MOSFET & IC Basics GATE Problems (Part - II)ย ยท When a gate voltage is applied, a depletion region...
Transcript of MOSFET & IC Basics GATE Problems (Part - II)ย ยท When a gate voltage is applied, a depletion region...
MOSFET & IC Basics โ GATE Problems (Part - II)
1. In MOSFET devices the N-channel type is better than the P โ Channel
type in the following respects.
(a) It has better immunity
(b) It is faster
(c) It is TTL compatible
(d) It has better drive
capability
[GATE 1988: 2 Marks]
Soln. In N โ Channel MOSFETs the charge carriers are electrons while in
P โ channel MOSFETs holes are the charge carriers.
The mobility of electrons is always greater than the mobility of holes.
i.e. ๐๐ > ๐๐
Thus, N โ Channel MOSFETs are faster
Option (b)
2. In a MOSFET, the polarity of the inversion layer is the same as that of the
(a) Charge on the GATE โ EC โ electrode
(b) Minority carries in the drain
(c) Majority carriers in the substrate
(d) Majority carries in the source
[GATE 1989: 2 Marks]
Soln. In a MOSFET the polarity of inversion layer is the same as that of
majority carriers in the source.
For example, for N โ MOSFETs the source is of N โ type and
inversion layer formed is of electrons.
Option (d)
3. Which of the following effects can be caused by a rise in the
temperature?
(a) Increase in MOSFET current (IDS)
(b) Increase in BJT current (IC)
(c) Decrease in MOSFET current (IDS)
(d) Decrease in BJT current (IC)
[GATE 1990: 2 Marks]
Soln. The current equation for BJT is
๐ฐ๐ช = ๐ท ๐ฐ๐ + (๐ + ๐ท)๐ฐ๐ช๐ถ
As temperature increases, the leakage current ICO increases, so the
current IC increases in BJT.
Since mobility decreases as temperature increases. So in MOSFETs
current IDS decreases with rise in temperature
Option (b) and (c)
4. When the gate โ to โ source voltage (VGS) of a MOSFET with threshold
voltage of 400 mV, working in saturation is 900 mV, the drain current is
observed to be 1 mA. Neglecting the channel width modulation effect and
assuming that the MOSFET is operating at saturation, the drain current
for an applied VGS of 1400 mV is
(a) 0.5 mA
(b) 2.0 mA
(c) 3.5 mA
(d) 4.0 mA
[GATE 2003: 2 Marks]
Soln. Given,
๐ฝ๐ป = ๐๐๐ ๐๐ฝ = ๐. ๐ ๐ฝ
Voltage applied at gate
๐ฝ๐ฎ๐บ = ๐๐๐ ๐๐ฝ =0.9 V
๐ฐ๐ซ๐บ = ๐ ๐๐จ
Find the drain current for
๐ฝ๐ฎ๐บ = ๐๐๐๐ ๐๐ฝ
MOSFET is operating in saturation
๐ฐ๐ซ๐บ = ๐ฒ (๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐
๐๐, ๐ ร ๐๐โ๐ = ๐ฒ (๐. ๐ โ ๐. ๐)๐
๐๐, ๐ฒ =๐๐โ๐
(๐.๐)๐= ๐ ร ๐๐โ๐ ๐จ
๐ฝ๐
For VGS = 1.4 V
๐ฐ๐ซ๐บ = ๐ฒ (๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐
= ๐ ร ๐๐โ๐ (๐. ๐ โ ๐. ๐)๐
๐ฐ๐ซ๐บ = ๐ ๐๐จ
Option (d)
5. The drain of an n โ channel MOSFET is shorted to the gate so that
๐๐บ๐ = ๐๐ท๐ . The threshold voltage (VT) of MOSFET is 1 V. If the drain
current (ID) is 1 mA for VGS = 2 V, then for ๐๐บ๐ = 3 ๐, ID is
(a) 2 mA
(b) 3 mA
(c) 9 mA
(d) 4 mA
[GATE 2004: 2 Marks]
Soln. Given,
๐ฝ๐ฎ๐บ = ๐ฝ๐ซ๐บ
Then the device is in saturation.
So, ๐ฐ๐ซ๐บ = ๐ฒ (๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐
For ๐ฐ๐ซ๐บ = ๐ ๐๐จ , ๐ฝ๐ฎ๐บ = ๐๐ฝ , ๐ฝ๐ป = ๐๐ฝ
๐ = ๐ฒ (๐ โ ๐)๐
๐๐, ๐ฒ = ๐ ๐๐จ ๐ฝ๐โ
Again,
๐ฐ๐ซ = ๐ฒ (๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐
For ๐ฝ๐ฎ๐บ = ๐ ๐ฝ
๐ฐ๐ซ = ๐ ร (๐ โ ๐)๐
๐ฐ๐ซ = ๐ ๐๐จ
Option (d)
6. An n โ channel depletion MOSFET has following two points on its
๐ผ๐ท โ ๐๐บ๐ curve
(i) ๐๐บ๐ = 0 ๐๐ก ๐ผ๐ท = 12 ๐๐ด ๐๐๐
(ii) ๐๐บ๐ = โ6 ๐ ๐๐ก ๐ผ๐ท = 0
Which of the following Q point will give the highest transconductance
gain for small signals?
