Voltage Gain of Cascode Amplifier

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Lecture 11 ANNOUNCEMENTS • Prob. 5 of Pre‐Lab #5 has been clarified. (Download new version.) • HW#6 has been posted. • Review session: 3‐5PM Friday (10/5) in 306 Soda (HP Auditorium) • Midterm #1 (Thursday 10/11): • Material of Lectures 1‐10 (HW# 2‐6; Chapters 2, 4, 5) • 2 pgs of notes (double‐sided, 8.5”×11” ), calculator allowed OUTLINE • Review of BJT Amplflifiers • Cascode Stage Reading: Chapter 9.1 EE105 Fall 2007 Lecture 11, Slide 1Prof. Liu, UC Berkeley Review: BJT Amplifier Design • A BJT amplifier circuit should be designed to 1. ensure that the BJT operates in the active mode, 2. allow the desired level of DC current to flow, and 3. couple to a small‐signal input source and to an output “load”. Æ Proper “DC biasing” is required! (DC analysis using large‐signal BJT model) • Key amplifier parameters: (AC analysis using small‐signal BJT model)

– Voltage gain Av ≡ vout/vin – Input resistance Rin ≡ resistance seen between the input node and ground (with output terminal floating) – OtOutpu t resitistance Rout ≡ resitistance seen bbtetween the outttput node and ground (with input terminal grounded) EE105 Fall 2007 Lecture 11, Slide 2Prof. Liu, UC Berkeley Large‐Signal vs. Small‐Signal Models • The large‐signal model is used to determine the DC oppgerating point (VBE, VCE, IB, IC) of the BJT.

• The small‐signal model is used to determine how the output responds to an input signal.

EE105 Fall 2007 Lecture 11, Slide 3Prof. Liu, UC Berkeley Small‐Signal Models for Independent Sources

• The voltage across an independent voltage source does not vary with time. (Its small‐signal voltage is always zero.) It is regarded as a short circuit for small‐signal analysis.

Large-Signal Small-Signal Model Model

• The current through an independent current source does not vary with time. (Its small‐signal current is always zero.) It is regarddded as an open ciiitrcuit for small‐silignal analliysis.

EE105 Fall 2007 Lecture 11, Slide 4Prof. Liu, UC Berkeley Comparison of Amplifier Topologies

Common Emitter Common Base Emitter Follower

• Large Av < 0 • Large Av > 0 • 0 < Av ≤ 1 ‐ Degraded by RE ‐Degraded by RE and RS ‐ Degraded by RB/(β+1) ‐ Degraded by RB/(β+1) ‐ Degraded by RB/(β+1) • Large Rin • Moderate Rin • Small Rin (due to RE(β+1)) ‐ Increased by RB ‐ Increased by RB/(β+1) • Small Rout ‐ Increased by RE(β+1) ‐ Decreased by RE ‐ Effect of source • Rout ≅ RC • Rout ≅ RC impedance is reduced by β+1 • r degrades A , R • r degrades A , R o v otout o v otout ‐ DdDecreased by RE but impedance seen but impedance seen looking into the collector looking into the collector • ro decreases Av, Rin, can be “boosted” by can be “boosted” by and Rout emitter degeneration emitter degeneration

EE105 Fall 2007 Lecture 11, Slide 5Prof. Liu, UC Berkeley Common Emitter Stage

VA = ∞ VA < ∞ v − R out ≈ C 1 R vin B + RE + g m β +1

Rin = RB + rπ + (β +1)RE

Rout = RC Rout ≈ RC ||(rO[1+ gm(RE || rπ )])

EE105 Fall 2007 Lecture 11, Slide 6Prof. Liu, UC Berkeley Common Base Stage

VA = ∞ v R R out ≈ C ⋅ E 1 R vin B RS + RE + RS RE + gm β +1

⎛ 1 R ⎞ ⎜ B V < ∞ Rin ≈ ⎜ + ⎟ RE A ⎝ gm β +1⎠

Rout = RC Rout ≈ RC ||(rO[1+ gm(RE || rπ )])

EE105 Fall 2007 Lecture 11, Slide 7Prof. Liu, UC Berkeley Emitter Follower

V < ∞ VA = ∞ A

v R v R || r out = E out = E O 1 R 1 R vin S vin S RE + + RE || rO + + gm β +1 gm β +1

Rin = rπ + (1+ β )RE Rin = rπ + (β +1)(RE || rO ) ⎛ 1 R ⎞ ⎛ R 1 ⎞ R =⎜ + s ||R R = ⎜ s + || R || r out ⎜ g β +1 E out ⎜ E O ⎝ m ⎠ ⎝ β +1 gm ⎠ EE105 Fall 2007 Lecture 11, Slide 8Prof. Liu, UC Berkeley Ideal Current Source

Circuit Symbol I-V Characteristic Equivalent Circuit

• An ideal current source has infinite output impedance. How can we increase the output impedance of a BJT that is used as a current source?

