domingo, 27 de junio de 2010

The Transistor Amplifier.

Transistors are used in a great variety of circuits. Fortunately, we can divide the ways in which they are used into two fairly simple classes: amplifiers and switches.

Transistors switches form the basis of all modern electronic digital computers. This particular lab doesn't deal with digital electronics. Here we will look at an example of using a bipolar transistor in an amplifier.

Figure 6 illustrates a typical single-transistor amplifier circuit. This arrangement is often called the common emitter amplifier because the input voltage to the transistor appears between the base & emitter, and the output voltage appears between the collector & emitter — i.e. the emitter terminal is shared by (or 'common to') the input and output.

dia6.gif - 15Kb

Note, and  are the voltages between each of the transistor base, collector, and emitter terminals and the 'ground' (zero volts). They aren't the same thing as  or  which are the voltages from base-to-emitter and collector-to-emitter! The diagram also shows the input and output signal AC voltages,  and . Thesearen't equal to  and  because the 0·1F capacitors block any d.c. connection between these potentials. (If you're puzzled by all this, ask a demonstrator.)

In order to build a working amplifier you have to choose suitable values for resistors, , and . For now, assume that  (i.e. it is a piece of wire). We will want to choose a value for  later, but for now we'll worry about everything else.

Anyone who has been confused by reading an electronics textbook will suspect that choosing the 'right' values for the resistors is quite complicated. However, it is possible to select satisfactory values using some simple rules. It is worth bearing in mind again that electronics is a practical subject which shares some things with cookery! (Transistors can get hot, too...) In particular, there are situations (and this is one) where there isn't always a single 'correct' solution for the resistor values you need. It is possible to make a working amplifier using a wide range of resistor values. For a theorist or mathematician this can be depressing — there isn't one 'right' answer. For the rest of us it's good news as it means there are a wide range of values which are 'OK'. It also means that some simple approximations aren't likely to lead to serious problems.

Experience with bipolar transistors has taught engineers that — 9 times out of 10 — a good start is to make just three assumptions and use them as 'rules' unless we know better:—
  1. The base-emitter voltage will always be about 0·6 Volts (or 0·6 for a PNP transistor).
  2. The current gain (the  value) will be a few hundred.
  3. The large  value means that , so we can assume that 

If you look at your transistor's characteristic curves you should see that, although  does depend upon , over most of the measured range it is around 0·6 Volts or so. The  of your transistor will probably be somewhere in the 200 — 600 range. So these approximations are a moderately good place to start in the absence of any better information.

The resistors in the amplifier circuit will determine the steady bias voltages and currents, , etc. The capacitors in the circuit are used to control the effects of a.c. signals. Start off by ignoring the capacitors as they don't affect the way the actual transistor operates. We can therefore work out all the resistor values, etc, without bothering about them.

There are various ways to decide what values to choose for the bias resistors. They all give roughly similar results, and the following simple argument is about as good as any other.

For the circuit to work as an amplifier we need to make the collector voltage, , move up and down in response to any input signal variations. These changes in collector voltage are coupled out through the capacitor to provide the output voltage signals, . This means that — in the absence of any input signal — the transistor should have a 'moderate' set of applied bias voltages/currents to give  'room' to move up and down under the influence of any input.

The circuit is driven by a +15V power line and the collector-emitter voltage is applied via the two series resistors,  & . In the absence of any good reason for making some other choice we might just as well assume that the available voltage should be shared equally between , and the transistor. We therefore want about 5 volts across , 5 volts across , and 5 volts between the collector and emitter. This means that the amplifier should have,  V,  V, and  V. 

Darlington transistor

Circuit diagram of a Darlington pair using NPN transistors
In electronics, the Darlington transistor (often called a Darlington pair) is a compound structure consisting of two bipolar transistors (either integrated or separated devices) connected in such a way that the current amplified by the first transistor is amplified further by the second one. This configuration gives a much higher current gain (written β, hfe, or hFE) than each transistor taken separately and, in the case of integrated devices, can take less space than two individual transistors because they can use a shared collector. Integrated Darlington pairs come packaged singly in transistor-like packages or as an array of devices (usually eight) in an integrated circuit.
The Darlington configuration was invented by Bell Laboratories engineer Sidney Darlington in 1953. He patented the idea of having two or three transistors on a single chip, sharing a collector.
A similar configuration but with transistors of opposite type (NPN and PNP) is the Sziklai pair, sometimes called the "complementary Darlington."


