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### Analogue

1. 1. Small-Signal Analysis Small-Signal Analysis An ExampleHOW is it done?1. Consider each current and voltage in the circuit as the sum of two parts: R vin a BIAS (or QUIESCENT) component, and a SIGNAL component V D + vd VIN ID + id2. Analyse the circuit in two stages, considering first the bias components and then the signals. These separately satisfy Kirchhoff’s Laws: FULL CIRCUIT Full Circuit in Full Circuit Small-Signal Quiescent state with Signals Equivalent Circuit I1 I2 + i2 I1 + i1 i1 I2 i2 R R rd = VT/ID I4 I4 + i4 i4 VIN vin vd I3 I3 + i3 i3 VD ID id  In = 0 and  (In + in ) = 0  in = 0 BIAS CIRCUIT SMALL-SIGNAL EQUIVALENT CIRCUIT3. At each stage, use suitable approximations to describe non-linear circuit elements: • KVL in BIAS CCT gives: STAGE 1: Simple approximations to large-signal behaviour should VIN = ID R + VD (2.1) be adequate (if circuit is well-designed) so ID  (VIN - 0.7)/R, assuming VD  0.7 and VIN >> 0.7 e.g. VD  0.7 V for a forward biased diode • KVL in SMALL-SIGNAL EQUIVALENT CCT gives: STAGE 2: Provided signals are small enough, approximate LINEAR equations can be used vin = id R + vd (2.2) e.g. id = gdvd for a forward biased diode and we also know that id = vd/rd, so NOTE: gd depends on the bias conditions vd = vin rd/(rd + R)WHY this approach? where rd = 1/gd is the DYNAMIC RESISTANCE of the diode• Because linear approximations make the analysis easier, and often give us all the information we need about small signal behaviourEE1 Analogue Electronics 2010/2011 - Part 2 ASH 1 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 2
2. 2. Small-Signal Equivalent Models Small-Signal Equivalent Models 2-Terminal Devices 2-Terminal Devices• Small-Signal Equivalent Circuit is formed by replacing each circuit element with its SMALL-SIGNAL EQUIVALENT MODEL (SSEM) Device Symbol Defining equations Small-signal equiv. model• The SSEM describes schematically the relationship between the signal voltages and currents at the device terminals DC voltage source V = const rd = dV/dI = 0 short-circuit• DYNAMIC RESISTANCE I = const DC current source For 2-Terminal devices where the I-V relationship depends only on the rd  open-circuit instantaneous values of I and V, the SSEM is a resistor with value rd, where: Resistor V=IR rd = [dV/dI]OP (2.3) R rd = R Here []OP indicates that the derivative is evaluated at the operating point Junction diode I = I0 exp(V/VT) (forward biased) rd = VT /I Examples: resistor (rd = R); diode at low frequency Capacitor Q=CV• DYNAMIC CAPACITANCE C cd = C For devices where the current depends on the rate of change of the voltage across the terminals, we can describe this dependence in terms NB Second order and high frequency effects neglected of a dynamic capacitance cd: cd = [dQ/dV]OP (2.4) This is similar to the usual capacitor equation (i.e. C = Q/V), except that we are now allowing the stored charge Q to be a non-linear function of the applied voltage Examples: capacitor (cd = C); diode at high freq (cd appears in parallel with rd)EE1 Analogue Electronics 2010/2011 - Part 2 ASH 3 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 4
