INTRINSIC GAIN METHOD EX-1

FINDING GAIN OF DEGENERATED & CASCODE AMPLIFIER 

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In this post, we shall discuss two basic amplifiers. 

Q1) Find the gain.

Solution: 

From the signal flow, it is evident that the amplifier is in CS mode.

Let's directly write the gain expression. What is the Intrinsic Gain Method? 

$GAIN=V_{IN} \times \frac{R_{LOAD}}{R_{LOAD}+R_{OUT}}$

Now, ${R_{LOAD}=R_{D}}$ ; ${R_{OUT}=r_{0}+M \times R_{S}}$, where $M= 1+g_{m}r_{0}$

How did I write the above expression? Go check.

$GAIN= -g_{m}r_{0} \times \frac {R_{D}}{R_{D}+(r_{0}+M \times R_{S})} \approx -g_{m}r_{0}\times \frac {R_{D}}{M \times R_{S}} \approx -\frac{R_{D}}{R_{S}}$

Let's try to solve it by another method called, $G_{m}r_{out}$ method and match the results with the above result.

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Another Method: 

We know the equivalent transconductance of the source-degenerated CS is $G_{M}= - \frac {g_{m}}{1+g_{m}R_{S}}$

$R_{eq}=R_{LOAD} || R_{OUT} \approx R_{D}$

$GAIN=G_{M}R_{eq}=- \frac {g_{m}}{1+g_{m}R_{S}} \times R_{S}$

Considering, $g_{m}R_{S}>>1$

$GAIN \approx -\frac{R_D}{R_S}$, which matches with the previous result.

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Q2) Find the gain.

           

Solution: Let's draw the schematic for a small signal.

The above diagram shows that M1 and M2 are in CS and CG mode respectively. 

$GAIN= -g_{m1}r_{01} \times M_{2} \times \frac {R_{LOAD}}{R_{LOAD}+R_{OUT}}$

 where, $R_{LOAD}= r_{03}+M_3 \times r_{04} \approx M_3 \times r_{04}$

             $R_{OUT}= r_{02}+M_2 \times r_{01} \approx M_2 \times r_{01}$ 

Now,

The $GAIN= -g_{m1} \times \frac {M_{2} \times r_{01} \times R_{LOAD}}{R_{LOAD}+R_{OUT}}$ =$-g_{m1}(R_{LOAD}|| R_{OUT})$

Let's also try with  $G_{m}r_{out}$ method.

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Another Method: 

$\frac{I_{SC}}{V_{IN}}= G_{M}=-g_{m1} \frac {r_{01}}{\frac{1}{g_{m2}}+r_{01}} \approx -g_{m1}$

$GAIN=G_MR_{eq}=-g_m \left(R_{LOAD}||R_{OUT} \right)$, which matches with above result.

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We will solve more complex circuits in the upcoming posts.

Check previous posts of the series we started "INTRINSIC GAIN METHOD".

PART1: INTRINSIC GAIN METHOD

PART2: INTRINSIC GAIN METHOD PARAMETERS 

PART3: FINDING EQUIVALENT IMPEDANCE

For part 5 follow our blog and don't forget to comment

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MOSFET INTRINSIC GAIN

This post will be dedicated to find out the intrinsic gain for 3 different MOSFET amplifier Configurations.

First Consider the small signal model of MOSFET.

Here Body effect on incremental MOSFET model is neglected. 

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Common Source:

Here biasing is not shown and only MOSFET in common source is analyzed.

       $V_{out}=  -g_{m}V_{gs}r_{o}$

   => $V_{out}= -g_{m}V_{in}r_{o}$

 => $\frac{V_{out}}{V_{in}}= -g_{m}r_{o}$ 

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Common Drain:

$V_{out} = g_{m}V_{gs}r_{o}$

   

=>  $V_{out} = g_{m}(V_{in}-V_{out}r_{o})$

=>  $V_{o}(1+g_{m}r_{o}) = V_{in}g_{m}r_{o}$

=> $\frac{V_{out}}{V_{in}} = \frac{g_{m}r_{o}}{M}$ 

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Common Gate:

Vin Voltage source does not draw any current due to unavailability of path to ground. All $g_{m}V_{gs}$ current flows through $r_{o}$

$V_{gs}=V_{in}$

=>$V_{out}= V_{in}- g_{m}(-V_{in})r_{o}$

=>$\frac {V_{out}}{V_{in}}=(1+g_{m}r_{o})$

=>$\frac {V_{out}}{V_{in}} = M$ >

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Extra: If anywhere you want to consider body effect on intrinsic gain expression (Body grounded, where source has some incremental signal nature). 

just Put $M=1+(g_{m}+g_{mb})r_{o}$

This is 2nd post of the series "Finding MOSFET gain". In 3rd post we will learn the intuitive way to find impedances of any MOSFET circuit. From the 4th post onwards we will take different amplifier circuits and try find out their gains and impedances.

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Post 2

FINDING MOSFET CIRCUIT GAIN USING INTRINSIC GAIN METHOD

Here in this post, I'll help build your basics of the MOSFET incremental model and try to find some parameter values, which will be used extensively to find out the gain of any MOSFET circuit.

There are three method to find out the gain of the amplifier

• Conventional Method (using the small-signal method)
• Intrinsic gain method approach
• \(G_{m}, R_{out}\)  method

Here in this tutorial, We will only be discussing the second method in this post. For this you only need to remember three expressions along with a voltage divider equation and some basic methods of finding equivalent impedance (from inspection) at any node. 

That's all!

(There will be a separate tutorial dedicated to the last method.)

First, we shall define the intrinsic gain of different configurations (CS, CD, CG) and find their value once.

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Let's start-------

                   Configuration                     Intrinsic Gain                  

                 Common Source                    $-g_{m}r_{0}$

              

                 Common Drain                          $\frac{g_{m}r_{0}}{M}$

                  Common Gate                            $ M $

But what is $M$ in the expressions? 
Well, I call it Mirroring Factor. For the time being just remember its expression  $M= (1+g_{m}r_{0})$. ( It will be explained in the subsequent blog.)

Other than above 3 expressions you have to remember the Thevenin's equation.

$Actual \, Gain = Intrinsic \, Gain \times \frac{R_{L}}{R_{L}+R_{out}}$ 

$R_{L}$= Actual load seen by the circuit

$R_{out}$= Output impedance seen into the circuit >

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Without justifying the above equation, let's just jump into one of the simplest MOSFET amplifier circuit.

Observation: Common Source Configuration

So, $\frac{v_{out}}{v_{in}}= A_{Intrinsic}\times \frac{R_{L}}{R_{L}+R_{out}}$

            $ = -g_{m}r_{0} \times \frac{R_{D}}{R_{D}+r_{0}}$

            $ = -g_{m} \times \frac{r_{0}R_{D}}{r_{0}+R_{D}}$

            $ = -g_{m}(r_{0}||R_{D})$

This is a very well known result, doesn't seem surprising to us.

You will appreciate this, as we are going to solve quite a complex amplifier circuit. I promise, you will never have any fear while dealing with MOSFET Circuits ever again. 

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In the upcoming lectures, We will discuss more about the basics of intrinsic Gain. So get ready!

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