Explain Vceq & Icq in Collector-Base Bias Config.

[ranju]

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..

 

[NascentOxygen]

the base resistance in collector to base bias configuration is given by
Rb=Vcc-Vbe-IcRl / Ib

Isn't this equation needing a set of paentheses? Here, borrow a spare pair of mine: ( )

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..

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.

 

[ranju]

that means V_CEq = V_cc-V_be-I_cR_L...?

 

[Jony130]

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

 

[NascentOxygen]

that means V_CEq = V_cc-V_be-I_cR_L...?

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.

 

[ranju]

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_CR_L / I_B
we can neglect V_BE in comparison to V_CC , so is V_CEQ=V_CC-I_CR_L...??

 

[NascentOxygen]

This is the given circuit with which I was referring to ..
Here in the solution , they took R_B=V_CEQ / I_B

Approximately.

and if we apply KVL we ger R_B= V_CC-V_BE-I_CR_L / I_B

Approximately. But what happened to those extra parentheses I leant you?

we can neglect V_BE in comparison to V_CC

There is no good reason for doing so, biasing need not be quite so rough and ready.

so is V_CEQ=V_CC-I_CR_L...??

Approximately. So what is being neglected in making this approximation?

 

[ranju]

so is that a generalised formula for CE voltage at quiscient point...this what I was asking...