CHAPTER 6
Other
Transistor
Circuits
OBJECTIVES
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
gain.
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 base current is 10mA / 70 = 0.14mA
The Thevenin’s equivalent of Rb1 & Rb2 is
RTH = (Rb1  Rb2) / (Rb1 + Rb2) = 1.5k
and 0.14mA  1.5k =0.2V, so Vb = 6.6 - .2 = 6.4V
But 5% of 6.7V is 0.34V. The minimum Vb is 6.36V,
so Vb = 6.4V seems OK.
Biasing Example
Let’s find the power dissipation of the transistor:
PD = Vce  Ic = (12V – 6V)  0.30A = 1.8 Watts
Most likely, this guy needs a heat-sink.
Input Impedance
Let’s find Zin for the emitter follower that we just
biased:
Zin is the parallel combination of the biasing
resistors together with the impedance “looking into”
the base: Zin = Rb1 || Rb2 || Re
But Rb1 || Rb2 = RTH, which we calculated to be
RTH = 1.5k and Re = 70  600 = 42k
Since Re is so much bigger than RTH, we can say:
Zin  RTH = 1.5k
Output Impedance
Zout is the parallel combination of Re and the
equivalent base resistance divided by beta
Output Impedance
Now let’s find Zout for the same emitter follower we
biased:
Zout = (RTH / ) || Re = (1.5k / 70) || 600
But, RTH /  = 1.5k / 70 = 21 Ohms
So, for all practical purposes, Zout  20 Ohms
Let’s just check that:
Actual Zout = (21  600) / (21 + 600) = 20.3 Ohms
Voltage Gain
We should find that Av is close to 1
Voltage Gain
The equation for the voltage gain of an emitter follower is:
Av = re / (re + r’e)
where r’e = 25mV / Ie, and re is Re in parallel with the load
being driven by the emitter follower.
Let’s find Av for the circuit we biased:
R’e = 25mV / 10mA = 2.5 Ohms
Re = Re = 600 Ohms
Av = 600 / (600 +2.5) = 600 / 602.5 = 0.996
0.996 is close enough to 1 for most purposes.
Power Gain
Power gain (Ap) is output power divided by input
power: Ap = Pout / Pin
Since P = V2 / R,
Pout = (Vout)2 / Rout and Pin = (Vin)2 / Rin
Some algebra, and: Ap = (Vout / Vin)2  (Rin / Rout)
For a buffer, Vout = Vin, so Ap = Rin / Rout
For the emitter follower we’ve been using, Rin = Zin
and Rout = Zout, so its power gain is:
Ap = Rin / Rout = 1.5k / 20 = 75
Darlington Pairs
For a Darlington pair, Ic / IB = 1  2  2
Darlington Pairs
Some things to know about Darlington pairs:
• Since the collectors are tied together, the transistors
can not saturate. When used as a switch, Vce  Ic
can generate a lot of heat when Ic is big.
• A transistor’s beta is often lower for very low values
of Ic. So the beta of Q1 may be a lot less than the
beta of Q2.
• The Vbe of a Darlington pair is 2  0.7 = 1.4V
• The equivalent fT of the pair is lower than the fT of
either transistor.
Common Base Amplifiers
Work at higher frequencies than a common emitter can
Common Base Amplifier
The Miller Effect
Common emitter amplifiers lose gain at higher
frequencies because of what’s known as the
Miller Effect.
The capacitance of the collector-base junction (CCB)
looks bigger at the base than it really is. It looks like
CM = Av  CCB where CM is the Miller Capacitance.
It’s caused by negative feedback from the output
(collector) back to the input (base).
An RC low-pass filter is formed by CM and the
resistance of the signal source driving the base.
Common Base Amplifier
The base is at signal ground in a common base
amplifier (but not necessarily at DC ground). So CCB
can only shunt some signal to ground, not back to the
input. That eliminates the Miller Effect.
The trade-off is that Zin is very low, on the order of 50
Ohms. But in a high-frequency amplifier, it’s usually
required that Zin and Zout be around 50 Ohms.
Comparison of Configurations
All three configurations have their place in circuits
the base is at signal ground
Differential Amplifiers
The “diff amp” is commonly used in linear ICs
Differential vs. Single-Ended
•
All the amplifiers we have seen so far share one
characteristic: they have only one input. They are single-ended
amplifiers. A signal is applied from that one input to ground.
•
Differential amplifiers have two inputs, commonly referred
to as the “plus input” and the “minus input”. A signal is applied
across the two inputs.
•
Signals applied simultaneously to both inputs with respect
to ground are called “common mode” signals.
Differential Amps & CMR
• Suppose there is +10 mV (with respect to ground)
on one input of a differential amplifier and –10 mV
(with respect to ground) on the other input. Then the
differential signal is 20 mV. If the diff amp has a gain
of 10, the output will be 10  20 mV = 200 mV.
• Now suppose that +100 mV (with respect to ground)
is applied to both inputs at the same time. That’s a
common mode signal. Since the differential voltage
is 0, and the output will be zero.
• Since common mode signals produce no output,
differential amplifiers have “common mode rejection”
(CMR). CMR is very important, as we will see.
Constant-Current Source
• Referring back to figure 6-17, the purpose of Q3 is
to act as a “constant-current source” for the
differential pair formed by Q1 and Q2.
• By definition, a constant-current source will conduct
the same amount of current irrespective of the
voltage across it.
• Since dynamic resistance (RD) is RD = V / I, the
RD of a constant-current source is infinite.
• The constant-current source in the emitter circuit of
Q1 and Q2 prevents common-mode signals from
causing base current. There’s the CMR.
Troubleshooting
The troubleshooting techniques for single-ended
amplifiers:
– Measure DC bias levels
– Trace signals
– Isolate stages
also work for differential amplifiers.
In addition, the technician should look for imbalances
in what should be equal DC voltage and current
levels.
Diff Amps & Noise
• One of the main problems in using amplifiers is the
presence of electrical noise and interference. A
common form of “man-made” noise is “hum” from
the 60 Hz power line.
• Noise gets mixed in with low-level signals, and when
the signals are amplified, so is the noise. Like death
and taxes, it’s hard to avoid noise.
• Most man-made noise appears on all inputs with
respect to ground. It’s common mode, and will be
rejected by a diff-amp. So small differential signals
can be “extracted” from large common mode noise.