ELE 2110A Electronic Circuits Week 6: Small Signal Analysis for Single Stage BJT amplifiers ELE2110A © 2008 Lecture 06 - 1 Topics to cover … z Family of single-stage BJT amplifiers z Small signal analysis – Common-Emitter Amplifier z Common-Emitter Amplifier with Emitter R z Common-Base Amplifier Reading Assignment: Chap 14.1 – 14.5 of Jaeger and Blalock , or Chap 5.7 of Sedra & Smith ELE2110A © 2008 Lecture 06 - 2 Signal Injection v i = I exp BE C S V T ⎛ ⎜ ⎜ ⎜⎜ ⎝ ⎞ ⎟ ⎟ ⎟⎟ ⎠ z Transistor is a VCCS: z To cause changes in current, vBE must be changed – Base or emitter terminals can be used as i/p terminal – Collector is not used as an i/p terminal because even if Early voltage is considered, collector voltage has negligible effect on terminal currents ELE2110A © 2008 Lecture 06 - 3 Signal Extraction z Large changes in iC or iE create large voltage drops across collector and emitter resistors – Collector or emitter can be used as o/p terminal – Base terminal is not used as o/p terminal due to small iB ELE2110A © 2008 Lecture 06 - 4 BJT Amplifier Family z Constraints for signal injection and extraction yield three families of amplifiers Input Output – Common-Emitter (C-E): – Common-Base (C-B): – Common-Collector (C-C): Base Emitter Base Collector Collector Emitter – All circuit examples in the following slides use the four-resistor bias circuits to establish Q-point for the various amplifiers. ELE2110A © 2008 Lecture 06 - 5 Topics to cover … z Family of single-stage BJT amplifiers z Small signal analysis – Common-Emitter Amplifier z Common-Emitter Amplifier with Emitter R z Common-Base Amplifier ELE2110A © 2008 Lecture 06 - 6 Using Small Signal Models in BJT ckt Analysis Steps for using small-signal models: z Determine the DC operating point of the BJT – in particular, the collector current z z Calculate small-signal model parameters gm, rπ, & re for this DC operating point Eliminate DC sources – Replace DC voltage sources with short circuits – Replace DC current sources with open circuits z Replace BJT with an equivalent small-signal model – Choose most convenient one depending on surrounding circuitry z Analyze ELE2110A © 2008 Lecture 06 - 7 Common Emitter Amplifier z z z z Biased using the four-resistor network AC input is injected to base through a coupling capacitor AC output is taken from collector Emitter is ac-grounded ELE2110A © 2008 Lecture 06 - 8 DC operating point z All capacitors in original amplifier circuits are replaced by open circuits, disconnecting vI, RI, and R3 from circuit z Q-point can be found from dc equivalent circuit by using DC model for BJT ELE2110A © 2008 Lecture 06 - 9 AC Equivalent z ELE2110A © 2008 Replace all ∞ capacitors by short circuits, ∞ inductors by open circuits z Set all independent DC sources to 0, i.e., replace DC voltage sources by short circuits and DC current sources by open circuits z Replace transistor by small-signal model z Use small-signal AC equivalent to analyze AC characteristics of amplifier Lecture 06 - 10 Simplified AC Equivalent R B = R R = 10 k Ω 30 k Ω 1 2 R = R ELE2110A © 2008 C R 3 = 4 . 