MOSFET Basic FET Amplifiers The MOSFET Amplifier

Name: MOSFET Basic FET Amplifiers The MOSFET Amplifier
Uploaded: 2017-08-05T16:59:31+00:00
Duration: PTM32S11
Channel: Ross Marsh
Description: MOSFET Basic FET Amplifiers The MOSFET Amplifier

MOSFET Basic FET Amplifiers The MOSFET Amplifier
The figure shows a basic MOSFET amplifier where the dc source VGSQ is used to provide the bias voltage between gate and source. The ac voltage source vi which represents the signal source is connected in series with the bias voltage VGSQ.

The figure shows the corresponding characteristic, load line and Q-point for the amplifier.

The Q-point is dc load line and Q-point are functions of vGS, VDD and RD and the transistor parameters.

The sinusoidal input voltage (vi) causes variations in the gate-to-source voltage (vGS), drain current (iD) and drain-to-source voltage (vDS). Sinusoidal input voltage (vi) Sinusoidal variation in the drain current (iD) Sinusoidal variation in the drain-to-source voltage (vDS)

The total vGS is the sum of VGSQ and vi i.e; As vi increases, the instantaneous value of vGS increases, and the bias point moves up the load line.

Larger value of vGS means larger drain currrent iD and smaller value of vDS. The negative swing of vi causes vGS to decrease below the quiescent value, and the bias point moves down the load line.

A smaller value of vGS means smaller drain currrent iD and larger value of vDS. Thus this configuration causes phase inversion

Phase inversion between input and output voltages

For the MOSFET to operate as a linear amplifier, it must be biased in the saturation region and the instantaneous drain current and drain-to-source voltage must also be confined to this region.

As long as the amplifier operates in the linear region, symmetrical input signal will produce symmetrical output. If the limit is exceeded, the output signal will be clipped and distortion will occur

The total instantaneous value of the gate-to-source voltage is given by; Considering only ac component; Hence; dc component ac component

The total instantaneous drain current is; Substituting for vGS; or

The term is the dc component of the drain current The term is the time-varying component of the drain current which is linearly related to the input signal vgs. The term is the component of the drain current which is proportional to the square of the input signal vgs.

The term produces undesirable harmonics or non-linear distortion in the output voltage. To minimize these harmonics, we must have; This is the small-signal condition that must be satisfied for linear amplifier.

If the small-signal condition is satisfied, the term containing vgs2 may be neglected and the equation for iD becomes; The equation may be written as; where;  the dc component

MOSFET Basic FET Amplifiers The MOSFET Amplifier and;
 the ac or signal component From the above equation, we can write;  the transconductance relating the output current to the input voltage

The transconductance can also be written as; which gives us;

MOSFET Basic FET Amplifiers The MOSFET Amplifier AC Equivalent Circuit
In the ac equivalent circuit, all dc sources are set to zero and VDD is considered to be at signal ground.

Small-signal Equivalent Circuit If we neglect the effect of the output resistance ro, the MOSFET and its small-signal equivalent circuit are as follows (in terms of instantaneous ac values):

Small-signal Equivalent Circuit The small-signal equivalent circuit in terms of phasor components (note the notation used):

Small-signal Equivalent Circuit The small-signal equivalent circuit with ro taken into account.

Small-signal Equivalent Circuit

MOSFET Basic FET Amplifiers The MOSFET Amplifier Example 1
The MOSFET in the figure has the following parameters; The transistor is biased for IDQ = 0.4 mA. (a) Draw the ac equivalent circuit. (b) Draw the small-signal equivalent circuit. (c) Determine the small-signal voltage gain Av.

MOSFET Basic FET Amplifiers The MOSFET Amplifier Example 1 – Solution
For the ac equivalent circuit, all dc voltage sources are set to zero;

MOSFET Basic FET Amplifiers The MOSFET Amplifier Example 1 – Solution
Since  = 0, the output resistance ro is omitted in the small-signal equivalent circuit;

Example 1 – Solution (cont’d) (c)

Example 1 – Solution (cont’d) At the output terminal; At the input terminal; Substituting for Vgs; The small-signal voltage gain is;

Example 1 – Solution (cont’d) Substituting values;

MOSFET Basic Amplifier Configurations Common-source
R1 and R2 establish the transistor biasing. CC1 couples the signal to the input acts as an open circuit for dc voltage. At signal frequency, this capacitor is considered as a short circuit. The series combination vi and RSi represents the Thevenin’s equivalent of the signal source. The source is grounded and is common to both input and output – hence the name common-source

Since the coupling capacitor is considered as a short circuit at signal frequency and all dc sources are set to zero under ac condition, the ac equivalent circuit for the amplifier may drawn as follows;

The parallel combination of R1 and R2 may be replaced by Ri where; Ri is the input resistance of the amplifier.

The small-signal equivalent circuit may drawn as shown below; In the above circuit, the output resistance of the transistor, ro is finite and taken into account.

Since RD and the output resistance of the transistor, ro are in parallel, the small-signal equivalent circuit may be redrawn as shown below.

