Ankitha Miryala 310_Lab6

advertisement
Ankitha Miryala
Section 7
PSU ID: 950603219
TA: Alexander
Cocking
Formal Design Report
EE 310 Electronic Circuit Design
Fall 2013
Experiment 6
Project: Amplifier Design Using an Active
Load
Introduction:
In this experiment we will design an amplifier circuit, which will provide
a gain. We will be using the CD4007 chip that we used in the
experiment 4 and 5. Amplifiers are an important part of analog
circuits. MOSFETs can be configured in such a way that a small ac
input signal gives a significantly larger ac output signal. One common
way of doing this is to use PMOS transistors to construct a current
mirror to act as an active load for an amplifying NMOS transistor. The
main purpose of this experiment is to observe the voltage gains of the
various amplifier configurations and also observe how the body effect
changes the voltage gain. Basically we will be constructing an AC
amplifier using CMOS devices and using a current source as an active
load. First a basic common-emitter circuit is constructed and
examined and the measured voltage transfer characteristic is
compared to the expected results. Then the common-gate
configuration is examined and compared to the expected values.
Here we also take into account the body effect and finally the body
effect current is characterized using a specially designed circuit.
Table listing of the parameters for the MOSFETs from Lab 4:
Device
K(πœ‡A/V2)
πœ† (per mV)
VA (volt)
NMOS1
150
7.6
131.58
NMOS2
420
7.8
128.21
NMOS3
420
6.8
147.06
PMOS1
336
0.0420
23809.52
PMOS2
331
0.0419
24449.88
PMOS3
334
0.0421
23752.97
CIRCUIT SCHEMATICS
Task1 - Current Mirror/Active Load Schematic
Figure 1 - Current Mirror/Active Load
The 50 k resistor in series with the ammeter insures that VSD2 in this test is +5V; the
same value as it will be in the actual amplifier circuit (Figures 3 and 4).
Task 2 – Common-Source Amplifier Circuit Schematic
Task 5 – Common-Gate Amplifier Circuit Schematic
Figure 5: Schematic of ac equivalent circuit of the circuit shown in Figure 2, if the source Vi
is moved to node Y and node X is grounded.
Task 7 – Directly Measuring the Body Effect Schematic
Figure 7: Schematic of circuit in Figure 2 if vi is connected to node Z and nodes X and Y are both
grounded.
DATA AND GRAPHS
Task 1 Theoretical calculation of VGS2, R3
𝑖𝐷 = 𝐾𝑝 (𝑉𝑆𝐺 − 𝑉𝑇𝑃 )
𝑉𝑆𝐺 = √𝑖𝐷 ⁄𝐾𝑝 − 𝑉𝑇𝑃
𝑖𝑅𝐸𝐹 = 100πœ‡π΄ = 𝑖𝐷
𝑉𝑆𝐺 = √100πœ‡π΄⁄0.331π‘šπ΄ – 1.53𝑉
𝑉𝑆𝐺 = 2.08𝑉 : Theoretical value
By KVL,
𝑅3 =
𝑉𝐷𝐷 − 𝑉𝑆𝑆 − 𝑉𝑆𝐺
𝑖𝑅𝐸𝐹
𝑅3 =
10 − 0 − 2.08
100πœ‡π΄
𝑅3 = 79.2𝐾٠: Theoretical value
Measured Value of 𝑅3
𝑅3 = 90.8𝐾Ω
Task 2 Theoretical calculation of VGS1, R1, R2
Here we have to calculate 𝑉𝐺𝑆1 to make Io = 100πœ‡π΄
Assume 𝑅𝑠 𝐼𝐷 = 𝑉𝑆𝐷 β‰ͺ 1
𝑉𝐺𝑆 = (
𝑅2
) 10𝑉
𝑅1 + 𝑅2
And
𝑉𝐺𝑆 = √𝑖𝐷 ⁄𝐾𝑛 + 𝑉𝑇𝑛
𝑉𝐺𝑆 = √100πœ‡π΄⁄0.42π‘š + 1.14𝑉
𝑉𝐺𝑆 =1.628V
𝑅1 =
𝑉𝐷𝐷 − 𝑉𝑆𝑆 − 𝑉𝐺𝑆
10πœ‡π΄
𝑅1 =
10 − 0 − 1.628
10πœ‡π΄
𝑅1 = 837.2𝐾Ω
𝑅2 =
𝑉𝐺𝑆
10πœ‡π΄
𝑅2 =
1.628𝑉
10πœ‡π΄
𝑅2 = 162.8𝐾Ω
Check:
𝑉𝐺𝑆 = (
162.8𝐾Ω
) 10𝑉
837.2𝐾٠+ 162.8𝐾Ω
𝑉𝐺𝑆 = 1.628𝑉 Hence true.
