POWER SUPPLY ANALYSIS AND DESIGN EXERCISE MR uf. = L sec. (FL ) N sec. sec = L pri. sec 2 V FL VRIP

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Piers Eaton
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1 ELETRON 3 POWER UPPLY NLY ND DEGN EXERE No.1 Power upply nalysis MR H z 5 REGULTOR 0 to 166N u in 1 u 0. 1 u o L O D ) Using chade curves, detere the following: in (NL), in (L), (NL), (L), (L), (L), (peak), (rms), (ave), (surge), in (rms) and in (pp). B) s the transformer correct for the job? Explain. Note that he transformer can run at its imum rated current without any problems there is safety margin built in. ) erify all of step results using mcap. The transformer parameters you have to specify for Microcap are: L, L, coupling coef. ssume L is such that the transformer magnetising current is 10% of its full load current. N N L ( NL) L L ,7 10,75 ( mag) 0,1 (L ) N 115 ω (mag ) 10π 37, m 8,H L N L N 1 0,1 4 37, m N 10, 75 8, ( 10,75) 71 mh ssug a coupling coefficient of 99.5%, the transformer µap parameter values should read: 8., 71m, in the value field. Don't forget to add a series resistor in the ondary winding to simulate the transformer losses. D) Detere the imum heat sink thermal resistance allowed if the voltage regulator has T J () 150 o, θ J 1,5 o /W and θ 0, o /W assume T 5 o. No. Power upply Design Design a D power supply of 10 for a 0 to 5 load. pecify all of the components to be used. Don't forget to specify the heat sinks required by the voltage regulator, the PNP transistor and the power MOETs. The LM113 is a precision reference used to provide a more accurate current limit because BE of the PNP is not known accurately. et the current limit above 5, say around 6.5, to guarantee voltage regulation right up to 5. Do this problem without chade curves using R ( ( NL) ( L) ) ( L) for diodes L ( D ) t RP ( PP ) ( PP ) E ( pk ) L ( D ) D L ( D ) 0.6 E RP L Hz LM 113 LM 317T L O D 0 to 5 10 Power upply Exercise Rev. 5//003. auriol Page 1

No.1 ) Transformer : (NL) 10,7 (L) 10 at (L) 4 R s (10,710)/4 0,175Ω % L ( D) NL ( D) irst iteration ssume an initial value for in (L) in (NL) 10.7 * 1,41 13.

9 econd iteration ssume in(l) 11.9 and redo R L 11.9 / 5.64Ω R s /R L 0.175 / 5.64 * 100 3.1% ωr L 10π * 5000µ * 5.58 10.5 We read in (L)/ in (NL) 0.85 therefore in (L) 0.85 * 13.13 11.

2 No.1 ) Transformer : (NL) 10,7 (L) 10 at (L) 4 R s (10,710)/4 0,175Ω % L ( D) NL ( D) irst iteration ssume an initial value for in (L) in (NL) 10.7 * 1, Let in (L) 10 * 1,41 1,14 R L 1,14 / 6,07Ω R s /R L / 6.07 *100.88% ωr L 10π * 5000µ * We read in (L)/ in (NL) 0.86 in (L) 0.86 * econd iteration ssume in(l) 11.9 and redo R L 11.9 / 5.64Ω R s /R L / 5.64 * % ωr L 10π * 5000µ * We read in (L)/ in (NL) 0.85 therefore in (L) 0.85 * final ( avg) ( peak) ( avg) rom the diode current curves, we read at nωr L * 10π * 5000µ * (11.16 / ) 1 and R /nr L 1.57% (rms).75 * (avg).75 * 1.75 (surge) NL /R s / P (peak) 9 * (avg) 9 * 1 9 p Power upply Exercise Rev. 5//003. auriol Page

