Analog Integrated Circuits (Jieh Tsorng Wu)

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Unformatted text preview: alanced, the current-source noise represents a common-mode signal and will produce no differential output. Noise 11-36 Analog ICs; Jieh-Tsorng Wu Effect of Ideal Feedback on Noise Performance vi2 ii2 vi2 v1 a vo ii2 v1 a vo f × vo f × vo • For ideal feedback systems, the equivalent input noise generators can be moved unchanged outside the feedback loop and the feedback has no effect on the circuit noise performance. Noise 11-37 Analog ICs; Jieh-Tsorng Wu Effect of Input Series Feedback Feedback on Noise Performance vi2 a vi2 v1 ii2 a RE vo a RF ii2 vf2 v1 RE a vo RF 2 ve vf2 = 4kT RF ∆f 2 ve = 4kT RE ∆f RF ve RE vf vi = vi a + ii aR + + RF + RE RF + RE ⇒ Noise vi2 = vi2 + ii2 R 2 + 4kT R∆f a a 11-38 R = RF RE ii ≈ ii a ii2 ≈ ii2 a Analog ICs; Jieh-Tsorng Wu Effect of Input Shunt Feedback Feedback on Noise Performance if2 RF vi2 a ii2 a v1 vo a vi ≈ vi a Noise vi2 ≈ vi2 a v1 ii2 if2 = 4kT ⇒ RF vi2 a vo 1 ∆f RF ii = ii a + vi a RF ii2 = ii2 + a 11-39 + if vi2 a 2 RF + 4kT 1 ∆f RF Analog ICs; Jieh-Tsorng Wu Effect of Feedback on Noise Performance To analyze the noise performance of a practical feedback system, first use the loading approximation according to its feedback configuration to find the loading for the input port due to the feedback network. For series feedback at the input vi2 = vi2 + ii2 |Zf b|2 + 4kT Rf b∆f a a ii2 ≈ ii2 a For shunt feedback at the input vi2 ≈ vi2 a ii2 = ii2 + a vi2 a |Zf b|2 + 4kT 1 ∆f Rf b where Zf b is the loading of the feedback network for the input port, and Rf b represents the resistive part (thermal noise) of the loading. Noise 11-40 Analog ICs; Jieh-Tsorng Wu Effect of Cµ on Noise Performance B rb 2 vb 2 ib Cµ B v1 rc C Ccs rπ Cπ g m v1 ro 2 ic E • Note that the collector-base capacitor Cµ represents single-stage shunt feedback, and thus does not significantly affect the equivalent input noise generators of a transistor, even if Miller effect is dominant. The capacitor itself contributes no noise. Also, in calculating ii2, the term vi2 /|Zf b|2 can be neglected, since |Zf b| = 1/|ωCµ | is quite a large at frequencies of interest. Noise 11-41 Analog ICs; Jieh-Tsorng Wu Single-Stage Amplifier with Local Feedback if2 if2 RF vi2 RF 2 Vo vi2 1 RF vi2 RF Vi RE ii2 1 RE ii2 2 RE ii2 RE 2 ve vi2 1 vi2 2 vi2 1 ≈ 4kT rb + 2gm ∆f 1 ≈ 4kT rb + + RE 2gm ∆f ∆f ii2 1 ii2 2 ii2 ∆f Noise ≈ 2qIB ∆f ≈ 4kT rb + 1 + RE 2gm 4kT ≈ 2qIB + ∆f RF ≈ 2qIB 11-42 Analog ICs; Jieh-Tsorng Wu Operational Amplifier Noise Model ii2 − a vi2 a ii2 + a • With FET input stage, the current noises can often be ignored at low frequencies since their values are small. Noise 11-43 Analog ICs; Jieh-Tsorng Wu A Low-Pass Filter Example Cf Cf 2 i1 ii2 − a Rf Vi Vo R1 if2 Rf Vo R1 R2 R2 vi2 a ii2 + a 2 v2 2 vo1 = 2 ia− + 2 i1 + if2 Rf 1 + j 2πf Rf Cf 2 2 vo2 = vi2 a + 2 2 ia+R2 + 2 v2 Rf /R1 1+ 1 + j 2πf Rf Cf 2 2 2 2 voT = vo1 + vo2 Noise 11-44 Analog ICs; Jieh-Tsorng Wu A Current Amplifier Example io Q2 Q1 20 k is 500 5k io vi2 a Q2 Q1 if2 ii2 a 5.5 k Noise 20 k 5k || 500 11-45 Analog ICs; Jieh-Tsorng Wu A Current Amplifier Example • Neglect flicker noise and assume IC1 = 0.5 mA IC2 = 1 mA rb1 = rb2 = 100 Ω β1 = β2 = 100 fT 1 = 300 MHz fT 2 = 500 MHz • For both first and second stages, the driving signals are high-impedance current sources, thus we need to consider only equivalent noise current generators. • The equivalent noise current from the 2nd stage is approximately 2qIB2 + 4kT 1 = 2q (10 µA + 2.6 µA) 20 kΩ which can be neglected when compared to 2qIC1 = 2q × 500 µA. Noise 11-46 Analog ICs; Jieh-Tsorng Wu A Current Amplifier Example • The equivalent input noise current for the amplifier is ii2 ∆f = ii2 a vi2 a + + 4kT 5.5 kΩ (5.5 kΩ)2 ∆f IC 4kT 4kT 1 + rb1 + + = 2q IB + 2gm1 5.5 kΩ (5.5 kΩ)2 |β1|2 ∆f = 2q 5 µA + 500 µA |β1 |2 + 2q × 0.2 µA + 2q × 9.1 µA 500 × 10−6 A2/Hz = 2q 14.3 + |β1|2 • We know that β (j f ) = Noise βo 1+ βf j fo T1 ⇒ 1 1 = 2 2 |β1(j f )| βo1 11-47 2 1+ βo1f 2 fT21 Analog ICs; Jieh-Tsorng Wu A Current Amplifier Example • The current gain of the amplifier is AI ≈ 11 and is constant up to B = 100 MHz = fT 1/3. The total output noise is B 2 ioT = 0 ii2 A2 d f = A2 × I ∆f I B 2q 14.3 + 0 = A2 × 2q × 10−6 14.3f + I 500 2 βo1 500 × 10−6d f |β1|2 3 f+ 500 f f2 3 T1 B 0 = A2 × 2q × 10−6 × (14.3B + 18.6B) = A2 × 1.05 × 10−15 A I I 2 • The equivalent input noise current is ii2 = T Noise 2 ioT A2 I ⇒ 11-48 ii T = 32.4 nA rms Analog ICs; Jieh-Tsorng Wu Feedback and Frequency Compensation Jieh-Tsorng Wu ES A December 5, 2002 1896 National Chiao-Tung University Department of Electronics Engineering Feedback Si Se So a Sf b f So = a · Se Sf b = f · So So Se = Si − Sf b a a 1 = = ≈ if T Closed-Loop Gain = A ≡ Si 1 + af 1+T f δA 1 δa Gain Sensitivity = = · A 1+T a Loop Gain = T ≡ a × f Feedback 12-2 1 Analog ICs; Jieh-Tsorng Wu Effect of Negative Feedback on Distortion If a is a nonlinear amplifier 3 2 So = a1Se + a2Se + a3Se + · · · T...
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This note was uploaded on 03/26/2013 for the course EE 260 taught by Professor Choma during the Winter '09 term at USC.

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