Finding the (Q) points for a BJT?????

[shanecy]

The question gives the following schematic and says to find the Q point for the Voltage(Collector to Emitter) and the Q point for the Collector Current. It gives beta as 50.

I'm using two standard equations from my book to find the two Q points. I followed step by step instructions from my book and I'm being told I'm wrong. The person doesn't support their answer with any work but only says that I cannot use Beta (B) or the two equations from my book "see below" to find the Q points. However, wether i'm finding the Q point or not the two equations are the same just with the subscript Q added.

Could someone work this and perhaps tell me if I'm wrong and what I am not taking into consideration. The schematic below is all that is given with B being 50. The 2N3904 was just added in by my simulator but the actual problem just says it's an NPN BJT transistor.

FYI my work/answers is in the first attachment with the two equations the book is using. 

I'm also attaching in the last attachment additional equations I used in my strategy for solving this problem.

Also the voltage from base to emitter is assumed to be .7V.

[fmaimon]

First consider Ib negligible. So Vb = 20 * 10 / (10 + 68) = 2.564V

Vbe = Vb - Ve = 0.7V, so Ve = 2.564 - 0.7 = 1.864V

Then we get Ie = Ve / Re = 1.864 / 1.1 = 1.695 mA

Ie ~= Ic, so Vrc = 1.695 * 6.2 = 10.51V

Vce = 20 - Vrc - Ve = 20 - 10.51 - 1.864 = 7.63 V


If you want more precision, you can iterate these results with the other data, like beta. First you calculate the Ib = Ic / beta, it's impact on Vb and Ic, and recalculte all of above. Do it until the results don't change significantly.

[shanecy]

Could you explain why implementing my solution was incorrect. Why can't I use the formulas above?