enee313 homework #11 solutions - umd
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ENEE313 Homework #11 Solutions
1. (11.1) Qualitative MOSFET questions
(a) The cross section for an N-MOSFET under zero gate voltage is shown below:
In the above case, all terminal voltages are zero (grounded).
In the next graph, when the gate voltage is large enough to establish a channel, electrons
are pulled from the source into the channel.
(b) Inversion layer: The inversion layer in an N-MOSFET is a conducting channel which contains
electrons that are attracted by a positive gate-source voltage. This gate-source voltage
forward biases the source (n) β substrate (p) junction, causing electrons diffusing across the
junction, and the associated E field keeps them near the surface. The surface of the P-type
substrate is βinvertedβ to contain mobile electrons. These channel electrons will flow to the
drain under a positive drain-source voltage. No gate current is required to maintain the
inversion layer since the gate oxide (insulator) blocks any carrier flow. Thus, the gate
voltage controls the formation of an inversion layer, which in turn forms the flow of
electrons from the source to the drain once an additional voltage is applied at the drain
terminal. In other words, the MOSFET is a voltage-controlled current source, and the
inversion layer is the key concept connecting the input (gate control voltage) and response
(drain output current).
Carrier types: Mobile carriers in N-channel MOSFETs are electrons. Mobile carriers in P-
channel MOSFETs are holes.
(c) The higher the gate voltage is applied, the more change in the substrate (body) surface
potential, and the deeper the inversion gets into. This results in higher electron
concentration in the inversion layer (channel). Since the channel current is mainly drift
current under the drain-source field, it depends on the carrier concentration by:
π½πβπππππ = π½πππππ‘ β ππβπππππ
The higher the electron concentration, the higher the channel current.
2. (11.2) Qualitative description of MOSFET operation
The MOSFET is usually considered to be a three-terminal device. The three terminals are the
gate, source, and drain. (The MOSFET also has a fourth terminal called the body, but the
body is usually shorted to the source or grounded, it thus plays a secondary role, and we
wonβt worry about it too much in this context.)
One of the key features of the MOSFET is the gate oxide. This is a very thin insulating layer
that separates the gate terminal from the rest of the device. Because of this oxide, when we
apply a DC voltage to the gate, no gate current will flow. Despite many nuances, the general
operation goes as follows.
A voltage applied to the gate terminal and the electric field associated with it control the
current that flows between the drain and the source. This is where the MOSFET gets its
name as a βfield-effect transistorβ. For an N-channel device, we apply a positive voltage
from the gate to the source. The resulting electric field penetrates the oxide and then
forward biases the source-substrate n-p junction near the top of the device close to the
surface (between the oxide and the silicon). Because of the insulating oxide layer, there is
virtually no gate current. This is in contrast to a BJT where there is a DC base current and a
very specific ratio between the base (input) and collector (output) currents.
The electrons that come out of the source then form the conducting channel along the
surface. A positive voltage on the drain terminal reverse biases the drain-substrate n-p
junction near the surface, and the drain field parallel to the surface pulls channel electrons
into the drain, which give rise to the drain current. The drain current is proportional to the
concentration of electrons in the channel, which is controlled mainly by VGS.
3. (11.3) Reason for MOSFET being a field effect device
A MOSFET has a very thin layer of silicon dioxide (SiO2) or other types of dielectric materials
which separates the gate terminal and the rest of the device. Because such oxide layer is
insulating (SiO2 is an insulator), during normal device operation, there is no DC current
flowing through the gate terminal (the input terminal) under DC gate voltages. Thus, the DC
input resistance of a MOSFET is infinity.
In an N-MOSFET, a conducting channel forms when an applied gate voltage and its
associated E field pulls electrons from the source into the substrate near the surface.
Meanwhile, there is no input current at the gate terminal. Therefore, the MOSFET is
completely controlled by the electric field, and the device is a field-effect device. Another
applied voltage at the drain and its associated E field attracts the channel electrons so they
flow into the drain, forming conduction current. Similar things happen with holes in P-
MOSFETs.
On the other hand, in a BJT, the conduction current is formed when the emitter-base
junction is forward biased. Since there is a forward current from the base to the emitter,
the device is not completely controlled by field.
4. (11.4) Threshold voltage
The threshold voltage is the required gate-source voltage ππΊπ to be applied to:
1. Induce surface inversion, and
2. Form a channel at the surface of the substrate.
For an N-MOSFET, in the p-type substrate, there are π0 = ππ΄ holes and π0 =ππ2
ππ΄ electrons
per unit volume. Also, π0 βͺ π0. We know that the substrate potential is:
ππ = βππ ln (π0
ππ) = βππ ln (
ππ΄
ππ)
Which equivalently gives:
π0 = ππ΄ = πππβππ
ππ
When the surface potential is changed to ππ = βππ by applying a gate voltage equal to the
threshold voltage, the electron concentrations is:
π|ππ = πππ
ππ ππ = πππ
βππ
ππ = π0
This shows that the electron concentration at the surface is equal to the deep-substrate
hole concentration where the effect from the gate field diminishes. So, the type of the
material has been inverted.