Task 3: Record method used to adjust R1/R2
Here we are adjusting the Q-point of the device
To do this we use potentiometers to adjust the resistances to
get the desired value of current and output voltage
R1/R2 was adjusted using 𝑅2 = 250𝐾٠potentiometer
connected in series with R1 and adjusted until the DC Q point
was at approximately 5V (i.e. halfway between the supply
voltage and ground)
We choose
𝑅2 = 250𝐾٠pot to adjust the voltage
𝑅3 = 100𝐾٠pot to adjust the current
We have to adjust pots so that 𝑉𝑂𝐷𝐢 = 5𝑉 and πΌπ‘œ = 100πœ‡π΄
Here, we altered R2 so that the output voltage 𝑉𝑂_𝐷𝐢 = 5𝑉
Record actual R1, R2, Vo
From DMM we measured the respective values as below:
𝑅1 = 848𝐾٠(we used a 820𝐾٠and a 18𝐾٠in series)
𝑅2 = 178𝐾٠(we altered R2 so that the output voltage)
𝑉𝑂_𝐷𝐢 = 5𝑉
Measure VGS1, VGS2, ID1
𝑉𝐺𝑆1 = 1.52𝑉 of NMOS
𝑉𝑆𝐺2 = −1.96𝑉 of PMOS
πΌπ‘œ = 𝐼𝐷1 = 103πœ‡π΄
schematic) = 184kohm
Multimeter
XMM1 = Vo = 5.204 V
XMM2 = -V_GS2 = 1.952V
XMM3 *A/V = 97.467 uA
XMM4 = V_GS1 = 1.677V
The oscilloscope
CH1 = Vo
CH2 = Vi
Av = Vo/Vi = 76 v/v
Error%(V_GS1) = (1.677-1.45)/ 1.677 = 13.5%
Error%(V_GS2) = (2.02- 1.952)/1.952 = 3.5%
Error%(Id) = (97.46-93)/97.46 =4.6%
The main differences between simulated results and the actual value of the lab should
result from the different parameters used by the MULTISM and those of my own chips.
In addition, the multimeter can also interfere with the results.
Task 4 Small signal equivalent circuit & calculation of Av, Rin,
Rout
As we are done with the circuit biasing we will now add an AC
signal at node X in the figure for an AC amplifier. This is
respectively done in the small-signal model as well
π‘Ÿπ‘œ1 =
1
1
=
= 1.316𝑀𝛺
πœ†πΌπ·π‘„ 0.0076 ∗ 100𝑒 − 6
π‘Ÿπ‘œ2
1
1
=
=
= 238.1π‘˜π›Ί
πœ†πΌπ·π‘„ 0.0420 ∗ 100𝑒 − 6
𝑉𝐷𝑆 = 5.2𝑉 from DMM
π‘”π‘š = 2√𝐾𝑛 ∗ 𝐼𝐷𝑄 (1 + πœ‚π‘‰π·π‘† ) =
2√
150πœ‡π΄
𝑣2
∗ 100πœ‡π΄(1 + 0.0076 ∗ 5.2) = 249.74πœ‡S
π‘£π‘œ = −π‘”π‘š ∗ 𝑣𝑔𝑠 (π‘Ÿπ‘œ1 ||π‘Ÿπ‘œ2 )
𝑣𝑖 = 𝑣𝑔𝑠
π‘£π‘œ
= −π‘”π‘š ∗ (π‘Ÿπ‘œ1 ||π‘Ÿπ‘œ2 )
𝑣𝑖
π‘£π‘œ
𝑣𝑖
=249.74πœ‡S∗ (201.6π‘˜π›Ί) = −50.35𝑉/𝑉 is
AV(THEORITICAL)
To find the values of RIN and ROUT, one must reduce to the circuit by
removing any sources and adjusting components as shown above. Each
resistor can then be calculated by obtaining the correct equivalent
equations.
To find the values of Rin and Rout we should remove all the sources to get
the equivalent resistances at both the ends.
π‘…π‘œπ‘’π‘‘ = π‘Ÿπ‘œ1 ||π‘Ÿπ‘œ2 = 201.6π‘˜π›Ί
𝑅𝑖𝑛 = 𝑅1 ||𝑅2 = 837.2π‘˜π›Ί||162.8π‘˜π›Ί = 136.3π‘˜π›Ί
Plot of Vi and Vo & measurement of Av
The plot reveals values for VO and Vi which using the equation
AV = VO / Vi allows to determine the voltage gain. One thing to
keep in mind is both VO and Vi are positive number on the plot
but because they are out of phase the voltage gain is
automatically negative.
