MOSFET - Differential Amplifier (Small Signal Analysis)

Summary

This video from the ALL ABOUT ELECTRONICS YouTube channel delves into the small signal analysis of a differential amplifier, focusing on determining its gain. The presenter explains that for linearity in the input-output relationship, the differential input should be small. Using small signal approximation, the video derives the differential gain formula as -gm * (Rd ∥ R0), considering the channel length modulation. An example illustrates the low gain of the amplifier and discusses limitations in increasing gain due to power consumption and supply voltage constraints. The video concludes with a teaser for the next episode, which will explore methods to enhance the differential amplifier's gain.

Takeaways

• 🔍 The video covers the small signal analysis of a differential amplifier, focusing on finding its gain.
• 📊 A previous video discussed large signal analysis, providing the operating range and relationship between differential input and output.
• 🎚️ For linear amplification, the differential input must be small to ensure a linear relationship between input and output.
• 🔧 The video explains the small signal approximation condition and how it simplifies the differential amplifier's analysis.
• 🔗 Small signal analysis treats DC sources and common mode inputs as zero, focusing on the differential components.
• 🛠️ The video replaces MOSFETs with their small signal models to analyze the circuit's behavior more easily.
• ⚖️ By using Kirchhoff's laws and assuming matched MOSFETs, the analysis is simplified, treating node x as a virtual ground.
• 📉 The differential gain of the amplifier is derived as -gm * Rd, showing the relationship between the output and input.
• 📏 Channel length modulation and its effect on the differential gain are considered, adjusting the gain formula to include output resistance.
• 📈 The video concludes by discussing ways to improve the gain, such as increasing the drain resistor or trans-conductance, while also acknowledging the limitations of these methods.

Q & A

What is the purpose of small signal analysis in the context of a differential amplifier?

The purpose of small signal analysis is to linearize the relationship between the differential input and output of a differential amplifier. This allows for easier calculation of the amplifier's gain when the differential input is very small.

Why is the input voltage Vin1 - Vin2 considered to be small in small signal analysis?

The input voltage Vin1 - Vin2 is considered small in small signal analysis because the relationship between the differential input and output is nonlinear for large inputs. To maintain a linear relationship and use the amplifier effectively, the differential input must be small.

How does the small signal approximation affect the current through the MOSFETs?

Under the small signal approximation, the differential current (Id1 - Id2) simplifies, allowing us to neglect higher-order terms. This approximation makes it easier to analyze the circuit and find the small signal currents that contribute to the output.

What is the role of the transconductance (gm) in determining the gain of the differential amplifier?

The transconductance (gm) is directly proportional to the differential gain of the amplifier. The differential gain is given by -gm * Rd, where gm represents the relationship between the gate-to-source voltage (Vgs) and the small signal drain current (Id).

Why can node X be treated as a virtual ground in the small signal analysis?

Node X can be treated as a virtual ground because, under small signal conditions, the voltage at this node does not change due to the symmetry of the circuit and the matching of the two MOSFETs. This simplifies the analysis of the circuit.

What happens when the effect of channel length modulation is included in the small signal model?

When the effect of channel length modulation is included, the output resistance of the MOSFET (R0) is introduced in parallel with Rd. This modifies the differential gain equation to -gm * (Rd ∥ R0), slightly reducing the gain.

What is the impact of increasing the drain resistor (Rd) on the differential gain?

Increasing the drain resistor (Rd) increases the differential gain of the amplifier. However, this also increases the voltage drop across the resistor, potentially pushing the MOSFET out of saturation, which may require a higher supply voltage.

How can the transconductance (gm) be increased to improve the differential gain?

The transconductance (gm) can be increased by increasing the bias current (Id) or the width-to-length ratio (W/L) of the MOSFET. However, increasing Id leads to higher power consumption, and increasing W/L results in a larger area occupied by the circuit.

What is the effect of increasing the bias current (Id) on the power consumption of the differential amplifier?

