High Frequency BJT Model Cascode BJT Amplifier 1
Gain of 10 Amplifier Non-ideal Transistor C in R 1 V CC R 2 v s Gain starts dropping at > 1MHz. Why! Because of internal transistor capacitances that we have ignored in our low frequency and mid-band models. 2
Sketch of Typical Voltage Gain Response for a CE Amplifier A v db Low Frequency Band Due to external blocking and bypass capacitors. Internal Cs o.c. Midband ALL capacitances are neglected, i.e. External Cs s.c. Internal Cs o.c. 3dB 20 log 10 A v db High Frequency Band Due to BJT parasitic capacitors C π and C µ. External Cs s.c. f L BW = f H f L f H f H GBP= A v mid BW f Hz (log scale) 3
High Frequency Small-signal Model (Fwd. Act.) r x SPICE CJC = C µ0 CJE = C je0 TF = τ F v be Two capacitors and a resistor added. A base to emitter capacitor, C π A base to collector capacitor, C µ A resistor, r x, representing the base terminal resistance (r x << r π ) = =C de C je C de 2C je 0 C de C de = F g m RB = r x 0 1 V m CB V 0c τ F = forward-base transit time 4
High Frequency Small-signal Model (IC) 5
High Frequency Small-signal Model The transistor parasitic capacitances have a strong effect on circuit high frequency performance! They attenuate base signals, decreasing v be since their reactance approaches zero (short circuit) at high frequencies. As we will see later C µ is the principal cause of this gain loss at high frequencies. At the base C µ looks like a capacitor of value k C µ connected between base and emitter, where k > 1 and may be >> 1. r x This phenomenon is called the Miller Effect. 6
Prototype Common Emitter Circuit V CC C in v o b c v o v s v s r v be g m v be V B C byp e At high frequencies low frequency capacitors are short circuits High frequency small-signal ac model 7
Multisim Simulation 50 b 2 pf c v o v s 50k v r be C g m v be Mid-band gain 2.5 k 40mS v be 12 pf e 5.1k A v mid = g m = 204 Half-gain point 8
Introducing the Miller Effect b c v o v s r C v be g m v be e The feedback connection of between base and collector causes it to appear in the amplifier like a large capacitor 1 K has been inserted Between the base and emitter terminals. This phenomenon is called the Miller effect and the capacitive multiplier 1 K acting on equals the CE amplifier mid-band gain, i.e. K= A v mid = g m. NOTE: CB and CC amplifiers do not suffer from the Miller effect, since in these amplifiers, one side of is connected directly to ground. 9
Millers Theorem v s r b r i e -i g m v be c v o <=> V be I Z I + + A v mid V be - - V o A v mid = v o v s v o v be = g m V o V be =A v mid = g m 1 Z = 2 f 10
I =I 1 Z I 2 = I + + V 1 A v V 1 V 2 - - ESE319 Introduction to Microelectronics Millers Theorem <=> I 1 =I I 2 = I + + Z 1 Z 2 V 1 V 2 =A v V 1 - - I =I 1 = V 1 V 2 Z = V 1 A v V 1 = V 1 => Z Z 1 A v I =I 2 = V V 2 1 V 2 V 1 A 2 v = = V 2 Z Z Z 1 1 A v => Z 1 = V 1 I 1 = V 1 I = Z 1 = 1 j 2 f 1 A v Z 2 = V 2 I 2 = V 2 I = Z 1 A v Z 1 1 Z A v Z 2 = V 2 I 2 = V 2 I 1 j 2 f if A v >> 1 Ignored in practical circuits 11