3 k Ω 100 k Ω Lecture 06 - 11 Small Signal Performance Parameters In AC analysis, we are interested to find z z z Voltage gain Input resistance Output resistance and sometimes z Current gain z Input signal range ELE2110A © 2008 Lecture 06 - 12 Voltage Gain Terminal voltage gain between base and collector is: v v Avt = v c = v o = − g m R L b be Overall voltage gain from source vi to output voltage across R3 is: ⎞ ⎛v v ⎟ be ⎟ = A ⎜⎜ be vt ⎜ v v ⎟ i ⎠ i ⎝ ⎡ ⎤ R rπ ⎢ ⎥ B ⎥ ∴ Av = − g m R ⎢⎢ L ⎢ R + R r ⎥⎥ B π ⎦ I ⎣ v o ⎛⎜ v o Av = v = ⎜ v i ⎜⎝ be ⎞⎛ ⎟⎜ ⎟⎜ ⎟⎜ ⎠⎝ ( R L ⎞ ⎟ ⎟ ⎟ ⎠ ) = ro R R C 3 ELE2110A © 2008 Lecture 06 - 13 Input Resistance z The total resistance looking into the amplifier at coupling capacitor C1 represents total resistance of the amplifier presented to signal source v x = i x ( R rπ ) B v R = x = R rπ = R R rπ B 1 2 in ix ELE2110A © 2008 Lecture 06 - 14 Output Resistance z Output resistance is the total equivalent resistance looking into the output of the amplifier at coupling capacitor C3 To find Rout, input source is set to 0 and test source is applied at output vx vx + + gm v ix = But vbe=0. be R ro C vx = R ro ≅ R As ro>> RC. ∴ R out = C C ix z ELE2110A © 2008 Lecture 06 - 15 Example 1 Analysis: To find the Q-point, dc equivalent circuit is constructed. 105 I +V + (β +1)I (1.6×104) = 5 B BE B z z z Problem: Find voltage gain, input and output resistances. Given data: β= 65, VA =50 V Assumptions: Active-region operation, VBE=0.7 V, small signal operating conditions. ELE2110A © 2008 ∴ I = 3 .71µA B I = 65 I = 241µA B C I = 66 I = 245µA E B 5 −104 I −V − (1.6 ×104 ) I − (−5) = 0 E C CE ∴V = 3.67V CE Active region of operation is correct. Lecture 06 - 16 Example 1 (Cont.) Next we construct the ac equivalent and simplify it R = R rπ = 6.23kΩ in B I gm = C = 9.64×10− 3S V T βV rπ = T = 6.64kΩ I C V +V ro = A CE = 223kΩ I C ELE2110A © 2008 R out = R vo Av = v i r = 9 . 57 k Ω C o ⎡ ⎤ R ⎢ ⎥ in ⎢ ⎥ = − 84 .0 = − gm ( Rout R ) ⎢ 3 ⎢ R + R ⎥⎥ ⎢⎣ I in ⎥⎦ Lecture 06 - 17 Topics to cover … z Family of single-stage BJT amplifiers z Small signal analysis – Common-Emitter Amplifier z Common-Emitter Amplifier with Emitter R z Common-Base Amplifier ELE2110A © 2008 Lecture 06 - 18 Common-Emitter Amplifier with Emitter R AC equivalent: Input AC ground Output A resistor exists at the emitter of the ac equivalent circuit Also called emitter degenerated CE amplifier RE provides negative feedback ELE2110A © 2008 Lecture 06 - 19 Negative Feedback due to RE z z z z z z ELE2110A © 2008 Assume iC Ç for any reason iE (=iC/α) Ç vE Ç vBE È iC È : counter act the original change More discussion on negative feedback later in the course Lecture 06 - 20 Overall Voltage Gain Overall voltage gain from vi to vo can be expressed as: vo Av ≡ = v i Define terminal voltage gain as: And it can be observed that: ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞⎛ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎠⎝ ⎞ vo vb ⎟⎟ ⎟ v v ⎟⎟ b i⎠ vo Avt ≡ vb vb RB || Rin = vi RI + ( RB || Rin ) Thus, the task breaks down into two steps: z Find the terminal voltage gain z Find the input resistance Rin. ELE2110A © 2008 Lecture 06 - 21 Terminal Voltage Gain Use hybrid-π model (neglecting ro): Using rπ g m = β , we have g m rπ R gm R L L Avt = − ≅− 1+ g m R rπ + ( g m rπ +1)R E E for gmrπ = β >> 1 vb = irπ + ( β + 1)iRE = i(rπ + ( β + 1) RE ) vo = − βiRL βR vo L Avt ≡ = − v rπ + (β +1)R