12 - Differential Amplifiers Example: OpAmp ECSE 334...

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1 ECSE334 – Introduction to Microelectronics ECSE 334 Introduction to Microelectronics Roni Khazaka Lecture #12 ECSE334 – Introduction to Microelectronics 2 Khazaka - 2008 Differential Amplifiers Example: OpAmp Amplifies differential signal v D + - + - ( ) + = v v A v d o ECSE334 – Introduction to Microelectronics 3 Khazaka - 2008 Differential Large Signal Operation Q 2 R D v D1 v S Q 1 R D V DD I i D1 i D2 v O -+ v D2 v G1 v G2 v CM + - 2 id v 2 id v + - - + 2 id CM v v + 2 id CM v v ECSE334 – Introduction to Microelectronics 4 Khazaka - 2008 Differential Large Signal Operation Can show that (see S&S sec 7.1.3): 2 1 2 1 2 2 + = OV id id OV D V v v V I I i 2 2 2 1 2 2 = OV id id OV D V v v V I I i () L W k I V n OV = Overdrive voltage when the current is balanced (1/2 in each transistor) ECSE334 – Introduction to Microelectronics 5 Khazaka - 2008 Differential Large Signal Operation L W k I V n OV = Normalized current Higher slope Î Higher Gain Lower V OV Î Higher Gain Lower V OV Î Less Linearity Graph below, assumes constant bias current I ECSE334 – Introduction to Microelectronics 6 Khazaka - 2008 Differential Large Signal Operation L W k I V n OV = For a constant bias current I : Increase V OV Î Decrease Gain Increase V OV Î Increase linearity Î Tradeoff Gain vs. Linearity We can increase the gain by increasing the bias current I, but this is done at the expense of more power consumption.
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2 ECSE334 – Introduction to Microelectronics 7 Khazaka - 2008 Large Signal vs. Small Signal Small Signal Operation ECSE334 – Introduction to Microelectronics 8 Khazaka - 2008 Differential Large Signal Operation + 2 2 1 id OV D v V I I i 2 2 2 id OV D v V I I i Can use linear small signal model when: 1 2 2 << OV id V v 2 1 2 1 2 2 + = OV id id OV D V v v V I I i 2 2 2 1 2 2 = OV id id OV D V v v V I I i ECSE334 – Introduction to Microelectronics 9 Khazaka - 2008 Example mA 4 . 0 = I 2 mA/V 2 . 0 = = ox n n C k μ For V OV =0.2, 0.3, and 0.4V. Find W/L , and g m . Also, find maximum v id such that: 1 . 0 2 2 < OV id V v () L W k I V n OV = 2 OV n V k I L W = V 2 . 0 = OV V 50 V 2 . 0 mA/V 2 . 0 mA 4 . 0 2 2 = = L W mA/V 2 2 mA 4 . 0 50 mA/V 2 . 0 2 2 2 = × = = D ox n m I L W C g ECSE334 – Introduction to Microelectronics 10 Khazaka - 2008 Example mA 4 . 0 = I 2 mA/V 2 . 0 = = ox n n C k For V OV =0.2, 0.3, and 0.4V.
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12 - Differential Amplifiers Example: OpAmp ECSE 334...

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