Lab 6 - EE 420L
Authored
by Hongzhong Li
Today 3/20/15
lih12@unlv.nevada.edu
Lab
description:
This lab will utilize the ZVN3306A and ZVP3306A MOSFETs
and we will experiment the 3 topologies of amplifiers: Common-Drain,
Common-Source, and Common gate. Finally, we will experiement the
push-pull amplifier topology.
Experiment#1: Common-Drain Amplifiers
- In your lab report discuss the operation of these circuits.
- In
a common-drain amplifier, the transistor's gate terminal serves as the
input and the source is the output so the drain is common to both input
and output.
- The
circuit uses a voltage divider to allow a small DC voltage supply to
the gate terminal of the transistor. This allows the circuit to
transoform impedances and the output to stablizes and therefore achieve
a unity gain of approximately 1.
- Hand
calculate, and then verify your hand calculations with experimentation
and simulations, the gains and the input and output
resistances ensuring that your test signals are at a high enough
frequency that the caps have negligible impedance but not so high that
the gain is dropping off.
Hand Calculation
- Simulation Result
Transient Simulation | Using the schematic above, We
can see that outputn and outputp are the same as vin. That means both
of them has roughly 1V/V gain. |
|  |
Input Resistance | By
adding a 33k resistor before the input capacitor in the schematic.The
output of NMOS and PMOS becomes half of the input signal. |
|  |
Output Resistance
| By
adding a 56ohms and 93 ohms (NMOS and PMOS, respectively) resistor in
series of a 10uF decoupling capacitor to the output of the circuit.The
output of NMOS and PMOS becomes half of the original value
(100mV/2=50mV in this case). |
|  |
Experimental Result
- For the experimental design, the input is 100mV@10kHz. From the
following figure:
- The NMOS gain is equal to 124mV/128mV = 0.9688
- The PMOS gain is equal to 120mV/124mV = 0.9677.
| NMOS Gain | PMOS Gain |
 |  |
- In your lab report discuss, in your own words, how to measure the input resistance.
- In
order to measure the input resistance, add the input resistance before
the input cap. Then test the voltage of resisotr two nodes, calculate
the current flows through the added resistance. Thus the input
resistance is the gate voltage divided by the current.
| NMOS Input Resistance | PMOS Input Resistance |
 |  |
- Again, in your lab report discuss how to measure the output resistance.
- In order to measure the output resistance, add the output resistance
series with 10uF to the output ode. Then measure the output voltage and
calcualte the current flows through the added resistance. And also
measure the input voltage on the gate of the MOSFET. Thus the output
resistance is the difference of the input voltage and output voltage
divided by the current.
| NMOS Output Resistance | PMOS Output Resistance |
 |  |
| Data Analysis-NMOS |
| Hand Calcution | Simulation | Experimental |
| Gain | 0.947 | 1 | 0.9677 |
| Rin | 33.3k ohms | 33k ohms | 33k ohms |
| Rout | 55 ohms | 56 ohms | 62 ohms |
| Data Analysis-PMOS |
| Hand Calcution | Simulation | Experimental |
| Gain | 0.917 | 1 | 0.9677 |
| Rin | 33.3k ohms | 33k ohms | 33k ohms |
| Rout | 93 ohms | 56 ohms | 62 ohms |
Experiment#2: Common-Source Amplifiers
- In your lab report discuss the operation of these circuits.
- In
a common-source amplifier, the input signal enters the gate of the
MOSFET and exists through the drain. The source is therefore common to
both input and output.
- The
circuit modulate the amount of current going into the load by changing
the voltage across the output resistance and therefore achieve a
considerable gain in a limited range of frequency response.
- Hand
calculate, and then verify your hand calculations with experimentation
and simulations, the gains and the input and output
resistances ensuring that your test signals are at a high enough
frequency that the caps have negligible impedance but not so high that
the gain is dropping off.
