ESE319 Introduction to Microelectronics
Miller Effect Cascode BJT Amplifier
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 1 ESE319 Introduction to Microelectronics
Prototype Common Emitter Circuit
Ignore “low frequency” High frequency model capacitors
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 2 ESE319 Introduction to Microelectronics
Multisim Simulation
Mid-band gain
∣Av∣=g m RC=40mS∗5.1k=20446.2 dB
Half-gain point
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 3 ESE319 Introduction to Microelectronics Introducing the Miller Effect
The feedback connection of C between base and collector causes it to appear to 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 common emitter amplifier mid-band gain, i.e. K = − g m R C .
Common base and common collector 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 (based on P. V. Lopresti 2006) update 15Oct08 KRL 4 ESE319 Introduction to Microelectronics High Frequency CC and CB Models
ground B C B C
E E
Common Collector Common Base C is in parallel with R . C is in parallel with R . B C
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 5 ESE319 Introduction to Microelectronics Miller's Theorem I Z I 1=I I 2=I + + + +
V1 V2=K V1 V V K V <=> 1 ic2 2= 1 - - - -
V1−V2 V1−K V1 V1 V1 V1 Z I = = = => Z1= = = Z Z Z I 1 I 1−K 1−K
V2 V2 Z Z Z = = =− = ≈Z => 2 −I −I 1 1 2 −1 1− K K if K >> 1
Let's examine Miller's Theorem as it applies to the HF model for the BJT CE amplifier.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 6 ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis
I Determine effect of C : C B V C V Using phasor notation: o I =−g V I I RC m C RC or
g V Vo= −g m V I C RC m where E I C = V −Vo sC
Note: The current through C depends only on V ! I C = V g mV RC −I C RC s C
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 7 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis II From slide 7:
I C = V g mV RC −I C RC s C
Collect terms for I and V : C
1s RC C I C = 1g m RC sC V
1g R sC s 1g R C I m C V m C V C = = 1s RC C 1s RC C
Miller Capacitance C : C eq=1−KC =1g m RC eq
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 8 ESE319 Introduction to Microelectronics
Common Emitter Miller Effect Analysis III
C eq=1g m RC C
For our example circuit:
1g m RC=10.040⋅5100=205
Ceq=205⋅2 pF ≈410 pF
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 9 ESE319 Introduction to Microelectronics Apply Miller's Theorem to BJT CE Amplifier
Z Z Z Z 2= 1= 1 1−K 1− K 1 For the BJT CE Amplifier: Z= and K=−gm RC j C 1 1 1 => Z1= and Z 2= ≈ j 1g m RC C 1 j C j 1 C g m RC
C eq=1g m RC C Miller's Theorem => important simplification to the HF BJT CE Model
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 10 ESE319 Introduction to Microelectronics Simplified HF Model
' I C C Rsig B' + ' ' + + R =r ∥R ∥R .≈0 ' L o C L V sig V C V r C RL o Rsig B x B' C - g mV - ++ + r ' V sig RB V C r R R V (b) g V o C L sig - m - ' RB∥r Vsig=Vsig E ' R ∥r R R B sig L ' ' (a) Rsig=r ∥RB∥Rsig Vo −g m RL Thevenin ≈ V ' ' |Vo/Vsig| (dB) sig 1 j C in Rsig R I C sig B' C -6 dB/octave ++ + -20 dB/decade ' −3dB V C ' V V sig Ceq R o - g V L - 20log A m 10 M 1 f = H ' C in 2C in Rsig ' C =C C =C C 1g R 0 in eq m L f f (Hz log scale) H (c) (d)
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 11 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 meet all of the criteria above. The cascode amplifier will meet all of these criteria. a cascode is a combination of a common emitter stage cascaded with a common base stage. (In “olden days” the cascode amplifier was a cascade of grounded cathode and grounded grid vacuum tube stages – hence the name “cascode,” which has persisted in modern terminology.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 12 ESE319 Introduction to Microelectronics The Cascode Circuit CE Stage CB Stage iC1 ic2 i v-out B1 ib2 ie1 ic1 i E1 i iC2 ie2 b1 i B2 Rb
i E2 veg1 vcg2 RB=R2∥R3 Rin1= = =low ie1 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. Cb and Cbyp are to act as “open circuits” at dc and act as “short circuits” at all operating frequencies of interest, i.e. min .
