![EE105 Fall 2007 Lecture 13, Slide 2 Prof. Liu, UC Berkeley
Cascoding Cascode?
• Recall that the output impedance seen looking into the
collector of a BJT can be boosted by as much as a factor
of β, by using a BJT for emitter degeneration.
• If an extra BJT is used in the cascode configuration, the
maximum output impedance remains βro1.
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![EE105 Fall 2007 Lecture 13, Slide 23 Prof. Liu, UC Berkeley
Input Impedance of CE Stage
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Lecture 13
- 1. EE105 Fall 2007 Lecture 13, Slide 1 Prof. Liu, UC Berkeley Lecture 13 OUTLINE • Cascode Stage: final comments • Frequency Response – General considerations – High-frequency BJT model – Miller’s Theorem – Frequency response of CE stage Reading: Chapter 11.1-11.3 ANNOUNCEMENTS • Midterm #1 (Thursday 10/11, 3:30PM-5:00PM) location: • 106 Stanley Hall: Students with last names starting with A-L • 306 Soda Hall: Students with last names starting with M-Z • EECS Dept. policy re: academic dishonesty will be strictly followed! • HW#7 is posted online.
- 2. EE105 Fall 2007 Lecture 13, Slide 2 Prof. Liu, UC Berkeley Cascoding Cascode? • Recall that the output impedance seen looking into the collector of a BJT can be boosted by as much as a factor of β, by using a BJT for emitter degeneration. • If an extra BJT is used in the cascode configuration, the maximum output impedance remains βro1. ( ) 11211 121121 || ||)]||(1[ ooomout ooomout rrrrgR rrrrrgR βπ ππ ≤≈ ++= 1111max, 121121 ||)]||(1[ oomout ooomout rrrgR rrrrrgR β ββ π ππ =≈ ++≤
- 3. EE105 Fall 2007 Lecture 13, Slide 3 Prof. Liu, UC Berkeley Cascode Amplifier • Recall that voltage gain of a cascode amplifier is high, because Rout is high. • If the input is applied to the base of Q2 rather than the base of Q1, however, the voltage gain is not as high. – The resulting circuit is a CE amplifier with emitter degeneration, which has lower Gm. ( )21211 πrrgrgA omomv −≈ ( )212 2 1 oom m in o m rrg g v i G + =≡
- 4. EE105 Fall 2007 Lecture 13, Slide 4 Prof. Liu, UC Berkeley Review: Sinusoidal Analysis • Any voltage or current in a linear circuit with a sinusoidal source is a sinusoid of the same frequency (ω). – We only need to keep track of the amplitude and phase, when determining the response of a linear circuit to a sinusoidal source. • Any time-varying signal can be expressed as a sum of sinusoids of various frequencies (and phases). Applying the principle of superposition: – The current or voltage response in a linear circuit due to a time-varying input signal can be calculated as the sum of the sinusoidal responses for each sinusoidal component of the input signal.
- 5. EE105 Fall 2007 Lecture 13, Slide 5 Prof. Liu, UC Berkeley High Frequency “Roll-Off” in Av • Typically, an amplifier is designed to work over a limited range of frequencies. – At “high” frequencies, the gain of an amplifier decreases.
- 6. EE105 Fall 2007 Lecture 13, Slide 6 Prof. Liu, UC Berkeley Av Roll-Off due to CL • A capacitive load (CL) causes the gain to decrease at high frequencies. – The impedance of CL decreases at high frequencies, so that it shunts some of the output current to ground. −= L Cmv Cj RgA ω 1 ||
- 7. EE105 Fall 2007 Lecture 13, Slide 7 Prof. Liu, UC Berkeley Frequency Response of the CE Stage • At low frequency, the capacitor is effectively an open circuit, and Av vs. ω is flat. At high frequencies, the impedance of the capacitor decreases and hence the gain decreases. The “breakpoint” frequency is 1/(RCCL). 1222 + = ωLC Cm v CR Rg A
- 8. EE105 Fall 2007 Lecture 13, Slide 8 Prof. Liu, UC Berkeley Amplifier Figure of Merit (FOM) • The gain-bandwidth product is commonly used to benchmark amplifiers. – We wish to maximize both the gain and the bandwidth. • Power consumption is also an important attribute. – We wish to minimize the power consumption. ( ) LCCT CCC LC Cm CVV VI CR Rg 1 1 nConsumptioPower BandwidthGain = = × Operation at low T, low VCC, and with small CL superior FOM