3. 3. Small-Signal BJT Model - 1 Small-Signal BJT Model - 2 (VCCS version, neglecting ro) (ICCS version, neglecting ro)• Need an SSEM to describe the relationship between the signals ib, ic, vbe • Often we will want to express ic in terms of ib, using a current-controlled and vce in active mode current source (ICCS): Given large-signal relationships: IC = IS exp(VBE/VT) ib ic B ib  ib C IB = IC / = (IS /) exp(VBE/VT) rbe• IB-VBE relationship is diode-like  Use a dynamic resistance to represent the base-emitter junction E ie• IC-VBE relationship is also diode-like, but now the voltage between B & E is controlling the current between C & E NB  here (also called hfe) is the slope of the IC-IB characteristic at the  Use a voltage-controlled current source (VCCS) to represent the operating point. Strictly we should distinguish this from the  used collector signal current: earlier in the large-signal equations. However, the difference is usually small, so we shall ignore it SIMPLIFIED HYBRID- MODEL: • How are the parameters in the two models related? ib ic ib = vbe /rbe (2.5) Eliminating vbe from Equations 2.5 and 2.6 gives ic = gmrbeib, and we also B gmvbe C know that ic =  ib, so: vbe rbe ic = gm vbe (2.6)  = gm rbe (2.9) ie = ib + ic E ie • Some TYPICAL VALUES: For a BJT with  = 100, biased at IC = 1 mA: NOTE: rbe often referred to as r gm = IC /VT  40 mS• SMALL-SIGNAL PARAMETERS: rbe =  /gm  2.5 k rbe = [dVBE/dIB]OP = VT/IB (2.7) gm = [dIC/dVBE]OP = IC/VT (2.8)EE1 Analogue Electronics 2010/2011 - Part 2 ASH 5 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 6
4. 4. Small-Signal BJT Model - 3 C-E Amplifier Revisited The Early Effect Quiescent Analysis• For a real transistor, IC does show some dependence on VCE: IC VCC VCC IC VBIAS = VCC RB2 _________ (RB1 + RB2 ) RC RC RB1 RB = (RB1 //RB2 ) VCE COUT RB VBE CIN incr VBE V IN VOUT RB2 VBIAS RE CE RE VCE 0 FULL CIRCUIT BIAS CIRCUIT -VA• All curves, when extrapolated, cross the VCE axis at roughly the same • CIN, COUT and CE open-circuit at DC  omit from bias CCT point (-VA, 0). VA is the EARLY VOLTAGE, which normally lies in the range 50 to 100 V • KVL on input side of bias CCT gives:• Equation 1.2 becomes: VBIAS = IB RB + VBE + IE RE IC = IS.exp(VBE/VT).(1 + VCE/VA) (2.10) VBE  0.7 V, IE = (1 + )IB• Finite slope of the IC-VCE curve represents a SMALL-SIGNAL  IE  (VBIAS - 0.7)/[RE + RB/(1 + )] (2.12) OUTPUT RESISTANCE ro: ro = [VCE/IC]OP  VA/IC (2.11) NOTE: IE now shows some dependence on  as a result of the finite source impedance of VBIAS. However, we can minimise this by making RB << (1 + )RE.• So SSEM becomes: B gmvbe C gm = IC /VT • As before: rbe IC =  IE vbe ro ro = VA /IC rbe =  /gm and the quiescent voltages at the transistor terminals are: E VE = IERE, VB  VE + 0.7, VC = VCC - RC ICEE1 Analogue Electronics 2010/2011 - Part 2 ASH 7 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 8
5. 5. C-E Amplifier Revisited C-E Amplifier Revisited Small-Signal Analysis - 1 Small-Signal Analysis - 2• Assume capacitors are so large that they are effectively short-circuit at Small-Signal response can be characterised by three parameters: signal frequencies: • VOLTAGE GAIN: VCC Av = - gm (RC//ro) (2.14) RC RB1 [since vout = -gmvbe(RC//ro), and vbe = vin ] O/P I/P COUT CIN Similar to Equn 1.10, except that we are now taking the finite output resistance of the transistor into account RB2 e.g. If VA = 100 V and IC = 1 mA, then ro = VA /IC = 100 K. In this case RE CE the voltage gain with RC = 10 K is reduced from 400 (assuming ro infinite) to 364 FULL CIRCUIT • INPUT RESISTANCE: Ri = (RB1//RB2//rbe) = (RB//rbe) (2.15) Ri tells us about the loading effect imposed by the amplifier on the input gmvbe CIN COUT signal source. Usually it is dominated by the transistor input resistance vbe rbe ro rbe RB1 RB2 RC vin vout • OUTPUT RESISTANCE: RE CE Ro = (RC//ro) (2.16) Ro is the source impedance associated with the amplifier’s output SMALL-SIGNAL EQUIVALENT CCT voltage Also of interest: gmvbe • CURRENT GAIN Ai, which is the ratio of the short-circuit output vbe rbe ro current to the input current. This can be deduced from Av, Ri and Ro: RB1 RB2 RC vin vout Ai = AvRi/Ro SSEC assuming Large CapacitorsEE1 Analogue Electronics 2010/2011 - Part 2 ASH 9 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 10
