ANALOG INTEGRATED CIRCUITS (TWO STAGE AMPLIFIER) ECE 196 LAB 10 MENCHIE LABADAN BS ECE IV Submitted to: PROF. OLGA LABAJO GERASTA
ANALOG INTEGRATED
CIRCUITS
(TWO STAGE AMPLIFIER)
ECE 196 LAB 10
MENCHIE LABADAN
BS ECE IV
Submitted to:
PROF. OLGA LABAJO GERASTA
EXPERIMENT STEPSEXPERIMENT STEPSEXPERIMENT STEPSEXPERIMENT STEPS
TWO STAGE OP AMPTWO STAGE OP AMPTWO STAGE OP AMPTWO STAGE OP AMP
HSPICE CODE:
MOSFETS:
GAIN
GRAPH
DIFFERENTIAL OP AMPDIFFERENTIAL OP AMPDIFFERENTIAL OP AMPDIFFERENTIAL OP AMP
Mosfets mn6 and mn7 are removed. The new node of vout is at node x. The capacitor is now
connected at node x.
MOSFETS:
GRAPH
PERFORMANCE COMPARISON
Structure gain (dB) -3dB (MHz) G.B (MHz)
two stage amplifier 56.9 6.9734 7.2623
differential amplifier 43 0.48794 0.53044
Hence from the simulation results, the performance of the Two-Stage Operational amplifier is much
better than Differential Amplifier.
We have the following HPICE Codes
All the mosfets operate in saturation region as seen below and Id does not exceed 10uA,
The figure shows the graph of the phase margin and the gain.
The graph below shows the Vin Vout curve expressed in terms of time,
Hence the Phase Margin is,
PM = 180 – 177.86 = 2.14
As we connected the circuit as a unit gain buffer, we can see that the phase margin is less than 0
degrees. Hence we add the negative value of phase margin to 180 degrees to get our final phase
margin, which is 2.14. Hence the system is unstable. It is because the typical value of phase margin
must be larger than 45 degrees for it to be considered a stable system.
In my circuit, I added a compensation capacitor of 2.15pF. The HPICE code is,
Phase Margin = 180 – 135.07 = 44.93° ≈ 45°
After compensation, pole1 frequency decreased and pole 2 frequency increased making the phase
margin around 45° and also the system becomes more stable.
Effect on Phase Margin
Hspice code
at 2.15p PM=180-136.47=43.53
at 3pf PM=180-128.87=51.53
at 5pf PM=180-120.27=59.53
at 10pf PM= 180- 111.7=68.3
Conclusion: As we increase Cc the phase margin also increased
Effect on Slew Rate
Hspice code
At 2pF
At 3pF
At 5pF
At 10pF
Conclusion:As we increase Cc the slew rate decreases, because larger Cc need more time to charge
or discharge
Rc = 1/gm6 = 1/138.2344u = 7.234kΩ
First, we choose Rc =1.234k ohm to move the zero to infinite.
From the simulation, we can see that the gain does not increase after second pole, it means that the
zero is moved to infinite. And PM = 180-127.2= 52.8
We try a new value of Rc which is,
Rc =( 1/gm6)((Cc+Cl)/Cc) = (1/138.2344u) (2.15p+10p)/(2.15p) = 40.88kΩ
At Rc= 40.88kΩ
Phase Margin = 180 – 93.19 = 86.81
Hence the phase margin increased.
Input Common Range (ICMR) Typical Value – 0.7 to 3V
Vin and Vout are shorted.
The graph is,
ICMR = 1.77 - 0.572= 1.198. Hence it is still is in the range of the typical value for ICMR.
Output Voltage Swing
We have the hspice code,
The graph below shows the Vin and Vout curve for the output voltage swing,
Where Vout = -(100R/10R)Vin = -10Vin
R = 1 MΩ
COMMON MODE REJECTION RATIO
Common Mode Rejection Ratio (CMRR): the ability of ejecting
common mode signal.
HSPICE CODE:
two stage amp-CMRR_mench
.lib 'C:\synopsys\rf018.l' TT
.option post
mp3 a a vdd vdd pch l=1u w=5.5u
mp4 x a vdd vdd pch l=1u w=5.5u
mn1 x vin+ b gnd nch l=1u w=2.5u
mn2 a vin- b gnd nch l=1u w=2.5u
mn5 b c gnd gnd nch l=1u w=5u
mn8 c c gnd gnd nch l=1u w=5u
mp6 vout x vdd vdd pch l=1u w=33u
mn7 vout c gnd gnd nch l=1u w=12u
vdd vdd gnd 1.8
Cl vout gnd 10p
Iref vdd c 3.7u
vin+ vin+ gnd dc 0.9 ac 1
vin- vin- gnd dc 0.9 ac 0
.ac dec 10 100 10g
.print vdb(vout)
.alter
vin- vin- gnd dc 0.9 ac 1
.op
.end
CMRR = 88.3
βSLEW RATE AND SETTLING TIME
Slew Rate: the maximum rate of change dvo/dt is called slew
rate.
Settling Time: the time during output signal be stable shown as
left figure
We have the following HSPICE CODE for Vinp which has a peak voltage of 1V
The transient time and the slew rate is,
We have the following graph,
For a Vip of 1.4 V we have the following hspice code :
The transient time and slew rate is,
The graph is,
From the simulation, we can see that the slew rate will increase while the swing of input pulse
increases.
QUESTIONQUESTIONQUESTIONQUESTION
Ans
The common source is used for higher gain. If we use other structures, there’s a posiibility that it the
gain would be much lower.