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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.
The increase in the current is building up the magnetic field surrounding the coil. Energy is stored in that field. Consider the energy supplied by the voltage source during the
WHOLE COIL WINDING
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