Hand Calculation
Simulation Result
Transient Simulation | Using
the schematic above, We
can see that outputn and outputp are about 7 times and 5 times larger
than the input with a 180degree phase shift, respectively . |
| 
|
Input Resistance | By
adding a 33k resistor before the input capacitor in the schematic.The
output of NMOS and PMOS becomes half of the original output value in the AC simulation. |
| 
|
Output Resistance
| By
adding a 1kohms resistor resistor in
series of a 10uF decoupling capacitor to both NMOS and PMOS output of the circuit.The
output of NMOS and PMOS becomes half of the original value
(70mV/2=35mV for NMOS an 50mV/2=25mV for PMOS in this case). |
|  |
Experimental Result
- For the experimental design, the input is 100mV@10kHz. From the
following figure:
- The NMOS gain is equal to 1.15/200mV = 5.75 with 90 degree phase
- The PMOS gain is equal to 336mV/196mV
= 1.7, this discrepency is due to the experimental transconductance is
much smaller than the hand calculation value of 11mA/V.
| NMOS Gain | PMOS Gain |
| 
|
- In your lab report discuss, in your own words, how to measure the input resistance.
- In
order to measure the input resistance, add the input resistance before
the input cap. Then test the voltage of resisotr two nodes, calculate
the current flows through the added resistance. Thus the input
resistance is the gate voltage divided by the current.
| NMOS Input Resistance | PMOS Input Resistance |
| 
|
- Again, in your lab report discuss how to measure the output resistance.
- In
order to measure the output resistance, add the output resistance
series with 10uF to the output ode. Then measure the output voltage and
calcualte the current flows through the added resistance. And also
measure the input voltage on the gate of the MOSFET. Thus the output
resistance is the difference of the input voltage and output voltage
divided by the current.
| NMOS Output Resistance | PMOS Output Resistance |
| 
|
| Data Analysis-NMOS |
| Hand Calcution | Simulation | Experimental |
| Gain | -6.5 | -6.7 | -5.75 |
| Rin | 33.3k ohms | 33k ohms | 33k ohms |
| Rout | 1k ohms | 1k ohms | 1k ohms |
| Data Analysis-PMOS |
| Hand Calcution | Simulation | Experimental |
| Gain | -5.2 | -5.2 | -1.7 |
| Rin | 33.3k ohms | 33k ohms | 33k ohms |
| Rout | 1k ohms | 1k ohms | 1k ohms |
Experiment#3: Common-Gate Amplifiers
- In your lab report discuss the operation of these circuits.
- In
a common-gate amplifier, the source terminal of the transistor serves
as the input, the drain is the output and the gate is connected to
ground which is common to both input and output.
- The
circuit operate as a current buffer which takes the current at the
input which have a small resistance and replicates it at the output
which works well due to its high output resistance.
- Hand
calculate, and then verify your hand calculations with experimentation
and simulations, the gains and the input and output
resistances ensuring that your test signals are at a high enough
frequency that the caps have negligible impedance but not so high that
the gain is dropping off.
Hand Calculation
Simulation Result
Transient Simulation | Using the schematic above, We
can see that outputn and outputp are about 7 times and 5 times larger than the input with no phase shift, respectively . |
| 
|
Input Resistance | By
adding a 155ohms to NMOS and 193 ohms to PMOS before the input capacitor in the schematic.The
output of NMOS and PMOS becomes half of the original output value in the AC simulation. |
| 
|
Output Resistance
| By
adding a 1kohms resistor resistor in
series of a 10uF decoupling capacitor to both NMOS and PMOS output of the circuit.The
output of NMOS and PMOS becomes half of the original value |
| 
|
Experimental Result
- For the experimental design, the input is 100mV@10kHz. From the
following figure:
- The NMOS gain is equal to 780mV/160mV = 4.9 with no phase shift.