(a) ๐๐บ๐ = โ6 ๐
(b) ๐๐บ๐ = โ3 ๐
(c) ๐๐บ๐ = 0 ๐
(d) ๐๐บ๐ = 3 ๐
[GATE 2006: 2 Marks]
Soln. Transconductance (๐๐) =๐น ๐ฐ๐ซ
๐น ๐ฝ๐ฎ๐บ for ๐ฝ๐ซ๐บ = ๐๐๐๐๐๐๐๐
The characteristics is plotted.
VGS
12mA
-6V
๐ฐ๐ซ(๐๐จ)
0
As ID increases VGS also increases Larger VGS will have high
transconductance
Thus, Option (d)
7. In the C MOS inverter circuit shown if the transconductance parameters
of N MOS and P MOS transistor are
๐พ๐ = ๐พ๐ = ๐๐๐ถ๐๐
๐๐
๐ฟ๐= ๐๐๐ถ๐๐
๐๐
๐ฟ๐= 40 ๐๐ด ๐2โ
and threshold voltages are ๐๐ = 1๐ the current I is
NMOS
PMOS
5V
2.5V
(a) 0 A
(b) 25 ยตA
(c) 45 ยตA
(d) 90 ยตA
[GATE 2007: 2 Marks]
Soln. Given
๐ฒ๐ = ๐ฒ๐ = ๐๐ ๐๐จ ๐ฝ๐โ
๐ฝ๐ป = ๐ ๐ฝ
The device is in saturation. So the current is given by
๐ฐ๐ซ๐บ =๐ฒ๐
๐ (๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐
๐๐
๐(๐. ๐ โ ๐)๐ = ๐๐ ร (๐. ๐)๐ = ๐๐ ๐๐จ
Option (c)
8. Two identical N MOS transistors M1 and M2 are connected as shown
below. ๐๐๐๐๐ Chosen so that both transistor are in saturation. The
equivalents ๐๐ of the pair is defined to be ๐ ๐ผ๐๐ข๐ก
๐ ๐๐ at constant๐๐๐ข๐ก. The
equivalent ๐๐ of the pair is
M2
M1
๐ฝ๐๐๐
๐ฝ๐๐๐๐
๐ฐ๐๐๐
๐ฝ๐
(a) The sum of individual ๐๐ of two transistors.
(b) The product of individual ๐๐ of the transistors.
(c) Nearly equal to the ๐๐ of M1
(d) Nearly equal to ๐๐ ๐0โ of M2
[GATE 2008: 2 Marks]
Soln. As per the principle of transconductance
๐
๐๐=
๐
๐๐๐
=๐
๐๐๐
๐๐, ๐๐ =๐๐๐
๐๐๐
๐๐๐+๐๐๐
โ ๐๐๐
Option (c)
9. The measured Transconductance ๐๐ of an NMOS transistor operating in
the linear region is plotted against voltage VG at a Constant drain voltage
VD. Which of the following figures represents the expected dependence
of ๐๐ on VG .
(a) (b)
(d)(c)
gm gm
gm gm
VG
VG VG
VG V
[GATE 2008: 2 Marks]
Soln. Drain current ID in the linear region is given by
๐ฐ๐ซ = ๐ฒ๐ [(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐ฝ๐ซ๐บ โ๐ฝ๐ซ๐บ
๐
๐]
๐๐๐ ๐๐ =๐ ๐ฐ๐ซ
๐ ๐ฝ๐ฎ๐บ= ๐ฒ๐(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐ฝ๐ซ๐บ
๐บ๐, ๐๐ โ ๐ฝ๐ฎ๐บ
So, linear characteristic
Option (c)
10. Consider the following two situations about the internal conditions in an
n โ channel MOSFET operating in the active region.
S1: The inversion charge decreases from source to drain
S2 : The channel potential increases from source to drain
Which of the following is correct?
(a) Only S2 is true
(b) Both S1 and S2 are false
(c) Both S1 and S2 are true but S2 is not a reason for S1
(d) Both S1 and S2 are true and S2 is reason for S1
[GATE 2009: 2 Marks]
Soln. MOSFET is also considered as gate โ controlled resistor.