𝐴𝑣 =
−508π‘šπ‘‰
80π‘šπ‘‰
= −6.35 is AV(EXPERIMENTAL)
We tried hard during the lab but the oscilloscope results did not
give a gain close to 140 as it was supposed to be. The
discrepancy in the gain is because of our πœ† values of our PMOS
from lab 4.
Task 5 Small signal equivalent circuit & calculation of Av, Rin,
Rout
In task 5, we shorted X to ground and sent Vi from Y, making a
CG amplifier. We follow the same procedure of Task 4 except in
Task 5 the circuit is not a common source amplifier but a
common gate amplifier. This makes the AC small signal analysis
different by having the gate go to ground instead of source.
Also, we will have to take the body effect into consideration.
π‘…π‘œπ‘’π‘‘ = π‘Ÿπ‘œ1 ||π‘Ÿπ‘œ2 = 201.6π‘˜π›Ί
𝑅𝑖𝑛 = 51𝛺
𝐴𝑣 = (π‘”π‘š + π‘”π‘šπ‘ +
1
)𝑅
π‘Ÿπ‘œ1 π‘œπ‘’π‘‘
𝐴𝑣 = (249.74πœ‡π‘† + 62.44πœ‡π‘† + 0. .02)201.6π‘˜π›Ί
𝐴𝑣 = −629.355𝑉/𝑉 this is the 𝐴𝑣_𝐢𝐺
−1.36𝑉
𝐴𝑣 = 12.8π‘šπ‘‰ =-106.26
Task 6 Calculate π’ˆπ’Žπ’ƒ and 𝜼 from your measurements
π‘”π‘š = 2 ∗ √𝐾𝑛 (1 + πœ†πΌπ·π‘„ )
π‘”π‘š = 249.74πœ‡π‘†
π‘”π‘šπ‘ = η ∗ π‘”π‘š
Taking η = 0.25 as given in the lab instructions
π‘”π‘šπ‘ = 62.44πœ‡π‘†
|𝐴𝑣_𝐢𝑆 |
π‘”π‘š ∗ π‘…π‘œπ‘’π‘‘
𝟏
=
=
|𝐴𝑣_𝐢𝐺 | (π‘”π‘š + π‘”π‘šπ‘ ) ∗ π‘…π‘œπ‘’π‘‘ 𝟏 + η
50.35
106.26
η=
=
𝟏
𝟏+η
1
− 1 = 1.11
0.47388
π‘”π‘šπ‘ = η ∗ π‘”π‘š
π‘”π‘šπ‘ = 249.74πœ‡π‘† ∗ 1.11 = 277.27πœ‡π‘†
Task 7 Small signal analysis of test circuit
We conclude the experiment by designing a test circuit that
measures the body trans conductance and body effect
parameter directly.
We connect to Vi to node 2 (i.e.) connect node X to node Y, so
that we can get gmb directly.
Now we get rid of π‘”π‘š 𝑣𝑔𝑠 current so that we can fing the gmb.
AV(MEASURED) =
π‘£π‘œ
𝑣𝑖
=
1.06𝑉
25.6π‘šπ΄
=41.41V/V
π‘£π‘œ
= π‘”π‘šπ‘ ∗ (π‘Ÿπ‘œ1 ||π‘Ÿπ‘œ2 )
𝑣𝑖
π‘”π‘šπ‘ = (𝑣_π‘œ⁄𝑣_𝑖) / (π‘Ÿπ‘œ1 ||π‘Ÿπ‘œ2 )= 2.05*10-4
η=
π‘”π‘šπ‘
π‘”π‘š
=
2.05∗10−4
249.74∗10−6
= 0.822
AV (TEST) = ≅ 𝐴𝑉 𝐢𝐺 − |𝐴𝑉 𝐢𝑆 | = 99.156 − |−46.23| = 52.926
DISCUSSION
Reasoning behind task 7 test circuit (How does it work?)
The circuit used in TASK 7 is similar to the other circuits used in
this lab and works very similarly except that the gate and the
source are tied together here. Here, we connect node X to
node Y and measure the gain from the oscilloscope. From the
small signal viewpoint we eliminate the gmvgs so that we gave
only this
𝐴𝑣 = (π‘”π‘šπ‘ +
1
π‘Ÿπ‘œ1
)π‘…π‘œπ‘’π‘‘ , remaining so that we can directly
calculate π‘”π‘šπ‘ .
