VIEWS: 12 PAGES: 8 POSTED ON: 2/26/2012
The Common-Base Ampliﬁer Basic Circuit Fig. 1 shows the circuit diagram of a single stage common-base ampliﬁer. The object is to solve for the small-signal voltage gain, input resistance, and output resistance. Figure 1: Common-base ampliﬁer. DC Solution (a) Replace the capacitors with open circuits. Look out of the 3 BJT terminals and make Thévenin equivalent circuits as shown in Fig. 2. V + R2 + V − R1 VBB = RBB = R1 kR2 R1 + R2 VEE = V − REE = RE VCC = V + RCC = RC (b) Make an “educated guess” for VBE . Write the loop equation between the VBB and the VEE nodes. To solve for IC , this equation is IC IC VBB − VEE = IB RBB + VBE + IE REE = RBB + VBE + REE β α (c) Solve the loop equation for the currents. VBB − VEE − VBE IC = αIE = βIB = RBB /β + REE /α (d) Verify that VCB > 0 for the active mode. VCB = VC − VB = (VCC − IC RCC ) − (VBB − IB RBB ) = VCC − VBB − IC (RCC − RBB /β) 1 Figure 2: DC bias circuit. Small-Signal or AC Solutions (a) Redraw the circuit with V + = V − = 0 and all capacitors replaced with short circuits as shown in Fig. 3. Figure 3: Signal circuit. (b) Calculate gm , rπ , re , and r0 from the DC solution.. IC VT VT VA + VCE gm = rπ = re = r0 = VT IB IE IC (c) Replace the circuits looking out of the base and emitter with Thévenin equivalent circuits as shown in Fig. 4. RE vtb = 0 Rtb = 0 vte = vs Rte = Rs kRE Rs + RE 2 Figure 4: Signal circuit with Thévenin emitter circuit. Exact Solution (a) Replace the BJT in Fig. 4 with the Thévenin emitter circuit and the Norton collector circuit as shown in Fig. 5. Figure 5: Emitter and collector equivalent circuits. (b) Solve for ic(sc) . RE ic(sc) = −Gme vte = −Gme vs Rs + RE 1 αr0 + re0 rx 0 Gme = 0 0 re = + re Rte + re kr0 r0 + re 1+β (c) Solve for vo . RE vo = −ic(sc) ric kRC kRL = Gme vs ric kRC kRL Rs + RE 0 r0 + re kRte ric = 0 1 − αRte / (re + Rte ) 3 (d) Solve for the voltage gain. vo RE Av = = Gme ric kRC kRL vs Rs + RE (e) Solve for rin . 0 r0 + Rtc rin = R1 kR2 krie rie = re 0 re + r0 + Rtc / (1 + β) (f) Solve for rout . rout = ric kRC Example 1 For the CB ampliﬁer in Fig. 1, it is given that Rs = 100 Ω, R1 = 120 kΩ, R2 = 100 kΩ, RC = 4.3 kΩ, RE = 5.6 kΩ, R3 = 100 Ω, RL = 20 kΩ, V + = 15 V, V − = −15 V, VBE = 0.65 V, β = 99, α = 0.99, rx = 20 Ω, VA = 100 V and VT = 0.025 V. Solve for Av , rin , and rout . Solution. Because the dc bias circuit is the same as for the common-emitter ampliﬁer example, the dc bias values, re , gm , rπ , and r0 are the same. In the signal circuit, the Thévenin voltage and resistance seen looking out of the emitter are given by RE vte = vs = 0.9825vs Rte = Rs kRE = 98.25 Ω Rs + RE The Thévenin resistances seen looking out of the base and the collector are Rtb = 0 Rtc = RC kRL = 3.539 kΩ 0 Next, we calculate re , Gme , ric , and rie . Rtb + rx 1 αr0 + re 0 1 0 re = + re = 12.03 Ω Gme = 0 0 = S 1+β Rte + re kr0 r0 + re 111.4 0 r0 + re kRte r0 + Rtc 0 ric = 0 = 442.3 kΩ rie = re 0 = 12.83 Ω 1 − αRte / (re + Rte ) re + r0 + Rtc / (1 + β) The output voltage