5. (11.5) MOSFET calculations
Given quantities:
ππ΄ = 1017 cmβ3
ππ·βππππ¦ = 1019 cmβ3
π‘ππ₯ = 10 nm = 10 Γ 10β7 cm = 10β6 cm
πΏ = 0.5 ΞΌm = 5 Γ 10β5 cm
π = 20 ΞΌm = 2 Γ 10β3 cm
ππ = 500 cm2/V s
ππ = 0.0259 V
(a) Threshold voltage:
π0 = ππ ln (ππ΄ππ·βππππ¦
ππ2 )
π0 = 0.954 π
ππ = βππ ln (ππ΄
ππ)
ππ = β0.418 π
πΆππ₯ =πππ₯
π‘ππ₯=
3.9 Γ 8.85 Γ 10β14 F/cm
10β6 cm
πΆππ₯ = 3.45 Γ 10β7 F/cm2
So, the threshold voltage is:
πππ» =1
πΆππ₯[4ππππππ΄|ππ|]
12 + 2|ππ| β π0
πππ» = 0.363 π
(b) For the linear region, i.e., when ππ·π β€ ππΊπ β πππ», the drain current is:
πΌπ· = πππΆππ₯
π
πΏ[(ππΊπ β πππ»)ππ·π β
1
2ππ·π
2 )]
For the saturation region, i.e., when ππ·π β₯ ππΊπ β πππ», the drain current is:
πΌπ· =1
2πππΆππ₯
π
πΏ(ππΊπ β πππ»)2
The πΌπ·-ππ·π graph is shown below:
(c) The depletion region width π₯π under the gate is given by:
π₯π = [2πππ
πππ΄(ππ β ππ)]
1/2
When the surface potential ππ = 0 V, the equation becomes:
π₯π = [2πππ
πππ΄(βππ)]
12
The substrate potential ππ = βππ ln (ππ΄
ππ) = β0.4175 π is calculated in (a), so we get:
π₯π = 7.35 Γ 10β6 cm = 0.0735 ΞΌm
(d) When the surface potential ππ = 0 V, using the value for π₯π from (c) above, the electric
field in the oxide is:
πΈππ₯ =πππ΄π₯π
πππ₯
πΈππ₯ = 3.41 Γ 105 V/cm
6. (11.6) Deriving the MOSFET linear region current formula
When the drain-source voltage ππ·π is low, the drain terminal current is mainly drift current.
We start with the drain current density formula (Equation 11.3):
π½π·π,πππππ‘ = πππππβπΈπβ
To get the terminal current in Amperes, multiply the current density per unit area (A/cm2)
by the cross-section area of the conducting channel:
πΌπ· = π΄πβ π½π·π,πππππ‘
The cross sectional area is π΄πβ = ππ‘πβ where π is the channel width (perpendicular to
what we drew for the MOSFET βcross sectionβ in Question 1; the βcross sectional areaβ in
this context refers to a different plane.) and π‘πβ is the effective thickness of the channel
layer below the surface. Putting things together, we have:
πΌπ· = π΄πβ π½ = π΄πβπππππβπΈπβ = ππ‘πβπππππβπΈπβ
πΌπ· = π(πππβπ‘πβ)(πππΈπβ)
There are two important components in the above formula:
1. The channel charge density ππβ = πππβπ‘πβ represents the amount of mobile electrons
per unit surface area capable of forming drain current. Its unit is Coulomb/cm2. It is
mainly controlled by the gate voltage, which gives rise to surface inversion and pulls
electrons across the source-substrate junction, which form the channel.
2. The other term πππΈπβ represents the drift velocity, which is apparently controlled by the
channel field πΈπβ, which is parallel to the channel and the surface.
We continue and derive the low-field drain current formula (Equation 11.8) as follows.
First, treating the MOSFET like a capacitor, the charge stored in it is π = πΆπ. But we want
the result in Coulomb/cm2, so use πΆππ₯ =πππ₯
π‘ππ₯ in F/cm2. Also, we need to only count the
conducting (mobile) electrons in the channel and exclude the background charges due to
ionized acceptors; the threshold voltage will do the job. Thus,
ππβ = πππβπ‘πβ = πΆππ₯(ππΊπ β πππ»)
Next, find the channel field πΈπβ. From the definition of field and potential, πΈπβ = βππ
ππ₯.
Neglecting the minus sign (merely indicating direction), under a low drain voltage ππ·π, it is
approximated by:
πΈπβ βππ·π
πΏ
πΏ is the channel length.