When +ve gate voltage in an n โ channel MOSFET exceeds VT,
electrons are induced in p โ type substrate. This channel is also
connected to n+ source and drain regions. Channel looks like induced
n โ type resistor. With increase in gate voltage the channel becomes
more conducting. The drain current increase (linear region). When
more drain current flows, there is more ohmic voltage drop along the
channel. The channel potential varies near zero from source to the
potential at drain end.
The voltage difference between gate and channel reduces from VG
(near source) to (๐ฝ๐ฎ โ ๐ฝ๐ซ) near drain. When drain bias is increased
(๐ฝ๐ฎ โ ๐ฝ๐ซ) = ๐ฝ๐ป, the channel is piched off. When VD increase further
the drain current is in saturation.
So, S1 & S2 are true and S2 is reason for S1
Option (d) (Refer: Streetman)
11. The source of a silicon (๐๐ = 1010 ๐๐๐ ๐๐3) n โ channel MOS transistor
has an area 1 sq ยตm and a depth of 1 ยตm. If the dopant density in the
source is 1019/๐๐3 the no. of holes in the source region with above
volume is approx.
(a) 107
(b) 100
(c) 10
(d) 0
[GATE 2012: 2 Marks]
Soln. Given,
๐๐ = ๐๐๐๐ ๐๐๐โ
๐จ๐๐๐ (๐จ) = ๐ ร ๐๐โ๐๐/๐๐
๐ ๐๐๐๐ (๐ ) = ๐๐โ๐ ๐
๐ฝ๐๐๐๐๐ = ๐จ . ๐ = ๐ ร ๐๐โ๐ ร ๐๐๐ ๐๐
= ๐๐โ๐๐/๐๐
= ๐๐โ๐๐/๐๐๐
๐ต๐ซ = ๐ = ๐๐๐๐/๐๐๐
๐ =๐๐
๐
๐ต๐ซ=
๐๐๐๐
๐๐๐๐ = ๐๐ ๐๐๐โ
So,
Holes in volume V is
๐ฏ = ๐ . ๐ฝ = ๐๐โ๐๐ ร ๐๐
= ๐๐โ๐๐
Since not integer number โ ๐
Option (d)
12. A depletion type N โ channel MOSFET in biased in its linear region for
use as a voltage controlled resistor . Assume threshold voltage ๐๐๐ป =
0.5๐ , ๐๐บ๐ = 2.0๐ , ๐๐ท๐ = 5๐ , ๐ ๐ฟโ = 100, ๐ถ๐๐ = 10โ8 ๐น/๐๐2 and
๐๐ = 800 ๐๐2 ๐ โ ๐ โ . The value of the resistance of the voltage
controlled resistor (๐๐ ฮฉ) is __________
[GATE 2014: 2 Marks]
Soln. Given,
Depletion type MOSFET (N โ Channel)
In linear region
๐ฝ๐ป๐ฏ = ๐. ๐๐ฝ
๐ฝ๐ฎ๐บ = ๐. ๐๐ฝ
๐ฝ๐ซ๐บ = ๐๐ฝ
๐พ/๐ณ = ๐๐๐
๐ช๐ถ๐ฟ = ๐๐โ๐ ๐ญ ๐๐๐โ
The value of voltage controlled resistor is given by
๐๐ ๐ =๐
๐๐๐ช๐ถ๐ฟ๐พ๐ณ
(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)
๐๐ซ๐บ =๐
๐๐๐ ร ๐๐โ๐ ร ๐๐๐[๐ โ (โ๐. ๐)]
๐๐ซ๐บ = ๐๐๐ ๐
Answer: ๐๐๐ ๐
13. For the n โ channel MOS transistor shown in figure, the threshold
voltage VTH is 0.8V. Neglect channel length modulation effects. When
the drain voltage VD = 1.6, the drain current ID was found to be 0.5 mA. If
