ESE319 Introduction to Microelectronics
High Frequency BJT Model & Cascode BJT Amplifier
Kenneth R. Laker, update 01Oct14 KRL 1 ESE319 Introduction to Microelectronics
Gain of 10 Amplifier – Non-ideal Transistor
RC R1 C in V CC
RS R2 RE
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.
Kenneth R. Laker, update 01Oct14 KRL 2 ESE319 Introduction to Microelectronics Sketch of Typical Voltage Gain Response for a CE Amplifier ∣Av∣dB Low Midband High Frequency ALL capacitances are neglected, i.e. Frequency Band External C's s.c. Band Internal C's o.c. Due to external blocking and by- 3dB Due to BJT parasitic pass capacitors. capacitors Cπ and Cµ. Internal C's o.c. External C's s.c. 20log10∣Av∣dB
f Hz f L f H (log scale)
BW = f H − f L≈ f H GBP=∣Av−mid∣BW
Kenneth R. Laker, update 01Oct14 KRL 3 ESE319 Introduction to Microelectronics High Frequency Small-signal Model (Fwd. Act.)
C r x
C
v be
Two capacitors and a resistor added to mid-band small signal model.
● base-emitter capacitor, Cµ ● base-collector capacitor, Cπ ● resistor, , representing the base terminal resistance ( ); rx rx << rπ ignored in hand calculations.
Kenneth R. Laker, update 01Oct14 KRL 4 ESE319 Introduction to Microelectronics High Frequency Small-signal Model (Fwd. Act.) SPICE/MultiSim C r x CJC = C µ0 CJE = C je0 TF = τ C F RB = r v x be
C =C deC je≈C de2C je0≈C de C 0 C je0 C = C = I C m je m C g i V CB V EB de= m F= F Note C= dt 1 1 V T d v 0cb 0eb C ≈2C 0 C = base-charging (diffusion) cap Non-linear, voltage controlled de = forward-base transit time m = junction grading coefficient 0.2 to 0.5) τF
Kenneth R. Laker, update 01Oct14 KRL 5 ESE319 Introduction to Microelectronics High Frequency Small-signal Model (IC)
C π C µ
C = collector-substrate depletion capacitance, ignored in hand calcs. CS
Kenneth R. Laker, update 01Oct14 KRL 6 ESE319 Introduction to Microelectronics High Frequency Small-signal Model The transistor parasitic capacitances have a strong effect on circuit high frequency performance!
● C attenuates base signal, decreasing v since X → 0 (short circuit) π be Cπ at very high-frequencies.
●
● As we will see later; is principal cause of gain loss at high-frequencies. Cµ At the base base-collector looks like a capacitor of value between Cµ k Cµ base-emitter, where k > 1 and may be >> 1. C r x
C
● This phenomenon is called the Miller Effect.
Kenneth R. Laker, update 01Oct14 KRL 7 ESE319 Introduction to Microelectronics
Prototype Common Emitter Circuit
V CC C R C RS b c vo RS C in vo + v R r be g v B vs RB C m be R vs - C R C V B E byp e
At high frequencies High frequency large blocking and bypass small-signal ac model capacitors are “short circuits”
Kenneth R. Laker, update 01Oct14 KRL 8 ESE319 Introduction to Microelectronics
Multisim Simulation
C 2 pF RS 50 b c vo
v v RB r be g v Mid-band gain @ 100 kHz s C m be RC 50k 40mS v 2.5k 12 pF be 5.1k e
o Av−mid=−g m RC=−20446dB@ 180
Gain @ 8.69 MHz
Kenneth R. Laker, update 01Oct14 KRL 9 ESE319 Introduction to Microelectronics Introducing the Miller Effect C
RS b c vo
v v RB r be g m vbe s C RC
e
The feedback connection of C between base-collector causes it to appear in the amplifier like a large capacitor 1 − K C between the base-emitter terminals. This phenomenon is called the “Miller effect” and capacitor multiplier “1 – K ” acting on C equals the CE amplifier mid-band gain, i.e. .
K=Av−mid =−g m RC 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.
