Today’s topic:
1. Basic current mirrors
2. single-stage amplifiers
3. differential amplifiers
Chapter 3
The purpose of this Chapter is to discuss fundamental circuit building blocks.
A good knowledge of these basic building blocks is critical to understanding many
subjects to be discussed later.
CMOS current mirrors and gain stages are emphasized here, while the general smallsignal analysis method can be applied to BJT versions.
When analyzing transistor circuits to determine their small-signal behavior, it is
implicitly assumed that signals are small enough that linear approximations about an
operating point (Q point) accurately reflect how the circuit operates. These linear
approximations may be represented schematically by replacing transistors with
small-signal equivalent circuit, whose parameters (such as gm, rds etc) are
determined from the operating point.
A note on notations in the book and the lecture:
A general procedure for small-signal analysis:
A general procedure for small-signal analysis:
3.1 A simple current mirror
An ideal current mirror is a two port circuit that accepts an input current Iin and
produce an output current Iout=Iin.
Also, an ideal current mirror will have zero input resistance but high output resistance.
Both transistors are in active region (so the drain voltage of Q2 must be larger than Veff2).
If Q1 and Q2 have the same size, and finite rds is ignored, then Q1 and Q2 have the same
current (due to the same Vgs)
However, when rds are considered, the transistor with larger Vds will have larger current.
Chapter 3 Figure 01
A simple current mirror: input resistance
We can use the following small-signal equivalent circuit to compute the input
resistance of the current mirror. (Iin is open and low-frequency model is used).
Then, apply a test signal voltage Vy at node V1 and measure the current iy
Input resistance is given by rin = Vy/iy =
Q1 is sometimes referred to as a diode-connected transistor.
Chapter 3 Figure 02
A simple current mirror: output resistance
Using the model for the input resistance from previous slides leads to a simplified
small-signal equivalent circuit for the overall current mirror. .
Output resistance is given by rout=Vx/ix =rds2
Chapter 3 Figure 03
rout = rds2
Chapter 3 Figure 01
In this configuration, as long as gate terminal is
small-signal ground, the resistance looking into the
drain is rds2
A simple current mirror: output resistance
Using the model for the input resistance from previous slides leads to a simplified
small-signal equivalent circuit for the overall current mirror. .
Output resistance is given by rout=Vx/ix =rds2
Chapter 3 Figure 03
An example 3.1 (page 119)
Estimation using low-frequency small-signal
linear equivalent circuit
3.2 A common-source amplifier
A common use of simple current mirrors is to be active loads in a single-stage amplifier.
By using an active load, a high-impedance output load can be realized without using
excessively large resistors or a large power supply voltage (for example, a 100k resistor load
with 100μA bias current would need a power supply of at least 10V).
The common-source (CS) topology is the most popular gain stage, especially when high
input impedance is desired.
Chapter 3 Figure 04
Small-signal analysis of CS: DC gain Av
Assume that all transistors are in active region, the small-signal model of the CS amplifier
with active load is shown below:
(note the resistance R2 is made up of parallel of rds1 and rds2.)
Depending on device sizes, currents and technology used, the typical gain for a CS
amplifier could be between -5 to -100.
Av is about one half of the intrinsic gain of Q1:
To maximize the gain, it is desirable to minimize Veff1. For a fixed bias current, this is to
increase W?
Chapter 3 Figure 05
Small-signal analysis of CS: rin and rout
Chapter 3 Figure 04
Chapter 3 Figure 05
An example 3.2 and 3.3 (page 121)
3.3 A common-drain amplifier
Another general use of current mirrors is to supply the bias current for a common-drain or
source follower (SF) amplifier.
SF amplifiers are commonly used as voltage buffers.
Chapter 3 Figure 06
Small-signal analysis of SF amplifier: gain Av
In the small-signal model, note that the voltage-controlled current source that models the
body effect of MOS transistors is included as there is finite source-body voltage (unlike CS
amplifier), which may become a major limitation on the small-signal gain.
Note that the voltage-controlled current source modeling the body effect produces a
current that is proportional to the voltage across it, which makes it equivalent to a resistor
of 1/gs1. This allows to simplify the circuit to the right-hand side.
Chapter 3 Figure 07
Chapter 3 Figure 08
Small-signal analysis of SF: rin and rout
1/gm1 || rds1 || 1/gs1
Chapter 3 Figure 06
Chapter 3 Figure 07
Transform to a resistor
when calculating rout
Chapter 3 Figure 08
Example 3.4 (page 123)
Hence, it can be seen that body-effect is the major source of error causing the gain of the
SF amplifier to be less than 1.0.
3.4 A common-gate amplifier
A common-gate amplifier is used as a gain stage when small input impedance is desired.
Also, it is often used when the input signal is a current as small input impedance is desired.
Aside from its low input impedance, the common-gate amplifier is similar to a CS amplifier
as the input signal is across Gate-Source terminal and output taken from the Drain terminal.
Hence, in both amplifiers, the small signal gain equals the product of gm and total
impedance at the drain.
