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CHAPTER 5

CHAPTER 5. Transistor Circuits. OBJECTIVES. Describe and Analyze: Need for bias stability Common Emitter Amplifier Biasing RC-coupled Multistage Amplifiers Direct-Coupled Stages Troubleshooting. Introduction.

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CHAPTER 5

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  1. CHAPTER 5 Transistor Circuits

  2. OBJECTIVES Describe and Analyze: • Need for bias stability • Common Emitter Amplifier Biasing • RC-coupled Multistage Amplifiers • Direct-Coupled Stages • Troubleshooting

  3. Introduction The DC bias values for VCE and Ic are collectively called the “Q-Point”. Because a transistor’s beta varies 2 to 1 or more from device to device, biasing circuitry needs to be designed so that the Q-point is not a function of beta. Likewise, the gain of a transistor amplifier should not depend on beta. Gain should be set by the values of external components such as resistors.

  4. Beta Changes with Temperature Not only does it vary from device to device, beta is also strongly dependent on temperature.

  5. Voltage Divider Biasing • Choose Rb1 & Rb2 so that: Rb1 || Rb2 << Re for the worst-case value of beta • Vb is fixed by Rb1 and Rb2, and:Ve = Vb – 0.7V • Re >> r’e. Therefore Ic = Ie = Ve / Re

  6. Biasing Example For a circuit like the one on the previous slide, calculate Vb, Ve, Ie, Ic, Vc, and Vce given:  = 50 Vcc =12V, Rb1 = 100k, Rb2 = 20k, Rc = 4k, Re = 2k, Vb = [Rb2 / (Rb1 + Rb2)]  Vcc = 12V / 6 = 2 Volts Ve = Vb – 0.7 = 2 – 0.7 = 1.3 Ic = Ie = Ve / Re = 1.3V / 2k = 0.65 mA Vc = Vcc - Rc  Ic = 12V – 4k  1.3mA = 6.8V Vce = Vc – Ve = 6.8V – 1.3V = 5.5V r’e = 25mV / Ie = 25mV / 0.65mA = 38.5 Ohms Is Re >> r’e? Is 2000 >> 38.5 ? Yes!

  7. Input Impedance Zin will not depend on  if: Rb1 || Rb2 << Re

  8. Voltage Gain: Unbypassed Re Av = rc / Re where rc = Rc || RL Gain is stable but low

  9. Voltage Gain: Bypassed Re Av = rc / r’e where rc = Rc || RL. But r’e = 25mV / Ie Gain is high, but changes with the signal current

  10. Voltage Gain: Compromise A trade-off between high gain and gain stability

  11. Emitter Biasing Very stable Q-point, but requires two voltage supplies

  12. Emitter Bias Example For a circuit like that of the previous slide, calculate Ie, Ic, Ve, Vc, Vce given Vcc = +12V, Vee = -12V, RE + Re = 10k, Rc = 4.7k Since, effectively, Vb is zero, Ve = -0.7V Ie = (Ve – Vee) / Re =11.3V / 10k = 1.13mA Ic is about the same as Ie, so Ic = 1.13mA Vc = Vcc – Rc Ic = 12V – 4.7k  1.13mA = 6.7V Vce = Vc – Ve = 6.7V – 0.7V = 6.0 Volts

  13. Voltage-Mode Feedback Can never saturate or cut off. High gain. Limited Vce.

  14. RC-Coupled Stages Circuit is no longer used, but illustrates the principle.

  15. Choosing Capacitors Key Idea: Compared to the values of Zin and Zout, the reactances of the capacitors (Xc) should be negligible in the frequency range the input signals. • Xc = 1 / (2fC) • Xc << Zin and Xc << Zout

  16. Xc Compared with Zin or Zout What ratio of Z to Xc is required to say that Xc is negligible compared to Zin or Zout? Not as high as you might assume. Zin and Zout are determined by resistors. Let Zx be the sum of Xc and R. But remember, it’s a vector (phasor) sum: Zx = sqrt[ R2 + X2 ] Let Xc be about a third of R. That is, Xc = .3R Then Zx = sqrt[ R2 + .09R2 ] = R  sqrt(1.09) = 1.04R So there is only a 4% effect if Xc is as big as a third of Zin or Zout.

  17. A Numerical Example The first stage of a two-stage amplifier has an output impedance of 2k. The input impedance of the second stage is 4k. The frequency range is 50 Hz to 5000 Hz. Select a coupling capacitor. Since Zout < Zin, we will compare Xc to Zout to be conservative. Let Xc = .3  Zout = .3  2k = 600 Ohms. Xc is highest at the low end of the frequency range. Xc = 1 / 2f C => C = 1 / 2f Xc C = 1 / 6.28  50  600 = 5.3 uF A 10 uF electrolytic capacitor should do nicely.

  18. Direct Coupled Amplifiers Having PNP as well as NPN transistors allows us to do away with coupling capacitors

  19. Gain of a Multi-Stage Amp Suppose you have two single-stage amplifiers, each with a voltage gain of 20. If the stages are coupled together, will the gain be 20  20 = 400? Not necessarily. In fact, probably not! The problem is that Zin of stage two “loads down” the output of stage one. With a transistor amp, the Zin of the second stage is effectively in parallel with the Rc of the first stage. So the voltage gain (Av) will be: Av = (Rc || Zin) / Re

  20. Troubleshooting • Check the power-supplies, but keep your fingers off any high-voltage that may be present. • Check the DC bias levels with no signal applied. • Check for shorted capacitors. • Check for open capacitors. • Try signal tracing using amplifier’s “normal” input. • Try signal tracing with an injected signal. • Try disconnecting one stage from the next, but remember to use resistors to simulate Zout.

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