21 Two-Port Model: CE Amplifier ■ Use transconductance amplifier form for model (not mandatory) ■ , Rin = rπ Rout = ro || RC, Gm = gm by inspection iout + + v = G v = v in Rin rπ m in Rout ro RC out _ − (a) i RS out + + + v v vs in R = rπ G v R = r R out RL − _ in m in out o C − (b) ■ Loaded ratio of iout to vin R ro RC out Gm ------------------------ = gm ------------------------------ Rout + RL ro RC + RL Increasing RC seems desirable for increasing iout / vin ... but note that the DC collector current decreases for a fixed VCC --> gm decreases EECS 105 Fall 1998 Lecture 21 Common-Source Amplifier ■ Configuration is similar to common-emitter VDD VDD RD RD = + iOUT IOUT iout RS + + + v s v = V + v R VOUT − OUT OUT out L + _ VBIAS + − − V BIAS − (a) (b) ■ Bias: remove source and load resistances (b) EECS 105 Fall 1998 Lecture 21 Graphical Load-Line Analysis ■ Load line is given by: ()⁄ ID = VDD – VOUT RD ID (mA) 1.0 0.9 4 VBIAS (V) 0.5 4 3 3 0.25 2.5 2 ≤ = 0.10 2 VBIAS VTn 1 V 0 1 1 2 3 4 5 = VDS VOUT (V) (a) VOUT High- 5 V 1 Gain 2 Region 3 4 2.5 V VBIAS (b) EECS 105 Fall 1998 Lecture 21 Small-Signal Model of CS Amplifier ■ ≈ ⁄ Substitute parameters at operating point selected so that VOUT VDD 2 iout + + v vgs gmvgs ro RD out − − (a) i RS out + + + v v g v r R R v s − gs m gs o D L out − − (b) 1/2 Transconductance is proportional to ID unlike bipolar transistor EECS 105 Fall 1998 Lecture 21 Two-Port Model of Common-Source Amplifier ■ Use transconductance amplifier form for model (most natural choice) , Rin = infinty Rout = ro || RD, Gm = gm by inspection iout + + = vin Gmvin Rout ro RD vout − − (a) i RS out + + + v v = s in Gmvin Rout ro RD vout − RL − − (b) Infinite input resistance is ideal for a voltage input Output resistance increases with RD increasing, but DC drain current ID will 1/2 decrease and gm will decrease with ID EECS 105 Fall 1998 Lecture 21 Current-Source Supplies ■ A current source to supply current, rather than a resistor, allows a high DC current for the device with a large incremental (small-signal) resistance iSUP + + v v SUP iSUP SUP ISUP roc − − (a) iSUP 1 r roc ISUP oc vSUP (b) (c) EECS 105 Fall 1998 Lecture 21 Common-Source with Current Source Supply ■ RD is replaced with idealized current source with internal resistance VDD VDD iSUP ISUP = + iOUT IOUT iout + + RS v = V + v V + OUT OUT out RL OUT vs − + − VBIAS − + − V BIAS − (a) (b) ■ For DC bias analysis, the small-signal source (with RS) and the load resistor RL are eliminated, along with the internal resistance roc of the current source EECS 105 Fall 1998 Lecture 21 Graphical Analysis of CS Amplifier with Current-Source Supply ID (mA) 1.0 0.9 4 V VBIAS 0.5 3 V 4 3 2 0.25 2.5 V ≤ = 0.10 2 V VBIAS VTn 1 V 1 0 12 3 4 5 = VDS VOUT (V) (a) High- 1 VDD Gain 2 Region 3 4 0 0 2.5 V VBIAS (b) ■ The region of input bias voltage VBIAS for which the current source and the MOSFET are in their constant-current regions is extremely small .... EECS 105 Fall 1998 Lecture 21 Common-Source/Current-Source Supply Models ■ The small-signal model is identical to the resistor supply, except that the current source’s internal resistance roc replaces RD iout + + v r v gs gmvgs ro oc out − − ■ Two-port model in both transconductance and voltage amplifier forms (the latter by direct conversion from the former ... by applying procedure for finding Av) i i out Rout out + + + v R v + = − v in Gmvin out in Avvin GmRoutvin out − −−− (a) (b) EECS 105 Fall 1998 Lecture 21 p-Channel Common-Source Amplifier ■ Source of p-channel is tied to positive supply; current supply sinks ISUP to ground or to lower supply VDD VDD RS + v i = I + i s − OUT OUT out + V + BIAS − V + BIAS − + v = V + v R V i OUT OUT out L OUT SUP ISUP _ _ (a) (b) ■ Small-signal model: substitute p-channel model directly EECS 105 Fall 1998 Lecture 21 p-Channel CS Small-Signal Model ■ Source is at top, but circuit can be inverted to show correspondence with n- channel common-source amplifier s + − g v r r vsg m sg o oc vout − + g d iout (a) iout g d − + g v r r vsg m sg o oc vout + − s (b) iout g d + + g v r r vgs m gs o oc vout − − s (c) EECS 105 Fall 1998 Lecture 21.