1 dc circuit analysis mosfet common-source circuit the source terminal is at ground potential and is...
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![Page 1: 1 DC circuit analysis MOSFET Common-Source Circuit The source terminal is at ground potential and is comon to both the input and output. Coupling capacitor](https://reader035.vdocument.in/reader035/viewer/2022062805/5697bfe91a28abf838cb6ac8/html5/thumbnails/1.jpg)
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DC circuit analysisMOSFETCommon-Source Circuit
The source terminal is at ground potential and is comon to both the input and output.
Coupling capacitor acts as an open circuit to dc but allows the signal voltage to be coupled to the gate
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DC circuit analysisMOSFETCommon-Source Circuit
The dc equivalent circuit;
Assume that the MOSFET is biased
in the saturation region i.e;
andTNGS Vv
DDGSG VRR
RVV
21
2
Applying the voltage divider rule;
2TNGSnD VVKI
DDDDD RIVI
Also;
TNGSDSDS Vvvv sat
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DC circuit analysisMOSFETCommon-Source Circuit
The dc equivalent circuit;
If the calculated vDS is greater than vDS(sat), the MOSFET is biased in the saturation region as we have initially assumed, and our analysis is correct.
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DC circuit analysisMOSFETCommon-Source Circuit
The dc equivalent circuit;
22 DSDSTNGSnD vvVvKi
If the calculated vDS is smaller than vDS(sat), the MOSFET is biased in the triode region and we have to recalculate the current using the equation for the triode region namely;
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DC circuit analysisMOSFETCommon-Source Circuit
The dc equivalent circuit;
The power dissipated in the MOSFET is;
DSDT VIP
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DC circuit analysisMOSFETCommon-Source Circuit Example 1
Calculate the drain current, ID and drain-to-source voltage, VDS and the power dissipated in the transistor PT.
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DC circuit analysisMOSFETCommon-Source Circuit Example 1 – Solution
V 64.3
10160280
160
21
2
DDGSG VRR
RVV
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DC circuit analysisMOSFETCommon-Source Circuit Example 1 – Solution (cont’d)
Assuming that the transistor is in saturation mode;
mA 669.0
264.325.0 2
2
TNGSnD VVKI
V 31.3
10669.010
DDDDDS RIVV
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DC circuit analysisMOSFETCommon-Source Circuit
V 64.1
264.3
sat
TNSGDS VVV
Since VDS > VDS(sat), the transistor is indeed in saturation mode as we have initially assumed.
Example 1 – Solution (cont’d)
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DC circuit analysisMOSFETCommon-Source Circuit
mW 21.2
31.3669.0
DSDT VIP
The power dissipated in the transistor;
Example 1 – Solution (cont’d)
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DC circuit analysisMOSFETCommon-Source Circuit
Design the circuit such that;
V 4
mA 2.1
k 20021
SDQ
DQ
V
I
RR
Example 2
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DC circuit analysisMOSFETCommon-Source Circuit
DDQSDQ RIV 10V 4
Hence;
V 6DDQRI
k 5mA 2.1
V 6V 6
DQD I
R
Example 2 – Solution
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DC circuit analysisMOSFETCommon-Source Circuit
2TPSGQpDQ VVKI
Substituting values;
22.14.02.1 SGQV
34.0
2.12.1 2 SGQV
V 932.22.13 SGQV
Example 2 – Solution (cont’d)
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DC circuit analysisMOSFETCommon-Source Circuit
V 932.21021
1
RR
R
2932.021
1 RR
R
But;
k 20021
21
RR
RR
Example 2 – Solution (cont’d)
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11/02/2008
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DC circuit analysisMOSFETCommon-Source Circuit
Solving the equations gives us;
k 2831R
k 6822R
Example 2 – Solution (cont’d)
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DC circuit analysisMOSFETCommon-Source Circuit Example 2 – Solution (cont’d)
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DC circuit analysisMOSFETCommon-Source Circuit Example 3
For the circuit in figure; VTN = 1 V
and Kn = 0.5 mA/V2.
Determine VGS, ID and VDS.