Common Emitter Miller Effect Analysis b V be I c V o Determine effect of : Using phasor notation: V s r V be g m V be I RC or I RC = g m V be I C V o = g m V be I e where Note: The current through depends only on! V be I =V be V o j I =V be g m V be I C R j V o 12
Common Emitter Miller Effect Analysis II From slide 7: I =V be g m V be I C R j b V be I C c V o Collect terms for and : I V be V s r v be g m V be I RC 1 j I =1g m R j V be e I = 1g m R j 1 j V be 1g m R j V be = j C eq V be 1 Miller Capacitance C eq : C eq =1 A v =1g m 13
I C I C b V be c V o b V be c V o I RC I RC V s r e g m V be V s r C eq e g m V be C eq =1 A v =1g m 14
Common Emitter Miller Effect Analysis III C eq =1g m R For our example circuit: 1g m =10.040 5100=205 C eq =205 2 pf 410 pf 15
Simplified HF Model b V be I c V o V s r v C be g m V be r o R L R L =r o R L e V s =V s r r b V be I c V o =r V s r v be g m V be R L Thevenin Equiv. e 16
Simplified HF Model b V be I c V o R L =r o R L =r V s r g m V be R L R V B r s =V s r e Millers Theorem b c V o I C V be V s r C eq g m V be RL C eq =1g m C tot = C eq C tot e 17
Simplified HF Model V s b I r C eq g m V be RL V be c V o R L =r o R L =r e V s =V s r r C tot C tot = 1g m R V be = 1/ j C tot 1/ j C tot V s V o V s db A v f = V o V s A v mid = g m R L g m R L = g m R L 1 j 2 f C tot f 1 j and f H = 1 2C tot f H 18
Frequency-dependent beta h fe jωc µ V π I C = (g m jωc µ )V π short-circuit current V =V be The relationship i c = βi b does not apply at high frequencies f > f H! Using the relationship i c = f(v π ) find the new relationship between i b and i c. For i b (using phasor notation (I x & V x ) for frequency domain analysis): = I 1 @ node B: b j C r V where r x 0 (ignore r x ) 19
Frequency-dependent h fe or beta jωc µ V I C = (g m jωc µ )V π π = I 1 b j C r C V @ node C: I c = g m j V (ignore r o ) Leads to a new relationship between the I b and I c : h fe = I c I b = g j C m 1 j C r 20
Frequency Response of h fe h fe = g m j 1 j C r multiplying N&D by r π h fe = g j r m 1 j r factor N to isolate g m 1 j g g m r m h fe = 1 j r g m = I C V T For small ω s r = V T I C = low : low g m 1 1 10 and: low r 1 1 10 Note: low r = low g m low g m We have: h fe =g m r = 21
Frequency Response of h fe cont. 1 j C = g g m r 1 j m z h fe = 1 j r 1 j g m r 1 j = 1 f f z j f f r = g m g m => f z f h fe db 20log 10 f f z f Hence, the lower break frequency or 3dB frequency is f β f = 1 2 r = g m 2 the upper: f z = where f z 10 f 1 2 / g m = g m 2 22
Frequency Response of h fe cont. Using Bode plot concepts, for the range where: f f h fe =g m r = For the range where: f f f z s.t. 1 j f / f z 1 We consider the frequency-dependent numerator term to be 1 and focus on the response of the denominator: f f f z h fe g m r f 1 j f = f 1 j f 23
Frequency Response of h fe cont. Neglecting numerator term: h fe = g m r f 1 j f = f 1 j f And for f / f >>1 (but < f / f z ): h fe f f = f f Unity gain bandwidth: h fe =1 f f f T = T 2 = f f = f T =1 f T = f BJT unity-gain frequency or GBP 24
Frequency Response of h fe cont. =100 r =2500 =12 pf =2 pf g m =40 10 3 S 12 10 3 1 = = 10 r 122 2.5 =28.57 106 rps f = 2 = 28.57 6.28 106 Hz=4.55 MHz f T = f =455 MHz z = g m = 40 10 3 10 12 2 Hz=20 10 9 rps f z = z 2 =3.18 109 Hz=3180 MHz 25