Using this equivalent circuit, at the output terminal; At the input terminal; Hence the voltage gain is;

Boundary between saturation and triode region For maximum symmetrical swing, the Q-point must be near the middle of the saturation region Triode region Saturation region

MOSFET Basic Amplifier Configurations Common-source Example 1
The transistor in the figure has the following parameters; Find the following; (a) gm and ro; (b) Av; (c) Ri and Ro.

Example 1 – Solution The circuit under dc condition (a)

Example 1 – Solution (cont’d) (b)

Example 1 – Solution (cont’d) (c)

MOSFET Basic Amplifier Configurations Common-source with RS
The source resistor RS tends to stabilize the Q-point against variations in transistor parameters. However, the inclusion of this resistor reduces the amplifier gain.

In this case, the body and substrate are not at the same potential. Therefore the body effect should be taken into account in the small-signal analysis, but in the following example this effect will be neglected.

MOSFET Basic Amplifier Configurations Common-source with RS Example 2
The transistor in the figure has the following parameters; (a) Determine the Q-point of the amplifier. (b) Find the small-signal voltage gain

Example 2 – Solution The circuit under dc condition The gate voltage (w.r.t. ground);

Example 2 – Solution (cont’d) The circuit under dc condition Substituting for ID and rearranging, we have; or;

Example 2 – Solution (cont’d) The circuit under dc condition Solving the quadratic equation, gives us; Substituting for VGS in; we obtain;

Example 2 – Solution (cont’d) The circuit under dc condition

Example 2 – Solution (cont’d) The Q-point: The transconductance is;

Example 2 – Solution (cont’d) The ac equivalent circuit may be drawn as follows;

Example 2 – Solution (cont’d) The small-signal equivalent circuit may be drawn as follows;

Example 2 – Solution (cont’d) The small-signal voltage gain is;

MOSFET Basic Amplifier Configurations Common-source with CS
The transistor is biased with a constant current source, IQ. The bypass capacitor, Cs minimizes the loss in small-signal voltage gain but maintains the stability of the Q-point. For small-signal analysis, Cs is considered as a short circuit.

MOSFET Basic Amplifier Configurations Common-source with CS Example 3
The parameters for the transistor in the figure are; (a) Determine the Q-point (c) Calculate the small-signal voltage gain Av.

Example 3 – Solution The circuit under dc condition Since IG is assumed to be zero, G is at the ground potential and IS = ID.

Example 3 – Solution (cont’d) The circuit under dc condition Substituting for ID; or; Solving the equation for VSG, we obtain;

Example 3 – Solution (cont’d) The circuit under dc condition

Example 3 – Solution (cont’d) The circuit under dc condition (a) The Q-point is;

Example 3 – Solution (cont’d) The ac equivalent circuit

Example 3 – Solution (cont’d) The small-signal equivalent circuit may drawn as follows; Note that ro is included in the equivalent circuit because the value of  is given.

Example 3 – Solution (cont’d) (b) The small-signal voltage gain is;

MOSFET Basic Amplifier Configurations Common-drain
Also known as source-follower The output is taken from the source. The drain is connected to VDD which is the signal ground and becomes the common terminal for the input and output, hence the name common-drain.

MOSFET Basic Amplifier Configurations Common-drain Q-point
Determined by means of DC analysis Voltage gain, Av; Input resistance, Ri; Output resistance, Ro; Determine by means of AC (small-signal) analysis

The small-signal analysis may be performed using the following small-signal equivalent circuit.

MOSFET Basic Amplifier Configurations Common-drain – Voltage gain Av
The small-signal equivalent circuit may be redrawn as follows;

MOSFET Basic Amplifier Configurations Common-drain – Voltage gain Av

At the input side; where

and Combining the two equations, gives us; Hence;

The small-signal voltage gain is;

The small-signal voltage gain also can be written as; Note that there is no phase shift

MOSFET Basic Amplifier Configurations
Common-drain – Input resistance Ri The input resistance of the common-drain amplifier is;

Common-drain – Output resistance Ro To calculate the output resistance, set all the independent voltage source to zero and apply a test voltage Vx at the output terminal as shown below;

Common-drain – Output resistance Ro Applying KVL at this node gives the equation

Common-drain – Output resistance Ro Since there is no current in the input side, we have; Substituting for Vgs, we have;

Common-drain – Output resistance Ro Or; The output resistance is;

Common-drain – Output resistance Ro Therefore;

MOSFET Basic Amplifier Configurations Common-drain Example 4
The transistor parameters in the figure are as follows; Design the circuit such that; Determine the small-signal voltage gain

Example 4 – Solution The dc equivalent circuit;

Example 4 – Solution (cont’d) The dc equivalent circuit;

Example 4 – Solution (cont’d) The dc equivalent circuit; If

Example 4 – Solution (cont’d) The dc equivalent circuit; If Unrealistic – negative VGS.

Example 4 – Solution (cont’d) The circuit with all the designed values;

Example 4 – Solution (cont’d) The transconductance is; The output resistance of the transistor;

Example 4 – Solution (cont’d) The small-signal voltage of the amplifier is;

MOSFET Basic Amplifier Configurations Common-drain Example 5
The transistor in the figure has the following parameters; Design the circuit such that IDQ = 3 mA. Determine the open-circuit small-signal voltage gain and output resistance What value of RL will result in a 10% reduction in gain?

