| ??? 03/24/06 12:36 Read: times |
#112959 - (.2*bulb resistance Responding to: ???'s previous message |
Terry Lingle said:
bulb voltage will be approximatly (.2*bulb resistance at that current) that resistance is fillament temperature dependant and must be determined from the dreaded data sheet for the bulb selected. When the current flow is .18A (i.e.,the current required for the bulb) the resistance would be 66.66R So when i calculate the bulb voltage it gives = .2 * 66.66 = 13.332 V. so i dont know how connecting a 12V supply could achieve this. Did i took it wrongly? Terry Lingle said:
In that case the transistor will see the remaining voltage across its c-E junction thus the potential dissapation issue. iam some what familiar in calculating the Vce and Ve. But since it is a bulb load i could not asses how much would be the drop accross the bulb, which i would need to know to calculate the dissiation accross Vce. I think the bulb i used was from a local manufacturer. I couldnt get any data regarding it other than the rating. So as you say going through a datasheet of a particular bulb would prove useful. can any one get me one. Terry said: It seems that you are a little light in the fundamentals of electronics and could benifit from some extra curricular work. Can anybody recomend a suitable basic text easily available in India The books i refer gives not much examples on driving different types of loads nor it is that descriptive at some cases. i think the book "Art of Electronics" would be useful as a reference material which i would try to get it in few days. I have found it being preferred during some discussion here and also heard with my friends also. Thanking you, Suresh. |