Kenneth R. Laker, update 01Oct14 KRL 10 ESE319 Introduction to Microelectronics
Miller's Theorem 1 Z= j 2 f C C R S b c v I Z −I o + + i -i r vbe g v V vs RB C m be R be V o C <=> Av−mid V be - - e
RB∥r ≫RS V o =Av−mid=−g m RC V be vo vo Av−mid= ≈ =−g m RC vs vbe
Kenneth R. Laker, update 01Oct14 KRL 11 ESE319 Introduction to Microelectronics Miller's Theorem I I I 2=−I = 1 Z I 1=I I 2=−I + + + +
V 1 Av V 1 V 2 V 1 Z Z V 2=Av V 1 <=> 1 2 - - - -
V 1 V 1 Z V 1−V 2 V 1− Av V 1 V 1 Z 1= = = I =I 1= = = => Z Z Z I 1 I 1−Av
1−Av 1 Z 1= 1 j 2 f C 1−Av V 2− V 2 V 2−V 1 Av V 2 V V Z −I =I = = = 2 2 2 Z 2= = = ≈Z if A >> 1 Z Z Z => I −I 1 v 2 1− 1 A 1− v Ignored in A V V v 2 2 1 practical Z 2= = ≈ I 2 −I j 2 f C circuits
Kenneth R. Laker, update 01Oct14 KRL 12 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis
Determine effect of C : I C C Using phasor notation:
RS V be I =−g V I b c RC m be C V o I or RC V be V RB r g m V be V o=I R RC = −g m V beI C RC s C RC C where e 1 I V V C = be− o j 2 f C
I C = V beg m V be RC−I C RC j 2 f C Note: The current through C
depends only on V be ! 1g m RC j2 f C I C = V be 1 j2 f R C C Kenneth R. Laker, update 01Oct14 KRL 13 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis II I C C V RS b be c From slide 13: V o
I R 1g R j2 f C C m C vbe V R r g m V be I C = V be s B C RC 1 j2 f R C C e
1g m RC j2 f C I C = V be≈ 1g m RC j2 f C V be= j2 f C eq V be 1 j2 f R C C 2 f RC C ≪1
Miller Capacitance C : C =1g R C =1−A C eq eq m C v
Kenneth R. Laker, update 01Oct14 KRL 14 ESE319 Introduction to Microelectronics
I I C C C C V R V RS be S b be c b c V o V o I RC I R g V C V s RB r m be r g V C RC V s RB C m be R C C C eq e e
C eq=1−AvC =1g m RC C
Kenneth R. Laker, update 01Oct14 KRL 15 ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis III
C eq=1g m RC C
For our example circuit (C = 2 pF): µ
1g m RC=10.040⋅5100=205
C eq=205⋅2 pF≈410 pF
Kenneth R. Laker, update 01Oct14 KRL 16 ESE319 Introduction to Microelectronics Simplified HF Model
C
RS V be I b C c V o
vbe g V ' V s RB r C m be R r R R ro RC RL L= o∥ C∥ L
e
C ' RB∥r ' RS V be I V s=V s b C c V o RB∥rRS
' ' v RS =r∥ RB∥RS r be g V ' V s RB C m be RL
e Thevenin Equiv.
