1 DC circuit analysis MOSFET Common-Source Circuit The source terminal is at ground potential and is...

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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

2


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

3



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.

4



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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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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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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mW 21.2

31.3669.0

DSDT VIP

The power dissipated in the transistor;


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Design the circuit such that;

V 4

mA 2.1

k 20021

SDQ

DQ

V

I

RR

Example 2

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DDQSDQ RIV 10V 4

Hence;

V 6DDQRI

k 5mA 2.1

V 6V 6

DQD I

R

Example 2 – Solution

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2TPSGQpDQ VVKI

Substituting values;

22.14.02.1 SGQV

34.0

2.12.1 2 SGQV

V 932.22.13 SGQV


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V 932.21021

1

RR

R

2932.021

1 RR

R

But;

k 20021

21

RR

RR


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11/02/2008

16


Solving the equations gives us;

k 2831R

k 6822R


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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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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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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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V 29.2

235.15

5

DDD RIV

65.329.2 SDDS VVV

V 94.5DSV

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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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The two points are used to plot a load line on the iD – vDS characteristic curves as shown.

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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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If vGS is less than VTN, the drain current is zero and the transistor is in the cutoff mode.

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If vGS is just greater than VTN, the transistor turns on.

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As vGS increases, the Q-point moves up the load line.

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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;

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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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