A Darlington pair behaves like a single transistor with a high current gain (approximately the product of the gains of the two transistors). In fact, integrated devices have three leads (B, C and E), broadly equivalent to those of a standard transistor.
A general relation between the compound current gain and the individual gains is given by:
\beta_\mathrm{Darlington} = \beta_1 \cdot \beta_2 + \beta_1 + \beta_2
If β1 and β2 are high enough (hundreds), this relation can be approximated with:
\beta_\mathrm{Darlington} \approx \beta_1 \cdot \beta_2
A typical modern device has a current gain of 1000 or more, so that only a small base current is needed to make the pair switch on. However, this high current gain comes with several drawbacks.
One drawback is an approximate doubling of base-emitter voltage. Since there are two junctions between the base and emitter of the Darlington transistor, the equivalent base-emitter voltage is the sum of both base-emitter voltages:
V_{BE} = V_{BE1} + V_{BE2} \approx 2V_{BE1}\!
For silicon-based technology, where each VBEi is about 0.65 V when the device is operating in the active or saturated region, the necessary base-emitter voltage of the pair is 1.3 V.

Another drawback of the Darlington pair is its increased saturation voltage. The output transistor is not allowed to saturate (i.e. its base-collector junction must remain reverse-biased) because its collector-emitter voltage is now equal to the sum of its own base-emitter voltage and the collector-emitter voltage of the first transistor, both positive quantities in normal operation. (In symbols, VCE2 = VBE2 + VCE1, so VC2 > VB2 always.) Thus the saturation voltage of a Darlington transistor is one VBE (about 0.65 V in silicon) higher than a single transistor saturation voltage, which is typically 0.1 - 0.2 V in silicon. For equal collector currents, this drawback translates to an increase in the dissipated power for the Darlington transistor over a single transistor.

Another problem is a reduction in switching speed, because the first transistor cannot actively inhibit the base current of the second one, making the device slow to switch off. To alleviate this, the second transistor often has a resistor of a few hundred ohms connected between its base and emitter terminals.

This resistor provides a low impedance discharge path for the charge accumulated on the base-emitter junction, allowing a faster transistor turn-off.

The Darlington pair has more phase shift at high frequencies than a single transistor and hence can more easily become unstable with negative feedback (i.e., systems that use this configuration can have poor phase margin due to the extra transistor delay).

Darlington pairs are available as integrated packages or can be made from two discrete transistors; Q1 (the left-hand transistor in the diagram) can be a low power type, but normally Q2 (on the right) will need to be high power. The maximum collector current IC(max) of the pair is that of Q2. A typical integrated power device is the 2N6282, which includes a switch-off resistor and has a current gain of 2400 at IC=10A.

A Darlington pair can be sensitive enough to respond to the current passed by skin contact even at safe voltages. Thus it can form the input stage of a touch-sensitive switch.

Common Source Amplifier with Source Degeneration.

The small-signal amplification performance of the CS amplifier discussed in the previous lecture can be improved by including a series resistance in the source circuit. (This is very similar – if not identical – to the effect of adding emitter degeneration to the BJT CE amplifier.) This so-called CS amplifier with source degeneration circuit is shown in


We have a choice of small-signal models to use for the MOSFET. A T model will simplify the analysis, on one hand, by allowing us to incorporate the effects of RS by simply adding this value to 1/gm in the small-signal model, if we ignore ro. This small-signal circuit is shown in


On the other hand, using the T model makes the analysis more difficult when ro is included. (The hybrid ð model is better at easily including the effects of ro.) However, ro in the MOSFET amplifier is large so we can reasonably ignore its effects for now in the expectation of making the analysis more tractable. 

Small-Signal Amplifier Characteristics

We'll now calculate the following small-signal quantities for this MOSFET common source amplifier with source degeneration: Rin, Av, Gv, Gi, and Rout. 