Kenneth R. Laker, update 01Oct14 KRL 17 ESE319 Introduction to Microelectronics Simplified HF Model
C ' R' r R R RS I L= o∥ C∥ L b C c V o ' R =r ∥ R ∥R V be S B S ' RB∥r ' r g V ' V s RB C m be V s=V s RL RB∥rRS
e Miller's Theorem
' RS b c V o I V be C ' r g V ' V s RB m be C C eq RL
C eq=1g m RC C e C C C tot = eq C tot
Kenneth R. Laker, update 01Oct14 KRL 18 ESE319 Introduction to Microelectronics Simplified HF Model
R' S b c V V o ' I be RL=ro∥RC∥RL C ' r g V ' ' V s RB m be C C eq RL RS =r∥ RB∥RS
' RB∥r e V s=V s ≈V s RB∥r RS C tot
C tot =C 1g m RC C V o 1 j2 f C dB / tot V V be= V s ' ∣ s∣ 1/ j2 f C totRS '
' ' V o −g m RL −g m RL Av f = ≈ ' = V s 1 j 2 f C R f tot S 1 j f H ' 1 Av−mid=−g m RL and f H = ' 2C tot RS
Kenneth R. Laker, update 01Oct14 KRL 19 ESE319 Introduction to Microelectronics
Multisim Simulation
C 2 pF RS 50 b c vo
v v RB r be g v s C m be RC 50k 40mS v 2.5k 12 pF be 5.1k Mid-band gain @ 100 kHz e
o Av−mid=−g m RC=−20446dB@ 180
C tot=12 pF205⋅2 pF ≈422 pF 1 1 f H ≈ f −3dB= ' ≈ ≈7.54 MHz Gain @ 8.69 MHz 2C tot RS 2 422 pF 50
Kenneth R. Laker, update 01Oct14 KRL 20 ESE319 Introduction to Microelectronics Frequency-dependent “beta” j2π f C V I = (g – j2π f C )V 0 µ b C m µ b V b short-circuit g V m b current
The relationship ic = βib does not apply at high frequencies f ≥ fH! Using the relationship – i = f (V ) – find the new relationship c b between i and i . For i (using phasor notation (I =I (jf) & V =V (jf)) b c b x x x x for frequency domain analysis): 1 @ node B': I = j2 f C C V where r x≈0 (ignore r ) b r b x
Kenneth R. Laker, update 01Oct14 KRL 21 ESE319 Introduction to Microelectronics Frequency-dependent “beta” j2π f C V I = (g – j2π f C )V µ b C m µ b V b ∞
g m V b
1 I = j2 f C C V @ node C: I =g − j2 f C V b r b c m b (ignore r ) o Leads to a new relationship between the Ib and Ic:
I c g m− j2 f C β(jf) → h jf = = fe I b 1 j2 f C C r
Kenneth R. Laker, update 01Oct14 KRL 22 ESE319 Introduction to Microelectronics
Frequency Response of “beta”
g m− j2 f C jf = I C V T 1 g m= r= j2 f C C V T I C r multiplying N&D by C rπ 1 For f = f low s.t. 2 f ≪1≤ low g 10 g m− j2 f C r m jf = 1 j2 f C C r 1 and 2 f C C r ≪1≤ low 10 factor N to isolate gm C 1− j2 f g m r jf =g m r = g m jf = 1 j2 f C C r
Kenneth R. Laker, update 01Oct14 KRL 23 ESE319 Introduction to Microelectronics
Frequency Response of “beta” cont.
C f f f dB 1 j2 f g r 1− j 1− j − m f f g m z z jf = = g r = 20log10 1 j2 f C C r f m f 1 j 1 j f f f f f z
Hence, the lower break frequency or frequency is – 3dB fβ
1 g m 1 g m f = = the upper: f = = 2C C r 2C C z 2C / g m 2C
where f z≫ f
Kenneth R. Laker, update 01Oct14 KRL 24 ESE319 Introduction to Microelectronics
Frequency Response of “beta” cont.
Using Bode plot concepts, for the range where: f f
jf =g m r= 1− j f / f ≈1 For the range where: f f f z s.t. ∣ z∣
We consider the frequency-dependent numerator term to be 1 and focus on the response of the denominator:
f f f z g m r jf ≈ = f f 1 j 1 j f f
Kenneth R. Laker, update 01Oct14 KRL 25 ESE319 Introduction to Microelectronics
Frequency Response of “beta” cont.
g m r Neglecting numerator term: jf = = f f 1 j 1 j f f f f f And for f / f >>1 (but < / z ): ∣ jf ∣≈ = f f f
f Unity gain bandwidth: jf =1⇒ ¿¿| f f =1⇒ f T = f ∣ ∣ f = T
T BJT unity-gain fre- f T = = f quency or GBP 2
Kenneth R. Laker, update 01Oct14 KRL 26 ESE319 Introduction to Microelectronics
Frequency Response of “beta” cont.