Increasing the bias current (Id) increases the transconductance (gm), which improves the gain. However, this also increases the power consumption of the amplifier and may require a higher supply voltage to maintain MOSFET saturation.

Why is the differential gain of a typical differential amplifier considered low?

The differential gain of a typical differential amplifier is considered low due to practical limitations in increasing the drain resistor (Rd), bias current (Id), and the width-to-length ratio (W/L). These factors restrict the achievable gain while maintaining reasonable power consumption and circuit area.

Outlines

00:00

📚 Introduction to Small Signal Analysis of Differential Amplifier

The paragraph introduces the topic of small signal analysis of a differential amplifier on the ALL ABOUT ELECTRONICS YouTube channel. It explains the purpose of the video, which is to analyze the gain of the differential amplifier using small signal analysis. The presenter reviews the large signal analysis from a previous video, where the operating range and the relationship between differential input and output were discussed. The importance of small signal approximation is emphasized, where the differential input should be very small to linearize the input-output relationship. The conditions for this approximation are also discussed, leading to a simplified equation that describes the linear relationship between the differential output and input. The video then sets the stage for further analysis by considering the differential amplifier's input components and the assumptions made during small signal analysis, such as treating DC sources as zero and replacing MOSFETs with their small signal models.

05:00

🔍 Deep Dive into Differential Amplifier's Small Signal Model

This paragraph delves deeper into the small signal model of the differential amplifier. It explains how to replace MOSFETs with their small signal models and how to simplify the analysis by assuming that the node between the gate and ground acts as a ground terminal for small differential inputs. The paragraph uses Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to prove that the node behaves as a ground under small signal conditions. The analysis then focuses on how the currents through the drain resistors relate to the small signal voltages, leading to expressions for the output voltages Vo1 and Vo2 in terms of the transconductance (gm) and the drain resistor (Rd). The paragraph concludes by deriving the expression for the differential gain of the amplifier, which is the ratio of the differential output voltage to the differential input voltage.

10:09

🔬 Impact of Channel Length Modulation on Differential Gain

The third paragraph discusses the impact of channel length modulation on the differential gain of the amplifier. It introduces the concept of the output resistance (R0) of the MOSFET and how it affects the differential gain when λ (the channel length modulation coefficient) is non-zero. The paragraph explains that the output resistance of the MOSFET is considered in the small signal equivalent circuit, and it modifies the expression for the differential gain to include the parallel combination of Rd and R0. The paragraph also provides an example to illustrate the calculation of the differential gain, including the determination of the output resistance and transconductance using given device parameters. It concludes by highlighting the limitations in increasing the gain of the differential amplifier due to practical constraints such as power consumption and supply voltage.

15:12

🚀 Strategies to Improve Differential Amplifier Gain

In the final paragraph, the presenter discusses strategies to improve the differential gain of the amplifier. It outlines the factors that limit the gain, such as the need to maintain the MOSFET in saturation and the constraints on increasing the drain resistor or transconductance. The paragraph explains that increasing the drain resistor or transconductance can improve the gain but is limited by the need to operate at low voltages and control power consumption. It also mentions the trade-off between increasing the W/L ratio to improve transconductance and the resulting increase in the amplifier's area. The paragraph concludes by previewing the next video, which will explore methods to increase the gain of the differential amplifier and design high gain differential amplifiers.

Keywords

💡Differential Amplifier

A differential amplifier is a type of electronic amplifier that amplifies the difference in voltage between two input signals while rejecting common-mode signals. In the video, the differential amplifier is the central topic, with the script detailing how to analyze its small signal behavior and calculate its gain.

💡Small Signal Analysis

Small signal analysis is a method used in electronics to study the behavior of circuits under the assumption that the input signal variations are small enough not to significantly affect the operating point of the circuit. The video uses this analysis to find the gain of the differential amplifier, considering the differential input to be very small.

💡Operating Range

The operating range refers to the set of conditions under which a circuit operates correctly. In the context of the video, the operating range of the differential amplifier is discussed in relation to the large signal analysis, which is a prerequisite for the small signal analysis.