6. 6. Emitter Degeneration - 1 Emitter Degeneration - 2 • Overall Ri now given by:What happens if we leave out the bypass capacitor? Ri = RB//[rbe + (1 + ) RE] (2.19) VCC • and vout = - ibRC, so Av given by: RC RB1 Av = -  RC /[rbe + (1 + ) RE] (2.20) COUT IMPORTANT CASES: CIN AV  - gmRC as RE  0 vIN vOUT RB2 RE Av  -  RC /RE if (1 + )RE >> rbe (these are the results we obtained in Part 1) FULL CIRCUIT Summary of C-E Amplifier Parameters ib  ib C-E with Emitter Grounded C-E vbe ro Degeneration (and ro = ) rbe vin RB1 RB2 RC vout ie Ri RB//rbe (RB//[rbe + (1 + )RE]) ve RE Ro RC//ro RC SSEC assuming Large Capacitors -  RC _______________ Av - gm (RC//ro ) NB Things get messy if ro is included, so we’ll neglect it [rbe + (1 + )RE]• KVL on input side gives: vin = vbe + ve = ib rbe + ie RE RE provides negative or DEGENERATIVE FEEDBACK which: and ie = (1 + )ib • Reduces gain  vin = [rbe + (1 + ) RE] ib (2.18) • Improves linearity (only a fraction of the input voltage appears across the base-emitter junction) So, RE increases the apparent input resistance at the base of the transistor from rbe to [rbe + (1 + ) RE] • Improves the high-frequency performance of the amplifierEE1 Analogue Electronics 2010/2011 - Part 2 ASH 11 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 12
7. 7. Amplifier Macromodels• Given Av, Ri and Ro, we can replace an entire amplifier by its Small- Signal Equivalent Model: Ro Input vin Ri Av vin ~ Output We call this a MACROMODEL, because it describes the external characteristics of the amplifier without providing detailed information about the circuit configuration• We can use the macromodel to determine how the amplifier will interact with other circuits at its input or output. e.g. Source Amplifier Load RS Ro vS ~ vin Ri Av vin ~ RL vL The observed voltage gain vL/vS is reduced by the potential dividers at the input and output, giving: vL/vS = [Ri /(Ri + RS)].Av.[RL /(RL + Ro)] (2.21) So, the IN-CIRCUIT GAIN DEPENDS ON THE SOURCE AND LOAD RESISTANCES! This is one of the reasons we are interested in Ri and Ro• Macromodels are extremely useful when it comes to analysing more complex circuits, because they allow us to reduce the number of circuit elements to a minimumEE1 Analogue Electronics 2010/2011 - Part 2 ASH 13 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 14
8. 8. AC- and DC-Coupled Circuits - 1 AC- and DC-Coupled Circuits - 2• DISCRETE CIRCUITS e.g. for amplifier on Page 11: • AC-coupling may be used to provide DC isolation between successive stages. In this case, each stage requires its own bias circuitry |vL/vS| This approach is often used in high-frequency applications, where there is no requirement for operation at signal frequencies near zero LOW-FREQ MID-BAND HIGH-FREQ• INTEGRATED CIRCUITS e.g. Operational Amplifiers • Stages inside chip are DC-COUPLED, so their bias conditions are CIN(RS + Ri ) = 1 interrelated NB LOG-LOG plot • Bias condition for entire circuit is established by external negative feedback COUT(Ro + RL) = 1 • DC-coupling gives extra flexibility of operation down to zero signal  frequency NB Graph assumes Av, Ri and Ro are constant throughout low-freq• AC-COUPLING - Effect on Frequency Response and mid-band  not directly applicable to bypassed C-E amplifier Source Amplifier Load • In this course we will consider only the MID-BAND, where the RS CIN Ro COUT vS ~ vin Ri Av vin ~ RL vL coupling capacitors are effectively short-circuit, and high-frequency ZS Zo effects can be ignored For a DC-Coupled circuit, the mid-band extends down to zero ZS = RS + 1/j CIN Zo = Ro + 1/j COUT frequency • Gain falls off at high frequency due to junction capacitances of • HIGH-PASS FILTERS at input and output transistors + parasitic effects. • Equation 2.21 becomes: vL/vS = [Ri /(Ri + ZS)].Av.[RL /(RL + Zo)] (2.22) Cont’d ...EE1 Analogue Electronics 2010/2011 - Part 2 ASH 15 EE1 Analogue Electronics 2010/2011 - Part 2 ASH 16