−3 =100 r=2500 C =12 pF C =2 pF g m=40⋅10 S
12 −3 1 10 ⋅10 6 = = =28.57⋅10 rps 12 2 2.5 C C r ⋅
28.57 6 f f 455 MHz f = = 10 Hz=4.55 MHz T = = 2 6.28
−3 12 g m 40⋅10 ⋅10 9 z= = Hz=20⋅10 rps C 2 f = z =3.18⋅109 Hz=3180 MHz z 2
Kenneth R. Laker, update 01Oct14 KRL 27 ESE319 Introduction to Microelectronics
“beta” Bode Plot |β(jf)| (dB)(dB) |β(jf)| vs. Frequency
f (MHz) f f β T
Kenneth R. Laker, update 01Oct14 KRL 28 ESE319 Introduction to Microelectronics
Multisim Simulation
Vb v-pi I 1 Ohm c I b V c 1 Ohm
v-pi
mS
Insert 1 ohm resistors – we want to measure a current ratio.
I c g m− j2 f C jf = = I b 1 j2 f C C r
Kenneth R. Laker, update 01Oct14 KRL 29 ESE319 Introduction to Microelectronics Simulation Results
Theory: f f 455 MHz Low frequency |β(jf)| T = =
Unity Gain frequency about 440 MHz.
Kenneth R. Laker, update 01Oct14 KRL 30 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.)
Kenneth R. Laker, update 01Oct14 KRL 31 ESE319 Introduction to Microelectronics The Cascode Amplifier CB Stage CE Stage CB Stage iC1 R R i 1 C c2 Q1 C byp i B1 C1 v-out vo B1 vO R ib2 Q1 Rs S i i Q2 e1 c1 i E1 E1 CE Stage i R iC2 ie2 b1 R C 2i V CC v R Rs S in B2 C2 s R C R E B2 Q2 B E2 i E2 v v v s R R R R R e1 c2 low r 3 R B= 2∥ 3 in1= = = ≈ e1 E i e1 ic2 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 near “short circuits” at all operating frequencies f f min .
Kenneth R. Laker, update 01Oct14 KRL 32 ESE319 Introduction to Microelectronics CE and CB Amplifier Feature Review CE Amplifier CB Amplifier Voltage Gain (A ) moderate (-R /R ) High (R /(R + r )) V C E C s e Important for Current Gain (A ) High (β) low (about 1) I Cascode
Input Resistance High (R || R ) low (r ) B β E e
Output Resistance High (R ||r ) High (R ||r ) C o C o
Kenneth R. Laker, update 01Oct14 KRL 33 ESE319 Introduction to Microelectronics Cascode Mid-Band Small Signal Model
ic1 CB B1 vo CB Stage iC1 Stage i C1 R R b1 vv g v 1 C r1 be11 gm v1 C i C1 v-out m1 be1 byp B1 B1 vO Q1 E1 B2 ie1 i E1 E1 CE Stage i Rin1=low R iC2 C2 c2 R C 2i V CC R ib2 R RS B2 C2 Rs S C s in vv 2 g vv r be2 m2m be22 B2 Q2 2 i v v E2 E2 s E2 s R CE R ie2 R 3 R B E E Stage
RB=R2∥R3 RB=R2∥R3
ASSUME: Q1 = Q2; e.g. β's identical, and both Forward Active
Kenneth R. Laker, update 01Oct14 KRL 34 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis
ic1 1. Show reduction in Miller effect CB B1 vo 2. Evaluate small-signal voltage gain Stage C1 ib1 vv r be11 ggm vv 1 1 m1 be1 OBSERVATIONS E1 a. The emitter current of the CB Stage is B2 ie1 R =low C2 ic2 in1 the collector current of the CE Stage. (This R ib2 R Rs S C also holds for the dc bias currents.) vv 2 g vv r be2 m2m be22 v 2 i i s E2 e1= c2 CE R ie2 R B E b. The base current of the CB Stage is: Stage i i i = e1 = c2 b1 1 1 g m1=g m2=g m c. Hence, both stages have about same r r r e1= e2= e collector current i c1 ≈ i c2 and same g , r , r . m e r r r π 1= 2= I C1≈ I C2
Kenneth R. Laker, update 01Oct14 KRL 35 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis cont. The input resistance R to the CB Stage is in1 ic1 B1 vo the small-signal “r ” for the CB Stage. CB e1 Stage C1 ib1 v ie1 ic2 r v be11 g v gmv1 i = = m be1 b1 1 1 E1 R r B2 ie1 in 1≈ e1 The CE output voltage, the voltage drop i C2 c2 from Q2 collector to ground, is: R ib2 R Rs S vv C be2 2 g mvv 2 r r r m be2 v vc2=ve1=−r ib1=− ie1=− ic2 s E2 1 1 CE R ie2 R B E Therefore, the CB Stage input resistance is: Stage ve1 r Rin1= = =r e1 −ie1 1 v R r r A c2 in1 e 1 e Miller Effect vCE−Stage= ≈− =− => C eq=1 C 2C vs RE RE RE Much reduced
Kenneth R. Laker, update 01Oct14 KRL 36 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis - cont.