Chapter 3 Figure 09
Chapter 3 Figure 10
Chapter 3 Figure 11
Chapter 3 Figure 11
Comparison of the impedances
Chapter 3 Figure 11
Chapter 3 Figure 08
Compare to Slide 16
If RL=rds2=rds1, rin=2/gm1 for low frequency. If RL is even larger, then rin is more than 2/gm1. 21
Small-signal analysis of CG: rout
Chapter 3 Figure 11
Ix = Vx/RL – (gm1+gs1)Vs1 + (Vx-Vs1)/rds1
-(gm1+gs1)Vs1 + (Vx-Vs1)/rds1 = Vs1/Rs
rout 
[1  ( g m  g s1 ) RS 
]rds1 RL
RL  rds1
If RL=rds2=rds1 and Rs=0, then not surprisingly rout=rds1/2
Example 3.5 (page 127)
3.5 Source-degenerated current mirrors
Simple current mirrors presented before have relatively small output impedance.
To increase this output impedance, a source degenerated current mirror can be used.
Chapter 3 Figure 13
Chapter 3 Figure 12
Small-signal model to find rout
Note that in the small-signal model to compute rout, (1) the gate voltage for both Q1 and Q2
is 0 as no small-signal current flows to the gate, (2) we neglect the body effect.
gm1Vgs1 = Vgs1/rds1
Compare to Slides 22
If body effect included
Chapter 3 Figure 13
Example 3.6 (page 128)
Therefore, compared to the simple two-transistor current mirror, the output impedance of
source-degenerated current mirror has output impedance increased by a factor equal to
Note also that such a formula can often be applied to moderately complicated circuits to
quickly estimate the impedance looking into a node.
3.6 Cascode current mirrors
Cascode current mirrors can be used to further increase the output impedance.
Note that rout’ looking into the drain of Q2 is simply rds2.
The rout looking into the drain of Q4, can be derived from the formula for sourcedegenerated current mirrors, by considering Q4 as a current source with a sourcedegenerated resistor of rds2 from Q2.
Thus, the output impedance is increased by a factor of
gm4rds4, which is the gain of a single transistor. Such a large
increase can be important to realize single-stage amplifiers
with large gains.
The drawback in using a cascode current mirror is the
reduction of the maximum output voltage swing.
Chapter 3 Figure 14
Both are small-signal
ground voltage 0
Chapter 3 Figure 14
Minimum output voltage
What is the minimum voltage at Vout to maintain Q2 and Q4 in active region? If we assume
all transistors have the same sizes and currents, and therefore the same Vgsi=Veffi+Vtni,
where i=1,2,3,4, then
So, the Vds2 for Q2 is larger than the minimum needed
which is Veff2.
Since the smallest output voltage VD4 can be VDS2+Veff2
before Q4 goes into triode region, so the minimum
allowed voltage for Vout is
Chapter 3 Figure 14
The loss of signal swing is a serious drawback when
modern analog IC are used with low power supply of
1V. (Later, we will see how to address this issue).
Example 3.7 (page 130)
So the minimum Vout is then 2Veff+ Vtn=0.98V,
which 0.77V larger than the simple current
mirror while rout is increased by 30 times.
Chapter 3 Figure 14
3.7 Cascode gain stage
In modern IC design, a commonly used configuration for a single-stage amplifier is a
cascode amplifier. This configuration consists of a CS transistor feeding into a CG one.
In (a) below, both CS Q1 and cascode transistor Q2 are NMOS (telescopic-cascode
In (b), CS Q1 is NMOS and cascode transistor Q2 is PMOS (folded-cascode amplifier).
Two major reasons for cascode stages: (1) they provide large gain when current sources
are realized with cascode current mirrors; (2) they limit the voltage across Q1, minimizing
short channel effects.
The main drawback of cascode stages is that the output voltage swing is reduced in order
to keep both Q1 and Q2 in active region when compared to the CS amplifier.
Chapter 3 Figure 15
Small-signal model
The impedance looking into source of Q2
The impedance looking into drain of Q2
Chapter 3 Figure 16
Example 3.9 (page 133)
Chapter 3 Figure 16
The gain in this case is only a factor of 2 larger than that for a
common-source amplifier.
Also, almost all of the gain is across Q2.
Example 3.10 revised (page 135)
Chapter 3 Figure 16
The gain and output resistance is dramatically improved
compared to the previous example when a cascode current
mirror is used for Ibias.
3.8 Differential pair and gain stage
A differential pair have identically-sized and –biased transistors Q1 and Q2.
It is usually used as the input stage of a Operational Amplifier.
Small-signal model using T model
To simplify the analysis, we ignore the output impedance of the transistors temporarily.
Defining the differential input voltage as
Chapter 3 Figure 18
Differential pair with resistive load
Note that rds is neglected.
Chapter 3 Figure 17
Differential pair with current mirror
As with the CS amplifier, we can replace the resistor by current mirrors as an
active load. Then, a complete differential-input, single-ended output gain
stage can be realized. This circuit is typically the first gain stage in a classical
two-stage integrated OpAmp to be discussed later.
From small-signal analysis of the differential pair,
How to determine the small-signal output
resistance rout?
Chapter 3 Figure 19
Computing rout
T model was used for Q1, Q2 and diode-connected Q3 was replaced by an
equivalent resistance and hybrid-pi model for Q4.
Then assuming the effect of rds1 can be ignored (since it is much larger than rs1), then
Chapter 3 Figure 20
Computing rout
Chapter 3 Figure 20
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