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Transcript
EE210 Digital Electronics Class Lecture 5 May 24, 2008 Solution to: QUIZ #1 HW #1 In This Class Bipolar JunctionTransistors (BJTs) We Will Discuss Following: 5.10 The Basic BJT Digital Logic Inverter Bipolar Junction Transistors (BJTs) Bipolar Junction Transistors (BJTs) Mode EBJ CBJ Cutoff Reverse Reverse Active Forward Reverse Saturation Forward Forward 5.3 BJT as Amplifier and Switch • Common-Emitter CKT • Input vI = vBE (bias + signal) at Base and Emitter • vO = vCE Collector and Ground • RC has Two Functions: Establish voltage at C and Convert iC to vO or vCE • vO = vCE = VCC - RC iC BJT as Amplifier and Switch (Cont…) • When vBE = vI < 0.5 V Cut Off iC ≈ 0 vO = vCC Segment XY • When vI > 0.5 V BJT Active & Cond., iC Inc. and vO Dec., vO = VCC - RC ISevBE/VT • Exp. Term Gives Rise to Steep Slope of YZ Segment BJT as Amplifier and Switch (Cont…) • Active mode ends When vO falls by 0.4 V below that of the vI. • CBJ Turns ON and BJT is in Saturation region at point Z. • In Saturation vO = vCE = VCEsat ≈ 0.1 – 0.2 V. • ICsat = (VCC – VCEsat)/RC • In Saturation BJT exhibit very small RCEsat b/w C and E 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 iC ≈ 0 and Thus acts as an Open Switch. 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 For: RB=10 kΩ, RC=1k Ω, β = 50, and VCC=5V VTC → 3 StraightLine Segments Correspond to BJT Cutoff, Active and Saturation Regions 1. vI=VOL=VCEsat= 0.2V vO = VOH = VCC = 5V 2. vI=VIL BJT starts to turn ON, Thus VIL=0.7V 3. VOL< vI < VIH : Active. Operate as Amp. Av ≈ - βRC/RB = - 50*1/10 = - 5 V/V 4. vI=VIH BJT enters into Saturation. VIH is value of vI at which BJT is at the Edge of Saturation. IB = ((VCC-VCEsat)/RC)/β = 0.096 mA VIH = IBRB + VBE =1.66V 5. vI=VOH=5V Deep Saturation vO = VCEsat ≈ 0.2V and Βforced= 11 6. Noise Margins: NMH=VOH-VIH=5-1.66=3.34V NML=VIL-VOL=0.7-0.2=0.5V Margins Vastly Different Inverter Less Than Ideal 7. The Gain in Transition Region (Slope) computed from X and Y points: Av= - (5 - 0.2)/(1.66 - 0.7) = - 5 V/V 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 • 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) Home Work No. 3 (Due May 12, 2007) 1. 2. 3. 4. Problem 3. 6 Problem 3.10 Problem 5.168 Problem 5.171 In Next Class We Will Discuss: Chap 4 MOS Field-Effect Transistors 4.1 Device Structure and Physical Operation 4.2 Current-Voltage Characteristics