ic1 Now, find the CE collector current in terms of the input voltage v : i i s Recall c1≈ c2 ib1 v vbe1 g mvvbe1 v m be1 i ≈ s b2 R R r 1 R ie1 S∥ B E RS Rin 1≈re1 ic2 v v ib2 s s vvbe2 g mvvbe2 i =i ≈ ≈ be2 m be2 c2 b2 RS∥RBr1 RE 1 RE
ie2 for bias insensitivity: 1RE ≫RS∥RBr
v v R ic1≈ie1=ic2≈ie2 vss o − C i ≈ vo≈−ic2 RC A c2 R v= ≈ OBSERVATIONS: E v s RE 1. Voltage gain A is about the same as a stand-along CE Amplifier. v 2. HF cutoff is much higher then a CE Amplifier due to the reduced C . eq Kenneth R. Laker, update 01Oct14 KRL 37 ESE319 Introduction to Microelectronics
V o ' dB R S V V s b c o ∣ ∣ I V be C ' r g V ' V s RB m be C C eq RL e
C tot
Common Emitter Stage
C tot =C 1g m RC C 1 Cascode Stage f H = ' ⇒ f H Cascode ≫ f H CE 2C tot RS r e C tot =C 1 C RE
.2C
Kenneth R. Laker, update 01Oct14 KRL 38 ESE319 Introduction to Microelectronics Cascode Biasing
1. Choose IE1 – make it relatively large to
reduce R in 1 = r e = V T / I E1 to push out HF I break frequencies. o.c. C1 I C 1 vO 2. Choose R for suitable voltage swing B Q1 C R =low V and R for desired gain. I in1 C1 E E1 I RS C in C2 Q2 3. Choose bias resistor string such that I E2 its current I is about 0.1 of the collector o.c. 1 current I . C1
1 4. Given RE, IE2 and V = 0.7 V calc. R3. I =I =I = I ⇒ I =2 I ≈I BE2 E2 C2 E1 C1 C1 E2 E2 5. Need to also determine R & R . 1 2
Kenneth R. Laker, update 01Oct14 KRL 39 ESE319 Introduction to Microelectronics Cascode Biasing - cont.
I I11 Since the CE-Stage gain is very small: v V O a. The collector swing of Q2 will be small. B1 Q1 b. The Q2 collector bias V = V - 0.7 V. Rin 1≈re1 C2 B1 V V C2 6. Set V − V = 1 V ⇒ V =1V B2 Q2 B1 B2 CE2 Since V = 1 V > V Q2 forward active. CE2 CE2sat
7. Next determine R . Its drop V = 1 V 2 R2 V CE2=V C2−V R e=V C2−V B2−0.7V with the known current. .=V B1−0.7V −V R e V −V R = B1 B2 .=V B1−0.7V −V B20.7V 2 I 1 .=V B1−V B2=1V
Kenneth R. Laker, update 01Oct14 KRL 40 ESE319 Introduction to Microelectronics Cascode Biasing - cont.
