How to Tackle a Recipe. I.There are 3 commonly used recipe formats: Standard Action Narrative.
Module 4-3 Frequency Response annotatedee105/fa17/lectures/Module_4-3... · source terminal of...
Transcript of Module 4-3 Frequency Response annotatedee105/fa17/lectures/Module_4-3... · source terminal of...
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EE105 – Fall 2015Microelectronic Devices and Circuits
Frequency Response
Prof. Ming C. Wu
511 Sutardja Dai Hall (SDH)
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Amplifier Frequency Response:Lower and Upper Cutoff Frequency
• Midband gain Amid and upper and lower cutoff frequencies ωH and ωL that define bandwidth of an amplifier are often of more interest than the complete transfer function
• Coupling and bypass capacitors (~ 𝝁𝝁F) determine ωL
• Transistor (and stray) capacitances (~ pF) determine ωH
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Lower Cutoff Frequency (ωL) Approximation:Short-Circuit Time Constant (SCTC) Method
1. Identify all coupling and bypass capacitors
2. Pick one capacitor (𝑪𝑪𝒊𝒊) at a time, replace all others with short circuits
3. Replace independent voltage source with short, and independent current source with open
4. Calculate the resistance (𝑹𝑹𝒊𝒊𝑺𝑺) in parallel with 𝑪𝑪𝒊𝒊
5. Calculate the time constant, 𝟏𝟏𝑹𝑹𝒊𝒊𝑺𝑺𝑪𝑪𝒊𝒊
6. Repeat this for each of the n capacitor
7. The low cut-off frequency can be approximated by
ωL ≅1
RiSCii=1
n
∑
Note: this is an approximation. The real low cut-off is slightly lower
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Lower Cutoff Frequency (ωL)Using SCTC Method for CS Amplifier
SCTC Method:
fL ≅1
2π1
RiSCii=1
n
∑
For the Common-Source Amplifier:
fL ≅1
2π1
R1SC1
+1
R2SC2
+1
R3SC3
#
$%
&
'(
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Lower Cutoff Frequency (ωL)Using SCTC Method for CS Amplifier
Using the SCTC method:
fL ≅1
2π1
R1SC1
+1
R2SC2
+1
R3SC3
"
#$
%
&'
For C1:
R1S = RI + RG RiG( ) = RI + RG
For C3 :
R2S = RS RiS = RS1gm
For C2 :
R3S = R3 + RD RiD( ) = R3 + RD ro( )
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Design: How Do We Choose the Coupling and Bypass Capacitor Values?
• Since the impedance of a capacitor increases with decreasing frequency, coupling/bypass capacitors reduce amplifier gain at low frequencies.
• To choose capacitor values, the short-circuit time constant (SCTC) method is used: each capacitor is considered separately with all other capacitors replaced by short circuits.
• To be able to neglect a capacitor at a given frequency, the magnitude of the capacitor’s impedance must be much smaller than the equivalent resistance appearing at its terminals at that frequency
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Coupling and Bypass Capacitor Design Common-Source Amplifiers
For the C-S Amplifier:
Rin = RG RinCS Rout = RD Rout
CS
For coupling capacitor C1 :
C1 >>1
ω RI + Rin( )For coupling capacitor C3 :
C2 >>1
ω R3 + Rout( )
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Lower Cutoff Frequency fLDominant Pole Design
• Instead of having the lower cutoff frequency set by the interaction of several poles, it can be set by the pole associated with just one of the capacitors. The other capacitors are then chosen to have their pole frequencies much below fL.
• The capacitor associated with the emitter or source part of the circuit tends to be the largest due to low resistance presented by emitter or source terminal of transistor and is commonly used to set fL.
• Values of other capacitors are increased by a factor of 10 to push their corresponding poles to much lower frequencies.
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Capacitively Coupled CS Amplifier
Find the pole frequency associated with each coupling/bypass capacitor:
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Figure 10.7 Sketch of the low-frequency magnitude response of a CS amplifier for which the three
pole frequencies are sufficiently separated for their effects to appear distinct.
