Full-wave rectification, bulk capacitor calculations Chris Basso January 2009

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1 ull-wave recificaion, bulk capacior calculaions Chris Basso January 9 This shor paper shows how o calculae he bulk capacior value based on ripple specificaions and evaluae he rms curren ha crosses i. oal Vbulk ou Vbulk D6 N47 D5 N47 bulk Vmains D7 N47 6 D8 N47 CBulk u C = 5 B Curren 5/V(Vbulk) igure : a classical full-wave recificaion. The bulk capacior curren can be esimaed as follows: d V dvc bulk C, cos bulk Cbulk Cbulk CbulkV () d d Wha we need is o derive he ime a which he refueling of he capacior from V min o V sars. This is exacly where he inpu signal reaches V min (neglecing he diodes drops): V min V () The oal refueling ime from V min o V is herefore a fourh of he period (he disance from o he volage occurrence) minus (): V min V (3) 4 Now ha he capacior curren is known, we need o calculae he diode curren d,. This is nohing else han he capacior curren added wih he load curren. However, as shown by igure, he load curren is no a coninuous curren. This is because as he swiching power supply mainains a consan oupu power, he ripple superimposed on he bulk capacior permanenly changes he operaing poin of he power supply, impog a variable drawn curren. As shown in igure, his load curren load () is raher disored bu can be approximaed as a recified usoidal signal. Therefore, he average or dc curren drawn by he swiching converer is:

2 ou V V V (4) min min Solving his equaion is a complex process. Raher, we can calculae he maximum load curren ha occurs a he minimum bulk volage: load,max V ou (5) min The diode curren is herefore immediaely found by: (6) d, Cbulk, load,max d () 4.8 d = load lo ibulk, idiode, iload in amperes C bulk = c load () C bulk () V lo iload in amperes abs, vbulk in vols load,max V bulk () 6m 58. load ( ) load () 38m 44m 5m 56m 6m ime in seconds V min igure : bulk capacior refueling waveforms. As shown in igure, he capacior curren reaches zero when he diode curren equals ha of he load. This occurs exacly a he op of he usoidal inpu volage (where he derivaive is null). The ime a which his even akes place is known, his is calculaed in (3). Therefore, he diode curren down slope can be approximaed by: S diode d, load (7) Where load ( ) represens he minimum load curren when he e wave is a is op: load ou load,min (8) Vmax

3 Once he slope is known, we can calculae he ime aken o go from d, o, leading o he oal diode conducion ime c : c d, (9) S diode rom his value, i becomes possible o calculae he load average curren in a simpler manner han in (4): Where represens he mains frequency. () d, c T d, c We can show ha he rms curren in he capacior can be pu under he following form: () Cbulk, rms c The diode rms curren is derived as follows: d, rms () c Each pair of conducing diodes sees half of he load average curren: The oal inpu rms curren comes easily as: d, avg (3) d, rms, oal (4) c Design example: Suppose we need he following specificaions: V in is 85 V rms, 5 Hz V is herefore V V min is chosen o 5 V (argeed ripple) ou = 9 W is 86%. calculae he bulk capacior value:

4 V min V ou 4 C bulk µ V V min (5). calculae wih (): V min V.368 ms (6) 3. calculae he bulk capacior curren wih () Cbulk, bulk C V cos 3.84 A (7) 4. calculae he oal charging ime : V min V 3.63 ms (8) 4 5. calculae he load and minimum curren values : A (9) ou load,max.9 Vmin load 6. calculae he diode curren: ou,min.87 A () V max,,max A () d Cbulk, load 7. calculae he diode curren down slope from he value o : S diode d, load ka s () m 8. Derive he diode oal conducion ime c :

