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ESE319 Introduction to Microelectronics High Frequency BJT Model Cascode BJT Amplifier 2008 Kenneth R. Laker, update 08Oct12 KRL 1 ESE319 Introduction to Microelectronics Gain of 10 Amplifier – Non-ideal Transistor C in RS R1 RC V CC R2 RE vs Gain starts dropping at > 1MHz. Why! Because of internal transistor capacitances that we have ignored in our low frequency and mid-band models. 2008 Kenneth R. Laker, update 08Oct12 KRL 2 ESE319 Introduction to Microelectronics Sketch of Typical Voltage Gain Response for a CE Amplifier ∣Av∣dB Low Frequency Band High Frequency Band Midband ALL capacitances are neglected, i.e. External C's s.c. Internal C's o.c. Due to external 3 dB blocking and bypass capacitors. Internal C's o.c. Due to BJT parasitic capacitors Cπ and Cµ. External C's s.c. 20 log10∣Av∣ dB fL BW = f H − f L≈ f 2008 Kenneth R. Laker, update 08Oct12 KRL f H f Hz H GBP=∣ Av−mid∣ BW (log scale) 3 ESE319 Introduction to Microelectronics High Frequency Small-signal Model (Fwd. Act.) SPICE CJC = Cµ0 CJE = Cje0 C rx C TF = τF RB = rx vbe Two capacitors and a resistor added. A base to emitter capacitor, Cπ A base to collector capacitor, Cµ A resistor, rx, representing the base terminal resistance (rx << rπ) 2008 Kenneth R. Laker, update 08Oct12 KRL C = C 0 m V CB 1 V 0c C =C de C je≈C de 2 C je 0 ≈C de C de = F g m τF = forward-base transit time 4 ESE319 Introduction to Microelectronics High Frequency Small-signal Model (IC) 2008 Kenneth R. Laker, update 08Oct12 KRL 5 ESE319 Introduction to Microelectronics High Frequency Small-signal Model The transistor parasitic capacitances have a strong effect on circuit high frequency performance! They attenuate base signals, decreasing vbe since their reactance approaches zero (short circuit) at high frequencies. As we will see later Cµ is the principal cause of this gain loss at high frequencies. At the base Cµ looks like a capacitor of value k Cµ connected between base and emitter, where k > 1 and may be >> 1. C rx C This phenomenon is called the Miller Effect. 2008 Kenneth R. Laker, update 08Oct12 KRL 6 ESE319 Introduction to Microelectronics Prototype Common Emitter Circuit V CC RC R S C in vs C RS vo RB V B RE vs C byp At high frequencies “low frequency” capacitors are “short circuits” 2008 Kenneth R. Laker, update 08Oct12 KRL b RB r c C v be g m v be vo RC e High frequency small-signal ac model 7 ESE319 Introduction to Microelectronics Multisim Simulation C 2 pF RS 50 vs RB 50 k b c r C v be g m v be 40 mS v be 2.5 k 12 pF vo RC Mid-band gain 5.1 k e Av−mid =−g m R C =−204 Half-gain point 2008 Kenneth R. Laker, update 08Oct12 KRL 8 ESE319 Introduction to Microelectronics Introducing the Miller Effect C RS vs b RB r c C v be g m v be vo RC e The feedback connection of C between base and collector causes it to appear in the amplifier like a large capacitor 1− K C has been inserted Between the base and emitter terminals. This phenomenon is called the “Miller effect” and the capacitive multiplier “1 – K ” acting on C equals the CE amplifier mid-band gain, i.e. K = A v−mid =−g m.R C NOTE: CB and CC amplifiers do not suffer from the Miller effect, since in these amplifiers, one side of C is connected directly to ground. 