Other Transistor Circuits

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Describe and Analyze:
• Common Collector Amplifiers
• Common Base Amplifiers
• Darlington Pairs
• Current Sources
• Differential Amplifiers
• Troubleshooting

Common Collector Amplifiers

The common-collector amplifier, more commonly
called an emitter follower, is used as a “buffer”

Buffers Amps

The ideal buffer amplifier has unity voltage gain
(Av = 1), infinite input impedance (Zin = ), and zero
output impedance (Zout = 0). The power gain would
also be infinite (Ap = )
The “job” of a buffer amp is to prevent loading of a
signal source. If a high-impedance signal source is
connected to a low-impedance point in a circuit,
most of the signal will be lost in the source’s internal
resistance. The buffer goes in between the source
and the rest of the circuit.

The Emitter Follower Buffer
As we shall see, the emitter follower has a voltage
gain slightly less than one (Av  1), a high input
impedance (Zin  Re), and low output impedance
(Zout  Re || Rb / ). It has a reasonably high power
Emitter followers are used very often in linear circuits,
even in linear ICs. They are simple, yet effective.

Biasing the Emitter Follower
• Emitter followers typically use resistor divider biasing,
just like the common-emitter amplifier.
• Usually, the collector is tied directly to Vcc, so the
collector to emitter voltage is Vce = Vcc – Ve. If Vcc
is too high, then the transistor can get hot since the
power dissipated is PD = Vce  Ic. Remember that Ic
is basically equal to Ie = Ve / Re = (Vb – 0.7) / Re.

• Emitter followers sometimes use a collector resistor
to lower the Vce drop.

A Biasing Example
The Specifications:
Suppose we have a circuit like that of figure 6-1.
Vcc is 12 V, and we want Ve to be 6 V  5%.
Find Rb1 and Rb2 so that there is 2 mA of current
through Rb2 (Ib2 = 2mA). The minimum beta is 70.
The emitter resistor is 600 Ohms (Re = 600).
Also, find the power dissipation in the transistor.

Biasing Example (cont.)

Find Vb: Vb = Ve + 0.7V = 6.0V + 0.7V = 6.7V


Find Rb2: Rb2 = Vb / Ib2 = 6.7V / 2mA = 3.35k


Choose a standard resistor value: Let Rb2 = 3.3k
Find Rb1: [Rb2 / (Rb1 + Rb2)]  Vcc = Vb
So, Rb1 = Rb2  [(Vc – Vb) / Vb]
Rb1 = 3.3k  [(12V – 6.7V) / 6.7V] = 2.61k


Choose a standard resistor value: Let Rb2 = 2.7k

Biasing Example
Now let’s check to see if we got it right:

Vb = [Rb2 / (Rb1 + Rb2)]  Vcc
Vb = [3.3k / (3.3k + 2.7k)]  12V
Vb = (3.3 / 6.0)  12 = 6.6V
Maximum b...
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