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. Zin will not depend on  if: Rb1 || Rb2 << Re
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. Very stable Q-point, but requires two voltage supplies
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. Circuit is no longer used, but illustrates the principle.
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[ R R2 ] = 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.