Finally,
πΌπ· = π(πππβπ‘πβ)(πππΈπβ) = πππβ (ππ
ππ·π
πΏ) = π[πΆππ₯(ππΊπ β πππ»)] (ππ
ππ·π
πΏ)
πΌπ· = πππΆππ₯
π
πΏ(ππΊπ β πππ»)ππ·π
7. (11.7) MOSFET linear region and saturation region
In the case of an N-MOSFET, when a positive gate-source voltage is applied that exceeds the
threshold voltage, surface inversion occurs. Electrons from the source flow into the p-type
body region, forming a channel. In this situation, a positive drain-source voltage and its
associated E-field parallel to the channel pulls mobile electrons to the drain, forming drain
current. It has two operation regions as the following:
When the drain voltage is low ππ·π β€ ππΊπ β πππ», the drain current πΌπ· is linearly dependent
on the drain bias as shown in Question 6. Hence, we call it the linear region of operation.
For a relatively large ππ·π (still below ππΊπ β πππ»), the I-V characteristic curves transition
away from the βlinearβ shape as the MOSFET approaches saturation, but we still call this
region of operation the linear region. Increasing ππ·π is associated with linearly increasing
electric field (in the channelβs direction), and the pull of the channel electrons linearly
increases, until we enter the saturation region.
When the drain voltage is high ππ·π β₯ ππΊπ β πππ», surface inversion is barely maintained near
the drain, and the channel is considered to be depleted. We call the channel as βpinched
offβ under this condition, and the increase in the drain voltage and drain-source field does
not increase the conduction current any more. Thus, we call the channel current as
βsaturatedβ, and the MOSFET is operating in the saturation region. The depletion region
around the drain-body junction is reverse biased, and it is much larger than in the linear
operating region. The extra drain-source voltage in [ππ·π β (ππΊπ β πππ»)] is dropped across
the drain-body depletion region.
8. (11.8) MOSFET small signal model parameters
The two small signal parameters asked about, output resistance ππ and transconductance
ππ are defined by the small-signal change in the drain current β terminal voltage
relationship (Equation 11.67):
ΞπΌπ· =ππΌπ·
πππΊπΞππΊπ +
ππΌπ·
πππ·πΞππ·π
ΞπΌπ· = ππ ΞππΊπ + ππ Ξππ·π
ππ = ππβ1
When the MOSFET operates in the saturation region, we have the drain current formula
with channel modulation effect given as (Equation 11.66):
πΌπ· =πππΆππ₯π
2πΏ(ππΊπ β πππ»)2(1 + π ππ·π)
The small signal transconductance is defined as:
ππ =ππΌπ·
πππΊπ
ππ =πππΆππ₯π
πΏ(ππΊπ β πππ»)(1 + π ππ·π)
When the channel length modulation is negligible under many practical circumstances, we
may ignore the βπβ for the calculation of ππ. Using the large-signal drain current πΌπ·
(typically achieved by the bias circuit) and ignoring the small effect from the βπβ term, we
can re-write the last equation as:
ππ =2πΌπ·
(ππΊπ β πππ»)= β
2πππΆππ₯ππΌπ·
πΏ
The small-signal output conductance is defined as:
ππ =ππΌπ·
πππ·π
=π
πππ·π[πππΆππ₯π
2πΏ(ππΊπ β πππ»)2(1 + π ππ·π)]
ππ =πππΆππ₯π
2πΏ(ππΊπ β πππ»)2 π
The small-signal output resistance is the reciprocal of the above quantity:
ππ = ππβ1
Using the large-signal drain current πΌπ· (typically achieved by the bias circuit) and ignoring
the small effect from the β1 + πβ term, we can re-write it as:
ππ = ππβ1 = (ππΌπ·)β1
9. (11.9) MOSFET small signal equivalent circuit model parameter calculations
Given quantities are:
ππΊπ = 2 V
ππ·π = 3 V
From Question 5, we know πππ» = 0.363 V.
Since ππΊπ β πππ» = 2 V β 0.363 V = 1.637 V β€ ππ·π = 3 V, the MOSFET is operating in
saturation region. Therefore, the drain current is
πΌπ· =πππΆππ₯π
2πΏ(ππΊπ β πππ»)2 = 0.0093 A
The small-signal transconductance can be found in two ways:
ππ =πππΆππ₯π
πΏ(ππΊπ β πππ») = 0.0113 Ξ©β1
or
ππ =2πΌπ·
(ππΊπ β πππ»)= 0.0114 Ξ©β1
It is a bit larger due to rounding errors in our previous calculations.
The output transconductance can also be found in two ways:
ππ =πππΆππ₯π
2πΏ(ππΊπ β πππ»)2 π = 0.000462 Ξ©β1
Or with the alternative formula, ππ = ππΌπ· = 0.000465 Ξ©β1 which is a bit larger due to
rounding errors in our previous calculations.
So, the output resistance is:
ππ = ππβ1 = 2.17 kΞ©
The gate-source capacitance πΆππ and gate-drain capacitance πΆππ are given by Equations 11.72
and 11.73:
πΆππ =2
3πΆππ₯πΏπ = 2.30 Γ 10β14 F
πΆππ = πΆππ₯πΏππ = 3.45 Γ 10β15 F
The small signal equivalent circuit of this N-MOSFET is as follows:
GATE TERMINAL
SOURCE TERMINAL
DRAIN TERMINAL