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 13 ESE319 Introduction to Microelectronics Cascode Mid-Band Small Signal Model
ic1 1. Show reduction in Miller effect Vout 2. Evaluate small-signal voltage gain i b1 OBSERVATIONS a. The emitter current of the CB stage is ie1 i Rin1=low the collector current of the CE stage. (This i c2 b2 also holds for the dc bias current.)
ie1=ic2 Rb ie2 b. The base current of the CB stage is: i i i = e1 = c2 b1 1 1 c. Hence, both stages have about same i i collector current c1 ≈ c2 and same gm. g m1=g m2=g m
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 14 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis cont. The input resistance R to the CB stage in1 i c1 Vout is the small-signal “R ” for the CE stage C ib1 i i i = e1 = c2 b1 1 1 R =low ie1 in1 The CE output voltage, the voltage drop ic2 ib2 Rc from Q2 collector to ground, is: r 1 r 1 v =v =−r i =− i =− i cg2 eg1 1 b1 1 c2 1 e1 Rb ie2 Therefore, the CB Stage input resistance is:
veg1 r 1 Rin1= = =r e1 −ie1 1 v R r r A cg2 in1 e1 1 e1 vCE−Stage= ≈− =− => C eq=1 C 2C vsig RE RE RE
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 15 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis - cont. i c1 Vout Now, find the CE collector current in terms of
ib1 the input voltage Vsig: Recall ic1≈ic2 v i ≈ sig i b2 e1 Rsig∥RBr 21 RE ic2 ib2 vsig vsig ic2=ib2≈ ≈ Rsig∥RBr 21 RE 1 RE Rb ie2 for bias insensitivity: 1 RE ≫Rsig∥RBr 2 v i ≈ sig , => c2 R OBSERVATIONS: E 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
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 16 ESE319 Introduction to Microelectronics
Approximate Cascode HF Voltage Gain
R − C Vout RE Av= ≈ ' V sig 1 j C in Rsig
where r C =C C =C 1 e1 C C 2C in eq R ' E Rsig=Rsig∥RB≈Rsig
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 17 ESE319 Introduction to Microelectronics
Cascode Biasing
1. Choose IE1 – make it relatively large to I1 IC1 reduce R in 1 = r e1 = V T / I E1 to push out HF v-out break frequencies.
IE1 R =low in1 2. Choose RC for suitable voltage swing I C2 V and R for desired gain. C1G E
IE2 3. Choose bias resistor string such that its current I is about 0.1 of the collector 1 current I . C1
4. Given R , I and V = 0.7 V calculate E E2 BE2
R3.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 18 ESE319 Introduction to Microelectronics
Cascode Biasing - cont. Since the CE-Stage gain is very small: I 1 a. The collector swing of Q2 will be small. V v-out b. The Q2 collector bias V = V - 0.7 V. B1G C2G B1G 5. Set V − V ≈ 1 V ⇒ V ≈1V V B1G B2G CE2 C2G V B2G This will limit V V V V 0.3 V CB2 CB2= CE2− BE2= which will keep Q2 forward active.