VD is adjusted to be 2V by changing the values of R and VDD, the new
value of ID (in m A) is
R
D
S
๐ฝ๐ซ๐ซ
G
(a) 0.625
(b) 0.75
(c) 1.125
(d) 1.5
[GATE 2014: 2 Marks]
Soln. Given,
๐ฝ๐ป๐ฏ = ๐. ๐ ๐ฝ
๐ฝ๐ซ = ๐. ๐ ๐ฝ
๐ฐ๐ซ = ๐. ๐ ๐๐จ
For the given figure we notice that Gate is connected to drain
So, ๐ฝ๐ฎ๐บ = ๐ฝ๐ซ๐บ = ๐ฝ๐ซ
In saturation ID is given by
๐ฐ๐ซ๐บ = ๐ฒ(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐
๐บ๐, ๐ฐ๐ซ๐
๐ฐ๐ซ๐
=[๐ฝ๐ฎ๐บ๐โ๐ฝ๐ป]
๐
[๐ฝ๐ฎ๐บ๐โ๐ฝ๐ป]๐
๐บ๐, ๐ฐ๐ซ๐
๐ฐ๐ซ๐
=(๐ โ ๐. ๐)๐
(๐. ๐ โ ๐. ๐)๐= ๐. ๐๐
๐๐, ๐ฐ๐ซ๐= ๐. ๐๐ ร ๐. ๐
= ๐. ๐๐๐ ๐๐จ
Option (c)
14. For the MOSFET shown in the figure the threshold voltage |๐๐ก| = 2๐
and ๐พ =1
2๐๐ (
๐
๐ฟ) = 0.1๐๐ด/๐2. The value of ID (in mA) is ________
๐ฝ๐ซ๐ซ = +๐๐๐ฝ
๐น๐
๐น๐
๐๐ ๐ฒ๐
๐๐ ๐ฒ๐
๐ฝ๐บ๐บ = โ๐๐ฝ
๐ฐ๐ซ
[GATE 2014: 2 Marks]
Soln. The two MOSFET are in series so, same current will be flowing. For
the bottom MOSFET it is easier to find VGS
๐ฝ๐ฎ๐บ = ๐ฝ๐ฎ โ ๐ฝ๐บ = ๐โ ๐
= ๐๐ฝ
VGS is greater than threshold voltage,
So, MOSFET is in saturation
๐ฐ๐ซ =๐
๐ ๐๐ ๐ช๐ถ๐ฟ
๐พ
๐ณ[๐ฝ๐ฎ๐บ โ ๐ฝ๐ป]๐
= ๐. ๐ ๐๐จ/๐ฝ๐ [๐ โ ๐]๐
๐. ๐ ๐๐จ/๐ฝ๐ [๐]๐
๐ฐ๐ซ = ๐. ๐ ๐๐จ
Answer: 0.9 mA
15. The slope of the ๐ผ๐ท ๐ฃ๐ . ๐๐บ๐ curve of an n โ channel MOSFET in linear
region is 10โ3 ฮฉโ1 at ๐๐ท๐ = 0.1๐ . For the same device, neglecting
channel length modulation, the slope of the โ๐ผ๐ท ๐ฃ๐ ๐๐บ๐ curve
(๐๐ โ๐ด ๐โ ) under saturation region is approximately ________
[GATE 2014: 2 Marks]
Soln. Given,
Slope of ๐ฐ๐ซ ๐๐ . ๐ฝ๐ฎ๐บ curve for N โ MOSFET in linear region is
๐๐โ๐ ๐โ๐
๐. ๐. ๐ ๐ฐ๐ซ
๐ ๐ฝ๐ฎ๐บ= ๐๐โ๐
In linear region the drain current is given by
๐ฐ๐ซ = ๐ฒ๐ [(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐ฝ๐ซ๐บ โ๐
๐ ๐ฝ๐ซ๐บ
๐ ]
๐๐๐๐๐ ๐ฒ๐ = ๐๐ ๐ช๐ถ๐ฟ๐พ
๐ณ
๐ ๐ฐ๐ซ
๐ ๐ฝ๐ฎ๐บ= ๐ฒ๐ ๐ฝ๐ซ๐บ
๐๐, ๐ฒ๐ =
๐ ๐ฐ๐ซ๐ ๐ฝ๐ฎ๐บ
๐ฝ๐ซ๐บ=
๐๐โ๐
๐.๐= ๐๐โ๐
For saturation region
๐ฐ๐ซ =๐
๐ ๐ฒ๐ [๐ฝ๐ฎ๐บ โ ๐ฝ๐ป]๐
๐๐, โ๐ฐ๐ซ = โ๐ฒ๐
๐ [๐ฝ๐ฎ๐บ โ ๐ฝ๐ป]
๐ โ๐ฐ๐ซ
๐ ๐ฝ๐ฎ๐บ= โ
๐ฒ๐
๐
๐๐,๐ โ๐ฐ๐ซ
๐ ๐ฝ๐ฎ๐บ= โ
๐๐โ๐
๐= ๐. ๐๐๐๐ โ๐จ / ๐ฝ
Answer: ๐. ๐๐๐๐ โ๐จ / ๐ฝ
16. An ideal MOS capacitor has boron doping concentration of 1015 ๐๐โ3
in the substrate. When a gate voltage is applied, a depletion region of
width 0.5 ๐๐ is formed with a surface (channel) potential of 0.2 V. Given
that โ0= 8.854 ร 10โ14F/cm and the relative permittivity of silicon and
silicon dioxide are 12 and 4 respectively the peak electric filed in ๐ ๐๐โ