- The PMOS gain is equal to 400mV/156mV
= 2.56, this discrepency is due to the experimental transconductance is
much smaller than the hand calculation value of 11mA/V.
| NMOS Gain | PMOS Gain |
| 
|
- In your lab report discuss, in your own words, how to measure the input resistance.
- In
order to measure the input resistance, add the input resistance before
the input cap. Then test the voltage of resisotr two nodes, calculate
the current flows through the added resistance. Thus the input
resistance is the gate voltage divided by the current.
| NMOS Input Resistance | PMOS Input Resistance |
| 
|
- Again, in your lab report discuss how to measure the output resistance.
- In
order to measure the output resistance, add the output resistance
series with 10uF to the output ode. Then measure the output voltage and
calcualte the current flows through the added resistance. And also
measure the input voltage on the gate of the MOSFET. Thus the output
resistance is the difference of the input voltage and output voltage
divided by the current.
| NMOS Output Resistance | PMOS Output Resistance |
| 
|
| Data Analysis-NMOS |
| Hand Calcution | Simulation | Experimental |
| Gain | 6.5 | 6.2 | 4.9 |
| Rin | 155 ohms | 155 ohms | 162 ohms |
| Rout | 1k ohms | 1k ohms | 1k ohms |
| Data Analysis-PMOS |
| Hand Calcution | Simulation | Experimental |
| Gain | 5.2 | 4.8 | 2.56 |
| Rin | 190ohms | 190 ohms | 620 ohms |
| Rout | 1k ohms | 1k ohms | 1k ohms |
Experiment#4: Push-Pull Amplifiers
Hand Calculation
- Discuss the operation of this amplifier in your lab report including both DC and AC operation.
DC operation:
If the input current is positive, the gate of M1(PMOS) is charged up
and thus shuts off. At the same time, the gate of M2 (NMOS) goes up and
turns it on. If the input current is negative, M1(PMOS) turns on and
M2(NMOS) shuts off
AC operation:
This amplifier bias the current in the output, a positive AC input
current causes both the gate of M1 and M2 to go up which shuts M2 off
and turn M1 on. M1 and M2 are pushing or pulling a current to/from the
output. If the resistance of
Rf is increased, the gain will go up proportional to it.
- Do you expect this amplifier to be good at sourcing/sinking current? Why or why not?
This amplifier
is good at sourcing sinking current because it includes both a PMOS and
a NMOS, when the PMOS is on, it sources current and when the NMOS is
on, it will sink current.
- What happens to the gain if the 100k resistor is replaced with a 510k resistor? Why?
If you increase R1 to 510k then the gain will be 5
times larger because of the equation of the gain above.- Again compare your hand calculations to simulation and experimental results.
Note that the gain of this amplifier is large so the
output may saturate at VDD and Ground. To avoid this saturation you can
reduce the AC input voltage using a voltage divider.
Simulation Result
In
the LTspice simulation, the input amplitude is 1mV @10kHz. The
first result uses RF=100k and its gain is roughly 2k V/V. The transient
analysis shows that the opamp can be pushed to ground but only pulled
to 4.4V instead of VDD. The second uses RF=510k and its gain is
3.4k V/V instead of 5 times. This is because the output amplitude can
not be pulled above VDD. The transient analysis shows that output range
is from ground to VDD.
| With 100k resistor |
 |
| With 510k resistor |
 |
Experimental Result
In
the experimental design, a voltage divider has been used by two
resistors 10k and 10 ohms. The input signal is 100mV@10kHz. The real
input of the opamp is 0.1mV@10kHz
For 510k Resistor : the output pushes to 4.88V before it saturates.
For 100k Resistor: the gain is 168mV/0.1mV=1.68k
With 510k Resistor
| With 100k resistor |
| 
|
| Data Analysis: Push-Pull amplifier |
| Gain-Hand Calculation | Experimental |
|
100k
| 3k | 1.68k |
|
This concludes the lab, all the work have been back up and uploaded to the email for future reference.
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