5 c d, ms (3) S 4 diode 9. he average dc curren can now be found : d, c.6 A (4). from which we can ge he bulk capacior rms curren: C,,.7 bulk rms load avg A (5) c. hen each diode rms curren: A (6) d, rms.58 c. plus each diode average curren: 3. and finally, he oal rms inpu curren: A (7) d, avg.63 d, rms, oal.3 A (8) c Comparison beween he calculaed values and he simulaed values Below is an array comparing he resuls given by a simulaor and he one obained from he analyical derivaions (same inpu volage and minimum bulk volage): C bulk = µ, ou = 9 W Calculaed simulaed Error in,rms.3 A. A 6.% d, 5.9 A 5.67 A 4% C bulk, rms.84 A.7 A 7.6% d,avg.63 A.593 A 6% C bulk = 49 µ, ou = W Calculaed simulaed Error in,rms 4.9 A 4.66 A 5% d, 3. A.5 A 5.6% C bulk,rms 4. A 3.8 A 7.9%

6 d,avg.4 A.3 A 7.7% C bulk = 6 µ, ou = 5 W Calculaed simulaed Error in,rms.4 A.6 A 6.9% d, 33 A 3. A 9% C bulk,rms. A 9.5 A 7.4% d,avg 3.5 A 3.3 A 6% Selecing a normalized value for he bulk capacior f equaion (5) gives us a capacior value, we are likely going o selec a differen value based on he rms curren requiremen as well as he available normalized values. Le us suppose we have seleced a 5-µ capacior furher o a recommendaion of µ. How do we rescale all he calculaed parameers? The firs hing is o compue he new minimum bulk volage, V min. Unforunaely, he process is complex because he curren consumed by he load as V bulk goes down, permanenly changes ce he power supply keeps ou consan. There is no close form soluion and we mus use a numerical approach o obain he value we are looking for. Le s discover he seps. The energy accumulaed in he capacior during is charge equals ha released during is discharge ime d from V o V min. Therefore, we have: (9) ou.5cbulk V Vmin d The discharge ime d can be evaluaed by looking a he below figure and combining he various ime evens: V plo abs, vbulk in vols V min V in () d V bulk () ¼ T 4. plo ibulk in amperes. -. C bulk () m 34.m 36.9m 39.6m 4.3m ime in seconds igure 3: he charge/discharge cycle can be broken down in hree differen ime duraions.

7 has acually been already derived in (5). We can see from igure 3 ha he discharge ime d acually equals he ime o go from he volage o (/4 h of he inpu period) plus. Oherwise saed: d V min V min V V (3) 4 4 We can now subsiue d definiion ino (9) and obain he final equaion: V min V ou.5cbulk V Vmin (3) 4 This equaion does no have a closed-form soluion. One easy way o find is soluion is o plo boh sides of he equaion and check he crosg poin. Anoher opion is o plo he difference beween he righ and lef sides and check when i reaches zero as V min is incremened from o V : V min V (3) ou.5cbulk V Vmin 4 We have enered he daa ino Mahcad and we obained a curve showing he zeroing of he curve for V min = 68 V (igure 4). Having he minimum volage, we can calculae he ime needed o reach V min ug (): V min V.95 ms (33) Then he capacior charging ime (equaion (3)): V min V 3.49 ms (34) 4 Then go o poin 3. above and roll ou he calculaions. The following simulaed daa have been compared o he numerical values based on various capacior and power selecions:

8 .5 Y = f(v min ) V min (V) igure 4: he graph of equaion (3) shows a minimum volage of 68 V. C bulk = 5 µ, ou = 9 W Calculaed simulaed Error V min 67.8 V 69 V.7% in,rms. A.95 A.3% d, 6. A 6. A.9% C bulk,rms.97 A.78 A.6% d,avg.54 A.54 A % (yes!) C bulk = 33 µ, ou = W Calculaed simulaed Error V min 67.3 V 67.7 V.6% in,rms 4.67 A 4.69 A.4% d, 3.5 A 3. A 3% C bulk,rms 4 A 3.97 A.75% d,avg. A.3 A.6% C bulk = µ, ou = 5 W Calculaed simulaed Error V min V 76.5 V.7% in,rms.67 A A.75% d, 35.8 A A.64% C bulk,rms.8 A.4 A.% d,avg.85 A.98 A 4.3% As one can judge, he agreemen is raher good.

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