Electronic Circuits EE359A
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1 Electronic Circuits EE359A Bruce McNair B Lecture 12 1
2 MOSFET vs. BJT current-voltage characteristic i C ( v) i D ( v) v The drain current of a MOSFET follows a square law relationship The collector current of a BJT follows an exponential relationship This means that the BJT can control a current that varies over ~5 orders of magnitude, compared to MOSFET current that varies as v OV 2 ( V), or about 1 order of magnitude. 2
3 MOSFET vs. BJT design parameters Significant parameter MOSFET W/L BJT A E (E-B junction area) Range :1 3
4 MOSFET vs. BJT design tradeoffs Available parameters MOSFET I D, V OV, W, L BJT I C, V BE, I S Useful parameters Pick any 3 I C (V BE is related and not very adjustable) 4
5 Differential and Multistage Amplifiers Ch 8 5
6 Noise issues with single ended systems Noise, interference Ground loops, offset reference 6
7 Noise issues with single ended systems Noise, interference Large effective antenna 7
8 DC offset issues with single ended systems offset reference 8
9 Differential benefits + - Differential signal Independent of ground reference + - 9
10 Differential benefits Differential amplification lends itself to twisted pair wiring + - Alternate loops cancel Small antenna loops
11 MOS differential pair MOSFETs are biased for saturation mode operation (not triode region) Resistive loads (for now) Ideal current source 11
12 Common-mode input voltage Assume identical MOSFETs, identical drain resistors 12
13 Common-mode input voltage Range of common mode voltage is limited: I V + V + V + V V V + V R 2 V CS ss CS t OV CM DD t D 13
14 Differential input voltage For all current to flow through Q 1 : 1 W I = k v V 2 L ( ) 2 ' n GS1 t 14
15 Differential input voltage For all current to flow through Q 1 : This determines limits of v id : 1 W I = k v V 2 L ( ) 2 ' n GS1 t 2V v 2V OV id OV 15
16 Differential input voltage For all current to flow through Q 1 : This determines limits of v id : 1 W I = k v V 2 L ( ) 2 ' n GS1 t 2V v 2V OV id OV Both transistors are in saturation, even though one is not conducting 16
17 Large signal operation 1 W i = k v V 2 L ( ) 2 ' D1 n GS1 t 1 W i = k v V 2 L ( ) 2 ' D2 n GS2 t v = v v = v v id GS1 GS 2 G1 G2 17
18 Large signal operation 1 W i = k v V 2 L ( ) 2 ' D1 n GS1 t 1 W i = k v V 2 L ( ) 2 ' D2 n GS2 t v = v v = v v id GS1 GS 2 G1 G2 1 W i i = k v 2 L I = i + i ' D1 D2 n id D1 D2 18
19 Large signal operation 1 W i = k v V 2 L ( ) 2 ' D1 n GS1 t 1 W i = k v V 2 L ( ) 2 ' D2 n GS2 t i D1 v = v v = v v id GS1 GS 2 G1 G2 I I vid vid /2 = VOV 2 VOV 2 1 W i i = k v 2 L I = i + i ' D1 D2 n id D1 D2 i D1 I I vid vid /2 = 1 2 VOV 2 VOV 2 19
20 Large signal operation 20
21 Large signal operation Nonlinear operating characteristics 21
22 Large signal operation Limited linear operating range v id /2 << V OV 22
23 Large signal operation Linear operating range v id /2 << V OV increase V Ov by using smaller W/L (but g m and gain are decreased) 23
24 Small signal differential gain Differential output v = V + v 1 G1 CM 2 id v = V v 1 G2 CM 2 id 24
25 Small signal differential gain v = V + v 1 G1 CM 2 id v = V v 1 G2 CM 2 id 25
26 Small signal differential gain v = V + v 1 G1 CM 2 id v = V v 1 G2 CM 2 id 26
27 Small signal differential gain v = V + v 1 G1 CM 2 id v = V v 1 G2 CM 2 id v v = g =+ g v R 2 v R 2 A vod = g R v id o1 m D id o2 m D d m D id 27
28 Differential amplifier with current source loads v A = g r r ( ) od d m o1 o3 vid 28
29 Cascode differential amplifier v A = g R R ( ) ( ) on m3 03 o1 op m5 05 o7 ( ) od d m on op vid R = g r r R = g r r R op R on 29
30 Common-mode rejection ratio 30
31 Common-mode rejection ratio Consider mismatched R D : Q 1 has drain resistor R D Q 2 has resistor R D +DR D v v o1 o2 A cm R 2R D SS C+ΔR 2R SS v D icm v icm ΔR D R D ΔR D = = 2RSS 2RSS RD 31
32 Common-mode rejection ratio Consider mismatched g m : Q 1 : g m + 1 / 2 Δg m Q 2: g m - 1 / 2 Δg m ( 1 2) 1 ( 1 2) ( Δg ) R g g R v = v v = v + + m m D od o1 o2 icm gm gm RSS A A cm cm vod m v = 1 + 2g R R D Δg m 2RSS gm D icm m SS 32
33 Common-mode rejection ratio CMRR = A A d cm CMRR 2gmRSS = ΔRD RD CMRR 2gmRSS = Δgm gm 33
34 BJT differential pair Q 1 and Q 2 must not saturate 34
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