the base resistance in collector to base bias configuration is given by Rb=Vcc-Vbe-IcRl / Ib now , in a question only the value of Vceq is given & they found Rb as Rb=Vceq/Ib.!! what is Vceq over here..?? and another quantity Icq is used..what are these two quantities..please explain a bit..
Isn't this equation needing a set of paentheses? Here, borrow a spare pair of mine: ( ) Vceq would be the value of Vce at the quiescent point, the "Q point". Icq is the value of collector current at the Q point The Q point is the DC operating point that your bias arrangement sets. BTW, it is not correct to use lower-case subscripts for these DC bias points, you should be using upper-case. There is a strict convention to follow. V_{CEq} is the quiescent C-E voltage. I hope that answers your questions.
No, Vce = Vcc - Ic*Rc For this circuit http://www.cjseymour.plus.com/elec/intrbias/BJTCBIAS.jpg Try read this http://www.zen22142.zen.co.uk/Design/bjtbias.htm
Sketch your circuit. Mark on it a closed loop. You won't be able to draw any closed loop to confirm your quoted guess. So your guess must be wrong. Apply Kirchoff's voltage law to any complete loop on the schematic to find that loop's equation.
This is the given circuit with which I was referring to .. Here in the solution , they took R_{B}=V_{CEQ} / I_{B} and if we apply KVL we ger R_{B}= V_{CC}-V_{BE}-I_{C}R_{L }/ I_{B} we can neglect V_{BE} in comparison to V_{CC} , so is V_{CEQ}=V_{CC}-I_{C}R_{L}...??
Approximately. Approximately. But what happened to those extra parentheses I leant you? There is no good reason for doing so, biasing need not be quite so rough and ready. Approximately. So what is being neglected in making this approximation?