8. Biasing Transistor Amplifiers Lecture notes: Sec. 5 Sedra & - PowerPoint PPT Presentation

8. Biasing Transistor Amplifiers Lecture notes: Sec. 5 Sedra & Smith (6 th Ed): Sec. 5.4, 5.6 & 6.3-6.4 Sedra & Smith (5 th Ed): Sec. 4.4, 4.6 & 5.3-5.4 ECE 65, Winter2013, F. Najmabadi

Issues in developing a transistor amplifier: 1. Find the iv characteristics of the elements for the signal (which can be different than their characteristics equation for bias). o This will lead to different circuit configurations for bias versus signal 2. Compute circuit response to the signal o Focus on fundamental transistor amplifier configurations 3. How to establish a Bias point (bias is the state of the system when there is no signal). o Stable and robust bias point should be resilient to variations in µ n C ox (W/L),V t (or β for BJT) due to temperature and/or manufacturing variability. o Bias point details impact small signal response (e.g., gain of the amplifier). F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (2/29)

BJT biasing with Base Voltage (Fixed Bias) − = + BE KVL : V I R V BB B B BE − V V = 0 BB D I B R B − V V = β = β 0 BB D I I C B R B − = + CE KVL : V I R V CC C C CE β R = − − C ( ) V V V V 0 CE CC BB D R B * Typically V BB = V CC in order to reduce the need for additional reference voltages. F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (3/29)

Exercise 1: Find R C and R B such that BJT would be in active with V CE = 5V and I C = 25 mA. ( V CC = 15 V, Si BJT with β = 100 and V A = ∞ ). > = ≥ 0 = BJT is in Active since 0 and 5 0 . 7 V I V V C CE D = β = / 0 . 25 mA I I B C − = + = × − + 3 BE KVL : 1 5 0 . 25 10 0 . 7 I R V R B B BE B = 57 . 2 k R B − − = + = × + 3 CE KVL : 1 5 25 10 5 I R V R C C CE C = Ω 400 R C F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (4/29)

Exercise 2: Consider the circuit designed in Exercise 1 ( R C = 400 , R B = 57.2 k, V CC = 15 V ). Find the operating point of BJT if β = 200. Assume BJT is in Active : = > ≥ 0 . 7 V, 0 and 0 . 7 V V I V BE C CE − = + = × + 3 BE KVL : 1 5 57 . 2 10 0 . 7 I R V I B B BE B = 0 . 25 mA I B = β = 50 mA I I C B − − = + = × × + 3 CE KVL : 1 5 50 10 400 I R V V C C CE CE = − 5 V V CE Note, compared to Exercise 1: BJT in saturation!  I B is the same.  I C has increased.  V CE had decreased. F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (5/29)

Why biasing with base voltage (fixed bias) does not work?  Changes in BJT β changes the bias point drastically. o BJT can end up in saturation or in cut-off easily. − V V =  In fixed bias, I B is set through 0 BB D I B R B  BJT β then sets I C = β I B ( I C changes with β ). o CE circuit then sets V CE .  But, requirements for BJT in active are on I C and V CE and NOT on I B o I C > 0 , V CE > V D 0  To make bias point independent of changes in β , the bias circuit should “set” I C and NOT I B ! F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (6/29)

Biasing with Emitter Degeneration Requires a resistor in the emitter circuit! − = + + BE KVL : V I R V I R BB B B BE E E   R − = + B V V I  R  β + BB D 0 E E   1 << β + If : ( 1 ) R R B E − ≈ V V I R 0 BB D E E − V V Independent of β ! ≈ ≈ 0 BB D I I C E R E << β + Condition of means that the voltage drop across R B is small ( 1 ) R R B E and the bias voltage V BB – V D 0 appears across R E , setting I E and I C ≈ I E . Note that resistor R B is NOT necessary for good biasing but it may exist due to other considerations. F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (7/29)

Emitter resistor provides negative feedback! = + + V I R V I R BB B B BE E E ∝ / V BE V I e T B Negative Feedback: Independent of β ! β BE junction BE -KVL o If I C ≈ I E ↑ (because β ↑ ) , V BE ↓ I B ↓ I C ≈ I E ↓ β BE junction BE -KVL o If I C ≈ I E ↓ (because β ↓ ) , V BE ↑ I B ↑ I C ≈ I E ↑ F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (8/29)

Requirements for Biasing with Emitter Degeneration  Requires a resistor in the emitter circuit.  The bias voltage V BB – V D 0 should appear across R E to set I E ≈ I C : − = + V V I R I R BB BE B B E E << ⇒ << β + 1. ( 1 ) I R I R R R B B E E B E << β + ( 1 ) R R o We need to set to ensure min B E that this condition is always satisfied! 2. V BE ≈ V D 0 . In reality, V BE = V D 0 ± ∆ V BE with ∆ V BE ≈ 0.1 V >> o We need to set or 0.1 V I E R E ≥ 1 V I E R E F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (9/29)

