EE105 Fall 2015 Microelectronic Devices and Circuits Prof. Ming C. Wu wu@eecs.berkeley.edu 511 Sutardja Dai Hall (SDH) 11-1 Transistor Operating Mode in Amplifiers Transistors are biased in flat part of the I-V curves Saturation mode for MOSFET, and active mode for BJT MOSFET: channel pinched off at drain BJT: base-collector junction reverse-biased Drain (collector) voltage can vary without changing current 11-2 1
Voltage Transfer Curves i D = 1 2 k n (v GS n )2 V i C = I S e T Transistors are transimpedance amplifiers MOSFET: v GS controls i D BJT: controls i C Adding load resistors converts current to voltage à voltage amplifiers Drain (collector) voltage free to change in saturation (active) mode 11-3 Voltage Amplifier DC bias at Q point (Quiescent point) Small-signal input superimposed on DC bias voltage v DS = V DD 1 2 k v n ( V GS t ) 2 = dv DS = k n ( v GS ) dv GS Q Voltage Gain: = k n V OV 11-4 2
Maximum Input Signal Amplitude To keep the MOSFTET in Saturation region (linear part of I-V curve): v DS v OV = v GS at all time. This condition is met if v DS min = V DS v ds v GS max = V GS + Since v ds = V DS V GS + V DS V OV 1+ The input amplitude must be smaller than this value to ensure MOSFET stays in Saturation 11-5 BJT Amplifier v o = v CE = V CC i C V i C = I S e T = dv CE d Q = I C " = I % C $ ' # & Note and I C = V CC V CE V CE 0.3 for BJT in Active region max = V CC 0.3 11-6 3
Graphical Analysis with Load Line KVL: V DD = i D + v DS This is the equation for load line The intersection of load line and MOSFET I-V (with a specific v GS ) defines the bias point (Q) 11-7 Consideration for Bias Point Q 1 : too close to V DD Not enough room for positive signal swing Q 2 : too close to Triode region Not enough room for negative voltage swing 11-8 4
Small-Signal Model for MOSFET Hybrid-π Model i D (v GS, v DS ) = 1 2 k " n ( v GS ) 2 1+ v % DS $ ' # & At DC bias point, Q i D = I D + i D + i D v ds v GS v Q DS Q i D = I D + + 1 r o v ds The equivlent circuit is valid for both NMOS and PMOS. In PMOS, use absolute sign for all parameters: V GS, V t, V OV,, and replace k n with k p Transconductance: " = k n ( v GS ) 1+ v % DS $ ' k n v GS # & ( ) = k n V OV (Alternative: = 2I D V OV ) Output Resistance: r o = I D (in MOSFET: r o = 1 λi D ) 11-9 Applying Small-Signal Model v ds Previously, we used v DS = V DD 1 2 k v n ( V GS t ) 2 = dv DS = k n ( v GS ) dv GS Q Voltage Gain: = k n V OV With small-signal model, gain can be obtained much more easily: v ds ( r o ) ( ) = v ds r o Usually r o >>, 11-10 5
Small-Signal Model for BJT i C (, v CE ) = I S e At DC bias point, Q! 1+ v $ CE # & " % i C = I C + i C v be + i C v ce v Q CE Q i D = I D + + 1 r o v ds Transconductance: = I S e! 1+ v $ CE # & 1 = I C " % Output Resistance: ( r o ) 1 = I S e 11-11 1 = I C r o = I C Hybrid-π Model for BJT Similar to MOSFET, but with finite base resistance: i B = i C β Base-Emitter resistance = i B = i C i C i B = β 11-12 6
Applying Small-Signal Model v ce v be Previously, we used v o = v CE = V CC i C V i C = I S e T = dv CE d Q " = I % C $ ' # & With small-signal model, gain can be obtained much more easily: v ce v be ( r o ) ( ) = v ce r o v be Usually r o >>, 11-13 BJT Amplifier Example DC analysis (β=100, = ) V BB = I C β R BB + 0.7 I C = 2.3 ma = I C = 40 2.3 = 92 ma/v (ms) r o = AC analysis ( ) + R BB = 0.011 92mS 3kΩ = 3.04 V/V = β =1.09 kω 11-14 7
Systematic Procedure for Transistor Amplifier Analysis 1. Perform DC analysis (ignore small signal source) 2. Calculate small-signal parameters (,, r o, etc) 3. Generate AC small-signal equivalent circuit Replace DC voltage source by short circuit Replace DC current source by open circuit Replace transistor by hybrid-π model (or other model) 4. Perform circuit analysis to determine voltage gain or other amplifier performance parameters 11-15 MOSFET Amplifier Example: (1) Solve DC Bias Point Find voltage gain for the amplifier. The MOSFET has V t = 1.5V, k n = 0.25 ma/v 2 and = 50V Coupling capacitor is open circuit in DC, and short circuit for AC signal. To solve DC bias point, replace coupling capacitor withvopen circuit: V GS = V DS from bias connection R G V DD = I D +V DS = k n ( 2 V GS ) 2 +V DS 15 =1.25( V DS 1.5) 2 +V DS Solve quadratic equation: V DS = 4.4V I D =1.06 ma 11-16 8
MOSFET Amplifier Example: (2) Solve AC Small Signal Circuit Now replace coupling capacitor with short circuit, and replace MOSFET with hybrid-pi model with = k n V OV = 0.25 (4.4 1.5) = 0.725 ma/v r 0 = I D = 47 kω Next, simpliy resistance at output to R L ' = r o R L = 4.5 kω Do circuit analysis: $ ' v o = (i i )R & L % i i = v o & ' R G A v R L ' 1 1/ R G 1+ R L ' / R G For large R G, A v R L ' = 0.725* 4.5 = 3.3 V/V 11-17 MOSFET Amplifier Example: (3) Additional Parameters of Interest R in = i i = R in = = v o R G 10 MΩ = 2.3 MΩ 1+ 3.3 R G = A v 1 A v R G Maximum signal that can be applied while keeping MOS in Saturation: v DS min v GS max V V DS A v V GS + V DS V GS +V t 1+ A v = V t 1+ A v V DS = V GS here. 0.35 V 11-18 9
BJT Amplifier Example: (1) DC Bias Point: Find voltage gain for the amplifier. Assume β = 100, and neglect Early effect. Assume V BE = 0.7 V I B = V BB V BE R BB = I C = βi B = 2.3 ma 3 0.7 = 0.023 ma 100k V CE = V CC I C =10 2.3*3 = 3.1 V 11-19 BJT Amplifier Example: (2) AC equivalent circuit: = I C 2.3 ma = = 92 ma/v 25 mv = β =1.1 kω A v R BB + = 3.0 V/V 11-20 10