Dr. Nasim Zafar Electronics 1 - EEE 231 Fall Semester – 2012 COMSATS Institute of Information Technology Virtual campus Islamabad

Potential-Divider-Biasing Circuits: Examples and Exercises.. Lecture No: 19 Contents:  Base-Biased (Fixed Bias) Transistor Circuits.  Voltage-Divider-Bias transistor Circuits.  Examples and Exercises. Nasim Zafar2

References:  Microelectronic Circuits: Adel S. Sedra and Kenneth C. Smith.  Integrated Electronics : Jacob Millman and Christos Halkias (McGraw-Hill).  Introductory Electronic Devices and Circuits Robert T. Paynter  Electronic Devices : Thomas L. Floyd ( Prentice Hall ). Nasim Zafar3

Basic Circuits of BJT: NPN Transistor Nasim Zafar4 I E = I C + I B

Transistor Output Characteristics: 5Nasim Zafar

Transistor Output Characteristics: Load Line – Biasing and Stability:  Active region: – BJT acts as a signal amplifier. – B-E junction is forward biased and C-B junction is reverse biased.  Graphical construction for determining the dc collector current I C and the collector-to-emitter voltage V CE.  The requirement is to set the Q-point such that that it does not go into the saturation or cutoff regions when an a ac signal is applied.  Maximum signal swing depends on the bias voltage. Nasim Zafar6

The DC Operating Point: Biasing and Stability  Active region - Amplifier: BJT acts as a Signal Amplifier. 1. B-E Junction Forward Biased V BE ≈ 0.7 V for Si 2. B-C Junction Reverse Biased 3. KCL: I E = I C + I B C B E IBIB IEIE ICIC C B E IBIB IEIE ICIC 7Nasim Zafar

The DC Operating Point: Biasing and Stability Slope of the Load Line: V CC = V CE + V RC V CE = V CC - V RC V CE = V CC - I C R C 8Nasim Zafar

9 Current Equations in a BJT: NPN Transistor  Collector Current  Base Current  Emitter Current 9Nasim Zafar

1. Fixed-Biased Transistor Circuits. Base-Biased (Fixed Bias) Transistor Circuit: Single Power Supply 10Nasim Zafar

11 Base-Biased (Fixed Bias) Transistor Circuit: Advantage: Circuit simplicity. Disadvantage: Q-point shifts with temp. Applications: Switching circuits only. Circuit Recognition: A single resistor R B between the base terminal and V CC. No emitter resistor. Nasim Zafar Circuit Characteristics - 1:

12 Base-Biased (Fixed Bias) Transistor Circuit: Circuit Characteristics - 2: Load line equations: Q-point equations: Nasim Zafar

13 Base-Biased (Fixed Bias) Transistor Circuit: Q-point equations: Nasim Zafar 1. Base–Emitter Loop: V CC = V BE + I B R B

14 Base-Biased (Fixed Bias) Transistor Circuit:  = dc current gain = h FE Nasim Zafar 2. Collector–Emitter Loop: V CC = V CE + V RC V CE = V CC - I C R

15 Circuit 19.1; Example 19.1 Nasim Zafar

16 Example 19.2 Construct the DC Load line for circuit 19.1; shown in slide 12, and plot the Q-point from the values obtained in Example Determine whether the circuit is midpoint biased. Nasim Zafar The circuit is midpoint biased.

17 Example 19.3 (Q-point Shift.) The transistor of Circuit 19.1, has values of h FE = 100 when T = 25 °C and h FE = 150 when T = 100 °C. Determine the Q-point values of I C and V CE at both of these temperatures. Temp(°C)I B (mA)I C (mA)V CE (V) Nasim Zafar

3. 2. Voltage-Divider-Bias Circuits. 18Nasim Zafar

Voltage-Divider Bias Circuits: NPN Transistor.  Voltage-divider biasing is the most common form of transistor biasing used. A thorough understanding of the dc analysis of this circuit is essential for an electronic technician.  In the Circuit, R1 and R2 set up a voltage divider on the base. Notice the similarity to the emitter-biased circuit. Nasim Zafar19

20 Voltage-Divider Bias Characteristics-(1) Circuit Recognition: The voltage divider in the base circuit. Advantages: The circuit Q- point values are stable against changes in h FE. Disadvantages: Requires more components than most other biasing circuits. Applications: Used primarily to bias linear amplifier. Nasim Zafar

21 Voltage-Divider Bias Characteristics-(2) The Thevenin voltage: Nasim Zafar

22 Voltage-Divider Bias Characteristics-(3) Load line equations: Q-point equations (assume that h FE R E > 10R 2 ): Nasim Zafar

23 Circuit 19.2; Example 19.4 (a). Determine the values of I CQ and V CEQ for the circuit 19.2 shown in Fig below: Because I I E (or h FE >> 1), Nasim Zafar

24 Circuit 19.2; Example 19.4 (b). Verify that I 2 > 10 I B. Nasim Zafar

25 Example 19.5 A voltage-divider bias circuit has the following values: R 1 = 1.5 kW, R 2 = 680 W, R C = 260 W, R E = 240 W and V CC = 10 V. Assuming the transistor is a 2N3904, determine the value of I B for the circuit. Nasim Zafar

26 Load Line for Voltage Divider Bias Circuit. Example 19.5 Nasim Zafar

27 Which value of h FE do we use? Transistor specification sheet may list any combination of the following h FE : max. h FE, min. h FE, or typ. h FE. Use typical value if there is one. Otherwise, use Nasim Zafar