Emitter Degeneration Bias with a voltage divider Real Circuit = || R R R B 1 2 R = × Voltage Divider 2 V V + BB CC R R 1 2 F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (10/29)

Exercise 3: Find the bias point of the BJT (Si BJT with β = 200 and V A = ∞ ). = = = || 5 . 9 k || 34 k 5 . 03 k R R R 1 2 B 5 . 9 k R = × = × = 2 15 2 . 22 V V V + + BB CC 5 . 9 k 34 k R R 1 2 Assume BJT is in Active : = > ≥ 0 . 7 V, 0 and 0 . 7 V V I V BE C CE − = + + BE KVL : 2.22 I R V I R B B BE E E = × β + + + 3 2.22 5 . 03 10 /( 1 ) 0 . 7 510 I I E E = 2 . 84 mA I E Notes: = × β β + = /( 1 ) 2 . 82 mA I I 1. We need to solve the complete BE-KVL C E = β + = µ << β + /( 1 ) 14.1 A I I as we do not know if ( 1 ) R R B E B E 2. β >> 1 is a good approximation that − = + + CE KVL : 1 5 I R V I R reduces the amount of work. Answers C C CE E E = × − × + + × − × using β >> 1 approximation: 3 3 3 15 2 . 82 10 10 2 . 84 10 510 V CE = > = ≈ ≈ = µ 10.7 V 0.7 V 2.84 mA, 14.2 A V V I I I 0 CE D C E B = 10.7 V V CE F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (11/29)

Exercise 4: Design a BJT bias circuit (emitter degeneration with voltage divider) such that I C = 2.5 mA and V CE = 7. 5 V. ( V CC = 15 V Si BJT with β ranging from 50 to 200 and V A = ∞ ). Step 1: Find R C and R E − = + + CE KVL : 1 5 I R V I R C C CE E E − = × × + + 3 15 2 . 5 10 ( ) 7 . 5 R R C E + = 3.0 k R R C E Free to choose individual values R E & R C (we will see later that amplifier parameters sets the individual values) = Choose : 1.0 k R E = − = 3.0 k 2.0 k R R C E Circuit Prototype ≥ Check: 1 V I E R E − = × × = ≥ 3 3 2 . 5 10 10 2 . 5 1 V I E R E F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (12/29)

Exercise 4 (Cont’d): Design a BJT bias circuit (emitter degeneration with voltage divider) such that I C = 25 mA and V CE = 5 V. ( V CC = 15 V Si BJT with β ranging from 50 to 200 and V A = ∞ ). Step 2: Find R B and V BB << β + → ≤ β + = → = ( 1 ) 0.1 ( 1 ) 5.1 k 5.1 k R R R R R min min B E B E B Using relative error, ε = 10% Use largest R B (Will see later why) = + + V I R V I R BB B B BE E E ≈ + = + × × → = 3 3 - 0.7 2.5 10 10 3 . 20 V V V I R V 0 BB D C E BB Step 3: Find R 1 and R 2 Step 4: Find commercial R values: R R = = = 1 2 5 . 10 k || 5 . 10 k R R R R C = 2 k = = + B 1 2 23.9 k R R R 1 1 2 0 . 213 R E = 1 k 3.20 V R = R 1 = 24 k = = = 6.4 k R 2 BB 0 . 213 2 + 15 V R R R 2 = 6.4 k 1 2 CC F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (13/29)

Emitter-degeneration bias circuits Basic Arrangement Bias with one power supply Bias with two power supplies (voltage divider) = + + = + + = + + V I R V I R V I R V I R V I R V I R BB B B BE E E BB B B BE E E EE B B BE E E = + + − 0 I R V I R V B B BE E E EE F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (14/29)

MOS bias with Gate Voltage (Fixed Bias) W = 0 µ − 2 . 5 ( ) I C V V D n ox GS t L = − V V I R DS DD D D  This method is NOT desirable as µ n C ox ( W/L ) and V t are not “well- defined.” Bias point (i.e., I D and V DS ) can change drastically due to temperature and/or manufacturing variability. o See S&S Exercise 5.33 (S&S 5 th Ed: Exercise 4.19): Changing V t from 1 to 1.5 V leads to a 75% change in I D . F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (15/29)

MOS bias with Source Degeneration (Resistor R S provides negative feedback!) = − V V R I GS G S D W = µ − 2 0 . 5 ( ) I C V V D n ox GS t L Negative Feedback: I D Eq. GS KVL o If I D ↑ (because µ n C ox ( W/L ) ↑ or V t ↓ ) V GS ↓ I D ↓ I D Eq. GS KVL o If I D ↓ (because µ n C ox ( W/L ) ↓ or V t ↑ ) V GS ↑ I D ↑  Feedback is most effective if >> R I V S D GS − + = ⇒ ≈ 0 / V V R I I V R GS G S D D G S F. Najmabadi, ECE65, Winter 2013, Amplifier Biasing (16/29)