Loke said:i tried ady...but i still can't find it...3rd equation got 2 unknown Ib and Vbe...i just don't know how to solve it without the type of transistor given...can you help me to figure it out?
Loke said:ermmm...can you solve it with kirchhoffs law because i haven study schokley diode law... i cannot apply it on my work yet.
Loke said:Teacher only taught me silicon transistor,Vbe=0.7 and germanium transistor,Vbe=0.3 ...but i don't know can assume one ^^? ...and in this question both type of transistor is not given that's why i don't know how to proceed..:(
Loke said:ohh... thanks...i ask u 1 more question...if let say you don't assume Vbe=0.7...can you actually find an answer? I'm wondering.
Loke said:As long as the question does not provide the type of transistor,then we just assume it to be Vbe=0.7 ...am i right? i meant other than this question also?...
stevenb gave a good answer. Also, you can think of the base-to-emitter path as similar to a diode. As long as there is something providing voltage between the base and emitter, it will (usually) be about 0.7V.Loke said:As long as the question does not provide the type of transistor,then we just assume it to be Vbe=0.7 ...am i right? i meant other than this question also?...
Loke said:Determine IB, IC, IE, VCE and VCB in Figure 8 for the following values:
RB = 5 k ohm, RE = 500 ohm , VBB = 3.0 V, VCC = 20 V and Bdc = 80
problems:
-If i assume the Vbe in this question also = 0.7...i still can't solve it right away...can u help me figure it out?
-can i assume IC=IE...so IE=BdcIB ...any idea?
attempt at solution:
VBB-IBRB-VBE-IERE=0
VCC-VCE-IERE=0
VBE=0.7
IC=BdcIB