Q. A Silicon Transistor Whose Common Emitter Output Characteristics Are Shown In Dig Is Used In The Circuit With
V_{cc}=22.5v, R_{c}=5.6k, R_{e}=1k, R_{2}=10k, And R_{1}=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
I_{c}=bI_{b}.
V=10*22.5/100
R_{b}=(10*90)/100
Kirchoffs voltage law applied to the collector and base circuits respectively yields
-22.5 + 6.6I_{c} + I_{b}+ V_{ce}=0.
0.65-2.25+Ic+10.0Ib=0
Therefore
-1.60 + I_{c}+ 10 / 55I_{c}=0.
I_{c} = 1.36mA
I_{b} = 24.8mA
Also we know that
-22.5 + 6.6*1.36 + 0.025 + V_{ce}=0.
Hence
V_{ce}=13.5V.