- 9. EE105 Fall 2007 Lecture 13, Slide 9 Prof. Liu, UC Berkeley Bode Plot • The transfer function of a circuit can be written in the general form • Rules for generating a Bode magnitude vs. frequency plot: – As ω passes each zero frequency, the slope of |H(jω)| increases by 20dB/dec. – As ω passes each pole frequency, the slope of |H(jω)| decreases by 20dB/dec. + + + + = 21 21 0 11 11 )( pp zz jj jj AjH ω ω ω ω ω ω ω ω ω A0 is the low-frequency gain ωzj are “zero” frequencies ωpj are “pole” frequencies
- 10. EE105 Fall 2007 Lecture 13, Slide 10 Prof. Liu, UC Berkeley Bode Plot Example • This circuit has only one pole at ωp1=1/(RCCL); the slope of |Av|decreases from 0 to -20dB/dec at ωp1. • In general, if node j in the signal path has a small- signal resistance of Rj to ground and a capacitance Cj to ground, then it contributes a pole at frequency (RjCj)-1 LC p CR 1 1 =ω
- 11. EE105 Fall 2007 Lecture 13, Slide 11 Prof. Liu, UC Berkeley Pole Identification Example inS p CR 1 1 =ω LC p CR 1 2 =ω
- 12. EE105 Fall 2007 Lecture 13, Slide 12 Prof. Liu, UC Berkeley High-Frequency BJT Model • The BJT inherently has junction capacitances which affect its performance at high frequencies. Collector junction: depletion capacitance, Cµ Emitter junction: depletion capacitance, Cje, and also diffusion capacitance, Cb. jeb CCC +≡π
- 13. EE105 Fall 2007 Lecture 13, Slide 13 Prof. Liu, UC Berkeley BJT High-Frequency Model (cont’d) • In an integrated circuit, the BJTs are fabricated in the surface region of a Si wafer substrate; another junction exists between the collector and substrate, resulting in substrate junction capacitance, CCS. BJT cross-section BJT small-signal model
- 14. EE105 Fall 2007 Lecture 13, Slide 14 Prof. Liu, UC Berkeley Example: BJT Capacitances • The various junction capacitances within each BJT are explicitly shown in the circuit diagram on the right.
- 15. EE105 Fall 2007 Lecture 13, Slide 15 Prof. Liu, UC Berkeley Transit Frequency, fT • The “transit” or “cut-off” frequency, fT, is a measure of the intrinsic speed of a transistor, and is defined as the frequency where the current gain falls to 1. π π C g f m T =2 Conceptual set-up to measure fT in in in Z V I = inmout VgI = in m T inT minm in out C g Cj gZg I I =⇒ = == ω ω 1 1
- 16. EE105 Fall 2007 Lecture 13, Slide 16 Prof. Liu, UC Berkeley Dealing with a Floating Capacitance • Recall that a pole is computed by finding the resistance and capacitance between a node and GROUND. • It is not straightforward to compute the pole due to Cµ1 in the circuit below, because neither of its terminals is grounded.
- 17. EE105 Fall 2007 Lecture 13, Slide 17 Prof. Liu, UC Berkeley Miller’s Theorem • If Avis the voltage gain from node 1 to 2, then a floating impedance ZFcan be converted to two grounded impedances Z1and Z2: v FF F A Z VV V ZZ Z V Z VV − = − =⇒= − 1 1 21 1 1 1 121 v FF F A Z VV V ZZ Z V Z VV 11 1 21 2 2 2 221 − = − −=⇒−= −
- 18. EE105 Fall 2007 Lecture 13, Slide 18 Prof. Liu, UC Berkeley Miller Multiplication • Applying Miller’s theorem, we can convert a floating capacitance between the input and output nodes of an amplifier into two grounded capacitances. • The capacitance at the input node is larger than the original floating capacitance. vAA −≡0 ( ) F F v F CAjA Cj A Z Z 00 1 1 1 1 1 1 + = + = − = ω ω F F v F C A jA Cj A Z Z + = + = − = 0 0 2 11 1 11 1 11 ω ω
- 19. EE105 Fall 2007 Lecture 13, Slide 19 Prof. Liu, UC Berkeley Application of Miller’s Theorem ( ) FCmS inp CRgR + = 1 1 ,ω F Cm C outp C Rg R + = 1 1 1 ,ω
- 20. EE105 Fall 2007 Lecture 13, Slide 20 Prof. Liu, UC Berkeley Small-Signal Model for CE Stage
- 21. EE105 Fall 2007 Lecture 13, Slide 21 Prof. Liu, UC Berkeley … Applying Miller’s Theorem ( )( )µ ω CRgCR CminThev inp ++ = 1 1 , ++ = µ ω C Rg CR Cm outC outp 1 1 1 , Note that ωp,out > ωp,in
- 22. EE105 Fall 2007 Lecture 13, Slide 22 Prof. Liu, UC Berkeley Direct Analysis of CE Stage • Direct analysis yields slightly different pole locations and an extra zero: ( ) ( ) ( ) ( ) ( )outinXYoutXYinCThev outXYCinThevThevXYCm p outXYCinThevThevXYCm p XY m z CCCCCCRR CCRCRRCRg CCRCRRCRg C g ++ ++++ = ++++ = = 1 1 1 2 1 ω ω ω
- 23. EE105 Fall 2007 Lecture 13, Slide 23 Prof. Liu, UC Berkeley Input Impedance of CE Stage ( )[ ] π µπω r CRgCj Z Cm in || 1 1 ++ ≈