E b ELE2110A © 2008 Effect of RE: z For RE=0, Avt ≅ − gm R L – Upper limit of Avt z For gmRE>>1, Avt = − R / R L E – Avt becomes less dependent on gm which varies widely z Increasing RE decreases voltage gain! Lecture 06 - 22 Input Resistance To find Rin Æ to find Vb/i: vb = i (rπ + ( β + 1) RE ) v R = b = rπ + (β +1)R in E i = rπ + ( gmrπ +1)R E ≅ rπ (1+ gm R ) E for gmrπ = β >> 1 z Increasing RE increases input resistance! ELE2110A © 2008 Lecture 06 - 23 Output Resistance To find Rout: z Set vi to zero z Apply a test source at output z Rout = vx/ix Rth = RI // RB ve = (β +1 )iR E KVL at left mesh: i (Rth + rπ ) + ve = 0 ve (Rth + rπ ) + ve = 0 ( β + 1) RE ⇒ ve = 0 ∴ i = 0 ⇒ ix = βi = 0 ELE2110A © 2008 vx ∴Rout = =∞ ix Lecture 06 - 24 Output Resistance Now, we also include ro in our analysis: KVL along loop 1: vx = vr + ve = (ix − βi)ro + ve Voltage at E: ⎛⎛ ⎞ ⎞ ve = ix ⎜⎜ ⎜⎜ R + rπ ⎟⎟ // R ⎟⎟ E⎠ ⎠ ⎝ ⎝ th Current division at Emitter: i = −ix Put 2nd and 3rd equations into 1st equation: ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ∴ v x = ro i x + β i x R R ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛⎛ ⎞ E + i x ⎜⎜ ⎜⎜ R + rπ ⎟⎟ // R E ⎠ ⎝ ⎝ th + R + rπ E th ⎞ βR ⎟ E ⎟ ∴Rout ≡ ≅ ro 1+ ⎟ R R r + + ix E th π ⎟⎠ vx ELE2110A © 2008 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ R E R + R + rπ E th ⎞ ⎟ ⎟ ⎠ for large value of ro Lecture 06 - 25 Output Resistance βR ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ E Rout ≅ ro 1+ R + R + rπ E th Assuming (rπ + RE ) >> Rth and ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ro >> R , with β = gmrπ E gmrπ R ⎞⎟ E ⎟ = r ⎛⎜⎜1+ g (r // R ) ⎞⎟⎟ Rout ≅ ro 1+ m π E ⎠ R + rπ ⎟⎟ o ⎝ E ⎠ ⎛ ⎜ ⎜ ⎜⎜ ⎝ Effect of RE: z For RE = 0, Rout = ro z For RE = ∞, upper limit of Rout is obtained: Rout = (β +1)ro z Increasing RE increases Rout! ELE2110A © 2008 Lecture 06 - 26 Current Gain ii z Terminal current gain: z Overall current gain: ELE2110A © 2008 i A ≡ L = −β it i Ai ≡ RB iL iL i = = Ait RB + Rin is i is Lecture 06 - 27 Input Signal Range Remember, in deriving the linear small signal model, we assumed vbe << VT : iC = I C e vbe / VT ≅ I C (1 + vbe / VT ) IC = IC + v { V be DC 1T23 AC Typically, vbe is required to be less than 5mV! ELE2110A © 2008 Lecture 06 - 28 Input Signal Range vb = i (rπ + ( β + 1) RE ) v v b b v = irπ = rπ = be rπ + (β +1)R 1+ g m R + R / rπ E E E ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ R ⎞ ⎞⎟ ⎟ ⎟⎟ ⎟⎟ ⎟⎟ ⎠⎠ v ≤ 0.005 1+ g m R + E ≅ 0.005(1+ gm R )V E E β b If gm R >> 1 , vb can be increased beyond 5 mV limit. E I R I R E C ≅ E E >>1 gm R = E V V T T ∴I R >>V = 0.025V E E T z Condition for gmRE >>1 : z Increasing RE increases input signal range! ELE2110A © 2008 Lecture 06 - 29 Summary of Emitter Degenerated Common-Emitter Amplifier z Terminal voltage gain: Inverting and Large − gm R L 1+ g m R E ⎡ ⎢ ⎢ ⎢ ⎢ ⎣⎢ ⎤ R //R ⎥ B in ⎥ L − ⎥ 1+ g m R RI + RB //Rin ⎥⎦⎥ E gm R z Overall voltage gain: Inverting and Large z Input resistance: Large z Output resistance: Large ro ⎜⎜1+ gm (rπ RE ) ⎟⎟ ⎠ ⎝ z Terminal Current gain: Large z Effects of the resistor at Emitter: rπ (1+ gm R ) E ⎞ ⎛ −β – Voltage gain decreased, but more stabilized (less sensitive to gm) – Input signal range increased – Input and output resistance increased