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BJT Differential Amplifier: Analysis and Design, Study notes of Electrical and Electronics Engineering

University of Florida (UF)Electrical and Electronics Engineering

An in-depth analysis of the bjt differential amplifier circuit, including its review, small-signal analysis, and calculation of important parameters such as differential gain, common mode gain, and common mode rejection ratio (cmrr). The document also includes an example of finding ad, acm, and cmrr for a given circuit.

Typology: Study notes

Pre 2010

Uploaded on 03/18/2009

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koofers-user-u1v 🇺🇸

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1
L35. Differential Amplifier
•Review
– BJT differential pair with emitter resistance
– BJT differential pair with active load
Ri, Ro, Ad
• Reading
– Chapter 7.1 – 7.5 MOS/BJT Differential amp
BJT Diff Amp with Active Load
V
CC
VO
I (DC)
Q1Q2
QQ 43
2
I
(DC) 2
I
(DC)
2
V
g (ss) d
m2
V
g (ss) d
m
2
Vd2
Vd
−
2
I
(DC) 2
I
(DC)
P/N/P current
mirror
Active Load
N/P/N Diff.
Pair
• Uses high resistance seen at collector to replace RC
• Need BJT emitter at top and collector at bottom
– P/N/P current mirror as active load
pf3
pf4
pf5

Partial preview of the text

Download BJT Differential Amplifier: Analysis and Design and more Study notes Electrical and Electronics Engineering in PDF only on Docsity!

L35. Differential Amplifier

  • Review
    • BJT differential pair with emitter resistance
    • BJT differential pair with active load Ri , Ro , Ad
  • Reading
    • Chapter 7.1 – 7.5 MOS/BJT Differential amp

BJT Diff Amp with Active Load

VCC

V (^) O

I (DC)

Q 1 Q 2

Q 3 Q 4 2 (DC) I 2 (DC) I 2 (ss) gmVd 2 (ss) gmVd

2

Vd 2 − Vd 2 (DC) I 2 (DC) I

P/N/P current mirror Active Load

N/P/N Diff. Pair

  • Uses high resistance seen at collector to replace RC
  • Need BJT emitter at top and collector at bottom
    • P/N/P current mirror as active load

Small-signal Analysis

  • Circuit is not symmetric
    • cannot use half circuit technique
  • Determine gain from A (^) v=GmRo
    • G (^) m=short circuit transconductance
    • Ro =output resistance
  • To find Ro, we need to revisit the Common Base Amplifier - Find Input and Output Resistance of common base amplifier - Find Resistance of diode connected transistor

Input/output resistance of common-

base amplifier

R O

Small-signal Analysis

  • Ri = 2rπ=2(1+β)r (^) e
  • Ro(note: single-ended output) = r (^) o2//r (^) o
  • Differential (single-ended) gain:

Consider current mirror: i (^) o= g (^) m(v (^) d/2) – (g (^) m(-v (^) d/2)) = g (^) mv (^) d v (^) o = i (^) oRo Ad = v (^) o/v (^) d = g (^) mRo

CMMR

  • Common mode gain: See text for detailed derivation (7.5.4)

Acm = v (^) o/v (^) cm ≅ -r (^) o4/(β 3 R)

  • CMRR = |Ad |/|Acm| CMRR = g (^) m(r (^) o2//r (^) o4)(β 3 R)/r (^) o4 ≅ ½ β 3 g (^) mR

Example

  • Find Ad , Acm, and CMRR given a 1 mA current source with 100k incremental resistance. Given β=100, VA=100V for all BJTs.

Summary

  • BJT differential pair with active load Acm, CMRR
  • MOS differential pair