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DC circuit analysisMOSFETCommon-Source Circuit Example 3 – Solution
V 1
54060
4010
GV
D
DDS
I
RIV3105
5
D
SGGS
I
VVV3104
3104 GSD VI
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DC circuit analysisMOSFETCommon-Source Circuit Example 3 – Solution (cont’d)
12105.0
1105.0 23
23
2
GSGS
GS
TNGSnD
VV
V
VVKI
1228 2 GSGSGS VVV
V 646.27 GSV
Substituting for 3104 GSD VI
Hence;
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DC circuit analysisMOSFETCommon-Source Circuit Example 3 – Solution (cont’d)
12105.0 23 GSGSD VVI
1646.22646.2105.0 23
mA 35.1DI
V 65.3
135.15
5
SDS RIV
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DC circuit analysisMOSFETCommon-Source Circuit Example 3 – Solution (cont’d)
V 29.2
235.15
5
DDD RIV
65.329.2 SDDS VVV
V 94.5DSV
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DC circuit analysisMOSFETCommon-Source Circuit Example 3 – Solution (cont’d)
V 1
54060
4010
55521
2
RR
RVG
65.31 SGGS VVV
V 65.2GSV
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DC circuit analysisMOSFETLoad Line and Mode of Operation
205 D
DDDDDS
I
RIVV
or;
mA2020
5 DSD
VI
From the above equation, we obtain the following two points: when ID = 0, VDS = 5 V and when VDS = 0, ID = 5/20 mA
The load line equation is;
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DC circuit analysisMOSFETLoad Line and Mode of Operation
The two points are used to plot a load line on the iD – vDS characteristic curves as shown.
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DC circuit analysisMOSFETLoad Line and Mode of Operation
The Q-point is given by the drain dc current and drain-to-source voltage and is always on the load line.
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DC circuit analysisMOSFETLoad Line and Mode of Operation
If vGS is less than VTN, the drain current is zero and the transistor is in the cutoff mode.
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DC circuit analysisMOSFETLoad Line and Mode of Operation
If vGS is just greater than VTN, the transistor turns on.
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DC circuit analysisMOSFETLoad Line and Mode of Operation
As vGS increases, the Q-point moves up the load line.
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DC circuit analysisMOSFETLoad Line and Mode of Operation
The transition point is the boundary between saturation and non-saturation (triode) region.
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DC circuit analysisMOSFETConstant-Current Biasing
The figure shows a MOSFET amplifier where RS is replaced with a constant current source, IQ.
The gate-to-source voltage, VGS adjusts itself to correspond to the current IQ.
IQ causes the bias to be independent on the transistor parameters and hence stabilizes the operating (Q) point.
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DC circuit analysisMOSFETConstant-Current Biasing
The figure shows the corresponding dc equivalent circuit.
The gate terminal is at the ground potential because we assume zero gate current.
IQ sets the value of VGS, VD, VDS and hence the Q-point of the MOSFET.
The current source IQ can be constructed using MOSFETs but this topic will not be covered in this note.
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DC circuit analysisMOSFETConstant-Current Biasing Example 4
The MOSFET parameters in the figure are;
3/ ,A/V 80 V, 8.0 2' LWkV nTN
Determine VGS and VDS.
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15/02/08
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DC circuit analysisMOSFETConstant-Current Biasing Example 4 – Solution
The dc equivalent circuit is as follows;
2'
2 TNGSn
D VVL
WkI
Assuming the MOSFET is operating in the saturation mode;
Substituting values;
26
6 8.032
108010250
GSV
V 24.2GSV
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DC circuit analysisMOSFETConstant-Current Biasing Example 4 – Solution (cont’d)
The source potential;
V 24.224.20 GSGS VVV
Assuming zero gate current;
46 10102505
5
5
DQ
DDD
RI
RIV
V 5.2DV
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DC circuit analysisMOSFETConstant-Current Biasing Example 4 – Solution (cont’d)
The drain-to-source voltage;
24.25.2 SDDS VVV
V 74.4DSV
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DC circuit analysisMOSFETEXERCISES
Problems:
3.23;
3.25;
3.27 and
3.28
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QUIZ