written 8.2 years ago by | • modified 8.2 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 15
written 8.2 years ago by | • modified 8.2 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 15
written 8.2 years ago by |
Calculation for Rth and VG:
Calculation for VG:
VG=(VDDXR2)(R1+R2) = (16X270X103)(2.1X106+270X103) = 1.822 V
VG = 1.822 V
The reduce circuit as shown in fig 5.6.
Calculation for IDQ, VS and VGS:
IS=ID+IG
Since IG≈0
Hence IS≈ID
Applying KVL to gate-source node we get
VG−VGS−IDRS=0
VGS=VG−IDRS
Now, ID=IDSS(1−VGS/VP)2
By Substituting,
ID=8m(1−(1.822–1.5kID)(−4))2
After Simplification,
ID=16.9475X10(−3)−8.733ID+1.125X103I2D
Hence,
ID1=6.223X10(−3) and ID2=2.415X10(−3)
FET is in pinch off region and ID=0 at pinch-off region.
Hence, IDQ=ID2= 2.415mA
And VGSQ = -1.8V
VS=VG−VGS
= 1.822 – (-1.8)
=3.622
Hence,VS= 3.622
Calculation for VDSQ:
Now, Applying KVL to drain-source channel
VDD−VDS−IDQ(RS+RD)=0
16−2.415X10(−3)(2.4X103+1.5X103)=VDSQ
VDSQ = 7.306V