in the oxide region is _____________
[GATE 2014: 2 Marks]
Soln. Dopant is Boron, so p โ type substrate
Thus device is n MOSFET. Peak electric field in Si substrate
๐ฌ๐๐ =๐๐ฟ๐
๐พ๐
Where, ๐ฟ๐ โ ๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐
๐พ๐ โ ๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐ ๐๐๐ ๐๐
๐ฌ๐๐ =๐ ร ๐. ๐
๐. ๐๐๐= ๐. ๐ ๐๐โ
Electric field in oxide
๐ฌ๐ถ๐ฟ =โ๐
โ๐ถ๐ฟ . ๐ฌ๐๐
=๐๐
๐ร ๐. ๐๐ฝ ๐๐โ
= ๐. ๐ ๐ฝ ๐๐โ
Answer: ๐ฌ๐ถ๐ฟ = ๐. ๐๐ฝ ๐๐โ
17. For the MOSFET M1 shown in figure, assume
๐ ๐ฟโ = 2, ๐๐ท๐ท = 2.0๐, ๐๐ ๐ถ๐๐ = 100 ๐๐ ๐2โ ๐๐๐ ๐๐๐ป = 0.5๐. The
transistor M1 switches from saturation region to linear region when ๐๐๐
(in volts) is ___________
๐ฝ๐ซ๐ซ
R = ๐๐ ๐ฒ๐
๐ฝ๐๐๐
M1 ๐ฝ๐๐
[GATE 2014: 2 Marks]
Soln. For saturation region drain current is given by
๐ฐ๐ซ = ๐ฒ๐[๐ฝ๐ฎ๐บ โ ๐ฝ๐ป]๐
=๐
๐๐๐๐ช๐ถ๐ฟ
๐พ
๐ณ[๐ฝ๐ฎ๐บ โ ๐ฝ๐ป]๐
=๐
๐๐๐๐ช๐ถ๐ฟ
๐พ
๐ณ[๐ฝ๐]๐
๐๐๐๐๐, ๐ฝ๐ฎ๐บ โ ๐ฝ๐ป = ๐ฝ๐ซ๐บ = ๐ฝ๐
๐บ๐, ๐ฐ๐ซ =๐
๐ร ๐๐๐ ร ๐๐โ๐ ร ๐ ร ๐ฝ๐
๐
๐๐, ๐ฐ๐ซ = ๐๐๐ ร ๐๐โ๐ ๐ฝ๐๐ โ โ โ โ โ โ โ (๐)
Applying KVL in outer loop
๐ฝ๐ซ๐ซ = ๐ฐ๐ซ ร ๐๐๐ฒ + ๐ฝ๐ โ โ โ โ โ โ(๐)
From equation (1) and (2) we get
๐ = ๐๐๐ ร ๐๐โ๐ ๐ฝ๐๐ + ๐ฝ๐
๐๐, ๐ = ๐ฝ๐๐ + ๐ฝ๐
๐๐, ๐ฝ๐๐ + ๐ฝ๐ โ ๐ = ๐
๐ฝ๐ =โ๐ ยฑ โ๐ + ๐ ร ๐ ร ๐
๐ ร ๐=
โ๐ ยฑ ๐
๐
๐บ๐, ๐ฝ๐ = ๐๐ฝ ๐๐ โ ๐๐ฝ
Taking ๐ฝ๐ = ๐๐ฝ
๐ฝ๐ซ๐บ = ๐ฝ๐ฎ๐บ = ๐ฝ๐ป
๐๐, ๐ฝ๐ = ๐ฝ๐๐ โ ๐ฝ๐ป
๐๐, ๐ฝ๐๐ = ๐ + ๐. ๐ = ๐. ๐๐ฝ
Answer: 1.5V
18. In a MOS capacitor with an oxide layer thickness of 10 nm, the
maximum depletion layer thickness is 100 nm. The permittivities of the
semiconductor and the oxide layer are ๐๐ ๐๐๐ ๐๐๐ respectively.
Assuming ๐๐ /๐๐๐ = 3, the ratio of the maximum capacitance of the
minimum capacitance of this MOS capacitor is .
[GATE 2015: 2 Marks]
Soln. Given,
Oxide layer thickness (๐๐ถ๐ฟ) = ๐๐ ๐๐
Depletion layer thickness (๐๐ ๐๐) = ๐๐๐ ๐๐
๐บ๐
๐บ๐ถ๐ฟ= ๐
Metal
Si
tOX (10nm)
tdep =100
Both oxide layer capacitance (๐ช๐ถ๐ฟ) and depletion layer capacitances
are coming in series, so the maximum capacitance will be when
depletion layer capacitance (๐ช๐ ๐๐) will be absent
Maximum capacitance (๐ช๐๐๐) = ๐ช๐ถ๐ฟ
Minimum capacitance (๐ช๐๐๐) is
Given by
๐ช๐๐๐ =๐ช๐ถ๐ฟ ๐ช๐ ๐๐
๐ช๐ถ๐ฟ + ๐ช๐ ๐๐
Where ๐ช๐ถ๐ฟ ๐๐๐ ๐ช๐ ๐๐ are capacitances per unit area
๐ช๐๐๐
๐ช๐๐๐=
๐ช๐ถ๐ฟ
๐ช๐ถ๐ฟ ๐ช๐ ๐๐
๐ช๐ถ๐ฟ + ๐ช๐ ๐๐
=๐ช๐ถ๐ฟ + ๐ช๐ ๐๐
๐ช๐ ๐๐
= ๐ +๐ช๐ถ๐ฟ
๐ช๐ ๐๐= ๐ +
๐บ๐ถ๐ฟ
๐๐ถ๐ฟ๐บ๐บ
๐
= ๐ +๐บ๐ถ๐ฟ
๐บ๐ .