💡Differential Input

The differential input is the difference in voltage between two input signals in a differential amplifier. The video emphasizes that for the differential amplifier to function as an amplifier, the differential input should be very small, which linearizes the relationship between input and output.

💡Transconductance (gm)

Transconductance, often denoted as 'gm', is a measure of an amplifier's ability to convert an input voltage into an output current. It's a key parameter in the analysis of the differential amplifier's gain, as shown in the video where it's used in the formula for calculating differential gain.

💡Channel Length Modulation

Channel length modulation is a phenomenon in MOSFETs where the effective channel length changes with the voltage, affecting the drain current. The video mentions that the analysis initially neglects this effect but later includes it to provide a more accurate model of the differential amplifier's gain.

💡Common Mode Rejection Ratio (CMRR)

While not explicitly mentioned in the script, the concept of CMRR is inherently tied to differential amplifiers. It's a measure of an amplifier's ability to reject common-mode signals while amplifying differential signals. The video's focus on differential gain indirectly relates to CMRR, as a high CMRR is desirable in differential amplifiers.

💡Early Voltage (Va)

The Early voltage is a parameter used to model the output resistance of a MOSFET. In the video, the Early voltage is used to calculate the output resistance of the MOSFETs in the differential amplifier, which is crucial for determining the amplifier's gain with finite channel length modulation.

💡Drain Current (Id)

Drain current is the current that flows from the drain to the source of a MOSFET. The video discusses how the drain current changes in response to small signal inputs and how these changes are related to the transconductance, which is essential for the small signal analysis of the differential amplifier.

💡Drain Resistor (Rd)

The drain resistor is a resistor connected to the drain terminal of a MOSFET in an amplifier circuit. In the video, the value of the drain resistor is used in the calculation of the differential gain of the amplifier, highlighting its importance in determining the overall gain of the circuit.

💡Output Resistance (Ro)

Output resistance is the resistance that a device exhibits looking into its output. In the context of the video, the output resistance of the MOSFETs is considered when including the effects of channel length modulation, affecting the overall differential gain of the amplifier.

Highlights

Introduction to small signal analysis of the differential amplifier.

Objective to find the gain of the differential amplifier through small signal analysis.

Review of large signal analysis and operating range of the differential input from a previous video.

Explanation of the nonlinear relationship between differential input and output and the need for a small signal approximation.

Condition for small signal approximation where differential input is much less than 2 times Vov.

Simplification of the equation for Id1 - Id2 under small signal approximation.

Derivation of the linear relationship between differential output and input under small signal conditions.

Fixed parameters μn Cox (W/L) for a given bias current Iss in the small signal approximation.

Definition of the differential output Vo1 - Vo2 and its relation to the differential input.

Assumption of zero common mode input and ideal current source in small signal analysis.

Replacement of MOSFETs with their small signal model for analysis.

Identification of small signal currents Id1 and Id2 and their relation to Vgs1 and Vgs2.

Assumption of node x acting as a ground terminal for small differential input and its proof.

Application of KVL and KCL to prove node x acts as a ground for small differential input.

Derivation of the differential gain of the differential amplifier as - gm . Rd.

Inclusion of the channel length modulation effect on the differential gain.

Example calculation of differential gain with a bias current of 0.8 mA and given device parameters.

Calculation of output resistance R0 using the early voltage and drain current.

Calculation of trans-conductance gm using the formula √2 . Id . μn.Cox.(W/L).

Discussion on the limitations of increasing drain resistor or trans-conductance to improve gain.

Conclusion on the factors restricting the gain of the differential amplifier.

Teaser for the next video on increasing the gain of the differential amplifier and designing high gain differential amplifiers.

Transcripts

Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS.

So in this video, we will see the small signal analysis of the differential amplifier.

And with the help of the small signal analysis, we will find the gain of this differential

amplifier.