6. Next determine R . Its drop V = 1 V 2 R2 VCE2=VC2G−V Re=VC2G−V B2G−0.7V with the known current. V −V R B1G B2G .=V −V −V V 2= B1G BE1 B2G BE2 I 1
.≈V B1G−0.7V−V B2G0.7 V V −V 7. Then calculate R . R = CC B1G 1 1 .=V B1G−V B2G I 1
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 19 ESE319 Introduction to Microelectronics
Cascode Bias Example
I R VCC-ICRE-1.7 C C v-out
VCE1=ICRC–1–ICRE 1.0 =12 V =12 V VCE2=1
ICRE+0.7
ICRE
I E2≈I C2=I E1≈I C1 ⇒ I C1≈I E2 Cascode circuit Typical Bias Conditions
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 20 ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
1. Choose IE1 – make it a bit high to lower re1 V -I R -1.7 ICRC CC C E or r . Try I = 5 mA => r = 0.025 V / I = 5 . 1 E1 e1 E
VCE1=ICRC–1–ICRE 1.0 2. Set desired gain magnitude. For example =12 V if |A | = 10, then R /R = 10. V C E VCE2=1
I R +0.7 C E 3. Since the CE stage gain is very small, I R C E V can be small. Use V = V – V = 1 V. CE2 CE2 B1G B2G
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 21 ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
RC VCC=12 I C =5mA. ∣Av∣= =10 I R RE VCC-ICRE-1.7 C C
Determine RC for a 5 V drop across RC. VCE1=ICRC–1–ICRE 1.0 5V R 1000 C= −3 = VCE2=1 5⋅10 A
I R +0.7 RC RC C E R 100 I R E= = = C E 10 ∣Av∣
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 22 ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
I1 VCC=12 RC=1k I C =5mA. RE=100
I R V -I R -1.7 C C Make current through the string of bias CC C E resistors I = 1 mA. 1 V =I R –1–I R CE1 C C C E V 12 1.0 R R R CC 12k 1 2 3= = −3 = I 1 1⋅10 VCE2=1 We now calculate the bias voltages: ICRE+0.7 I R C E VCC −I C RE−1.7V=12V−0.5V−1.7V=9.8V
V B1B2=1.0V −3 V B2G=I C RE0.7=5⋅10 ⋅1000.7=1.2V
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 23 ESE319 Introduction to Microelectronics
Cascode Bias Example cont.
−3 V B2G=10 R3=1.2 V v-out R3=1.2 k −3 V B1G−V B2G=1⋅10 R2=1.0V
R2=1k
R1=12000−1200−1000=9.8k
R1=10 k
VCC=12 RC=1k V B2G=1.2V
I C =5mA. RE=100 V B1G−V B2G=1.0V
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 24 ESE319 Introduction to Microelectronics Multisim Results – Bias Example
IC1
VCC −7.3290.9710.339 12V−8.639 V Check I : I 3.36 mA C1 C1= = = RC 1000 IC1 about 3.4 mA. That's a little low.
Increase R3 to 1.5 k Ohms and re-simulate. 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 25 ESE319 Introduction to Microelectronics Improved Biasing
10 uF
10 uF
VCC −5.0480.9410.551 12V−6.540 V Check I : I 5.46mA C1 C1= = = RC 1000 That's better! Now measure the gain at a mid-band frequency with some “large” coupling capacitors, say 10 µF inserted. 2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 26 ESE319 Introduction to Microelectronics Single Frequency Gain
10 uF v-out
10 uF
Gain |A | of about 8.75 at 100 kHz. - OK for rough calculations. Some atten- v uation from low CB input impedance (RB = R2||R3)and some from 5 Ohm re.
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 27 ESE319 Introduction to Microelectronics Bode Plot for the Amplifier
18.84 dB
Low frequency break point with 10 µF. capacitors
18.84 dB
High frequency break point – internal capacitances only
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 28 ESE319 Introduction to Microelectronics Scope Plot – Near 5 V Swing on Output
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 29 ESE319 Introduction to Microelectronics
Determine Bypass Capacitors Low Frequency f ≤ f min v-out From CE stage determine C b 10 10 C b ≥ ≈ F 2 f min RB∥r bg 2 f min RB∥1 RE
RB=R2∥R3
From CB stage determine C byp 10 C ≥ F byp 2 f 1 R r min E 1 where R -> R S E
2008 Kenneth R. Laker (based on P. V. Lopresti 2006) update 15Oct08 KRL 30