Scilab f T 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); 26
h fe Bode Plot (db) f T 27
Multisim Simulation v-pi I c I b ms v-pi Insert 1 ohm resistors we want to measure a current ratio. h fe = I c = g m j I b 1 j C r 28
Simulation Results Low frequency h fe Theory: f T = f =455 MHz Unity Gain frequency about 440 MHz. 29
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.) 30
CB Stage CE Stage v s R s S C byp C in R 1 B1 ESE319 Introduction to Microelectronics The Cascode Amplifier i B1 i R E1 2 i B2 R 3 B2 i C1 C1 E2 Q1 E1 i C2 C2 Q2 i E2 v O v-out V CC =R 2 R 3 CE Stage ac equivalent circuit Comments: 1. R 1, R 2, R 3, and set the bias levels for both Q1 and Q2. 2. Determine for the desired voltage gain. 3. C in and C byp are to act as open circuits at dc and act as short circuits at all operating frequencies f f min. v s R s S i b2 i e2 Q2 i c2 i e1 CB Stage i b1 Q1 i c1 v o R in1 = v e1 i e1 = v c2 i c2 =low r e1 31
Cascode Mid-Band Small Signal Model CB Stage CE Stage v s R s S C byp C in R 1 B1 i B1 i R E1 2 i B2 R 3 i C1 C1 B2 E2 Q1 E1 i C2 C2 Q2 i E2 v O v-out V CC CB Stage CE Stage v s R s i b1 B2 B1 r 1 i e2 i e1 i b2 v r be2 2 2 v 1 be1 C2 R E E1 i c2 i c1 C1 g m 1 m1 v be1 g m m2 v be2 2 E2 R in1 =low v o =R 2 R 3 B =R 2 R 3 32
CB Stage CE Stage v s R s i b1 B2 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis B1 r 1 i e2 i e1 i b2 v r be2 2 2 v 1 be1 C2 R E i c2 g m1 =g m2 =g m r e1 =r e2 =r e r 1 =r 2 =r i c1 C1 g m 1 m1 v be1 g m m2 v be2 2 E2 E1 R in1 =low v o 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 b1 = i e1 1 = i c2 1 c. Hence, both stages have about same collector current i c1 i c2 and same g m, r e, r π. 33
CB Stage CE Stage v s R s ESE319 Introduction to Microelectronics Cascode Small Signal Analysis cont. i b1 B2 B1 r i e2 i e1 i b2 v r be2 2 v 1 be1 C2 R E i c2 i c1 C1 g 1 m v be1 g m v be2 2 E2 E1 R in 1 r e1 v o The input resistance R in1 to the CB Stage is the small-signal r e1 for the CB Stage, i.e. i b1 = i e1 1 = i c2 1 The CE output voltage, the voltage drop from Q2 collector to ground, is: v c2 =v e1 = r i b1 = r 1 i c2= r 1 i e1 Therefore, the CB Stage input resistance is: R in1 = v e1 i e1 = r 1 =r e1 A vce Stage = v c2 v s R in1 = r e 1 => C eq =1 r e 2 34
ESE319 Introduction to Microelectronics Cascode Small Signal Analysis - cont. i b1 i e1 i b2 vv be2 be2 v be1 i c2 i c1 g m v be1 be1 g be2 m v be2 R in 1 r e1 Now, find the CE collector current in terms of the input voltage v s : Recall i c1 i c2 i b2 i c2 = i b2 v s r 1 v s r 1 v s 1 i e2 for bias insensitivity: 1 r v s i A v = v o = R c1 i e1 =i c2 i e2 C i C2 v s v o = i c2 R OBSERVATIONS: E v s 1. Voltage gain A v is about the same as a stand-along CE Amplifier. 2. HF cutoff is much higher then a CE Amplifier due to the reduced C eq. 35