3 Ripple oltage Without hade curves, we have: Q t ( 1 10) in ( PP) PP L L 5000µ which always yields a larger value than the actual ripple voltage because we assumed that the capacitor discharges during an entire half cycle of the 60 Hz sinewave. rom chade curves, we read at R s /R L 3.1% and ωr L 10.5 in (rms) 6% of in (L) 0.06 * in (pp) * 1,73 * PP Transformer: ( pk ) L ( D ) or a bridge rectifier, we have (rms) 1,414 * (rms) * N N RP ( pk ) D ( no load ) P No.1 B) (rms) 3,89 from chade curve calculations which is less than the imum rated value of 4, therefore the 166N10 transformer is fine for the job. No.1 ) nitial surge current in transformer ondary winding at D ull Load L 8.H, L 71 mh, k 1.0 NOTE: The imum surge current is a function of the initial phase angle of and also depends on the mary and ondary inductance values for the same voltage ratio the inductance ratio will be the same and if we use inductances 10 times smaller, the initial surge current will go up. Because the inductance slows down the rate of increase of the current, it will never reach the imum value calculated, that is surge NL /R. Power upply Exercise Rev. 5//003. auriol Page 3

4 ettled Waveforms at D ull Load L 8.H, L 71 mh, k 1.0 ettled Waveforms and their RM alues at D ull Load omparison of theoretical values and microcap values L (D) Ripple (pk) (rms) (pk) (rms) (pk) (rms) Theo pp 14.3 p p 3.89 µap pp p p p 0.81 p 0.91 p p Power upply Exercise Rev. 5//003. auriol Page 4

5 nitial urge of Primary Winding urrent at D ull Load ettled Primary Winding urrent at D ull Load No.1 D) rom the µap simulation, we have in 11,48 D and in.3 pp 150' 1.5'/W 0.'/W J REGULTOR W 5' P reg ( )* 1.96W at full load θ o / W Power upply Exercise Rev. 5//003. auriol Page 5

No. ssumed 115 60 Hz Power upply Design 4 pp 7.67p or 19.57rms NOTE: from Hammond dc 0.6 x ac 1.67 10.5 rms ac 5.67 6.5 dc R g 1.67 Q3 only starting to conduct 5.4 4.9 typ. Q3 0. to 0.5 m s 4 7 0 16.

6 No. ssumed Hz Power upply Design 4 pp 7.67p or 19.57rms NOTE: from Hammond dc 0.6 x ac rms ac dc R g 1.67 Q3 only starting to conduct typ. Q3 0. to 0.5 m s Rs 1.9 Rs typ 3 m 0. to 0.5 Q1 Q % safety margin dropout 3 LM317T 50 u typ. 390 Let thermal limit of LM317 be 0.5.7k 0. to L O D 0 to 5 10 n LM113 (Zener) precision reference is used in order to obtain a more accurate current limit. ts imum current is rated at 50 m and one should ensure that it is not exceeded. z 0.7m 3 / h E H950L Transfer haracteristics The MOET G drive level is 3 typical at 55 o. G threshold ranges from 1 to with a median value of 1.5, therefore let us assume that imum and imum values of G will be ±0.5 away from the typical values. Hence the circuit must be able to provide a G value greater than 3.5 to ensure that the MOET s can deliver each to the load. s long as there is no current limiting in Q1 and Q, s R s < 1.9 and Q3 is O. s the load current increases, REG goes up and this in turn increases G of the MOET s because REG R g G s R s when Q3 is O. To ensure enough drive for G, let G 3 be reached with REG 0. but set the thermal limit of REG to RG 4.6 4Ω std P 1. W 0. R 4 Let R G be 4Ω, 3W G o 3 55 o R 0.64Ω P 5. 67W R 0.64 Let R G be 0.64Ω, 10W G Transformer: rom the circuit above, we need at least and n the Hammond 165 series, we have the (18,10) and the 165T (, 1). The 165T will be selected because the has a voltage a little too low. We therefore have: R (trans) (.3 ) / 1 5 mω NL ( D ).3 D 9.54 L ( D ) D 0.5 RP 7. 1 Linear Power upply Exercise Rev. 5//003. auriol Page 6

