| ??? 03/24/06 14:22 Modified: 03/24/06 14:26 Read: times |
#112978 - Constant current during inrush moment Responding to: ???'s previous message |
Suresh said:
Kai said, When the lamp is switched-on, current through it is limited to about 0.2A and up to about 2.0W will drop across the transistor, but only for a very short time. If the current through 3R0 resistor reachs about I = 0.6V / 3Ohm = 0.2A from reading the above note, i tried to calculate how you have said that 2W would be dissipated in the transistor (BD135) and, the 3R resistor would have drop of .6V. BUT couldnt get clear with that. also with "what would be the voltage drop in the bulb in that circuit" since i think it would not drop 12V fully on it. For the lamp shown above I have made the following inrush current test, remember?  So, this lamp shows an impedance of 12V / 0.18A = 66.7 Ohm, when being turned-on for a longer time, at least longer than the inrush phenomenon needs. At the inrush moment, on the other hand, flowing current is 8.8 times higher, as you can see from the scope plot. So, impedance then is 12V / (0.18A x 8.8) = 7.6 Ohm. The above constant current driver limits the current through the lamp to about 0.2A at any instant, because about 0.6V drops across the 3R0 resistor. So, when 0.2A is flowing through the cold lamp, and its impedance is only 7.6 Ohm at that instant, then only 0.2A x 7.6 Ohm = 1.5V is dropping across the lamp. The rest, namely 12V - 1.5V - 0.6V = 9.9V is dropping across the collector emitter junction of BD135. 9.9V times 0.2A gives a power of 2W dropping across the BD135 during the inrush moment. Kai |