V B1−V B2 1V R2= = I 1 I 1 V 8. Then calculate R . R = B2 V B1 3 3 Q1 I 1 Rin 1≈re1 where V B2=0.7V I E RE Q2 V CC Note: R1R2R3= I 1 9. Then calculate R . 1
V CC R1= −R2−R3 0.1 I C
Kenneth R. Laker, update 01Oct14 KRL 41 ESE319 Introduction to Microelectronics Cascode Bias Summary SPECIFIED: A , V , V (CB collector voltage); v CC C1 SPECIFIED: f => C (r ) => I (or I ) => r . H tot e E C e DETERMINE: R , R , R , R and R . C E 1 2 3 V R R C1 R = C SET: V −V =1V ⇒V =1V STEP1: C= E B1 B2 CE2 I C ∣Av∣
V B1−V B2 1V STEP2: R2= = V I 1 0.1 I C B1 Q1 Rin 1≈re1 V B2 0.7V I E RE STEP3: R3= = Q2 I 1 0.1 I C V CC V CC R1R2R3= = I 1 0.1 I C V CC STEP4: R1= −R2−R3 I C2=I E1≈I C1≈I E2=I C 0.1 I C
Kenneth R. Laker, update 01Oct14 KRL 42 ESE319 Introduction to Microelectronics
Cascode Bias Example
R R1 V -I R - C I R V CC C E C C C1 1.7 Q1 Q1 V V I R V I R VCE1CE1==ICCCRC−–1–IC CRCE− CE2− C E 1.0 Q2 =12 V Q2 =12 V VCE2=1
ICRE+0. RE 7 ICR
E
I E2≈I C2=I E1≈I C1 ⇒ I C1≈I E2 Cascode Amp Typical Bias Conditions
Kenneth R. Laker, update 01Oct14 KRL 43 ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
1. Choose IE1 – to set re. I 0.025V r 5mA Let re = 5Ω => E1 = / e = . Q1 V CE1=V CC−I C RC −1−I C RE
2. Set desired gain magnitude. For example Q2 if A = -10, then R /R = 10. V C E
3. Since the CE stage gain is very small, V can be small, i.e. V = V – V = 1 V. CE2 CE2 B1 B2
Kenneth R. Laker, update 01Oct14 KRL 44 ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
Specs: R r 5 C V CC=12V V C1=7V e= ∣Av∣= =10 RE =100 Q1 V CE1=V CC−I C RC −1−I C RE Determine R for V = 7V . C C1 Q2 IE1=0.025V /re=5mA
V CC−V C1 5V RC= = =1000 5⋅10−3 A 5⋅10−3 A R R R = C = C =100 E 10 ∣Av∣
Kenneth R. Laker, update 01Oct14 KRL 45 ESE319 Introduction to Microelectronics Cascode Bias Example cont.
V CC=12 RC=1k r e=5 RE=100 I 1 =100
R1 R C I R V -I R -1.7 C C Make current through the string of bias CC C E resistors I = 0.1 I = 0.5 mA. Q1 1 C VV =IVR –1–II RR 1 I R CE1CE1= CCCC− CC EC − − C E V R CC 12 2 1.0 R R R 24 k 1 2 3= = −4 = =12V I 1 5⋅10 Q2 VCE2=1 I R +0.7 Calculate the bias voltages (base side of Q1, Q2): R C E 3 RE ICRE V R1=V CC −1.7V −I C RE=12V −1.7V −0.5V =9.8V
V R2=V B1−V B2=1V −3 V R3=V B2=0.7I C RE=0.75⋅10 ⋅100=1.2V
Kenneth R. Laker, update 01Oct14 KRL 46 ESE319 Introduction to Microelectronics
Cascode Bias Example cont. −4 V B2=5⋅10 R3=1.2V R 2.4 k C B 3= V B1 Q1 −4 V B1−V B2=5⋅10 R2=1.0V
C in R 2 k RRSS V B2 2= Q2 Recall: R1R2R3=24 k
R1=24000−2.400−2000=19.6 k
Recall for ac: RB=R2∥R3
RB=2 k ∥2.4 k =1.1 k V CC=12 , RC=1k , V B2=1.2V ,
I C=5mA, RE=100, V B1−V B2=1.0V
Kenneth R. Laker, update 01Oct14 KRL 47 ESE319 Introduction to Microelectronics Completed Design 100 = 1 f = re=5⇒ I C=5mA H ' 2C tot RS C V C1=7V B V r B1 Q1 C =C 1 e C tot R RC E Av = =10 ∣ ∣ C in .=C 1.05C R RRSS V B2 E Q2 If C = 12 pF π C = 2 pF µ
C tot =14.1 pF R =19.6 k 1 f =225.8 MHz RC=1k Hcascode R2=2 k RE=100 For CE with |A | = 10 R3=2.4 k v f HCE =94 MHz NOTE: RB=R2∥R3=1.09 k ≪ R E=10 k
Kenneth R. Laker, update 01Oct14 KRL 48