1. Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop to get I C in terms of V o Change I C in terms of I B Equate the two equations to link V i with V o Bipolar Transistor Biasing Fixed Bias Biasing Circuit

2. = 4.8 = 4.3 V o (V) V i (V) 5 0.7 Cutoff Active 0.2 x 5 Saturation x

3. Biasing using Collector to Base Feedback Resistor Find R B and R C such that I E = 1mA, V CE = 2.3 V, V CC = 10 V and  =100. ICIC IEIE IBIB I C + I B = I E NOTE: Proposed to use branch current equations and node voltages

4. Biasing using Collector to Base Feedback Resistor (V C – V B ) / R B = I B but V C = V CE and V B = V BE = 0.7 V (2.3 – 0.7) / R B = (I E / (  +1) R B = 161.6 k  (V CC – V C ) / R C = I E R C = 7.7 k  I E = 1mA, V CE = 2.3 V, V CC = 10 V and  =100. VCVC VBVB

5. This is a very stable bias circuit. The currents and voltages are almost independent of variations in . Voltage Divider Biasing Circuit

6. Redrawing the input side of the network by changing it into Thevenin Equivalent R Th : the voltage source is replaced by a short-circuit equivalent Analysis

7. V Th : open-circuit Thevenin voltage is determined. Inserting the Thevenin equivalent circuit Analysis V TH Use voltage divider V TH

8. The Thevenin equivalent circuit Analysis

9. BJT Biasing in Amplifier Example Find V CE, I E, I C and I B given  =100, V CC =10V, R 1 = 56 k , R 2 = 12.2 k  R C = 2 k  and  R E = 0.4 k  V TH = R 2 /(R 1 + R 2 )V CC V TH = 12.2k/(56k+12.2k).(10) V TH = 1.79V R TH = R 1 // R 2 = 10 k 

10. BJT Biasing in Amplifier Circuits V TH = R TH I B + V BE + R E I E 1.79 = 10k I B + 0.7 + 0.4k (  +1)I B I B = 21.62  A I C =  I B = 100(21.62  )=2.16mA I E = I C + I B = 2.18mA V CC = R C I C + V CE + R E I E 10 = 2k(2.16m)+V CE +0.4(2.18m) V CE = 4.8 V

11. Basic Transistor Application

12. Digital Logic – NOT GATE  In the simple inverter circuit, if the input is approximately zero volts, the transistor is in cutoff and the output is high and equal to V CC.  If the input is high and equal to V CC, the transistor is driven into saturation, and the output is low and equal to V CE (sat).

13. Digital Logic – NOR Gate  If the two inputs are zero, both transistors Q 1 and Q 2 are in cutoff, and V O = 5 V.  When V 1 = 5 V and V 2 = 0, transistor Q 1 can be driven into saturation, and Q 2 remains in cutoff. With Q 1 in saturation, the output voltage V O = V CE (sat).  If V 1 = 0 and V 2 = 5 V, then Q 1 is in cutoff, and Q 2 can be driven in saturation, and V O = V CE (sat).

14.  If both inputs are high, meaning V 1 = V 2 = 5 V, then both transistors can be driven into saturation, and V O = V CE (sat).  In a positive logic system, meaning that the larger voltage is a logic 1 and the lower voltage is a logic 0, the circuit performs the NOR logic function.  The circuit is then a two-input bipolar NOR logic circuit.