CHAPTER 6: Other Transistor Circuits pdf

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 volt

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

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

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  R TH = 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

(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 (R D ) is R D = 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

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