Low Frequency Response
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High Frequency Response
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Capacitors in MOS Device
Cgs = (2 / 3)WLCox +Cov
Cgd =Cov
Csb =Cjsb(area+ perimeter) junctionCdb =Cjdb(area+ perimeter) junction
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(Simplified) High-Frequency Equivalent-Circuit Model for MOSFET
Capacitances between source/body, Csb, and betweendrain/body, Cdb, are neglected
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Intrinsic Response of FET:Unity-Gain Frequency, fT
Vgs =Ii
sCgs + sCgd
KCL: Io +Vgs
1/ sCgd
= gmVgs
Io = gmVgs − sCgdVgs ≈ gmVgs
= gmIi
sCgs + sCgd
AI =IoIi=
gmsCgs + sCgd
s = jω
IoIi=
gmω(Cgs +Cgd )
ωT =gm
Cgs +Cgd
fT : defined as frequency at which short-circuit current gain = 1fT : a figure-of-merit for transistor speed
Drain is grounded (short-circuit load)
As gate length reduces in advancedtechnology node, Cgs reduces and fTincreases
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Common-Source Voltage Amplifier
• Small-signal model:
• Csb is connected to ground on both sides, therefore can be ignored
• Can solve problem directly by nodal analysis or using 2-port models of transistor
• OK if circuit is “small” (1-2 nodes)
RS
We can find the complete transfer function of the circuit, but in many cases it’s good enough to get an estimate of the -3dB bandwidth
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CS Voltage Amp Small-Signal Model
Two Nodes! “Easy”
For now we will ignore Cdb to simplify the math
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Frequency Response
KCL at input and output nodes;; analysis is made complicated
VoutVin
=−gm ro || RL[ ] 1− jω /ωz( )1+ jω /ω p1( ) 1+ jω /ω p2( )
Low-frequency gain:
Zero: ωz =gm
Cgs +C gd
VoutVin
=−gm ro || RL⎡⎣ ⎤⎦ 1− j0( )1+ j0( ) 1+ j0( )
→−gm ro || RL⎡⎣ ⎤⎦
Two Poles
Zero
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Calculating the Poles
ω p1 ≈1
Rs Cgs + 1+ gm ′Rout( )Cgd + ′RoutCgd
ω p2 ≈′Rout / RS
RS Cgs + 1+ gm ′Rout( )Cgd + ′RoutCgd
These poles are calculated after doing some algebraic manipulations on the circuit. It’s hard to get any intuition from the above expressions.There must be an easier way!
Usually >> 1
Results of complete analysis: not exact and little insight
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Method: The Miller Effect
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The Miller Effect
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Using The Miller Effect
Effective input capacitance:
Cin =1
jωCMiller
=1
1− Av,Cgd
⎛
⎝⎜⎜
⎞
⎠⎟⎟
1jωCgd
⎛
⎝⎜⎜
⎞
⎠⎟⎟=
1jω 1− AvCgd( )Cgd⎡⎣ ⎤⎦
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CS Voltage Amp Small-Signal Model
Modified Small-Signal Model with Miller Effect:
Cgs+CMiller
• We can approximate the first pole by using Miller capacitance
• This gives us a good approximation of the -3dB bandwidth
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Comparison with “Exact Analysis”
Miller result (calculate RC time constant of input pole):
Exact result:
ω p1−1 = RS Cgs + 1+ gm ʹRout( )Cgd
⎡⎣ ⎤⎦+ ʹRoutCgd
ω p1−1 = RS Cgs + 1+ gm ʹRout( )Cgd⎡
⎣⎤⎦
As a result of the Miller effect there is a fundamental gain-bandwidth tradeoff
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Common Drain AmplifierCalculate Bandwidth of the
Common Drain (Source-Follower)
Procedure:
1. Replace current source with MOSFET-based current mirror
2. Draw small-signal model with capacitors (for simplicity, we will focus on Cgd and Cgs)
3. Find the DC small-signal gain4. Use the Miller effect to
calculate the input capacitance5. Calculate the dominant pole
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Small-Signal Model with Capacitors
• Find DC Gain• Find Miller capacitor for Cgs -- note that the gate-source capacitor is between the input and output!
RS
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Voltage Gain Across Cgs
voutro roc
= gmvgs = gm(vin − vout )
vout1ro roc
+ gm⎛
⎝
⎜⎜
⎞
⎠
⎟⎟= gmvin
voutvin
=gm1ro roc
+ gm⎛
⎝⎜⎜
⎞
⎠⎟⎟
=gm(ro roc )
1+ gm(ro roc )= AvCgs
Write KCL at output node:
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RS
Compute Miller Effected Capacitance
Now use the Miller Effect to compute Cin:Remember that Cgs is the capacitor from the input to the output
Cin =Cgd +CM
Cin =Cgd + (1− AvCgs )Cgs
Cin =Cgd + (1−gm (ro roc )1+ gm (ro roc )
)Cgs
Cin =Cgd + (1
1+ gm (ro roc ))Cgs
Cin ≈ Cgd
Miller Cap
(for large gm(ro//roc))
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Bandwidth of Source Follower
Input low-pass filter’s –3 dB frequency:
ω p−1 = RS Cgd +
Cgs
1+ gm (ro roc )
⎛
⎝⎜⎜
⎞
⎠⎟⎟
Substitute favorable values of RS, ro:
mS gR /1» ro >>1/ gm
ω p−1 ≈ 1/ gm( ) Cgd +
Cgs
1+BIG⎛
⎝⎜
⎞
⎠⎟ ≈ Cgd / gm
ω p ≈ gm /Cgd
Very high frequency!Model not valid at these high frequencies
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Some ExamplesCommon Source Amplifier:
AvCgd = Negative, large number (-100)
Common Drain Amplifier:
AvCgs = Slightly less than 1
CMiller = (1− AV ,Cgd )Cgd ≈100Cgd
CMiller = (1− AV ,Cgs )Cgs ! 0
Miller Multiplied Cap has detrimental impact on bandwidth
“Bootstrapped” cap has negligible impact on bandwidth!