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Q. A Silicon Transistor Whose Common Emitter Output Characteristics Are Shown In Dig Is Used In The Circuit With
Vcc=22.5v, Rc=5.6k, Re=1k, R2=10k, And R1=90k.
For This Transistor b=55.Find The Q Point.
In many cases transistor characteristics are not available but b is known .Then the calculation of Q point is carried out as shown.
We know that
Ic=bIb.
V=10*22.5/100
Rb=(10*90)/100
Kirchoffs voltage law applied to the collector and base circuits respectively yields
-22.5 + 6.6Ic + Ib+ Vce=0.
0.65-2.25+Ic+10.0Ib=0
Therefore
-1.60 + Ic+ 10 / 55Ic=0.
Ic = 1.36mA
Ib = 24.8mA
Also we know that
-22.5 + 6.6*1.36 + 0.025 + Vce=0.
Hence
Vce=13.5V.
Q. Afixed-biasmethod is illustrated in Figure. Assuming ICBO to be small compared to I BQ and I CQ , find RB such that the operating point corresponds to I CQ = 14 mA, V CEQ = 7
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