1/4 AN5369 application note www.dynexsemi.com AN5369 selection of scrs for parallel operation application note replaces september 2000 version AN5369-2.0 AN5369-2.1 july 2002 instantaneous forward current - (a) instantaneous forward volt drop - (v) i 2 i 1 scr 1 (v0, rt 1 ) scr 2 (v0, rt 2 ) v0 v1 v2 fig. 1 total current is divided between the two scrs selection of thyristors (and diodes) for parallel operation with non-reactor/resistor sharing. when thyristors are intended to be used in parallel without the use of reactors or resistors to force current sharing then the thyristors must be chosen in such a way that they will share the current in relation to their forward voltage drop at the operating current. this application note gives a simple way of doing this from the data presented in the device data sheet, based on a number of simplifying assumptions. the problems of differing turn-on performance, finger voltage etc. have not been considered here but are addressed in application note an4999 ?urn-on performance of thyristors in parallel? the treatise below is presented in terms of the fully turned on phase of a thyristor conduction cycle. the methodology can equally be applied to rectifier diodes in parallel. dynex semiconductor will append a 4 digit selection number to the device type to specify your particular sharing criteria. consider the case of two thyristors in parallel with voltage v 2 across them. in this situation the total current is divided between the two scrs such that scr 1 is carrying i 2 and scr 2 is carrying i 1 . at the current i 1 scr 1 has a forward voltage drop of v 1 . it is assumed here that all the variation in forward voltage drop between the thyristors is due to variation in the slope resistance and that the knee voltage v 0 is the same for both thyristors. scr 2 is assumed to be the data book limit case device and v 0 , rt 2 are the figures printed in the data book. now v 2 = v 0 + i 1 .rt 2 for scr 2 and v 2 = v 0 + i 2 .rt 1 for scr 1 i.e. i 1 .rt 2 = i 2 .rt 1 or rt 1 = i 1 .rt 2 /i 2 ..............................................................(1) at the current i 1 we have: v 1 = v 0 + i 1 .rt 1 v 2 = v 0 + i 1 .rt 2 v 2 -v 1 = i 1 .(rt 2 -rt 1 ) substituting for rt 1 from (1) we have v 2 -v 1 = i 1 .(rt 2 - i 1 .rt 2 /i 2 ) i.e. ? v = i 1 .rt 2 .(1 -i 1 /i 2 ).......................................................(2)
2/4 www.dynexsemi.com AN5369 application note suppose that there is a situation whereby n scrs are connected in parallel and are required to conduct a current i tot amps with a given cooling system. in the worst case n-1 scrs will conduct current i 1 and the remaining scr will conduct i 2 = i tot - (n-1).i 1 .....................?3) define a mis-sharing factor (x) such that i 2 = (1+ x).i 1 ....?..(4) note that different people define mis-sharing in different ways. some people would say that a 20% mis-sharing results in the current in two devices dividing 80/120. in the above definition this would be a mis-sharing factor of 0.5 because 120 = (1 + 0.5)*80 ! i tot = i 1 .(n-1) + (1+x).i 1 i.e. i 1 = i tot /(n + x) which is the lower current level.........(5) the maximum current i 2 is i tot - (n-1).i 1 this will be carried by the scr with slightly higher losses because the voltage across all devices is the same, so the device with the higher current has the higher losses. if the conduction losses of the data book device calculated at i 2 are within those that the cooling system can disperse then the design is feasible. substituting for i 1 /i 2 from (4) into (2) and for i 1 from (5) we get ? v = i tot .rt 2 . (x /((1+x)*(n+x)) at i 1 ?........(6) note that the value of ? v can be scaled by the factor i test /i 1 to the value of ? v at any test current point on the linear part of the forward volt drop curve. if ? v is too tight compared to the production spread indicated in the data sheet then the mis-sharing factor will have to be increased and hence the number of devices in parallel will have to increase. example 1 : consider 2 dcr1673sz28 thyristors in parallel which are required to conduct a total of 40ka peak and share current to within 20% of the average i.e. one thyristor can take 24ka while the other takes 16ka. for the dcr1673 rt is 0.093mw to find the mis-sharing factor x : 24ka=(1+x)*16ka i.e. x = 0.5 from equation (6) ? v = 40x10 3 * 0.093x10 -3 * 0.5/[(1+0.5)*(2+0.5)] = 496mv @ 16ka the production test point for the dcr1673 is 3ka so ? v =496 * 3/16 = 93mv @ 3ka example 2 : consider running a number of dcr803sg12 thyristors in parallel to conduct a total of 1200a average current half wave with a heat sink that gives a thermal resistance, junction to ambient, of 0.155?cw per device. the ambient temperature is 40?c and the desired junction temperature is 100?c . data sheet gives v0= 0.85v rt = 0.38mohms the maximum power that may be dissipated is (100-40)/0.155 = 387 watts. from the dissipation curves in the data sheet 387w is equivalent to 320a average 1/2 wave. this is the maximum current. assuming a mis-sharing factor of 0.2, the current in the other semiconductors is given by equation 4, namely, 320 = (1+0.2) * i 1 i.e. i 1 = 320/1.2 or 267a number of devices required in parallel for 1200a is given by equation 3 i.e. 320 = 1200 ?(n?) * 267 or n = 880/267 or 4.29 therefore 5 devices are needed in parallel for a mis-sharing factor of 0.2. because we have to use 5 thyristors instead of 4.29, the minimum current, given by equation equation 3, will therefore actually be: i 1 = (1200-320)/4 = 220a i.e. a mis-sharing factor (x) of (1-220/320) = 0.3125 not 0.2 as originally. therefore equation (6) gives ? v = 1200* 2*0 .00038*(0.3125/(1.3125*5.3125)) = 0.0289 volts or 29mv @ i 1 = 267a average or 377a peak. the datasheet v f curve shows that the total spread of v f at 377a for the dcr803 is ~50mv so that selecting devices to a 29mv band is reasonable and the use of two bands would utilise 100% of the production spread. if we reduce the number of devices in parallel from 5 to 4,the lower current level would be (1200-320)/3 = 293.33 a the mis-sharing factor (x) is therefore 1?293.33/320) = 0.0833 and therefore ? v = 12mv i.e. to use the whole distribution, 4 bands would be required. using multiple vf bands can cause problems with the supply of replacement devices and should be recognised before opting for the solution with the least devices in parallel.
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