calculate the Power loss is to use the voltage divider and find g with espect to as shown below: = Dg Dg = Don D'( Don ) D' 2 Knowing that: = Dg for the lossless Buck: = Dg/D for the BuckBoost: we can see that =1 Don for the lossless Buck case thus: η = η = Don ( Don ) D' 2 η = 0.99 = 99% η = 0.98 = 98% since η = Pout/Pin: Pin = Pout/η = 4000/.99 = W Pin = Pout/η = 4000/.98 = W Ploss = = 40.4 W Ploss = = W The BUK is better for this application!