So in the previous video, through a large signal analysis, we found the operating range

of this differential input,

and then, we also found the relationship between this differential input and the differential

output.

So, to get the exact relationship between the differential output and the differential

input, we just need to multiply this expression by the -Rd.

But anyway, as you can see this expression, here the relationship between this input and

output is nonlinear.

So, if we want to use this differential pair as an amplifier, then this input Vin1 - Vin 2

or this differential input should be very small.

So in that condition, the relationship between this input and output will become linear.

So in the previous video, we also found the condition for this small signal approximation.

That means whenever this differential input Vin1 - Vin2 is much less than 2 times

Vov, then under this small signal approximation, this term can be neglected.

And this Id1 - Id2 can be written like this.

Now if we further rearrange this term, and if we take this 4 outside, then it will become

2 and it will get cancelled with this 2.

So eventually, this is how the equation will look like.

And of course, this differential output Vo1 - Vo2 is equal to - Rd times this

Id1 - Id2.

That is equal to - Rd √ μn Cox (W/L) Iss Vin1-Vin2

So now if you see, then for the given differential pair, this μn Cox (W/L) is fixed.

That means for the given bias current Iss, this entire square root term will remain fixed.

And therefore, under this small signal approximation, the relationship between this differential

output and the differential input is now linear.

So under this small signal approximation, whenever this differential input is very very

small, then we can use this small signal analysis and with the help of it, we can find the gain

of this differential amplifier.

So now as you know, the input to this differential amplifier consists of two components.

That is the common mode component and the differential component.

So for the small signal analysis, all the DC sources and the common mode input will

be considered as zero.

That means here, this common mode input Vcm will act as zero, while this ideal current

source will also act as an open circuit.

Moreover, this voltage source Vdd will act as a short circuit and it will get connected

to the ground terminal.

And now for the small signal analysis, we need to replace these MOSFETs by their small

signal model.

So after the replacement, if we see the small signal equivalent circuit, then it will look

like this.

So here, this differential input is applied between the gate and the ground terminal,

while the drain resistor RD will appear between the drain and the ground terminal.

Now here, these currents Id1 and Id2 are the small signal currents, or in other words,

they are the change in the drain currents due to the differential inputs.

So this change in the drain current Id1 can be given as this gm1 . Vgs1.

Similarly, this current Id2 can be given as gm2 . Vgs2.

And here, this Vgs1 and Vgs2 are the change in the gate to source voltage due to these

differential inputs.

So now, let's say this node is equal to x and the voltage at this node is equal to Vx.

Now if these two MOSFETs are perfectly matched, then for a small signal differential input,

the voltage Vx will not change and this node x will act as a ground terminal.

And if that is the case, then the analysis of this circuit will become very easy.

So first of all, let's prove that this node indeed acts as a ground for this small differential

input.

And for that, let's apply the KVL on this left-hand side.

So if we apply the KVL, then we can write this voltage Vid1 - Vgs1 = Vx.

Similarly, if we apply the KVL on this right-hand side, then we can write this voltage Vid2 -Vgs2=Vx.

So if we equate these two expressions, then we can say that this Vid1 - Vgs1 = Vid2 - Vgs2.

Or we can say that this Vid1 - Vid2 = Vgs1 - Vgs2.

And let's say, this is the equation number 1.

So now, let's apply the KCL at this node.

So if you see, then at this node, this drain current Id1 and Id2 are the incoming drain

currents.

So we can say that this Id1 + Id2 = 0.

Now this Id1 = gm1 . Vgs1.

Similarly, this Id2 = gm2 . Vgs2.

So we can say that this gm1 . Vgs1 + gm2 . Vgs2 = 0.

Now if these two MOSFETs are perfectly matched, then this gm1 should be equal to gm2, right?

Let's say that is equal to gm.

So in that case, we can write this expression as gm .Vgs1 + Vgs2 = 0.

Now since the trans-connectance of the MOSFET cannot be 0, so we can say that this Vgs1

+ Vgs2 has to be 0.

