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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.