ELE2110A © 2008 Lecture 06 - 30 Topics to cover … z Family of single-stage BJT amplifiers z Small signal analysis – Common-Emitter Amplifier z Common-Emitter Amplifier with Emitter R z Common-Base Amplifier ELE2110A © 2008 Lecture 06 - 31 Common-Base Circuits Output AC/Small-signal equivalent: Load AC ground Input ELE2110A © 2008 RL = R3 || R7 Signal Source Lecture 06 - 32 Terminal Voltage Gain CB ≡ vo = + g R Avt m L ve z Apply test source to input (E) And use BJT small signal model: ELE2110A © 2008 z Non-inverting! Magnitude same as the CE amplifier with RE=0. Lecture 06 - 33 Input Resistance KCL at emitter: ve i = + g m ve rπ v CB = e = rπ = rπ //( 1 ) ≅ 1 Rin g g i rπ g m +1 m m z Rin is small (as gm is usually large)! (gm=IC/VT) ELE2110A © 2008 Lecture 06 - 34 Overall Voltage Gain For gm R << 1 , I CB Av = + gm RL This is the upper bound. For gm R >>1 , I R CB Av = + L R I Overall voltage gain is ⎤ ⎡ ⎞ ⎛ ⎞⎛ R // R ⎢ ⎜ v ⎟⎜ v ⎟ v in ⎥⎥ CB = o = ⎜ o ⎟⎜ e ⎟ = A ⎢ 4 Av ⎜ ⎟ ⎥ ⎢ v ⎜⎜⎝ ve ⎟⎟⎠⎜⎜ v ⎟⎟ vt ⎢ R + ( R // R ) ⎥ ⎢⎣ I i ⎝ i ⎠ 4 in ⎥⎦ ⎛ ⎞ ⎜ ⎟ R gm R ⎜ L 4 ⎟⎟ = ⎜ 1+ g m (R // R ) ⎜⎜ R + R ⎟⎟ 4 I ⎝ I 4⎠ gm R L ≅ for R4 >> RI 1+ g m R I ELE2110A © 2008 z z For large voltage gain, a very small RI is required! Not a good candidate for voltage amplifier Lecture 06 - 35 Example z z z z Problem: Find overall voltage gain. Given data: β=100, Q-point values: IC=245uA, VCE=3.64V, gm=9.8mS, rπ=10.2kΩ, rο=219kΩ. Assumptions: Small-signal operating conditions. Analysis: CB Rin ≅1/ gm =102Ω R = R R =18kΩ L 3 7 ACB vt = + gm RL = 176 ⎛ ⎞ CB ⎜ ⎟ R A ⎜ vt 4 ⎟⎟ = +8.59 = ACB ⎜ v 1+ gm ( R R ) ⎜⎜ R + R ⎟⎟ 4⎠ I 4 ⎝ I ELE2110A © 2008 Lecture 06 - 36 Input Signal Range ⎛ R4 ⎞ R4 // Rin vi ⎜⎜ ⎟⎟ veb = vi = RI + R4 // Rin 1 + g m ( RI // R4 ) ⎝ RI + R4 ⎠ v ≅ v (1+ gm R ) I i eb For small-signal operation, for R4 >> RI. v ≤ 0.005(1+ gm R )V I b Relative size of gm and RI determine signal-handling limit. ELE2110A © 2008 Lecture 06 - 37 Output Resistance Rout here is equivalent to the Rout of CE amplifier with RE=Rth and resistance at base equal to zero. z ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ∴RCB out = ro 1+ βR th R + rπ th ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Using β = gmrπ ⎞ ⎛ CB Rout ≅ ro ⎜⎜⎝1+ gm (rπ // Rth ) ⎟⎟⎠ z ELE2110A © 2008 Rout is large. Lecture 06 - 38 Current Gain i i CB = l = α ≅ +1 Ait ie z Terminal current gain: z Current gain from source to load: i ⎛⎜ i ⎞⎟⎛⎜ i ⎞⎟ R CB = l = ⎜ l ⎟⎜ e ⎟ = A ≅ A =1 for R >> R Ai ⎟ ⎟⎜ ⎜ it in R + R it i ⎜⎝ ie ⎟⎠⎜⎜ i ⎟⎟ in ⎝ ⎠ ELE2110A © 2008 Lecture 06 - 39 Summary of Common Base Amplifier z Terminal voltage gain: Non-inverting and Large Input resistance: Low Output resistance: High Current gain: close to Unity Input range: determined by gmRth z Excellent for use as a current buffer z z z z ELE2110A © 2008 Lecture 06 - 40 Current Buffer • Without a buffer: Io Rs Ii Io = RL RS I i << I i for RL >> Rs RS + RL Only a small portion of source current is delivered to load! • With a buffer: Current buffer Io Ii Rs Ri A=1 Ro RL Vo ⎛ RS ⎞ RO ⎜ Io = ⎜ I i ⎟⎟ A ⎝ RS + Ri ⎠ RL + RO ≅ AI i = I i for Ri << RS and RO >> RL • Requirement of a current buffer: - Low input resistance - High output resistance - Unity current gain ELE2110A © 2008 Lecture 06 - 41