28 Stability of Voltage Divider Bias Circuit:  The Q-point of voltage divider bias circuit is less dependent on h FE than that of the base bias (fixed bias).  For example, if I E is exactly 10 mA, the range of h FE is 100 to 300. Then I CQ hardly changes over the entire range of h FE. Nasim Zafar

Voltage-Divider Bias Circuit: Circuit-19.3; Problem 19.6 (a).  Find the operating point Q for this circuit.  The use of Thevenin equivalent circuit for the base makes the circuit simpler. Nasim Zafar29

Determination of V BB – The Thevenin Voltage V CC = I.(R 1 + R 2 ) -- Eq. (1) V Thev = I.R 2 Eq. (2) -- Eq. (3) Nasim Zafar30

Circuit-19.3; Problem 19.6 (a) Determination of V BB From Eq (3) V Thev = 2 Volts Nasim Zafar31

Circuit-19.3; Problem 19.6 (b). Determination of V RE Input Loop with R E V BB = V BE + V RE V RE = V BB – V BE V RE = 2V - 0.7V V RE = 1.3V Nasim Zafar32

Circuit-19.3; Problem 19.6 (c). Determination of I E V RE = I E R E Nasim Zafar33

Circuit-19.3; Problem 19.6 (d). Determination of V RC Since I E ≈ I C I E = 1.3mA Therefore, I C = 1.3mA V RC = I C R C = (1.3mA)(4x10 3 Ω) V RC = 5.2V Nasim Zafar34

Circuit-19.3; Problem 19.6 (e). Determination of V CE Output Loop V CC =V RC +V CE +V RE V CE = V CC -V RC -V RE V CE = 12V - 5.2V - 1.3V V CE = 5.5V Nasim Zafar35

Results of Problem 19.6 I E = I C = 1.3mA V RC = 5.2V V CE = 5.5V V RE = 1.3V V BB = 2V β dc was never used in a calculation. Hence, voltage-divider biased circuits are immune to changes in β dc. A single voltage source supplies both voltages, V CC and V BB Nasim Zafar36

Review of equations: In Review V RE = V BB – V BE I E ≈ I C Nasim Zafar37

Summary:  Voltage-divider biased circuits are immune to changes in β dc.  A single voltage source supplies both voltages, V CC and V BB  The circuit Q-point values are stable against changes in h FE.  Use of the Thevenin equivalent circuit for the base makes the circuit simpler.  Make the current in the voltage divider about 10 times IB, to simplify the analysis.  For design, solve for the resistor values (IC and VC specified). Nasim Zafar38

Nasim Zafar39

Circuit 19.4; Problem 19.7 (a). Given: V B = 3V and I = 0.2mA. IBIB I (a) RB1 and RB2 form a voltage divider. Assume I >> IB I = V CC /(R B1 + R B2 ) 0.2mA = 9 /(R B1 + R B2 ) 40 Nasim Zafar

Circuit 19.4; Problem 19.7 (b). Given: V B = 3V and I = 0.2mA. R B1 = 30K , and R B2 = 15K . IBIB I V B = V CC [R B2 /(R B1 + R B2 )] 3 = 9 [R B2 /(R B1 + R B2 )], Solve for R B1 and R B2. 41 Nasim Zafar (b) Determination of the Thevenin voltage:

Prob (c). Find the operating point The use of Thevenin equivalent circuit for the base makes the circuit simpler. V BB = V B = 3V R BB = R B1 || R B2 = 30K  15K  10K  42Nasim Zafar

Problem 19.7 (d). Write B-E loop and C-E loop B-E loop C-E loop B-E Voltage Loop: V BB = V RBB + V BE + V R E V BB = I B R BB + V BE + I E R E I E =2.09 mA C-E Voltage Loop: V CC = I C R C + V CE + I E R E V CE =4.8 V This is how all DC circuits are analyzed and designed! 43Nasim Zafar

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Example 19.7 Stage 2 C-E loop I E2 I C2 V CC = I E2 R E2 + V EC2 +I C2 R C2 15 = 2.8(2) + V EC (2.7) solve for V EC2 V CE2 = 1.84V 45Nasim Zafar

Example 19.7 C-E loop neglect IB2 because it is IB2 << IC1 I E1 I C1 V CC = I C1 R C1 + V CE1 +I E1 R E1 15 = 1.3(5) + V CE1 +1.3(3) VCE1= 4.87V 46Nasim Zafar

Example 19.7 Stage 2 B-E loop I B2 I E2 V CC = I E2 R E2 + V EB +I B2 R BB2 + V BB2 15 = I E2 (2K) +.7 +I B2 (5K) (3) Use I B2  I E2 /  solve for I E2 IE2 = 2.8mA 47Nasim Zafar

Example stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor. I C =  I B I C  I E V BE = 0.7 (npn) = -0.7 (pnp)  = 100 Find I C1, I C2, V CE1, V CE2 Use Thevenin circuits. 48Nasim Zafar

Example 19.7 R BB1 = R B1 ||R B2 = 33K V BB1 = V CC [R B2 /(R B1 +R B2 )] V BB1 = 15[50K/150K] = 5V Stage 1 B-E loop V BB1 = I B1 R BB1 + V BE +I E1 R E1 Use I B1  I E1 /  5 = I E1 33K / I E1 3K I E1 = 1.3mA I B1 I E1 49Nasim Zafar