๐
๐๐ถ๐ฟ
๐ช๐๐๐
๐ช๐๐๐= ๐ +
๐
๐ร
๐๐๐
๐๐= ๐. ๐๐
Answer: 4.33
19. In the circuit shown both enhancement mode NMOS transistor have the
following characteristics: ๐พ๐ = ๐๐๐ถ๐๐(๐ ๐ฟโ ) = 1๐๐ด/๐2; ๐๐๐ = 1๐. Assume that the channel length modulation parameter ฮป is zero and body
is shorted to source. The minimum supply voltage VDD (in volts) needed
to ensure that transistor M1 operates in saturation mode of operation is __
๐ฝ๐ซ๐ซ
M1
M2
2V
[GATE 2015: 2 Marks]
Soln. In the given circuit
M1 is to operate in saturation. Both MOSFETs are NMOS
๐ฒ๐ = ๐๐๐ช๐ถ๐ฟ(๐พ ๐ณโ ) = ๐ ๐๐จ/๐ฝ๐
๐ฝ๐ป๐ = ๐ ๐ฝ
Both MOSFETs are in series so same current will flow through then
For M1 : ๐ฝ๐ฎ๐บ๐= ๐๐ฝ (As per figure)
If M1 is assumed in saturation, then
๐ฐ๐ซ๐=
๐
๐๐ฒ๐[๐ฝ๐ฎ๐บ๐
โ ๐ฝ๐ป]๐
Minimum VDS required for M1 to operate in saturation
๐ฝ๐ซ๐บ๐= ๐ฝ๐ฎ๐บ๐
โ ๐ฝ๐ป = ๐ โ ๐ = ๐๐ฝ
For M2 : ๐ฝ๐ฎ๐บ๐โ ๐ฝ๐ซ๐ซ โ ๐ฝ๐ซ๐บ๐
= ๐ฝ๐ซ๐ซ โ ๐
๐ฐ๐ซ๐=
๐
๐ ๐ฒ๐[๐ฝ๐ฎ๐บ๐
โ ๐ฝ๐ป]๐
Since ๐ฐ๐ซ๐= ๐ฐ๐ซ๐
๐
๐ ๐ฒ๐[๐ฝ๐ฎ๐บ๐
โ ๐ฝ๐ป]๐
=๐
๐ ๐ฒ๐[๐ฝ๐ฎ๐บ๐
โ ๐ฝ๐ป]๐
๐๐, ๐ฝ๐ฝ๐ฎ๐โ ๐ฝ๐ป = ๐ฝ๐ฎ๐บ๐
โ ๐ฝ๐ป
๐ฝ๐ฎ๐บ๐= ๐ฝ๐ฎ๐บ๐
๐๐, ๐ฝ๐ฎ๐บ๐= ๐ฝ๐ซ๐ซ โ ๐
๐๐, ๐ฝ๐ซ๐ซ = ๐ + ๐ = ๐๐ฝ
Answer: ๐ฝ๐ซ๐ซ = ๐๐ฝ
20. The current in an enhancement mode NMOS transistor biased in
saturation mode was measured to be 1 mA at a drain source voltage of
5V. When the drain source voltage was increased to 6V while keeping
gate source voltage same. The drain current increased to 1.02 mA.
Assume that drain to source saturation voltage is much smaller than the
applied drain source voltage. The channel length modulation parameter ฮป
(๐๐ ๐โ1) is ______________.