7 ilter apacitor: We assumed rip 4 PP at a imum D of 6.5 at the unregulated supply O/P. ( 1 10 Hz) Q D t Let s use three 5000 µ capacitors in parallel. ( PP ) ( PP ) 4 µ L rip rip This will produce 3.6 PP imum ripple at 6.5 and NL ( D ).3 D 9.54 L ( D ) D 0.5 RP 7. 3 Diode Bridge: ssug D 6.5, we have (avg) 0.5 x (rms) 6.5 / and (rms) x (rms) 7.4 Maximum surge M (31.5 ) / 5 mω 1180 P 9.5 from circuit shown beside One should select diodes that can handle the above imum currents and P. The imum surge will never reach 1180 because of transformer inductance. elect diodes that can handle (avg) > 10 and P > 50 and simulate circuit to obtain actual surge current you must obtain winding inductance values from manufacturer or measure it to be able to simulate accurately Hz 165T p no load no load use and switch: ( NL) The Little use application data says to use nom / / for a sloblo rms fuse operating at 5 o ambient temperature if it operates at higher ambient temperatures, then the fusing current decreases and one must select a higher fuse rating. Let s use a fuse assug at 5 o. The switch must be able to handle.3 and the 115. LM113: This is a 1. precision reference with a imum rated current of 50 m.. t requires 0.5 m typical to operate, therefore let s use a bleeder resistor across the EB junction of Q3 to provide say 1.5 m to the LM113 when Q3 turns ON and BE R BE 0.7/1.5m 467 theo and use 470 Ω std The imum current will be m Z R h which is well below the imum of 50 m. BE E NOTE: the BD438 (PNP transistor Q3) data sheets specify h E 85 to 375 at 0.5 and E 1 PNP transistor BD , asssume 0.5 for safety. The worst case power dissipation for Q3 is then PQ 3 E W o θ θ / W.46 Use heat sink insulator with less than 3.5 o /W for safety Ω Q3 0.64Ω Q at 150 ' reg.46w 150' J 3.5'/W 5' Linear Power upply Exercise Rev. 5//003. auriol Page 7

8 LM317T: P θ reg θ (.4 10) W ' J 5'/W 6.W 5' / W o 4.9 typ. 4 Ω 5W 7.3 at full load 0 to Q3 0.64Ω Ω 0. to u Q1 Q 0.33 u 1N400 LM317T 0. to k L O D WHEN URRENT LMTED 0 to10 We have to select a heat sinkinsulator combination that has a thermal resistance of 11.1 o /W in order to current limit the LM317T at 0.5. Note that the MOET s go into current limiting before the LM317T does because that occurs when reg 0. typical which corresponds to G 3 typical at. The shortcircuit current of the LM317T will be o < Q3 O 6.W reg which means a shortcircuit current of 6.77 in the load. 1 T J <15 o Q3 ON Q3 ON T J 15 o 0.5 reg Regulator load regulation characteristics Power supply load regulation characteristics o 10 0 Q3 O 1 < P Q1 occurs at shortcircuit where D is 6. Q3 ON 1 Q3 ON LOD P θ Q1 MOET s H950L P ( ) θ 76.14W J Q D 76.14W ' 0.61'/W 5' / W Pick individual heat sinks with θ < 0.5 o /W for safety this includes insulator thermal resistance. o Linear Power upply Exercise Rev. 5//003. auriol Page 8

L4970A 10A SWITCHING REGULATOR

L4970A 10A SWITCHING REGULATOR 10A OUTPUT CURRENT.1 TO 40 OUTPUT OLTAGE RANGE 0 TO 90 DUTY CYCLE RANGE INTERNAL FEED-FORWARD LINE REGULA- TION INTERNAL CURRENT LIMITING PRECISE.1 ± 2 ON CHIP REFERENCE