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ELE1110C Tutorial 7

ELE1110C Tutorial 7. 24/10/2006 Sunkie, YU Xi. Outline. Principles of transistor examples. Principles of transistor. Structure of transistor (2 types) Three layers device NPN=> electrons as carriers PNP=> holes as carriers

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ELE1110C Tutorial 7

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  1. ELE1110CTutorial 7 24/10/2006 Sunkie, YU Xi

  2. Outline • Principles of transistor • examples

  3. Principles of transistor • Structure of transistor (2 types) • Three layers device • NPN=> electrons as carriers • PNP=> holes as carriers • Transistor is a voltage control current source e b c e b c N P N P N P

  4. Pay attention to the direction of arrows of emitter

  5. Base-emitter is forward biased • Lots of electrons diffused from emitter to base while holes from base to emitter • Base-collector is reverse biased • Attracts all injected electrons from emitter to the collector

  6. The emitter is much more heavily doped than the base, most of the current is contributed by electrons • Since is very small, • When there is a voltage between base and emitter (VBE), IE (or IC) is induced • Hence, it is voltage controls current source

  7. Transistor model—small signal model • Voltage controls current • IE vs. VBE • VBE– IE: • Just a Diode • EM Equation: • Small signal Model : • Linear Model Consider VBE and IC • Slope = trans-conductance gm = IC/ VBE = IC/VT,where VT is a constant = 25mV.

  8. A simple model for transistor • It is an easy way to do analysis • VC>VE for NPN, VE>VC for PNP

  9. Examples • Example 1 • Decide whether these transistors work

  10. Circuit A • NPN transistor • VC>VE & VBE=0.6V • conducting • Circuit B • NPN transistor • VC<VE & VBE=-0.6V • Non-conducting

  11. Circuit C • PNP transistor • VC>VE & VEB=-0.6V • Non-conducting • Circuit D • PNP transistor • VC<VE & VEB=0.6V • conducting

  12. Typical applications of transistors • Current source how to keep Iout a constant 5.6V 5V

  13. VB is fixed at 5.6V • VE=VB-0.6V=5V • IE=IC=VE/RE=1mA //RE=5.1k • We want the current to keep constant for a range of load at the collector • The question turns out to be when the transistor will not work

  14. Calculate the maximum voltage drop • The load can vary from 0 to Rmax

  15. Another two kinds of applications

  16. Example 2 • In the following circuit, if the base voltage is 5.6V

  17. What is VE? • What is IE (the current through the 500W resistor)? • What is IC (the current through RC)? • What is the output impedance of this circuit? • What value of RC would you use so that Vout = 10V. • What is the ac gain of this amplifier? • If b = 100, what is IB? What is the input impedance of this circuit? • What would be a suitable value for R1 and R2?

  18. In most cases, assume the transistor work • VE=5V • IE=VE/500=10mA • IC=IE=10mA • In this question DC output impedance Vout=20 Iclose=20/RC the output impedance R=Vout/Iclose=RC

  19. When Vout=10V, from the previous equation RC=(20-Vout)/IC= 1k • Given value of RC Gain = output signal voltage/ input signal voltage Let input signal = D vIN

  20. What is the function of R1 and R2? They are used to offer a biased voltage at the base. so it is easy to find the proportion of R1:R2 we don’t want to add too much current, so the larger the better. Typically, the scale is ( 10X design rule)

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