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Tim Sumner, Imperial College, Rm: 1009, x47552

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Presentation on theme: "Tim Sumner, Imperial College, Rm: 1009, x47552"— Presentation transcript:

1 Tim Sumner, Imperial College, Rm: 1009, x47552
Transistors 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

2 Review : What is a Transistor ?
Collector NPN Transistor Rules: VC > VE VB = VE Volts I C = I B   IC : Collector Current Base IB : Base Current IE : Emitter Current I E = I C + I B  I E = (1 + ) I B Emitter FOR MORE INFO... Relevant Books: Art of Electronics + Lab: Harowitz and Hill Electronic Circuits: E.C. Lowenberg, SCHAUM SERIES Electric Circuits: J.A. EDMINISTER SCHAUM SERIES 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

3 Some Transistor Applications I
+ V Current Source Follower: + V Load IC Vin IC Vin= 5.6 V Vout= Vin+ 0.6 Volts Vout= 5.0 V IE R = 5 K IE = 1 mA - V 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

4 Some Transistor Applications II
Push-Pull : Common Emitter Amplifier : + V + V RC >> Vout >> Qup Vout Vin = VB Vin = VE RE <<  IE = IC >> Qdown - V 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

5 Tim Sumner, Imperial College, Rm: 1009, x47552
The Emitter Follower I Given : Vcc = 15 Volts Ic = 0.5 mA;  = 100 f(100 Hz) = 3 db What about C1 ? Vcc Source R1 Ic Load C1 1F 10 K ~ 4.7 K R2 R3 Rbase= VB/IB = 1.64 M (2) RTH(divider) = 1/10 RB  RB = 164 K  R1 = R2= 328 K (1) R3=7.5 V/0.5 mA = 15K 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

6 The Emitter Follower II
(3) Think of an RC filter (high-pass) . 15K is parallel to 4.7 K and the whole think times   358 K ; parallel with 328 K  112 K; So f=(1/2RC)  F Vcc Source 328 K Ic Load C1 1F 10 K ~ 4.7 K 328K 15 K  = 100 f(100 Hz) = 3 db 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

7 Emitter Follower (a bit diff.)
Rin=Rload(1+) Rout=Rsource/(1+) 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

8 Common Emitter Amplifier
Gain = R5/R2 Rout=R5 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

9 Common Emitter Amplifier II
12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

10 The Ebers-Moll Model for BJT
12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

11 Tim Sumner, Imperial College, Rm: 1009, x47552
Temperature Effects 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

12 Temperature Effects and Current Mirrors
12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

13 Tim Sumner, Imperial College, Rm: 1009, x47552
Exercises Make a Current Source using a transistor Make common emitter follower Make a Common Emitter Amplifier 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

14 Making Gates using Transistors I
+5 Volts InPut A A .OR. B InPut B 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

15 Making Gates using Transistors II
InPut A A .OR. B InPut B A OUT 1 Here is what we have made 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

16 Exercise I Design and construct an AND gate
using transistors and resistors : InPut A A .AND. B InPut B A OUT 1 The AND gate TRUTH table 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

17 Tim Sumner, Imperial College, Rm: 1009, x47552
Exercise I continued Verify the truth table ! You can set an input to ‘1’ by connecting it to +5 V by a resistor of 1K and you set it to ‘0’ by grounding it. 5 V ‘0’ InPut B ‘0’ A .AND. B InPut A ‘1’ 5 V 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

18 Answer (almost): TTL CMOS
Not quite because this is a NAND gate but we do know by now how to invert using a transistor 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

19 Tim Sumner, Imperial College, Rm: 1009, x47552
Important conclusion You may understand by now that the inputs have no ability to set signals high or low. In contrast to this the outputs can drive signals to High or Low 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552

20 Tim Sumner, Imperial College, Rm: 1009, x47552
Remember: Inputs draw no current. 1K resistors to +5 V can set inputs to high or they can be set low by connecting to ground. Outputs can draw current and can force lines to low or high. 12/11/2018 Tim Sumner, Imperial College, Rm: 1009, x47552


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