9. 9. ANALOGUE ELECTRONICSPROBLEMS 21. You are asked to analyse a circuit containing a BJT with a  of 200 and an Early voltage of 100 V. Bias analysis shows that the transistor is in active mode, and that the quiescent collector current is 2 mA. Draw a small-signal equivalent model of the device, and assign values to the parameters rbe, gm and ro.2. (a) Calculate the collector bias current and the quiescent output voltage for the amplifier in Figure Q2. (b) Draw a small-signal equivalent circuit for the amplifier, and determine the small- signal parameters Ri (input resistance), Ro (output resistance) and Av (voltage gain). +5V 3 k 430 k O/P I/P  = 100 VA = 120 V Figure Q23. The figure below shows a common-emitter amplifier with a bypassed emitter resistor. The transistor has a  of 50 and an Early voltage of 100 V. Calculate the quiescent output voltage VOUT, and the small-signal voltage gain vout/vin for this circuit. You may assume that the bypass capacitor C is effectively short-circuit at signal frequencies. + 20 V 3.3 k VOUT + vout vin ~ 4V 1.1 k CEE1 Analogue Electronics 2010/2011 - Problems 2 ASH 1
10. 10. 4. (a) Calculate the quiescent collector current for the common-emitter amplifier in Figure Q4, and confirm that the transistor is biased roughly in the middle of the active region. (b) Draw a small-signal equivalent circuit of the amplifier, and determine the parameters Ri, Ro and Av, assuming the transistor has infinite small-signal output resistance. +20 V 10 k 330 k O/P I/P = 200 43 k 1.5 k Figure Q45. To achieve a higher voltage gain, two amplifier stages similar to the one in Question 4 are cascaded as shown below. Coupling capacitors are used to provide DC isolation between the signal source and the first stage, and between the two stages. C1 C2 Stage 1 Stage 2 vin ~ VOUT + vout Draw a small-signal equivalent circuit for the two-stage amplifier, in which each stage is represented by a macromodel. Hence calculate the overall small-signal voltage gain vout/vin, assuming the coupling capacitors are effectively short-circuit at signal frequencies.EE1 Analogue Electronics 2010/2011 - Problems 2 ASH 2
11. 11. 6. Derive an expression for the variation of vo/vi with frequency for the passive network in Figure Q6. Hence show that the cut-off frequency fc is given by 2fcC(R1 + R2) = 1. NOTE: fc is the frequency at which the low-freq and high-freq asymptotes of the frequency response meet. R1 C vi R2 vo Figure Q6 Using the above result, calculate the cut-off frequencies associated with the coupling capacitors in the amplifier of Question 5, assuming C1 = 47 nF and C2 = 220 nF. Hence sketch the variation of the overall voltage gain with frequency in the low-frequency and mid-band regions.7. Show that the small-signal voltage gain of the amplifier below is given by: Av = - (gm - 1/RB).(RC//ro//RB) where gm and ro are the usual hybrid- parameters of the transistor. VS RC RB O/P I/PEE1 Analogue Electronics 2010/2011 - Problems 2 ASH 3
12. 12. Answers1 rbe = 2.5 k, gm = 80 mS, ro = 50 k2 (a) IC = 1 mA, VOUT = 2 V; (b) Ri = 2.49 k, Ro = 2.93 k, Av = -1173 VOUT = 10.3 V, Av = -3544 (a) IC = 0.946 mA; (b) Ri = 33.8 k, Ro = 10 k, Av = -6.525 Av = 32.86 fc = 100 Hz, 16.5 Hz for C1, C2EE1 Analogue Electronics 2010/2011 - Problems 2 ASH 4