2008 Kenneth R. Laker, update 08Oct12 KRL 9 ESE319 Introduction to Microelectronics Miller's Theorem C RS vs b RB r c -i i g m v be C e R B∥r ≫ R S v o vo Av−mid = ≈ =−g m RC v s v be 2008 Kenneth R. Laker, update 08Oct12 KRL vo RC + <=> V be - I Z Av−mid V be −I + Vo - Vo =A v−mid =−g m RC V be 1 Z= 2 f C 10 ESE319 Introduction to Microelectronics I 2 =−I I =I1 Z + + Av V 1 V1 V2 - Miller's Theorem <=> - V 1−V 2 V 1− Av V 1 V1 I = I 1= = = => Z Z Z 1− Av 1 V 2− V 2 V 2 −V 1 Av V2 −I =I 2= = = => Z Z Z 1 1− Av 2008 Kenneth R. Laker, update 08Oct12 KRL I 2 =−I I 1= I + V1 Z1 Z2 + V 2 =Av V 1 - - V1 V1 Z Z 1= = = I1 I 1− Av Z 1= 1 j 2 f C 1− Av V2 V2 Z 2= = = I 2 −I Z ≈ Z if A >> 1 1 v 1− Av Ignored in V2 V2 1 practical Z 2= = ≈ I 2 − I j 2 f C circuits 11 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis C b V be RS Vs RB r IC Determine effect of C : c C V be g m V be Vo IR Using phasor notation: I R =−g m V be I C or V o =−g m V be I C RC C C RC e where I C = V be−V o j C Note: The current through C I C = V be g m V be RC − I C RC j C depends only on V be! Vo 2008 Kenneth R. Laker, update 08Oct12 KRL 12 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis II C From slide 7: I C = V be g m V be RC − I C RC j C b V be RS IC c Vs Collect terms for I C 1 j RC C I C = 1 g m R C j C V be and V be : RB r C IC = 1g m RC j C 1 j R C C v be g m V be IR e V be ≈ 1 g m R C j C V be = j C eq V be RC C ≪1 Miller Capacitance Ceq: C eq =1− Av C =1 g m RC C 2008 Kenneth R. Laker, update 08Oct12 KRL 13 Vo C RC ESE319 Introduction to Microelectronics C b V be RS Vs RB r IC C c g m V be C IR e b V be RS Vo Vs RB c C RC IC r C C eq C g m V be IR Vo C RC e C eq =1− Av C =1 g m RC C 2008 Kenneth R. Laker, update 08Oct12 KRL 14 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis III C eq = 1 g m RC C For our example circuit: 1g m RC =10.040⋅5100=205 C eq =205⋅2 pF ≈410 pF 2008 Kenneth R. Laker, update 08Oct12 KRL 15 ESE319 Introduction to Microelectronics Simplified HF Model C b V be RS RB Vs r IC C v be c Vo g m V be ro ' R L =r o∥ R C∥R L RC R L e C R B∥r V =V s R B∥r R S ' ' R S =r∥ RB∥R S b V be RS ' s ' Vs RB r IC C v be c g m V be Vo ' RL e Thevenin Equiv. 