That means this Vgs1 = - Vgs2.

Let's say this is the equation number 2.

Now here, since these differential inputs are complementary, so we can say that here

this Vid1 = - Vid2.

Let's say this is the equation number 3.

So let's use these two results and let's put the value of these two results in the equation

number 1.

So we have seen that as per the equation number 1, this Vid1 - Vid2 = Vgs1 - Vgs2.

So here, as per these two results, we can say that this Vid1 - (- Vid1) =

Vgs1 - (- Vgs1) = 2 . Vid1 = 2 . Vgs1.

or we can say that this Vid1 = Vgs1.

So earlier we have seen that this Vx = Vid1 - Vgs1.

Now whenever this Vid1 = Vgs1, so we can say that this Vx = 0.

So from this, we can say that for a small differential input, this node x will act as

a ground or equivalently, it can also be drawn like this.

So now, because this node acts as a virtual ground, so we can treat the two parts of these

circuits individually.

And hence, the analysis of this circuit becomes very easy.

So here, the current through this resistor Rd = Id1 or we can say that this

output voltage Vo1 = - Id1 . Rd = - gm1 . Vgs1 . Rd.

Now as we have just seen, this Vid1 = Vgs1.

So from this, we can say that this Vo1 = - gm1 . Rd . Vid1.

Similarly, for the second part, we can write this Vo2 = - gm2 . Rd . Vid2.

And here, we are assuming that both MOSFETs are identical.

That means here, this gm1 = gm2 and let's say that is equal to Gm.

So we can say that this Vo1 = - gm . Rd . Vid1, while this

Vo2 = - gm . Rd . Vid2.

That means Vo1 - Vo2 = - gm . Rd . (Vid1 - Vid2) or

we can say that this Vo1 - Vo2 / Vid1 - Vid2 = - gm .Rd.

So here, this is the ratio of the differential output to the differential input or in other

words, it represents the differential gain of this differential amplifier.

So we can say that this term is nothing but the differential gain.

That means the differential gain of this differential amplifier = - gm . Rd.

So in this way, using the small signal analysis, we found the gain of this differential amplifier.

Now so far in our discussion, we have neglected the effect of the channel length modulation.

That means we have assumed that this λ = 0.

But when this λ is non-zero or with the finite channel length modulation coefficient, the output

resistance of the MOSFET will also come into the picture.

So in such case, if we see this small signal equivalent circuit, then it will look like this.

Now here, due to the symmetry, this output resistance R0 1 should be equal to R0 2.

Let's say that is equal to R0.

So in that case, this R0 will be in parallel with the Rd.

And therefore, in that case, this differential gain will be equal to - gm Rd ∥ R0.

And of course, here due to the symmetry, we have assumed that this gm1 = gm2.

That means including the effect of the channel length modulation, this will be the expression

of the differential gain.

Alright, so now so far in our discussion, we have assumed that both MOSFETs are identical

and this current source is also ideal.

But even with these ideal assumptions, the gain of this differential amplifier used to

be very low.

So let's take one example, and through that example, let's understand why the gain of

this differential amplifier is very low.

So in this example, this differential amplifier is biased with 0.8 mA current, and here the

two MOSFETs are identical.

And these are the device parameters of two MOSFETs.

So here, with the help of these parameters, we will find the differential gain of this

amplifier.

Now here, since we have been given this early voltage, it means that these MOSFETs have

the finite output resistance.

And as we have seen, with the finite output resistance, this differential gain of this

amplifier can be given as - gm . Rd ∥ R0.

So here, the value of Rd is already known.

That means the only thing that we need to find is the value of this gm and the R0.

So first of all, let's find the value of this output resistance.

So as you know, in general, this output resistance of the MOSFET can be given as this Va /Id,

where the Va is the early voltage and the Id is the drain current.

So here, this Id is the bias current of these MOSFETs.

So for this MOSFET M1, this R01 = Va / Id1, while similarly,

for this second MOSFET, this R02 = Va / Id2.

