Lab 4
Part 1: Sizing Chart
We want to size a pmos CD transistor with a current source transistor with the following
specs:
(Specs)
As we used a current source we forced the value of the current and as the current is a strong function of VGS
,so we used a voltage source at the gate with 0 dc voltage ➔so VGS = VDS and VGS is determined by the
current and the transistor is in saturation .
(Pmos sizing)
PART 2: CD Amplifier
Simulate the OP point. Report a snapshot clearly showing the following parameters (add a
filter to your monitor).
(DC operating point for CD transistor)
Check that the transistor operates in saturation.
Yes , the transistor operates in saturation as the VGS = VDS then VGS – VTH will be of course less than VDS.
2. AC Analysis
Perform AC analysis (1Hz:10GHz, logarithmic, 20points/decade) to investigate the frequency
domain peaking.
(Bode plot magnitude)
As we can notice in the Bode plot ,there is a frequency domain peaking at a certain value
Which is
(frequency domain peaking)
So from this value we can predict that there is an underdamped system in which there complex conjugate
pole
Do you notice frequency domain peaking?
Yes , there exists frequency domain peaking at gain = 4.329 .
Analytically calculate quality factor (use approximate expressions). Is the system
underdamped or overdamped?
To calculate the quality factor we will use the expression sqrt(b2)/b1
To get b2 we will use the formula
From this formula we will find b2 = 2.5647f
To calculate b1 we will use the formula
We will find b1 = 31.8n
So to sum up we will find Q = sqrt(b2)/b1 which is equal 1.5925 , so that proves that the system is an
underdamped system as Q >0.5 and that there exists complex conjugate poles and as Q is also > 0.707
That will proves that there exists frequency domain peaking which shows that there exists Ringing (overshoot) in
step response (time domain), so we can predict that the overshoot value is close to e^(-pi/sqrt(4*Q^2-1))*100=35.38%
the overshoot value should be higher as we neglected the body effect in calculating Q.
(Optional) Perform parametric sweep: CL = 2p, 4p, 8p.
Report Bode plot magnitude overlaid on same plot.
Report the peaking vs CL.
Comment.
If we neglect the body effect and Cl had a large value we can see that the Q will have a inverse proportional
relation with CL which means that as long as CL increase Q decrease and the frequency domain peaking will
Decrease as in the graph and will became more flat.
(Optional) Perform parametric sweep: Rsig = 20k, 200k, 2M.
Report Bode plot magnitude overlaid on same plot.
Report the peaking vs Rsig.
Comment.
If we neglect the body effect and Cl had a large value we can see that the Q will have a direct proportional
relation with Rsig which means that as long as RSIG increase Q increase and the frequency domain peaking will
increase as in the graph and will have a larger peaking value.
3. Transient Analysis
Report Vin and Vout overlaid vs time.
Calculate the DC voltage difference (DC shift) between Vin and Vout.
From this graph we can see that the Dc shift between Vin and Vout in about 605mv
What is the relation between the DC shift and VGS?
the VGS is equal to the DC shift because our CD amplifier work as a follower , so the voltage at source node is
higher than the gate node and so on the current is a strong function in VGS so,it conserve the value.
How to shift the signal down instead of shifting it up?
We should use an Nmos transistor instead of Pmos.
Do you notice time domain ringing?
Yes , which was expected from above when we found that Q >0.5
Perform parametric sweep: CL = 2p, 4p, 8p.
• Report Vout vs time overlaid on same plot.
• Report the overshoot vs CL.
• Comment.
As we predicted from above the overshoot value at CP =2p should be around 37 ,the value calculated from
above was less because we neglected the body effect in our calculation of Q and also overshot relation Vs CP
is similar to peaking VS CP because overshot is a function of Q so as CP increase Q decrease and overshot
decrease.
6) (Optional) Perform parametric sweep: Rsig = 20k, 200k, 2M.
• Report Vout vs time overlaid on same plot.
• Report the overshoot vs Rsig.
• Comment.
As we predicted from above the overshoot value at Rsig should be around 37 ,the value calculated from
above was less because we neglected the body effect in our calculation of Q and also overshot relation Vs Rsig
is similar to peaking VS Rsig because overshot is a function of Q so as Rsig increase Q increase and overshot
increase.
4. Zout (Inductive Rise) (optional)
Plot the output impedance (magnitude and phase) vs frequency. Do you notice an inductive
rise?Why?
(Phase)
Yes there is a inductive effect because of cgs effect as Rsig//ro became larger than 1/gm at higher frequency
But because cgs is bootstrapped as CD is a buffer ,so the two poles are close to each other and so Zout doesn’t
Reach the Zout value of Rsig//ro which is 700k and cgd effect arise and Zout began to decrease to 0.
4) Does Zout fall at high frequency? Why?
Hint: Cgd appears in parallel with Rsig.
because cgs is bootstrapped as CD is a buffer ,so the two poles are close to each other and so Zout doesn’t
Reach the Zout value of Rsig//ro which is 700k and cgd effect arise and Zout began to decrease to 0.
5) Analytically calculate the zeros, poles, and magnitude at low/high frequency for Zout.
Compare with
simulation results in a table.
divide by 2pi don’t
forget.
Zout at low frequency =1/gm+gmb=7.47Kohm
Zout at high frequency=rsig//ro=753Kohm
Simulation
If we compare the analatical values with simulation we will find that the values are far from each other
because
The two poles are nearby and possibly complex conjugate
OCTC technique and Miller approx cannot be used
Pole 1 dominant 13.4828MHZ 13.841MHZ
Pole 2 115.850MHZ 103.7149MHZ
zero 1.17422MHZ 1.349456MHZ