2008 Kenneth R. Laker, update 08Oct12 KRL 16 ESE319 Introduction to Microelectronics Simplified HF Model C R ' S IC b ' c Vo V be ' Vs RB r g m V be C ' R S =r∥ RB∥R S R B∥r ' V s =V s R B∥r R S ' RL e Miller's Theorem ' RS ' Vs C eq =1g m RC C C tot =C C eq 2008 Kenneth R. Laker, update 08Oct12 KRL R L =r o∥ R C∥R L b RB c IC Cr Vo V be g m V be C eq ' RL e C tot 17 ESE319 Introduction to Microelectronics Simplified HF Model ' RS V ' s b RB IC Cr c V be Vo g m V be C eq ' R S =r∥ RB∥R S R B∥r ' V s =V s R B∥r R S ' RL e C tot C tot =C 1g m R C C ∣∣ Vo dB Vs 1/ j C tot V be = V s' 1/ j C tot R S ' ' ' R L =r o∥ R C∥R L ' Vo −g m R L −g m R L A v f = ≈ = V s 1 j 2 f C tot R'S f 1 j fH 1 ' f = Av−mid =−g m R L and H 2 C tot R'S 2008 Kenneth R. Laker, update 08Oct12 KRL 18 ESE319 Introduction to Microelectronics Frequency-dependent “beta” hfe jωCµVπ IC = (gm – jωCµ)Vπ short-circuit current V =V be The relationship ic = βib does not apply at high frequencies f > fH! Using the relationship – ic = f(Vπ ) – find the new relationship between ib and ic. For ib (using phasor notation (Ix & Vx) for frequency domain analysis): 1 where r x ≈0 (ignore rx) I = j C C V @ node B': b r 2008 Kenneth R. Laker, update 08Oct12 KRL 19 ESE319 Introduction to Microelectronics Frequency-dependent hfe or “beta” jωCµVπ 1 I b= j C C V r IC = (gm – jωCµ)Vπ @ node C: I c = g m − j C V (ignore ro) Leads to a new relationship between the Ib and Ic: g m− j C Ic h fe = = Ib 1 j C C r 2008 Kenneth R. Laker, update 08Oct12 KRL 20 ESE319 Introduction to Microelectronics Frequency Response of hfe h fe = g m − j C 1 j C C r multiplying N&D by rπ g m− j C r h fe = 1 j C C r factor N to isolate gm C 1− j g m r gm h fe = 1 j C C r 2008 Kenneth R. Laker, update 08Oct12 KRL IC g m= VT VT r = IC C 1 For small ω = ≪1 low : low s gm 10 and: low C C r ≪1 1 10 C Note: low C C r=low C C ≫low gm gm We have: h fe =g m r = 21 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. C f 1− j 1− j 1− j g m r z fz gm h fe = = g m r= 1 j C C r f 1 j 1 j f h fe dB 20log10 C f z≫ f C C r =C C ≫ => gm gm f f f z Hence, the lower break frequency or – 3dB frequency is fβ gm 1 f = = 2 C C r 2 C C the upper: gm 1 f z= = 2 C / g m 2 C where f z 10 f 2008 Kenneth R. Laker, update 08Oct12 KRL 22 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. Using Bode plot concepts, for the range where: f f h fe =g m r = f f f z s.t. ∣1− j f / f z∣≈1 For the range where: We consider the frequency-dependent numerator term to be 1 and focus on the response of the denominator: f f f z h fe ≈ gm r f 1 j f 2008 Kenneth R. Laker, update 08Oct12 KRL = f 1 j f 23 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. Neglecting numerator term: And for f / f >>1 (but < f / f z ): Unity gain bandwidth: h fe = g m r f 1 j f f 1 j f f = ∣h fe∣≈ f f f f ∣h fe∣=1⇒ f ¿ ¿| f = f =1⇒ f T = f T T f T= = f 2 2008 Kenneth R. Laker, update 08Oct12 KRL = BJT unity-gain frequency or GBP 24 ESE319 Introduction to Microelectronics Frequency Response of hfe cont. =100 r =2500 −3 C =12 pF C =2 pF g m =40⋅10 S 12 −3 1 10 ⋅10 6 = = =28.57⋅10 rps C C r 122⋅2.5 28.57 6 f = = 10 Hz=4.55 MHz 2 6.28 f T = f =455 MHz g m 40⋅10−3⋅1012 9 z= = Hz=20⋅10 rps C 2 z f z= =3.18⋅109 Hz=3180 MHz 2 2008 