6 introduction to transistor amplifiers concepts and
play

6. Introduction to Transistor Amplifiers: Concepts and Small-Signal - PowerPoint PPT Presentation

6. Introduction to Transistor Amplifiers: Concepts and Small-Signal Model Lecture notes: Sec. 5 Sedra & Smith (6 th Ed): Sec. 5.4, 5.6 & 6.3-6.4 Sedra & Smith (5 th Ed): Sec. 4.4, 4.6 & 5.3-5.4 ECE 65, Winter2013, F. Najmabadi


  1. 6. Introduction to Transistor Amplifiers: Concepts and Small-Signal Model Lecture notes: Sec. 5 Sedra & Smith (6 th Ed): Sec. 5.4, 5.6 & 6.3-6.4 Sedra & Smith (5 th Ed): Sec. 4.4, 4.6 & 5.3-5.4 ECE 65, Winter2013, F. Najmabadi

  2. Foundation of Transistor Amplifiers (1)  A voltage amplifier requires  MOS transfer function is NOT linear v o /v i = const. (2 examples below)  v o /v i can be negative (minus sign  In saturation, however, transfer represents a 180 o phase shift) function looks linear (but shifted) F. Najmabadi, ECE65, Winter 2013, Intro to Amps (2/32)

  3. Foundation of Transistor Amplifiers (2)  In saturation, transfer function appear to be linear Approximate the transfer function with a tangent line at point Q (with a slope of − A ): v DS − V DS = − A ( v GS − V GS ) v ds = − A v gs (linear relationship) for v ds = v DS − V DS and v gs = v GS − V GS F. Najmabadi, ECE65, Winter 2013, Intro to Amps (3/32)

  4. Foundation of Transistor Amplifiers (3)  Let us consider the response if NMOS remains in saturation at all times and v GS is a combination of a constant value ( V GS ) and a signal ( v gs ): = + v V v GS GS gs F. Najmabadi, ECE65, Winter 2013, Intro to Amps (4/32)

  5. The response to a combination of v GS = V GS + v gs can be found from the transfer function F. Najmabadi, ECE65, Winter 2013, Intro to Amps (5/32)

  6. Response to the signal appears to be linear  Response ( v o = v DS ) is also made of a constant part ( V DS ) and a signal response part ( v ds ).  Constant part of the response, V DS , is ONLY related to V GS , the constant part of the input (Q point on the transfer function of previous slide). o i.e., if v gs = 0 , then v ds = 0  The shape of the time varying portion of the response ( v ds ) is similar to v gs . o i.e., v ds is proportional to the input signal, v gs F. Najmabadi, ECE65, Winter 2013, Intro to Amps (6/32)

  7. Although the overall response is non-linear, the transfer function for the signal is linear! v ds Constant: Signal and Bias response v GS = V GS + v gs v DS = V DS + v ds v gs i D = I D + i d Approximately Linear Non-linear relationship among relationship among these parameters these parameters F. Najmabadi, ECE65, Winter 2013, Intro to Amps (7/32)

  8. Important Points and Definitions!  Signal: We want the response of the circuit to this input.  Bias: State of the system when there is no signal. o Bias is constant in time (may vary extremely slowly compared to signal) o Purpose of the bias is to ensure that MOS is in saturation at all times.  Response of the circuit (and its elements) to the signal is different than its response to the Bias (or to Bias + signal): o Signal iv characteristics of elements are different, i.e. relationships among v gs , v ds , i d is different from relationships among v GS , v DS , i D . o Signal transfer function of the circuit is different from the transfer function for total input (Bias + signal). Above observations & conclusions equally apply to a BJT in the active mode! F. Najmabadi, ECE65, Winter 2013, Intro to Amps (8/32)

  9. Issues in developing a transistor amplifier: 1. Find the iv characteristics of the elements for the signal (which can be different than their characteristics equation for bias). o This will lead to different circuit configurations for bias versus signal 2. Compute circuit response to the signal o Focus on fundamental transistor amplifier configurations 3. How to establish a Bias point (bias is the state of the system when there is no signal). o Stable and robust bias point should be resilient to variations in µ n C ox (W/L),V t (or β for BJT) due to temperature and/or manufacturing variability. o Bias point details impact small signal response (e.g., gain of the amplifier). F. Najmabadi, ECE65, Winter 2013, Intro to Amps (9/32)

  10. Signal Circuit 1) We will find signal iv characteristics of various elements. 2) In order to use circuit theory tools, we will use the signal iv characteristics of various elements to assign a circuit symbol. e.g., o We will see that the diode signal iv characteristics is linear so for signals, diode can be modeled as a “circuit theory” resistor. o In this manner, we will arrive at a signal circuit.

