Transcript
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CHAPTER 5
Transistor
Circuits
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OBJECTIVESDescribe and Analyze:
Need for bias stability
Common Emitter AmplifierBiasing
RC-coupled Multistage Amplifiers
Direct-Coupled Stages
Troubleshooting
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Introduction
The DC bias values for VCE and Ic are collectively
called the Q-Point. Because a transistors betavaries 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.
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Beta Changes with
Temperature
Not only does it vary from device to device,
beta is also strongly dependent on
temperature.
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Voltage DividerBiasing
Choose Rb1 & Rb2so that: Rb1 || Rb2> re. Therefore Ic= Ie = Ve / Re
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Biasing ExampleFor a circuit like the one on the previous slide, calculate Vb,
Ve, Ie, Ic, Vc, and Vce given: F = 50
Vcc=12V, Rb1 = 100k, Rb2= 20k, Rc = 4k, Re = 2k,
Vb = [Rb2/ (Rb1 + Rb2)] v Vcc= 12V / 6 = 2 Volts
Ve = Vb 0.7 = 2 0.7 = 1.3
Ic= Ie = Ve / Re = 1.3V / 2k = 0.65 mAVc= Vcc- Rcv Ic= 12V 4k v 1.3mA = 6.8V
Vce = Vc Ve = 6.8V 1.3V = 5.5V
re = 25mV / Ie = 25mV / 0.65mA = 38.5 Ohms
Is Re >> re? Is 2000 >> 38.5 ? Yes!
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Input Impedance
Zin will not depend on F if: Rb1 || Rb2
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Voltage Gain: Unbypassed Re
Av = rc/ Re where rc = Rc || RL Gain is stable but low
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Voltage Gain: Bypassed Re
Av = rc / re where rc = Rc|| RL.But re = 25mV / Ie
Gain is high, but changes with the signal current
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Voltage Gain: Compromise
A trade-off between high gain and gain stability
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EmitterBiasing
Very stable Q-point, but requires two voltage supplies
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EmitterBias 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
Icis about the same as Ie, so Ic= 1.13mA
Vc= Vcc Rcv Ic= 12V 4.7k v 1.13mA = 6.7V
Vce = Vc Ve = 6.7V 0.7V = 6.0 Volts
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Voltage-Mode Feedback
Can never saturate or cut off. High gain. Limited Vce.
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RC-Coupled Stages
Circuit is no longer used, but illustrates the principle.
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Choosing Capacitors
Key Idea:
Compared to the values ofZin and Zout, thereactances of the capacitors (Xc) should be negligible
in the frequency range the input signals.
Xc = 1 / (2TfC) Xc
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Xc Compared with Zin or ZoutWhat ratio ofZ to Xc is required to say that Xc is negligible
compared to Zin orZout? Not as high as you might assume.
Zin and Zout are determined by resistors. Let Zx be the sum ofXc and R. But remember, its 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 v sqrt(1.09) = 1.04R
So there is only a 4% effect if Xc is as big as a third ofZin or
Zout.
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A Numerical ExampleThe 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 acoupling capacitor.
Since Zout < Zin, we will compare Xc to Zout to be
conservative.
Let Xc = .3 v Zout = .3 v 2k = 600 Ohms.
Xc is highest at the low end of the frequency range.
Xc = 1 / 2TfC => C = 1 / 2TfXc
C = 1 / 6.28 v 50 v 600 = 5.3 uF
A 10 uF electrolytic capacitor should do nicely.
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Direct Coupled Amplifiers
Having PNP as well as NPN transistors allows us to do
away with coupling capacitors
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Gain of a Multi-Stage AmpSuppose you have two single-stage amplifiers, each
with a voltage gain of 20. If the stages are coupledtogether, will the gain be 20 v 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 ofthe second stage is effectively in parallel with the Rc of
the first stage. So the voltage gain (Av) will be:
Av = (Rc || Zin) / Re
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Troubleshooting
Check the power-supplies, but keep your fingers offany 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 amplifiers normal input.
Try signal tracing with an injected signal.
Try disconnecting one stage from the next, butremember to use resistors to simulate Zout.
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