[GATE 2015: 2 Marks]
Soln. Given,
N MOS transistor
Biased in saturation ๐ฐ๐ซ = ๐๐๐จ , ๐ฝ๐ซ๐บ = ๐๐ฝ
Drain current in saturation including the effect of channel length
modulation parameter
๐ฐ๐ซ =๐
๐ ๐๐๐ช๐ถ๐ฟ
๐พ
๐ณ(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐(๐ + ๐ ๐ฝ๐ซ๐บ)
So, we can write the ratio of two current
๐ฐ๐ซ๐
๐ฐ๐ซ๐
=๐ + ๐ ๐ฝ๐ซ๐บ๐
๐ + ๐ ๐ฝ๐ซ๐บ๐
๐๐, ๐. ๐๐
๐=
๐ + ๐๐
๐ + ๐๐
๐๐, ๐. ๐๐ + ๐. ๐๐ = ๐ + ๐๐
๐๐, ๐. ๐ ๐ = ๐. ๐๐
๐๐, ๐ =๐. ๐๐
๐. ๐= ๐. ๐๐๐๐ฝโ๐
21. Consider an n โ channel metal oxide semiconductor field effect transistor
(MOSFET) with gate to source voltage of 1.8V. Assume that ๐
๐ฟ= 4, ๐๐๐ถ๐๐ = 70 ร 10โ6 ๐ด๐โ2
The threshold voltage is 0.3V, and the channel length modulation
parameter is 0.09๐โ1. In the saturation region, the drain conduction (in
micro Siemens) is ________
[GATE 2016: 2 Marks]
Soln. Given,
N โ MOSFET
๐ฝ๐ฎ๐บ = ๐. ๐๐ฝ ,๐พ
๐ณ= ๐ , ๐๐๐ช๐ถ๐ฟ = ๐๐ ร ๐๐โ๐ ๐จ๐ฝโ๐
๐ฝ๐ป๐ฏ = ๐. ๐๐ฝ , ๐ = ๐. ๐๐๐ฝโ๐
Drain current in saturation including the effect of channel length
modulation is
๐ฐ๐ซ =๐
๐ ๐๐ ๐ช๐ถ๐ฟ (
๐พ
๐ณ) (๐ฝ๐ฎ โ ๐ฝ๐ป)๐(๐ + ๐๐ฝ๐ซ)
Channel conductance (๐๐ )
(๐๐ ) =๐ ๐ฐ๐ซ
๐ ๐ฝ๐ซ
๐ฝ๐ฎ = ๐๐๐๐๐๐๐๐
๐๐ =๐ ๐ฐ๐ซ
๐ ๐ฝ๐ซ=
๐
๐ ๐๐๐ช๐ถ๐ฟ (
๐พ
๐ณ) (๐ฝ๐ฎ โ ๐ฝ๐ป)๐ . ๐
So,
๐๐ =๐
๐ร ๐๐ ร ๐๐โ๐ ร ๐ ร (๐. ๐ โ ๐. ๐)๐ ร ๐. ๐๐
= ๐๐๐ ร ๐๐โ๐ ร ๐. ๐๐ ร ๐. ๐๐
= ๐๐. ๐๐ ร ๐๐โ๐
๐๐ = ๐๐. ๐๐ ๐ ๐บ๐๐๐๐๐๐
Drain conductance ๐๐ = ๐๐. ๐๐ ๐ ๐๐ข๐๐ฆ๐๐ง๐ฌ
22. A voltage VG is applied across the MOS capacitor with metal Gate and
p โ type silicon substrate at ๐ = 300 ๐พ. The inversion carrier density (in
number of units per unit area) for
๐๐บ = 0.8๐ ๐๐ 2 ร 1011๐๐โ2 ๐๐๐ ๐๐บ = 1.3๐, the inversion carrier density
is 4 ร 1011๐๐โ2. What is value of inversion carrier density of VG = 1.8V
(a) 4.5 ร 1011๐๐โ2
(b) 6.0 ร 1011๐๐โ2
(c) 7.2 ร 1011๐๐โ2
(d) 8.4 ร 1011๐๐โ2
[GATE 2016: 2 Marks]
Soln. Given,
For VG = 0.8V
Inversion layer carrier density = ๐ ร ๐๐๐๐๐๐โ๐
๐ญ๐๐ ๐ฝ๐ฎ = ๐. ๐๐ฝ , ๐ฐ๐๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐ ๐๐๐๐๐๐ = ๐ ร
๐๐๐๐๐๐๐
Find for VG = 1.8 V
For MOS capacitor
Charge (๐ธ) โ (๐ฝ๐ฎ โ ๐ฝ๐ป)
Charge ๐ธ๐ = ๐ฒ. (๐ฝ๐ฎ โ ๐ฝ๐ป)
Or, ๐ธ๐ = ๐๐จ ร ๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐ ๐๐๐๐๐๐
So,
๐ ร ๐๐๐๐ . ๐๐จ
๐ ร ๐๐๐๐ . ๐๐จ=
(๐. ๐ โ ๐ฝ๐ป)
(๐. ๐ โ ๐ฝ๐ป)
Or, VT = 0.3V
๐ต๐๐ ๐๐๐ ๐ฝ๐ฎ = ๐. ๐๐ฝ , ๐ธ๐ = ๐๐จ ๐๐
Where ๐๐ is inversion carrier density
๐๐ ๐ . ๐จ
๐ ร ๐๐๐๐ ๐ . ๐จ=
(๐. ๐ โ ๐. ๐)
(๐. ๐ โ ๐. ๐)
๐๐, ๐๐ = ๐ ร ๐๐๐๐/๐๐๐
Option (b)
23. Consider long channel N MOS transistor with source and body
connected together.