is given by RE vo = Gme (ric kRtc ) vte = Gme (ric kRtc ) vs = 30.97vs Rs + RE Thus the voltage gain is Av = 30.97 The input and output resistances are rin = R1 kR2 krib = 12.81 Ω rout = ric kRC = 4.259 kΩ Approximate Solutions These solutions assume that r0 = ∞ except in calculating ric . In this case, ic(sc) = i0 = αi0 = βib . c e 4 Figure 6: Simpliﬁed T model circuit. Simpliﬁed T Model Solution (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJT with the simpliﬁed T model as shown in Fig. 6. (b) Solve for i0 and ric . c ¡ 0 ¢ i0 ¡ 0 ¢ α 0 − vte = i0 re + Rte = c re + Rte =⇒ i0 = −vte 0 e c α re + Rte 0 r0 + re kRte ric = 0 1 − αRte / (re + Rte ) (c) Solve for vo . α RE α vo = −i0 ric kRC kRL = vte c 0 ric kRC kRL = vs 0 ric kRC kRL re + Rte Rs + RE re + Rte (d) Solve for the voltage gain. vo Rs α Av = = 0 +R ric kRC kRL vs Rs + RE re te (e) Solve for rie and rin . ve 0 0 − ve = i0 re =⇒ i0 = − e e 0 re ve 0 rie = = re −i0 e 0 rin = re kRE (f) Solve for rout . rout = ric kRC Example 2 For Example 1, use the simpliﬁed T-model solutions to calculate the values of Av , rin , and rout . 5 ¡ ¢ ¡ ¢ Av = 0.9825 × 8.978 × 10−3 × 3.511 × 103 = 30.97 rin = 12 Ω rout = 4.259 kΩ π Model Solution (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJT with the π model as shown in Fig. 7. Figure 7: Hybrid-π model circuit. (b) Solve for i0 and ric . c i0 c i0 i0 −vte 0 − vte = ib rx + vπ + i0 Rte = e rx + c + c Rte =⇒ i0 = c β gm α rx 1 Rte + + β gm α 0 r0 + re kRte ric = 0 1 − αRte / (re + Rte ) (c) Solve for vo . vte RE 1 vo = −i0 ric kRC kRL = r kR kR = vs r kR kR c rx 1 Rte ic C L Rs + RE rx 1 Rte ic C L + + + + β gm α β gm α (d) Solve for the voltage gain. vo RE 1 Av = = r kR kR vs Rs + RE rx 1 Rte ic C L + + β gm α (e) Solve for rout . rout = ric kRC 6 (f) Solve for rie and rin . i0 e 1+β 0 − ve = ib (rx + rπ ) = (rx + rπ ) =⇒ i0 = −ve e 1+β rx + rπ ve rx + rπ rie = 0 = −ie 1+β rin = rie kRE Example 3 For Example 1, use the π-model solutions to calculate the values of Av , rin , and rout . ¡ ¢ ¡ ¢ Av = 0.9825 × 8.978 × 10−3 × 3.539 × 103 = 30.97 rin = 12 Ω rout = 4.259 kΩ T Model Solution (a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the BJT with the T model as shown in Fig.??. Figure 8: T model circuit. (b) Solve for i0 and ric . c i0 c i0 −vte 0 − vte = ib rx + i0 (re + Rte ) = e rx + c (re + Rte ) =⇒ i0 = c β α rx re + Rte + β α 0 r0 + re kRte ric = 0 1 − αRte / (re + Rte ) (c) Solve for vo . vte RE 1 vo = −i0 ric kRC kRL = r kR kR = vs r kR kR c rx re + Rte ic C L Rs + RE rx re + Rte ic C L + + β α β α 7 (d) Solve for the voltage gain. vo RE 1 Av = = r kR kR vs Rs + RE rx re + Rte ic C L + β α (e) Solve for rie and rin . µ ¶ i0 e rx −ve 0 − ve = ib rx + i0 re e = rx + i0 re = i0 e e + re =⇒ i0 = e rx 1+β 1+β + re 1+β ve rx rie = = + re −i0 e 1+β rin = RE krie (f) Solve for rout . rout = ric kRC Example 4 For Example 1, use the T-model solutions to calculate the values of Av , rin , and rout . ¡ ¢ ¡ ¢ Av = 0.9825 × 8.978 × 10−3 × 3.539 × 103 = 30.97 rin = 12 Ω rout = 4.259 kΩ 8