Hello
I have been reading some material about IC thermal power calculations, but since this is my first time, please someone correct or confirm the following :
For a load of 360 ohm @ 12 volts, with 4 of 7 darlingtons active(worst case scenario), using ST ULN2003A (package DIP16) and the datasheet of June 2012.
each pin will take a load of 12/360 = 0,0333 amp.
according to datasheet at 100mA the V(ce) is of 0.9volts, so P = 0.9 * 0,0333, P = 0,02watts
for the total of the four pins, P = 0,08 watts.
Using the Tjmax of 150ºC , and RthJA 70ºC/W, and a maximum Air temp of 70º(read somewere that was the industrial value)
Qmax = (150-70)/(70) , Qmax = 1,14 watts.
Sp everything is fine by my calculations.
I have been reading some material about IC thermal power calculations, but since this is my first time, please someone correct or confirm the following :
For a load of 360 ohm @ 12 volts, with 4 of 7 darlingtons active(worst case scenario), using ST ULN2003A (package DIP16) and the datasheet of June 2012.
each pin will take a load of 12/360 = 0,0333 amp.
according to datasheet at 100mA the V(ce) is of 0.9volts, so P = 0.9 * 0,0333, P = 0,02watts
for the total of the four pins, P = 0,08 watts.
Using the Tjmax of 150ºC , and RthJA 70ºC/W, and a maximum Air temp of 70º(read somewere that was the industrial value)
Qmax = (150-70)/(70) , Qmax = 1,14 watts.
Sp everything is fine by my calculations.