Kenneth R. Laker, update 08Oct12 KRL 25 ESE319 Introduction to Microelectronics Scilab fT Plot //fT Bode Plot Beta=100; KdB= 20*log10(Beta); fz=3180; fp=4.55; f= 1:1:10000; term1=KdB*sign(f); //Constant array of len(f) term2=max(0,20*log10(f/fz)); //Zero for f < fz; term3=min(0,-20*log10(f/fp)); //Zero for f < fp; BodePlot=term1+term2+term3; plot(f,BodePlot); 2008 Kenneth R. Laker, update 08Oct12 KRL 26 ESE319 Introduction to Microelectronics hfe Bode Plot (dB) fT 2008 Kenneth R. Laker, update 08Oct12 KRL 27 ESE319 Introduction to Microelectronics Multisim Simulation v-pi Ic Ib v-pi mS Insert 1 ohm resistors – we want to measure a current ratio. g m− j C Ic h fe = = Ib 1 j C C r 2008 Kenneth R. Laker, update 08Oct12 KRL 28 ESE319 Introduction to Microelectronics Simulation Results Theory: Low frequency |hfe| f T = f =455 MHz Unity Gain frequency about 440 MHz. 2008 Kenneth R. Laker, update 08Oct12 KRL 29 ESE319 Introduction to Microelectronics The Cascode Amplifier ● A two transistor amplifier used to obtain simultaneously: 1. Reasonably high input impedance. 2. Reasonable voltage gain. 3. Wide bandwidth. None of the conventional single transistor designs will satisfy all of the criteria above. ● The cascode amplifier will satisfy all of these criteria. ● A cascode is a CE Stage cascaded with a CB Stage. ● (Historical Note: the cascode amplifier was a cascade of grounded cathode and grounded grid vacuum tube stages – hence the name “cascode,” which has remained in modern terminology.) 2008 Kenneth R. Laker, update 08Oct12 KRL 30 ESE319 Introduction to Microelectronics The Cascode Amplifier CB Stage C byp CE Stage RRs S vs R1 Cin B1 i C1 i B1 C1 R2 i E1 i B2 B2 R3 RC Q1 E2 v-out vO E1 i C2 C2 Q2 CB Stage CE Stage RRs S i c2 i b2 Q2 i e2 V CC v s RB RE Q1 i e1 RE R B =R 2∥R3 i c1 i b1 i E2 R in1= vo RC v e1 v c2 = =low≈r e1 i e1 i c2 ac equivalent circuit Comments: 1. R1, R2, R3, and RC set the bias levels for both Q1 and Q2. 2. Determine RE for the desired voltage gain. 3. Cin and Cbyp are to act as “open circuits” at dc and act as “short circuits” at all operating frequencies f f min . 2008 Kenneth R. Laker, update 08Oct12 KRL 31 ESE319 Introduction to Microelectronics Cascode Mid-Band Small Signal Model CB Stage C byp CE Stage RRs S vs R1 Cin B1 i C1 i B1 C1 R2 i E1 i B2 B2 R3 RC Q1 V CC RRs S vs i E2 RE r1 CE Stage r2 RB i e2 ggm vv 1 m1 be1 E1 i c2 R in1 =low C2 i b2 vvbe2 2 vo C1 vvbe1 1 i e1 B2 Q2 E2 i b1 v-out vO E1 i C2 C2 i c1 B1 CB Stage g m vv 2 m2 be2 E2 RE =R∥R 2∥R 3 RRB B=R 2 3 2008 Kenneth R. Laker, update 08Oct12 KRL 32 RC ESE319 Introduction to Microelectronics Cascode Small Signal Analysis i c1 B1 CB Stage i b1 r1 RRs S vs CE Stage i b2 vv 2 r be2 RB i e2 ggm vv 1 m1 be1 E1 i c2 R in1 =low C2 2 C1 vvbe1 1 i e1 B2 vo g m vv 2 m2 be2 E2 RE g m1=g m2=g m r e1=r e2=r e r 1=r 2 =r 2008 Kenneth R. Laker, update 08Oct12 KRL RC 1. Show