Assume that the electron mobility is dependent of VGS and VDS. Given,
๐๐ = 0.5 ๐ ๐ด ๐โ ๐๐๐ ๐๐ท๐ = 50 ๐๐ ๐๐๐ ๐๐บ๐ = 2๐
๐๐ = 8 ๐ ๐ด ๐โ ๐๐๐ ๐๐บ๐ = 2๐ ๐๐๐ ๐๐ท๐ = 0๐
Where
๐๐ =๐ ๐ผ๐ท
๐ ๐๐บ๐ ๐๐๐ ๐๐ =
๐ ๐ผ๐ท
๐ ๐๐ท๐
The threshold voltage (in volts) of the transistor is _________
[GATE 2016: 2 Marks]
Soln. Given long channel N MOS, it means it does not have the effect of
channel length modulation. Since VDS is 50 mV (small) it is operating
in linear region. The drain current is given by
๐ฐ๐ซ = ๐๐ ๐ช๐ถ๐ฟ (๐พ
๐ณ) [(๐ฝ๐ฎ๐บ โ ๐ฝ๐ป)๐ฝ๐ซ๐บ โ
๐
๐๐ฝ๐ซ๐บ
๐ ]
๐๐ =๐ ๐ฐ๐ซ
๐ ๐ฝ๐ฎ๐บ= ๐๐ ๐ช๐ถ๐ฟ (
๐พ
๐ณ) . ๐ฝ๐ซ๐บ
๐๐, ๐๐ ๐ช๐ถ๐ฟ
๐พ
๐ณ=
๐๐
๐ฝ๐ซ๐บ=
๐. ๐ ร ๐๐โ๐
๐๐ ร ๐๐โ๐= ๐๐ ร ๐๐โ๐
Now,
๐๐ =๐ ๐ฐ๐ซ
๐ ๐ฝ๐ซ๐บ= ๐๐ ๐ช๐ถ๐ฟ
๐พ
๐ณ[๐ฝ๐ฎ โ ๐ฝ๐ป]
๐๐, ๐ ร ๐๐โ๐ = ๐๐ ร ๐๐โ๐[๐ฝ๐ฎ๐บ โ ๐ฝ๐ป]
๐๐, ๐ฝ๐ฎ๐บ โ ๐ฝ๐ป =๐ ร ๐๐โ๐
๐๐ ร ๐๐โ๐
๐๐, ๐ฝ๐ป = ๐ฝ๐ฎ๐บ โ ๐. ๐
= ๐ โ ๐. ๐ = ๐. ๐๐ฝ
๐๐ง๐ฌ๐ฐ๐๐ซ: ๐ฝ๐ป = ๐. ๐๐ฝ
24. Figure I and II show two MOS capacitors of unit area. The capacitor in
Figure I has insulator material X of (thickness t1 = 1 nm and dielectric
constant โ1= 4) and y (of thickness ๐ก2 = 3๐๐ and dielectric
constantโ1= 20). The capacitor in figure II has only insulator material X
of thickness ๐ก๐๐ . If the capacitors are of equal capacitance, then the value
of ๐ก๐๐ (in nm) is________
Metal
ิ2 ิ1
Metal
ิ1
Si Si
t1 t2 teq
Figure IFigure II
[GATE 2016: 2 Marks]
Soln. MOS structure is like a parallel plate capacitor
The capacitance per unit area for this geometry is
๐ชโฒ =โ
๐
Where โ is permittivity of insulator
d is distance between plates
For area A,
๐ช =โ๐จ
๐
As per figure I the two capacitors formed by the two dielectrics are in
series
๐ช =๐ช๐ ๐ช๐
๐ช๐ + ๐ช๐
= [
๐๐ ร ๐๐โ๐ ร
๐๐๐ ร ๐๐โ๐
๐๐ ร ๐๐โ๐ +
๐๐๐ ร ๐๐โ๐
] โ๐
๐ช = ๐. ๐ ร ๐๐๐ โ๐
For the II case
๐ช =โ๐ โ๐
๐๐๐
๐๐, ๐๐๐=
โ๐ โ๐
๐ช
=โ๐ โ๐
๐. ๐ ร ๐๐๐ โ๐=
๐
๐. ๐ ร ๐๐๐
= ๐. ๐ ร ๐๐โ๐ ๐
๐๐๐= ๐. ๐ ๐๐
Ans.1.6 nm