written 8.8 years ago by | • modified 8.8 years ago |
Mumbai University > EXTC > Sem 3 > Analog Electronics 1
Marks: 10 M
Year: May 2014
written 8.8 years ago by | • modified 8.8 years ago |
Mumbai University > EXTC > Sem 3 > Analog Electronics 1
Marks: 10 M
Year: May 2014
written 8.8 years ago by |
Fig1 (a) determining Rth (b) determining VG
Rth=R1||R2
=140k||60k
=42kΩ
Calculation for VG
VG=VDD×R2R1+R2
=20×60k140k+60k
=6V
The reduced circuit as shown in Fig2
Fig2 thevenin’s equivalent circuit
IS=ID+IG
Since IG≈0
Hence IS≈ID
Applying KVL to gate-source node we get
VG−VGS−IDRS=0
VGS=VG−IDRS…………..(1)
Now, ID=IDSS(1−VGSVp)2…………(2)
Substituting eq(1) in eq(2)
ID=IDSS(1−VG−RS×IDVp)2
ID=8m(1−6−2k×ID−4)2
After simplification,
2ID=1m(10−2kID)2
0=0.1−40ID+4kI2D−2ID
Solving above quadratic equation
We get, ID1=3.65mA and ID2=6.85mA
Put above two values in eq(1)
ID2=6.85mA and VGS2=−7.7V since |VGS2|>|Vp| FET is in pinch off region and ID=0 at pinch-off region.
Hence, IDQ=ID1=3.65mA and VGSQ=−1.3V
Now, Applying KVL to drain-source channel
VDD−VDS−IDQ(RS+RD)=0
20−3.65×4.7=VDS
VDSQ=2.845V
IDQ=3.65mAandVGSQ=−1.3VVDSQ=2.845VVG=6V