reduction in Miller effect 2. Evaluate small-signal voltage gain OBSERVATIONS a. The emitter current of the CB Stage is the collector current of the CE Stage. (This also holds for the dc bias current.) i e1=i c2 b. The base current of the CB Stage is: i e1 i c2 i b1= = 1 1 c. Hence, both stages have about same collector current i c1≈ic2 and same gm, re, rπ. 33 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis cont. i b1 r RRs S vs RB i e2 ggm v 1 m be1 E1 R ≈r in 1 e1 i c2 C2 i b2 vv 2 r be2 vo C1 vvbe1 1 i e1 B2 CE Stage i c1 B1 CB Stage g mvv 2 m be2 E2 RE RC The input resistance Rin1 to the CB Stage is the small-signal “re1” for the CB Stage, i.e. i e1 i c2 i b1= = 1 1 The CE output voltage, the voltage drop from Q2 collector to ground, is: r r v c2=v e1 =−r i b1=− i c2=− i e1 1 1 Therefore, the CB Stage input resistance is: r v e1 Rin1= = =r e1 −i e1 1 v c2 R in1 re r AvCE −Stage = ≈− =− 1 => C eq =1 e C 2 C vs RE RE RE 2008 Kenneth R. Laker, update 08Oct12 KRL 34 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis - cont. i c1 i b1 v be1 i e1 RS i b2 vvbe2be2 gmmvvbe1 be1 i c2 Rin 1≈r e1 g mvv be2 m be2 i e2 i c1≈ie1 =i c2≈i e2 Now, find the CE collector current in terms of the input voltage vs: Recall i c1≈ic2 vs i b2≈ R S∥R B r 1 R E vs vs i c2 = ib2 ≈ ≈ R S∥R B r 1 R E 1 R E for bias insensitivity: 1 R E ≫ R S∥R B r v ss i C2≈ RE v o=−i c2 RC v o −R C Av = = vs RE OBSERVATIONS: 1. Voltage gain Av is about the same as a stand-along CE Amplifier. 2. HF cutoff is much higher then a CE Amplifier due to the reduced Ceq. 2008 Kenneth R. Laker, update 08Oct12 KRL 35 ESE319 Introduction to Microelectronics Cascode Biasing o.c. CB RS C in o.c. I1 IC1 Q1 IE1 vO R in1 =low IC2 Q2 IE2 1 2 I E2=I C2= I E1= I C1 ⇒ I C1≈ I E2 1 2008 Kenneth R. Laker, update 08Oct12 KRL 1. Choose IE1 – make it relatively large to reduce R in1=r e =V T / I E1 to push out HF break frequencies. 2. Choose RC for suitable voltage swing VC1 and RE for desired gain. 3. Choose bias resistor string such that its current I1 is about 0.1 of the collector current IC1. 4. Given RE, IE2 and VBE2 = 0.7 V calc. R3. 5. Need to also determine R1 & R2. 36 ESE319 Introduction to Microelectronics Cascode Biasing - cont. II11 VB1 VB2 Q1 vO Rin 1≈r e1 Q2 VC2 Since the CE-Stage gain is very small: a. The collector swing of Q2 will be small. b. The Q2 collector bias VC2= VB1 - 0.7 V. 6. Set V B1−V B2 =1V ⇒V CE2 =1 V This will limit VCB2 V CB2 =V CE2 −V BE2=0.3V which will keep Q2 forward active. V CE2=V C2 −V R e =V C2−V B2 −0.7V .=V B1−0.7V −V B2 0.7 V .