1. EE210 Digital Electronics Class Lecture 5 May 24, 2008

2. Solution to: QUIZ #1 HW #1

3. In This Class We Will Discuss Following: 5.10 The Basic BJT Digital Logic Inverter Bipolar JunctionTransistors (BJTs)

5. ModeEBJCBJ CutoffReverseReverse ActiveForwardReverse SaturationForwardForward

6. 5.3 BJT as Amplifier and Switch Common-Emitter CKT Input v I = v BE (bias + signal) at Base and Emitter v O = v CE Collector and Ground R C has Two Functions: Establish voltage at C and Convert i C to v O or v CE v O = v CE = V CC - R C i C

7. BJT as Amplifier and Switch (Cont…) When v BE = v I < 0.5 V Cut Off i C ≈ 0 v O = v CC Segment XY When v I > 0.5 V BJT Active & Cond., i C Inc. and v O Dec., v O = V CC - R C I S e vBE/VT Exp. Term Gives Rise to Steep Slope of YZ Segment

8. BJT as Amplifier and Switch (Cont…) Active mode ends When v O falls by 0.4 V below that of the v I. CBJ Turns ON and BJT is in Saturation region at point Z. In Saturation v O = v CE = V CEsat ≈ 0.1 – 0.2 V. I Csat = (V CC – V CEsat )/R C In Saturation BJT exhibit very small R CEsat b/w C and E

9. BJT as Amplifier and Switch (Cont…) Saturated BJT Provides Low Res. Path b/w Node C and Ground and Can be Thought of a Closed Switch. Cut Off BJT i C ≈ 0 and Thus acts as an Open Switch.

10. The Basic BJT Digital Logic Inverter We learned in Chap 1 that Logic Inverter is Most Fundamental Component of Digital System We will use this BJT Ckt to Realize Logic inverter Makes use of Cutoff and Saturation Modes of BJT to Work as Logic Inverter In These Modes the Power Dissipation is Low

11. For: R B =10 kΩ, R C =1k Ω, β = 50, and V CC =5V VTC → 3 Straight- Line Segments Correspond to BJT Cutoff, Active and Saturation Regions 1.v I =V OL =V CEsat = 0.2V v O = V OH = V CC = 5V 2. v I =V IL BJT starts to turn ON, Thus V IL =0.7V

12. 3.V OL < v I < V IH : Active. Operate as Amp. A v ≈ - βR C /R B = - 50*1/10 = - 5 V/V 4. v I =V IH BJT enters into Saturation. V IH is value of v I at which BJT is at the Edge of Saturation. I B = ((V CC -V CEsat )/R C )/β = 0.096 mA V IH = I B R B + V BE =1.66V 5. v I =V OH =5V Deep Saturation v O = V CEsat ≈ 0.2V and Β forced = 11 6. Noise Margins: NM H =V OH -V IH =5-1.66=3.34V NM L =V IL -V OL =0.7-0.2=0.5V Margins Vastly Different Inverter Less Than Ideal

13. 7. The Gain in Transition Region (Slope) computed from X and Y points: Av= - (5 - 0.2)/(1.66 - 0.7) = - 5 V/V

14. Saturated vs Nonsaturated BJT Digital Ckts Inverter we just discussed belongs to saturated variety of BJT Digital ckts – TTL Some TTL versions are in use, Generally Saturated Bipolar Digital ckts are no more Technology of Choice in Digital System Design The Reason Being the Speed of Operation LongTtime Delay to Turn OFF a Saturated BJT

15. Minority carrier distribution in base region of saturated BJT: Blue Triangle gradient gives the diffusion current across base Grey Rectangle Causes Transistor to be Driven Deeper into Saturation To Achieve High Speeds the BJT Should not Saturate – Current Mode Logic or ECL (Chap 11)

16. Home Work No. 3 (Due May 12, 2007) 1.Problem 3. 6 2.Problem 3.10 3.Problem 5.168 4.Problem 5.171

17. In Next Class We Will Discuss: Chap 4 MOS Field-Effect Transistors 4.1 Device Structure and Physical Operation 4.2 Current-Voltage Characteristics