• Input resistance, Rin. Referring to the small-signal equivalent circuit above in with I g=0 then

Rin = Rg

• Partial small-signal voltage gain, Av. We see at the output side of the small-signal circuit in 

V0= - gmVgs(RD||RL)

which is the same result (ignoring ro) as we found for the CS amplifier without source generation. At the gate, however, we find through voltage division that


This is a different result than for the CS amplifier in that vgs is only a fraction of vi here, whereas vgs =Vi without RS. Substituting (3) into (2), gives the partial small-signal AC voltage gain to be


• Overall small-signal voltage gain, Gv. As we did in the previous lecture, we can derive an expression for Gv in terms of Av. By definition, 


Applying voltage division at the input of the small-signal equivalent circuit in Fig. 4.44(b),

Substituting (6) into (5) we the overall small-signal AC voltage gain for this CS amplifier with source degeneration to be


• Overall small-signal current gain, Gi. Using current division at the output in the small-signal model above in


while at the input, 


Substituting (9) into (8) we find that the overall small-signal AC current gain is


• Output resistance, Rout. From the small-signal circuit in with vsig =0 then i must be zero leading to

Rout= RRD 


Adding RS has a number of effects on the CS amplifier. (Notice, though, that it doesn't affect the input and output resistances.) First, observe from (3) 


that we can employ RS as a tool to lower vgs relative to vi and lessen the effects of nonlinear distortion. This RS also has the effect of lowering the small-signal voltage gain, which we can directly see from (7). A major benefit, though, of using RS is that the small-signal voltage (and current) gain can be made much less dependent on the MOSFET device characteristics. (We saw a similar effect in the CE BJT amplifier with emitter degeneration.) We can see this here for the MOSFET CS amplifier using (7)

The key factor in this expression is the second one. In the case that gm RS >> 1 then


which is no longer dependent on gm. Conversely, without RS in the circuit ( Rs =0 ), we see from (7) that Gva gm . and is directly dependent on the physical properties of the transistor (and the biasing) because


in the case of an NMOS device. The "price" we pay for this desirable behavior in (12) – where Gv is not dependent on gm – is a reduced value for Gv. This Gv is largest when Rs =0 , as can be seen from (7). 

Example N32.1 (based on text exercises 4.32 and 4.33). Compute the small-signal voltage gain for the circuit below with Rs =0 , kn' W/ L' =1 mA/V2 and Vt = 1.5 V. For a 0.4-Vpp sinusoidal input voltage, what is the amplitude of the output signal? 


For the DC analysis, we see that VG =0 and ID= IS = 0.5 mA. (Why is VG = 0?) Consequently, 

VD =10 - RDID =10 -14 K •5m = 3V Assuming MOSFET operation in the saturation mode

Assuming MOSFET operation in the saturation mode


such that



VGS-1.5=+1 VGS=2.5 or 0.5V


Vs = - 2.5 V

for operation in the saturation mode. For the AC analysis, from (13) g<m= 10=(2.5-1.5)=1 mS Using this result in (7) with Rs =0 gives


For an input sinusoid with 0.4-Vpp amplitude, then

V0Gv . Vsig= V pp = 2.74 Vpp

Will the MOSFET remain in the saturation mode for the entire cycle of this output voltage? For operation in the saturation mode, vDG= vD >Vt = 1.5 V. On the negative swing of the output voltage, 


which is greater than Vt, so the MOSFET will not leave the saturation mode on the negative swings of the output voltage. On the positive swings, 


which is less than VDD = 10 V so the MOSFET will not cutoff and leave the saturation mode. (Interestingly, the MOSFET does leave the saturation mode on the negative swings for 15 RD= RL = 15 k beta, as used in the text exercises 4.32 and 4.33.) Lastly, imagine that for some reason the input voltage is increased by a factor of 3 (to 1.2 Vpp). What value of RS can be used to keep the output voltage unchanged? 

Lastly, imagine that for some reason the input voltage is increased by a factor of 3 (to 1.2 Vpp). What value of RS can be used to keep the output voltage unchanged? From (7), we can choose RS so that the so-called feedback factor 1+ gm Rs equals 3. The output voltage amplitude will then be unchanged with this increased input voltage.Hence, for


With R s=2 k beta the new overall small-signal AC voltage gain is from (7) 


The overall small-signal voltage gain has gone down, but the amplitude of the output voltage has stayed the same since the input voltage amplitude was increased. 

Freddy R Vallenilla R