Compare theoretical values to measurements
Percent Error =
|𝐸π‘₯π‘π‘’π‘Ÿπ‘–π‘šπ‘’π‘›π‘‘π‘Žπ‘™−π‘€π‘’π‘Žπ‘ π‘’π‘Ÿπ‘’π‘‘ |
𝐸π‘₯π‘π‘’π‘Ÿπ‘–π‘šπ‘’π‘›π‘‘π‘Žπ‘™
∗ 100%
Task 1 – R3
Measured = 90.8𝐾Ω
Experimental = 79.2𝐾Ω
Percent Error = 14.7%
Task 3 – R1
Measured = 848kΩ (Measured from Task 2)
Experimental = 837.2kΩ
Percent Error = 1.29 %
Task 3 – R2
Measured = 178 kΩ
Experimental = 168.23 kΩ
Percent Error = 5.83 %
Task 3 - VGS
Measured = 1.52 V
Experimental = 1.628 V
Percent Error = 6.63 %
Task 4 - AV
Measured = -50.35 V/V
Experimental = -6.35 V/V
Percent Error = 6.93%
Task 5 – AV
Measured = 629.355 V/V
Experimental = 106.26 V/V
Percent Error = 492.3%
Task 6 Vs Task 7 – gmb
Measured = 2.77 * 10 -4 S
Experimental 2.05× 10−4 S
Percent Error = 35.1%
Task 6 Vs Task 7 – η
Measured = 1.11
Experimental = 0.822
Percent Error = 35.04%
Amount of errors and reasons for errors
We see that the percentage errors for the resistances show
good results and the values are satisfactory. But the huge
discrepancy in the gain errors for both common gate and
common source is because our πœ† values from lab 4 were off by
a factor of 10 and therefore the inaccuracy. We tried very hard
during the lab and outside of class period but the oscilloscope
dint give us the appropriate gain for the common source. We
thoroughly understood the lab because of the many errors and
obstacles during the lab but our values were off because of the
chip parameter errors. We see that theoretical values did not
match with the experimental values but we were happy to get
satisfactory π‘”π‘šπ‘ and πœ‚ values.
Compare measurements to simulated values
VO
Simulated = 1.176 V
Experimental = 1.21 V
Percent Error = 3.3%
Vi
Simulated = 25.44 mV
Experimental = 27.8mV
Percent Error = 8.63%
AV
Simulated= 50.26 V/V
Experimental = 71.72 V/V
Percent Error = 28.2%
Amount of errors and reasons for errors
From above we can see that most of the values were within the
desired range. Errors may be due to the non-ideal experimental
conditions such as voltage supplied, DMM resolution and
human errors such as calculation approximations etc.
Answer questions related to different configurations
Why is the voltage gain for the common gate larger than the
voltage gain for the common source?
Answer: It Is due to how the circuits are for their AC small
equivalent circuits. Depending on the set up of the dependent
voltage source and location of the resistors we see that the
voltage gain for common gate would be larger than the
voltage gain for common source.
How does the calculated voltage gain compare with the
experimental value determined for the amplifier? Can you
appreciate the problem of loading effect when measuring
either a DC or AC voltage signal at the output node?
Answer: The calculated voltage gain is almost similar to the one
we got in experimental calculations. We could see the
problems of loading effect for both DC and AV voltage signal
at the output node, as the sensitivity was very high.
How do the experimental values of gmb and η obtained by this
test circuit compare with those previously determined? Is there
a reason why one of these experimental approaches should be
preferable from the standpoints of measurement accuracy
and simplicity?
Answer: We had 35.1% error in the experimental and theoretical
values for body effect parameter and body trans
conductance. This may be due to the errors noted above. We
could use the theoretical values from the standpoint of
accuracy and simplicity as there is less room for errors such as
human errors but the experimental value is the one that will be
in real conditions and the theoretical value is more for ideal
conditions.
Summary, Conclusions & Attachments
In this experiment we investigated the use of NMOS amplifying
transistor with complementary NMOS device configured as a
current mirror/active load. We found that the body effect
influences the overall small signal gain performance. This came
particularly came into play for common gate (CG) and
common drain (CD) amplifier configurations.
We saw that the body effect increases the amplifier’s voltage
gain.
We compared various results with the corresponding simulated
or theoretical value and we found that the errors was within the
acceptable margin. However, some results had a considerable
error and this may be due the reasons noted above.
The lab report pages are attached to this report, design
proposals were submitted in pre-labs at the starting of each
lab.
Download