o.c. C B C in o.c. I 1 ESE319 Introduction to Microelectronics I C1 I E1 Cascode Biasing Q1 R in1 =low Q2 I C2 I E2 v O 1. Choose I E1 make it relatively large to reduce R in1 =r e =V T / I E1 to push out HF break frequencies. 2. Choose for suitable voltage swing V C1 and for desired gain. 3. Choose bias resistor string such that its current I 1 is about 0.1 of the collector current I C1. 2 I E2 =I C2 = I E1 = 1 1 I C1 I C1 I E2 4. Given, I E2 and V BE2 = 0.7 V calc. R 3. 5. Need to also determine R 1 & R 2. 36
Cascode Biasing - cont. I 11 V B1 V B2 v O Q1 R in 1 r e1 Q2 V C2 Since the CE-Stage gain is very small: a. The collector swing of Q2 will be small. b. The Q2 collector bias V C2 = V B1-0.7 V. 6. Set V B1 V B2 =1V V CE2 =1V This will limit V CB2 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 0.7V.=V B1 V B2 7. Next determine R 2. Its drop V R2 = 1 V with the known current. R 2 = V B1 V B2 I 1 37
Cascode Biasing - cont. V B1 Q1 R r in 1 e1 Q2 R 2 = V B1 V B2 I 1 8. Then calculate R 3. Note: = 1V I 1 R 3 = V B2 I 1 where V B2 =0.7V I E R 1 R 2 R 3 = V CC I 1 9. Then calculate R 1. R 1 = V CC 0.1 I C R 2 R 3 38
Cascode Bias Summary SPECIFIED: A v, V CC, V C1 (CB collector voltage); SPECIFIED: I E (or I C ) directly or indirectly through BW. DETERMINE:,, R 1, R 2 and R 3. SET: V B1 V B2 =1V V CE2 =1V V B1 Q1 R r in 1 e1 Q2 I C2 =I E1 I C1 I E2 =I C STEP1: STEP2: STEP3: STEP4: = V C1 I C R 2 = V B1 V B2 I 1 R 3 = V B2 I 1 R 1 R 2 R 3 = V CC I 1 = A v = 1V 0.1 I C = 0.7V I E 0.1 I C = V CC 0.1 I C R 1 = V CC 0.1 I C R 2 R 3 39
Cascode Bias Example R 1 Q1 Q2 V C1 V CC -I C -1.7 I C Q1 V CE1 CE1 =V =I CCC I 1 I C CE V CE2 I C 1.0 =12 V Q2 =12 V V CE2 =1 I C +0.7 I C Cascode Amp I E2 I C2 =I E1 I C1 I C1 I E2 Typical Bias Conditions 40
Cascode Bias Example cont. Q1 Q2 V CE1 =V CC I C 1 I C 1. Choose I E1 to set r e. Try I E1 = 5 ma => r e =0.025V / I E =5. 2. Set desired gain magnitude. For example if A V = -10, then / = 10. 3. Since the CE stage gain is very small, V CE2 can be small, i.e. V CE2 = V B1 V B2 = 1 V. 41
Q1 Q2 Cascode Bias Example cont. Specs: V CC =12V =100 V CE1 =V CC I C 1 I C V C1 =7V I C =5 ma Determine for V C1 = 7V. A v = =10 = V CC V C1 5 10 3 A = 5V 5 10 3 A =1000 = A v = 10 =100 42
Cascode Bias Example cont. I 1 V CC =12 I C =5 ma. =1 k =100 R 1 V CC -I C -1.7 R 2 R 3 1.0 I C +0.7 Q1 Q2 I C V CE1 =V =I C CC I 1 I C C 1 I C V CE2 =1 I C =12V Make current through the string of bias resistors I 1 = 0.1 I C = 0.5 ma. R 1 R 2 R 3 = V CC I 1 = 12 5 10 4=24 k Calculate the bias voltages (base side of Q1, Q2): V R1 =V CC I C 1.7V =12V 0.5V 1.7V =9.8 V V R2 =V B1 V B2 =1V V R3 =V B2 =I C 0.7=5 10 3 1000.7=1.2 V 43
Cascode Bias Example cont. V B2 =5 10 4 R 3 =1.2V C B V B1 Q1 R 3 =2.4 k C in V B2 Q2 V B1 V B2 =5 10 4 R 2 =1.0V R 2 =2 k Recall: R 1 R 2 R 3 =24 k R 1 =24000 2.400 2000=19.6 k V CC =12, =1k, V B2 =1.2V, I C =5 ma, =100, V B1 V B2 =1.0V 44
=100 r e =5 I C =5mA V C1 =7 V A v = =10 ESE319 Introduction to Microelectronics Completed Design S C B C in R 1 =19.6 k R 2 =2 k NOTE: R 3 =2.4 k V B1 V B2 Q1 Q2 =1k =100 =R 2 R 3 =1.09 k =10 k f H = 1 2C tot C tot = 1 r e C R E.= 1.05 If C π = 12 pf C µ = 2 pf C tot =14.1 pf f Hcascode =225.8 MHz For CE with A v = 10 f HCE =94 MHz 45