Example 5 – Solution The circuit under dc condition Since ID = 3 mA; Because G is at ground potential, we have; The equation for the drain current is;

Example 5 – Solution (cont’d) Substituting values; Or; The above equation gives us two values; or

Example 5 – Solution (cont’d) For (unrealistic because VSG < |VTP| For (acceptable)

Example 5 – Solution (cont’d) The transconductance is;

Example 5 – Solution (cont’d) The ac equivalent circuit is as follows;

Example 5 – Solution (cont’d)

Example 5 – Solution (cont’d) (As calculated previously ) (Since RL =   open circuit)

Example 5 – Solution (cont’d) The open-circuit small-signal voltage gain is; The voltage gain after a reduction of 10% is;

Example 5 – Solution (cont’d) Substituting for Av in the equation; we have;

Example 5 – Solution (cont’d) Substituting the value for gm gives us; Solving the above equation, we obtaine; This equation may be written as;

Example 5 – Solution (cont’d) Or; Substituting values;

Example 5 – Solution (cont’d) Solving the equation for RL gives us;

Example 5 – Solution (cont’d) Using the equivalent circuit above, the output resistance Ro is given by the expression;

Example 5 – Solution (cont’d) The output resistance is;

MOSFET Basic Amplifier Configurations Common-gate
The input signal is applied to the source and gate terminal is grounded. The output is taken between the drain and gate terminals. The gate is therefore common to both the input and output and hence the name “common-gate”

CC1 and CC2 are the coupling capacitors for input and output respectively. In the following figure, the bias current is obtained from the current source IQ.

The dc analysis is the same as that of previous configurations.

Common-gate – small-signal voltage gain The gains can be determined using the following equivalent circuit;

Common-gate – small-signal voltage gain Applying KVL at the input, we obtain; Substituting for Ii, we obtain;

Common-gate – small-signal voltage gain Re-arranging, gives us;

Common-gate – small-signal voltage gain Substituting for Vgs in the expression for Vo, we obtain; The small-signal voltage gain is;

Common-gate – small-signal current gain To determine the current gain, the input voltage source is transform into a current source;

Common-gate – small-signal current gain At the input node; At the output;

Common-gate – small-signal current gain Rearranging the equation; gives us; Substituting for Vgs in the equation for Io, we have;

MOSFET Basic Amplifier Configurations The current gain is unity if
Common-gate – small-signal current gain Thus the current gain is; The current gain is unity if

MOSFET Basic Amplifier Configurations Common-gate – input resistance
Since; From the equivalent circuit; the input resistance becomes;

MOSFET Basic Amplifier Configurations Common-gate – output resistance
The output resistance may be found by setting Vi = 0

When Vi is set to zero; which only holds if

When Vgs = 0, the controlled current source gmVgs is open

The equivalent circuit becomes as shown below; Hence the output resistance is;

MOSFET Basic Amplifier Configurations Common-gate Example 6
The transistor parameters in the following figure are; (a) Draw the small-signal equivalent (b) Determine the small-signal voltage gain Av. (c) Find the input resistance Ri.

MOSFET Basic Amplifier Configurations Common-gate Example 6 – Solution
DC analysis May be performed using the dc equivalent circuit shown.

Example 6 – Solution (cont’d) From the dc equivalent circuit; Rearranging, we have;

Example 6 – Solution (cont’d) Substituting for ID; or; Solving the equation, we have;

Example 6 – Solution (cont’d) The transconductance is;

Example 6 – Solution (cont’d) The ac equivalent circuit is as follows;

Example 6 – Solution (cont’d) Replacing the transistor with its small-signal model, we have the following circuit;

Example 6 – Solution (cont’d)

Example 6 – Solution (cont’d)  The small-signal voltage gain is

Example 6 – Solution (cont’d) The input resistance is;

MOSFET Basic Amplifier Configurations Exercises
Common-source configuration Problems 4.19, 4.23 and 4.25 Common-drain configuration Problems 4.34, 4.39 and 4.41 Common-gate configuration Problems 4.44, 4.45 and 4.46

MOSFET Basic Amplifier Configurations Assignment
The common-source amplifier in Fig A1 was designed for a sinusoidal input voltage range vi of 0.1  0.2 V peak-to-peak. With this input voltage range, the amplifier is expected to produce an output voltage range of 1.3  2.6 V peak-to-peak (sinusoidal). When measured with an input voltage of 0.2 V peak-to-peak at 1 kHz, the output voltage was distorted as shown in Fig. A2. Explain the most probable cause for the distortion. In an attempt to rectify the problem, the values of some of the resistors in Fig. A1 were changed. After modification, the distortion was as shown in Fig. A3. Explain the most probable cause for the distortion.

MOSFET Basic Amplifier Configurations Assignment (cont’d) Fig. A1

MOSFET Basic FET Amplifiers The MOSFET Amplifier

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