EE105 Fall 2007 Lecture 11, Slide 9Prof. Liu, UC Berkeley Boosting the Output Impedance • Recall that emitter degeneration boosts the impedance seen looking into the collector. – This improves the gain of the CE or CB amplifier. However, headroom is reduced.

Rout = [1+ g m (RE || rπ )]rO + RE || rπ

EE105 Fall 2007 Lecture 11, Slide 10 Prof. Liu, UC Berkeley Cascode Stage

• In order to relax the trade‐off between output impedance and voltage headroom, we can use a transistor instead of a degeneration resistor:

Rout = [1+ gm (rO2 || rπ1 )]rO1 + rO2 || rπ1

Rout ≈ g m1rO1 ()rO2 || rπ1

IC 2 = I E1 ≅ IC 2 if β1 >>1

• VCE for Q2 can be as low as ~0.4V (“soft saturation”)

EE105 Fall 2007 Lecture 11, Slide 11 Prof. Liu, UC Berkeley Maximum Cascode Output Impedance

• The maximum output impedance of a cascode is

limited by rπ1.

If rO2 >> rπ1 : R ≈ g r r = β r out,max m1 O1 π 1 1 O1

EE105 Fall 2007 Lecture 11, Slide 12 Prof. Liu, UC Berkeley PNP Cascode Stage

Rout = [1+ gm1(rO2 || r 1)]rO1 + rO2 || rπ1

Rout ≈ gm1rO1(rO2 || rπ1 )

EE105 Fall 2007 Lecture 11, Slide 13 Prof. Liu, UC Berkeley False Cascodes

• When the emitter of Q1 is connected to the emitter of Q2, it’ s not a cascode since Q2 is a diode‐connected device instead of a current source.

⎡ ⎛ 1 ⎞⎤ 1 Rout = ⎢1 + g m1 ⎜ || rO 2 || r 1 ⎟⎥rO1 + || rO 2 || rπ 1 ⎣ ⎝ g m 2 ⎠⎦ g m 2

⎛ g m1 ⎞ 1 Rout ≈ ⎜1 + ⎟rO1 + ≈ 2rO1 ⎝ g m 2 ⎠ g m 2 EE105 Fall 2007 Lecture 11, Slide 14 Prof. Liu, UC Berkeley Short‐Circuit Transconductance

• The short‐circuit transconductance of a circuit is a measure of its strength in converting an input voltage signal into an output current signal.

iout Gm ≡ vin vout =0

EE105 Fall 2007 Lecture 11, Slide 15 Prof. Liu, UC Berkeley Voltage Gain of a Linear Circuit

• By representing a linear circuit with its Norton

equivalent, the relationship between Vout and Vin can be expressed by the product of Gm and Rout. Norton Equivalent Circuit

Computation of vout = −iout Rout = −Gm vin Rout short-circuit output current: vout vin = −Gm Rout

EE105 Fall 2007 Lecture 11, Slide 16 Prof. Liu, UC Berkeley Example: Determination of Voltage Gain

Determination of Gm Determination of Rout

iout vx Gm ≡ = gm1 Rout ≡ = ro1 vin ix vout =0

Av = −gm1rO1

EE105 Fall 2007 Lecture 11, Slide 17 Prof. Liu, UC Berkeley Comparison of CE and Cascode Stages

• Since the output impedance of the cascode is higher than that of a CE stage, its voltage gain is also higher.

vout = −gm1vinrO1

VA Av = −gm1rO1 = − Av ≈ −gm1rO2 gm2 (rO1 rπ 2 ) VT

EE105 Fall 2007 Lecture 11, Slide 18 Prof. Liu, UC Berkeley Voltage Gain of Cascode Amplifier

• Since rO is much larger than 1/gm, most of IC,Q1 flows into diode‐connected Q2. Using Rout as before, AV is easily calculated.

iout = gm1vin ⇒ Gm = gm1

π Av = −Gm Rout

= −g m1[(1+ g m2 ()rO1 || r 2 )rO 2 + rO1 || rπ 2 ]

≈ −g m1[g m2 (rO1 || rπ 2 )rO 2 ]

EE105 Fall 2007 Lecture 11, Slide 19 Prof. Liu, UC Berkeley Practical Cascode Stage • No current source is ideal; the output impedance is finite.

Rout ≈ rO3 || g m2rO2 (rO1 || rπ 2 )

EE105 Fall 2007 Lecture 11, Slide 20 Prof. Liu, UC Berkeley Improved Cascode Stage

• In order to preserve the high output impedance, a cascode PNP current source is used.

Rout ≈ [gm3rO3 (rO4 || r 3 )]|| [gm2rO2 (rO1 || rπ 2 )]

Av = −gm1Rout

EE105 Fall 2007 Lecture 11, Slide 21 Prof. Liu, UC Berkeley