6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 1 Lecture 19 Transistor Amplifiers (I) Common Source Amplifier November 15, 2005 Contents: 1. Amplifier fundamentals 2. Common source amplifier 3. Common source amplifier with current source supply Reading assignment: Howe and Sodini, Ch. 8, 8.1 8.6 Announcements: Quiz 2: 11/16, 7:30 9:30 PM, open book, must bring calculator; lectures #10 18. Quiz 2 TA Review Session: 11/15, 7:30 9:30 PM,
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 2 Key questions What are the key figures of merit of an amplifier? How can one make a voltage amplifier with a single MOSFET and a resistor? How can this amplifier be improved?
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 3 1. Amplifier fundamentals Goal of amplifiers: signal amplification. V v OUT v IN R L vout output signal v IN V input signal Features of amplifier: Output signal is faithful replica of input signal but amplified in magnitude. Active device is at the heart of amplifier. Need linear transfer characteristics for distortion not to be introduced.
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 4 Signal could be represented by current or voltage four broad types of amplifiers: voltage v s R L v amplifier out i out transconductance amplifier R L i s transresistance amplifier RL v out i out i s current amplifier R L
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 5 More realistic transfer characteristics: v OUT Q output signal v IN input signal Transfer characteristics linear over limited range of voltages: amplifier saturation. Amplifier saturation limits signal swing. Signal swing also depends on choice of bias point, Q (also called quiescent point or operating point). Other features desired in amplifiers: Low power consumption. Wide frequency response [will discuss in a few days]. Robust to process and temperature variations. Inexpensive: must minimize use of unusual components, must be small (in Si area)
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 6 2. Common Source Amplifier Consider the following circuit: V =V DD R D i R signal source i D signal load v OUT R L V GG V =V SS Consider it first unloaded by R L. How does it work? V GG, R D and W/L of MOSFET selected to bias transistor in saturation and obtain desired output bias point (i.e. V OU T = 0). v GS i D i R v out A v = v out < 0; output out of phase from input, but if amplifier well designed, A v > 1. [watch notation: v OU T (t) = V OU T v out (t)]
0 VSS V DD V OUT V SS 0 VT V DD V SS V GG V SS 6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 7 Load line view of amplifier: I R =I D load line V DD V SS R D V GG V SS =V DD V SS V GG V SS V GG V SS =V T Transfer characteristics of amplifier: V OUT V DD Want: Bias point calculation; small signal gain; limits to signal swing frequency response [in a few days]
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 8 Bias point: choice of V GG, W/L, and R D to keep transistor in saturation and to get proper quiescent V OUT. Assume MOSFET is in saturation: W I D = µ n C ox (V GG V SS V T ) 2 2L I R = V DD V OUT R D If we select V OUT =0: W I D = I R = µ n C ox (V GG V SS V T ) 2 = 2L V DD R D Then: 2V DD V GG = W V SS V T R D L µ n C ox
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 9 Small signal voltage gain: draw small signal equivalent circuit model: R D G v in v gs gm v gs D r o v out S v in gm v in r o //R D v out v out = g m v in (r o //R D ) Then unloaded voltage gain: v out A vo = = g m (r o //R D ) v in
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 10 Signal swing: V DD signal source R D v OUT V GG V SS Upswing: limited by transistor going into cut off: v out,max = V DD Downswing: limited by MOSFET entering linear regime: V DS,sat = V GS V T or v out,min V SS = V GG V SS V T Then: v out,min = V GG V T
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 11 Effect of input/output loading: V DD signal source R D i R i D i L v OUT R L V GG Bias point not affected because selected V OU T =0. Signal swing: V SS Upswing limited by resistive divider: R L v out,max = V DD R L R D Downswing not affected by loading Voltage gain: input loading ( ): no effect because gate does not draw current; output loading (R L ): R L detracts from voltage gain because it draws current. A v = g m (r o //R D //R L ) <g m (r o //R D )
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 12 Generic view of loading effect on small signal operation: Two port network view of small signal equivalent circuit model of voltage amplifier: R in is input resistance R out is output resistance A vo is unloaded voltage gain R out v in R in Avo v in v out R L input loading unloaded circuit output loading Voltage divider at input: Voltage divider at output: v in = R inrin A vo v in v out = R LRout R L Loaded voltage gain: v out R in R L A v = = A vo R in R L R out
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 13 Calculation of input resistance, R in : load amplifier with R L apply test voltage (or current) at input, measure test current (or voltage) For common source amplifier: i t v t v gs g m v gs r o //R D R L v t i t i t =0 R in = = No effect of loading at input.
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 14 Calculation of output resistance, R out : load amplifier at input with apply test voltage (or current) at output, measure test current (or voltage) For common source amplifier: v gs g m v gs r o //R D i t vt v gs =0 g m v gs =0 v t = i t (r o //R D ) R out = v t i t = r o //R D
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 15 Two port network view of common source amplifier: Rout v in R in Avo v in v out R L input loading unloaded circuit output loading A v = v out Rin R L R L = A vo = g m (r o //R D ) R in R L R out RL r o //R D Or: A v = g m (r o //R D //R L )
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 16 Design issues of common source amplifier (unloaded): Examine bias dependence: A vo = g m (r o //R D ) g m R D Rewrite A vo in the following way: W V DD V DD A vo g m R D = 2 µ n C ox I D L I D ID Then, to get high A vo : V DD I D Both approaches imply R D = Consequences of high R D : V DD I D large R D consumes a lot of Si real state large R D eventually compromises frequency response Also, it would be nice not to use any resistors at all! Need better circuit.
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 17 3. Common source amplifier with current source supply V DD i SUP signal source RS i D signal load v OUT R L V GG V SS Loadline view: i SUP =I D load line V GG V SS =V DD V SS I SUP V GG V SS 0 V SS V DD V OUT V GG V SS =V T
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 18 Current source characterized by high output resistance: r oc. Then, unloaded voltage gain of common source stage: A vo = g m (r o //r oc ) significantly higher than amplifier with resistive supply. Can implement current source supply by means of p channel MOSFET: V DD V B i SUP signal source i D v OUT V GG V SS
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 19 Relationship between circuit figures of merit and device parameters Remember: g m = W 2 µ n C ox I D L 1 L r o λn I D ID Then: Circuit Parameters Device A vo R in R out Parameters g m (r o //r oc ) r o //r oc I SU P W µ n C ox L adjustments are made to V GG so none of the other parameters change CS amp with current supply source is good voltage amplifier (R in high and A v high), but R out high too voltage gain degraded if R L r o //r oc.
6.012 Microelectronic Devices and Circuits Fall 2005 Lecture 19 20 Key conclusions Figures of merit of an amplifier: gain signal swing power consumption frequency response robustness to process and temperature variations Common source amplifier with resistive supply: tradeoff between gain and cost and frequency response. Trade off resolved by using common source amplifier with current source supply. Two port network computation of voltage gain, input resistance and output resistance of amplifier.