So here, this Id1 and Id2 are the drain currents of these two MOSFETs under the equilibrium

condition.

So as we have seen in the earlier videos, under the equilibrium condition, when the

differential input is zero and only common mode input is present, at that time, due to

the symmetry of the circuit, this 0.8 mA current will get equally divided between the two MOSFETs.

That means here, this Id1 = Id2 = 0.4 mA.

And here, we already know the value of this early voltage.

And therefore, this R01 = R02 = 20V / 0.4 mA, which is

equal to 50 kΩ.

That means here, the output resistance of these two MOSFETs is equal to 50 kΩ.

And let's call it as R0.

So similarly, now let's find the trans-conductance.

So as you know, this trans-conductance gm = √2.Id.μn.Cox.(W/L).

So here, this gm1= gm2.

So in general, let's call it as gm.

Now here, this Id1 and the Id2 are 0.4 mA.

Moreover, we have been also given the values of this μn . Cox as well as the W / L.

So here, this W / L is 20, while this μn . Cox = 0.2 mA / V².

So with the help of it, we can find the value of this trans-conductance.

That means here, this gm = √ 2 . 0.4 mA .0.2 mA per V².20

That is equal to 1.788 mA / V.

So in this way, we got the value of this trans-conductance as well as the output resistance.

So we know that this differential gain can be given as - gm . Rd ∥ R0.

That is equal to minus 1.788 . 5 kΩ ∥ 50 kΩ.

Now this parallel combination is equal to 4.55 kΩ.

That means this differential gain is equal to minus 1.788 times this 4.55 kΩ.

That is equal to - 8.13.

So here, this negative sign indicates that this differential output is 180 degree phase

apart from the differential input.

But if we just consider the magnitude, then roughly we can say that this differential

gain is equal to 8.

So as you can see, for this differential amplifier, this differential gain is very low.

So now let's see how we can improve this differential gain.

So as per the equation, this differential gain can be given as this gm . Rd ∥ R0.

So if we want to improve the differential gain, then either we need to increase the

value of this drain resistor or we need to improve the trans-conductance.

So if we increase the value of this drain resistor, then for the given bias current,

the drop across this resistor will also increase.

That means now, the voltage which is available at the drain terminal will also reduce.

So to keep the MOSFET in the saturation, we need to increase this supply voltage.

And nowadays, since the integrated circuits are operating at low voltages, so it is not

always possible to increase the supply voltage.

That means it is not possible to increase the value of this drain resistor beyond the

certain limit.

Now the second way is, if we increase the value of this trans-conductance, then also

we can improve the gain.

So as we have seen, this trans-conductance gm= √2 . Id . μn. Cox. (W/L).

So as per this equation, we can increase the trans-conductance either by increasing this

drain current or by increasing the W / L ratio.

So to increase the drain current, we need to increase this bias current.

But as we increase the bias current, then the power consumption of the circuit will

also increase.

Moreover, with the increase in the bias current, the drop across this drain resistor will also

increase.

And hence, to keep the MOSFETs in the saturation, we also need to increase the supply voltage.

So overall, by increasing the bias current, the power consumption of the circuit will

increase.

Therefore, to keep the power consumption in the certain limit, the bias current of the

circuit cannot be increased beyond the certain limit.

Then the second way is, by increasing the W / L ratio, we can also increase the trans-conductance.

And in a way, we can also increase the differential gain.

But as we increase the W / L ratio, then the overall area which is occupied by the

amplifier will also increase.

So because of all these reasons, the trans-conductance of the MOSFET cannot be increased beyond the

certain limit.

So all these factors restrict the gain of this differential amplifier.

So in the next video, we will see that how we can increase the gain of this differential

amplifier.

And we will also see how to design the high gain differential amplifiers.

But I hope in this video, you understood how to find the gain of this differential amplifier

using the small signal analysis.

So if you have any question or suggestion, then do let me know here in the comment section

below.

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