Hello and good day for all, I need advice in finding Vout and ac voltage gain in n-channel MOSFET common source amplifier.
Here i get :
Since the source is grounded, the gate-to-source voltage is
VGS = (R2 / (R1 + R2)) x Vdd
VGS = (33k / (51k + 33k)) x 15
VGS = 5.893V
ID = k [ VGS – VGS(th) ]2
Since: VGS = 7.5V, VGS(th) = 5V, ID = 5mA
Therefore : ID = k [ VGS – VGS(th) ]2
k = ID / [ VGS – VGS(th) ]2
k = 5mA / [7.5V - 5V]2
k = 0.8mAV-2
The Drain Current ID at VGS = 5.89V is ID = 0.8mAV-2 [ 5.89V – 5V ]2
ID = 0.8mAV-2 [ 5.89V – 5V ]2
ID = 0.634mA
The Drain Voltage is VD = VDS = 15V – [0.634mA x 3.3k]
VD = VDS = 15V – [0.634mA x 3.3k]
VDs = 12.90V
The Drain to source resistor is rDS = VDS / ID
rDS = 12.90V / 0.634mA
rDS = 20.35k
Rin = R1||R2
Rin = 51k x 33k / 51k + 33k
Rin = 1683k / 84k
Rin = 20.04k
ro = VM / ID
ro = 50V / 0.634mA
ro = 78.9k
Vout = −gmVgs(ro || RD) ----------------------> It is VGS same with Vgs?
Vout = -6mS Vgs ( 78.9k x 3.3k / 78.9 + 3.3k )
Vout = -6mS Vgs ( 3.17k )
Vout = -19.02 Vgs
Here i get :
Since the source is grounded, the gate-to-source voltage is
VGS = (R2 / (R1 + R2)) x Vdd
VGS = (33k / (51k + 33k)) x 15
VGS = 5.893V
ID = k [ VGS – VGS(th) ]2
Since: VGS = 7.5V, VGS(th) = 5V, ID = 5mA
Therefore : ID = k [ VGS – VGS(th) ]2
k = ID / [ VGS – VGS(th) ]2
k = 5mA / [7.5V - 5V]2
k = 0.8mAV-2
The Drain Current ID at VGS = 5.89V is ID = 0.8mAV-2 [ 5.89V – 5V ]2
ID = 0.8mAV-2 [ 5.89V – 5V ]2
ID = 0.634mA
The Drain Voltage is VD = VDS = 15V – [0.634mA x 3.3k]
VD = VDS = 15V – [0.634mA x 3.3k]
VDs = 12.90V
The Drain to source resistor is rDS = VDS / ID
rDS = 12.90V / 0.634mA
rDS = 20.35k
Rin = R1||R2
Rin = 51k x 33k / 51k + 33k
Rin = 1683k / 84k
Rin = 20.04k
ro = VM / ID
ro = 50V / 0.634mA
ro = 78.9k
Vout = −gmVgs(ro || RD) ----------------------> It is VGS same with Vgs?
Vout = -6mS Vgs ( 78.9k x 3.3k / 78.9 + 3.3k )
Vout = -6mS Vgs ( 3.17k )
Vout = -19.02 Vgs