=V B1−V B2 2008 Kenneth R. Laker, update 08Oct12 KRL 7. Next determine R2. Its drop VR2 = 1 V V B1−V B2 with the known current. R2= I1 37 ESE319 Introduction to Microelectronics Cascode Biasing - cont. V B1−V B2 1 V R2= = I1 I1 V B1 Q1 Rin 1≈r e1 Q2 8. Then calculate R3. V B2 R 3= I1 where V B2 =0.7V I E R E V CC Note: R 1R 2 R 3= I1 9. Then calculate R1. V CC R 1= −R 2−R 3 0.1 I C 2008 Kenneth R. Laker, update 08Oct12 KRL 38 ESE319 Introduction to Microelectronics Cascode Bias Summary SPECIFIED: Av, VCC, VC1 (CB collector voltage); SPECIFIED: IE (or IC) directly or indirectly through BW. DETERMINE: RC, RE, R1, R2 and R3. V C1 STEP1: R C = SET: V B1−V B2 =1V ⇒ V CE2 =1V IC V B1 Q1 Rin 1≈r e1 Q2 I C2= I E1≈ I C1≈ I E2 =I C 2008 Kenneth R. Laker, update 08Oct12 KRL RE= RC ∣A v∣ V B1−V B2 1V STEP2: R 2 = = I1 0.1 I C V B2 0.7 V I E R E STEP3: R 3= = I1 0.1 I C V CC V CC R 1R 2 R 3= = I 1 0.1 I C V STEP4: R 1= CC −R 2−R 3 0.1 I C 39 ESE319 Introduction to Microelectronics Cascode Bias Example R1 VCC-ICRE-1.7 V C1 RC I R C C Q1 Q1 VVCE1 =V =ICCC RC−I –1–I R −V CE2 −I C R E CR CE1 C CE 1.0 Q2 =12 V Q2 V =1 CE2 ICRE+0.7 Cascode Amp 2008 Kenneth R. Laker, update 08Oct12 KRL RE =12 V ICRE I E2≈ I C2= I E1≈ I C1 ⇒ I C1≈ I E2 Typical Bias Conditions 40 ESE319 Introduction to Microelectronics Cascode Bias Example cont. 1. Choose IE1 – to set re. Q1 Try IE1 = 5 mA => r e =0.025 V / I E =5 . V CE1=V CC − I C RC −1− I C R E Q2 2. Set desired gain magnitude. For example if AV = -10, then RC/RE = 10. 3. Since the CE stage gain is very small, VCE2 can be small, i.e. VCE2 = VB1 – VB2 = 1 V. 2008 Kenneth R. Laker, update 08Oct12 KRL 41 ESE319 Introduction to Microelectronics Cascode Bias Example cont. Specs: V CC =12 V V C1=7 V I C =5 mA =100 Q1 RC ∣Av∣= R =10 E V CE1=V CC − I C RC −1− I C R E Q2 2008 Kenneth R. Laker, update 08Oct12 KRL Determine RC for VC1 = 7V . V CC −V C1 5V RC = = =1000 −3 −3 5⋅10 A 5⋅10 A RC RC RE= = =100 ∣Av∣ 10 42 ESE319 Introduction to Microelectronics Cascode Bias Example cont. I1 V CC =12 R1 RC I R C C VCC-ICRE-1.7 Q1 Make current through the string of bias resistors I1 = 0.1 IC = 0.5 mA. =I RC−I –1–ICCREC −1− I C R E VVCE1 CE1=VC CC R2 1.0 Q2 VCE2=1 ICRE+0.7 R3 RC =1 k I C =5 mA. R E =100 RE I R C E =12V V CC 12 R1 R 2 R3= = =24 k −4 I 1 5⋅10 Calculate the bias voltages (base side of Q1, Q2): V R1=V CC −I C R E −1.7V =12 V −0.5V −1.7V =9.8 V V R2=V B1−V B2=1V V R3=V B2 =I C R E 0.7=5⋅10−3⋅1000.7=1.2 V 2008 Kenneth R. Laker, update 08Oct12 KRL 43 ESE319 Introduction to Microelectronics Cascode Bias Example cont. V B2 =5⋅10−4 R3=1.2V CB RRSS C in V B1 V B2 R3 =2.4 k Q1 V B1−V B2=5⋅10−4 R 2=1.0V Q2 R 2=2 k Recall: R1 R 2 R3=24 k R1 =24000−2.400−2000=19.6 k V CC =12 , RC =1 k , V B2 =1.2 V , I C =5 mA , R E =100 , V B1−V B2=1.0V 2008 Kenneth R. Laker, update 08Oct12 KRL 44 ESE319 Introduction to Microelectronics Completed Design =100 f H= r e =5 ⇒ I C =5 mA V C1=7 V RC ∣Av∣= R =10 E CB V B1 RRSS C in V B2 Q1 Q2 1 2 C tot R'S re C tot =C 1 C RE .=C 1.05 C If Cπ = 12 pF Cµ = 2 pF R1 =19.6 k R 2=2 k R3 =2.4 k C tot =14.1 pF f Hcascode=225.8 MHz RC =1 k R E =100 For CE with |Av| = 10 NOTE: R B=R 2∥R 3=1.09 k ≪ R E =10 k 2008 Kenneth R. Laker, update 08Oct12 KRL f HCE =94 MHz 45