  11. Bias and Signal Circuits Bias & Signal Bias Signal only = (Bias + Signal) - Bias + ? MOS : , , , V V I MOS : v , v , i , MOS : v , v , i , GS DS D GS DS D gs ds d = + ( v V v ,...) GS GS gs = + R : V , I R : v , i : R v V v D R R D r r D R R r = + ..... ..... i I i R R r ..... F. Najmabadi, ECE65, Winter 2013, Intro to Amps (11/32)

  12. Finding signal circuit elements -- Resistor Voltage Current iv Equation Resistor v R i R v R = R i R Bias + Signal: V R I R V R = R I R Bias: v r = v R − V R i r = i R − I R Signal: ?? = − = − = − v = v v V Ri RI R ( i I ) Ri r R R R R R R r r  A resistor remains as a resistor in the signal circuit. F. Najmabadi, ECE65, Winter 2013, Intro to Amps (12/32)

  13. Finding signal circuit elements -- Capacitor Voltage Current iv Equation Capacitor v C i C i C = C dv C /dt Bias + Signal: V C I C I C = C dV C /dt Bias: v c = v C − V C i c = i C − I C Signal: ?? − ( ) dv dV d v V = − = − = dv C C C C i i I C C C c = c i C c C C dt dt dt dt  A capacitor remains as a capacitor in the signal circuit. o Since V C = const., I C = 0 , i.e., A capacitor acts as an open circuit for bias circuit. F. Najmabadi, ECE65, Winter 2013, Intro to Amps (13/32)

  14. Finding signal circuit elements – IVS & ICS Independent Voltage Current iv Equation voltage source v IVS i IVS v IVS = V DD = const Bias + Signal: V IVS I IVS V IVS = V DD = const Bias: v ivs = v IVS − V IVS i ivs = i IVS − I IVS Signal: ?? = ≠ = − = − = v v V V V 0 v 0 , i 0 ivs ivs ivs IVS IVS DD DD  An independent voltage source becomes a short circuit! Similarly:  An independent current source becomes an open circuit! Exercise: Show that dependent sources remain as dependent sources F. Najmabadi, ECE65, Winter 2013, Intro to Amps (14/32)

  15. Summary of signal circuit elements  Resistors& capacitors: The Same o Capacitor act as open circuit in the bias circuit.  Independent voltage source (e.g., V DD ) : Effectively grounded  Independent current source: Effectively open circuit  Dependent sources: The Same  Non-linear Elements: Different! o Diodes & transistors ? F. Najmabadi, ECE65, Winter 2013, Intro to Amps (15/32)

  16. Diode Signal Response v D   v   + = D Bias Signal : exp i I   D s i D   nV T V D   V   = D Bias : exp I I   I D D s   nV T v d     + V v V     ? = − = − D d D Signal : i i I I exp I exp     i d d D D s s     nV nV T T       V v     = × −   D d i I exp exp 1     d s     nV  nV  T T  A different iv equation!     v   = × −  iv equation is non-linear!   d i I exp 1   d D    nV   Related to bias value, I D ! T F. Najmabadi, ECE65, Winter 2013, Intro to Amps (16/32)

  17. Diode small-signal model: v d     ? v   = × − i d   d i I exp 1   d D   nV   T 2       v v 1 v       = + + + d d d Taylor Series Exapnsion : exp 1 ....             nV nV 2 ! nV T T T     v v v     << ≈ + d d d If 1 : exp 1         nV nV nV T T T       v I ≈ × +   − =    d  D i I 1 1 v     d D D     nV nV   T T nV = = T v i r i d d d d I D F. Najmabadi, ECE65, Winter 2013, Intro to Amps (17/32)

  18. Formal derivation of small signal model  Signal + Bias for element A ( i A , v A ) : i A = f ( v A )  Bias for element A ( I A , V A ) : I A = f ( V A )  Signal for element A ( i a , v a ) : i a = g ( v a ) = i f ( v ) A A ( 2 ) ( ) f ( V ) ( ) (Taylor Series = + ⋅ − + ⋅ − + 2 ( 1 ) A f ( V ) f ( V ) v V v V ... A A A A A A 2 ! Expansion) ( 2 ) f ( V ) = + ⋅ + ⋅ + 2 ( 1 ) A f ( V ) f ( V ) v v ... A A a a 2 ! ≈ + ⋅ ( 1 ) f ( V ) f ( V ) v A A a Small signal means: = + = + ⋅ ( 1 ) i i I I f ( V ) v A a A A A a ( 2 ) f ( V ) ⋅ >> ⋅ 2 ( 1 ) A f ( V ) v v A a a = = ⋅ ( 1 ) 2 ! i g ( v ) f ( V ) v a a A a ( 1 ) ( ) f V << ⋅ A v 2 a ( 2 ) f ( V ) A F. Najmabadi, ECE65, Winter 2013, Intro to Amps (18/32)

Recommend


More recommend