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The term amplifier as used in this chapter means a circuit (or stage) using a single active device rather than a complete system such as an integrated circuit operational amplifier. An amplifier is a device for increasing the power of a signal. This is accomplished by taking energy from a power supply and controlling the output to duplicate the shape of the input signal but with a larger (voltage or current) amplitude. In this sense, an amplifier may be thought of as modulating the voltage or current of the power supply to produce its output.

The basic amplifier, figure 9.1, has two ports and is characterized by its gain, input impedance and output impedance. An ideal amplifier has infinite input impedance (R_{in} = ∞), zero output impedance (R_{out} = 0) and infinite gain (A_{vo} = ∞) and infinite bandwidth if desired.

Figure 9.1 Basic Amplifier Model

The transistor, as we have seen in the previous chapter, is a three-terminal device. Representing the basic amplifier as a two port network as in figure 9.1, there would need to be two input and two output terminals for a total of four. This means one of the transistor terminals must be common to both the input and output circuits. This leads to the names common emitter, etc. for the three basic types of amplifiers. The easiest way to determine if a device is connected as common emitter/source, common collector/drain, or common base/gate is to examine where the input signal enters and the output signal leaves. The remaining terminal is what is thus common to both input and output. In this chapter we will primarily be using n-type transistors (NPN, NMOS) in the example circuits. The same basic amplifier stages can just as easily be implemented using p-type transistors (PNP, PMOS). When larger multi-stage amplifiers are assembled, both types of transistors are often interspersed with each other.

Building-block amplifier stages:

- Inverting voltage amplifier (also called Common emitter or Common source amplifier)

- Current Follower (also called Common base or Common gate or cascode)

- Voltage Follower (also called Common collector or Common drain amplifier)

- Series feedback (more commonly: emitter/source degeneration)

- Shunt feedback

The common emitter/source amplifier is one of three basic single-stage amplifier topologies. The BJT and MOS versions function as an inverting voltage amplifier and are shown in figure 9.2. The base or gate terminal of the transistor serves as the input, the collector or drain is the output, and the emitter or source is common to both input and output (it may be tied to the ground reference or the power supply rail), which gives rise to its common name.

Figure 9.2: Basic n-type inverting voltage amplifier circuit (neglecting biasing details)

The common emitter or source amplifier may be viewed as a transconductance amplifier (i.e. voltage in, current out) or as a voltage amplifier (voltage in, voltage out). As a transconductance amplifier, the small signal input voltage, v_{be} for a BJT or v_{gs} for a FET, times the device transconductance g_{m}, modulates the amount of current flowing through the transistor, i_{c} or i_{d}. By passing this varying current through the output load resistance, R_{L} it will be converted back into a voltage V_{out}. However, the transistor’s small signal output resistance, r_{o}, is not typically high enough for a reasonable transconductance amplifier (ideally infinite). Nor is the output load, R_{L}, low enough for a decent voltage amplifier (ideally zero). Another major drawback is the amplifier’s limited high-frequency response due in part to the built in collector base or drain gate capacitance inherent to the transistor. More on how this capacitance effects the frequency response in a later section of this chapter. Therefore, in practice the output often is routed through either a voltage follower (common collector or drain stage), or a current follower (common base or gate stage), to obtain more favorable output and frequency characteristics. This latter combination is called a cascode amplifier as we will see later in the chapter on multi-stage amplifiers.

In comparison to the BJT common emitter amplifier, the FET common source amplifier has higher input impedance. The generally lower g_{m}of the FET vs. the BJT at equal current levels leads to lower voltage gain for the MOS version.

In order for the common emitter or source amplifier to provide the largest output voltage swing, the voltage at the Base or Gate terminal of the transistor is offset in such a way that the transistor is nominally operating halfway between its cut-off and saturation points. Note the NMOS (a) and NPN (b) characteristic curves in figure 9.2.1. This allows the amplifier stage to more accurately reproduce the positive and negative halves of the input signal superimposed upon the DC Bias voltage. Without this offsetting Bias Voltage only the positive half of the input waveform would be amplified.

(a) NMOS

(b) NPN

Figure 9.2.1 (a) I_{D} vs. V_{DS}curves and (b) I_{C} vs. V_{CE} curves

The red line superimposed on the two sets of curves represents the DC load line of a 400 ohm R_{L}. To maximize the output swing it is desirable to set the operating point of the transistor, with a zero input signal, at a drain or collector voltage of one half the supply voltage, which would be 4 volts in this case. Finding the corresponding drain or collector current along the load line gives us the target current level. This is around 10mA for R_{L} equal to 400 ohms. The next step is to determine the corresponding V_{GS} or I_{B} for a 10mA I_{D} or I_{C}. In the NMOS example each curve represents a different V_{GS} from 0.9 volts to 1.5 volts in 0.1 volt steps. The NMOS device used in this example has a transconductance of about 40mA/V. The I_{D} equal to 10mA point on the load line falls between the 1.4V and 1.3V curves or a V_{GS} of 1.32V. In the NPN example each curve represents a different I_{B} from 10uA to 100uA in 10uA steps. The 50uA curve happens to cross the load line at I_{C} =10mA. The β of the transistor must therefore be about 200. The task now is to somehow provide this DC offset or bias at the Gate or Base of the transistor.

The first bias technique we will explore is called voltage divider bias and is shown in figure 9.2.2. If we choose the correct resistor values for R_{1} and R_{2} that will result in a collector or drain current such that one half of the supply voltage, V+ appears across R_{L} we should have our desired value of V_{GS} or V_{BE} (I_{B}) for biasing with no signal input. For the MOS case we know that no current flows into the gate so the simple voltage divider ratio can be used to pick R_{1} and R_{2}. If V+ = 8V and we want V_{GS} to equal 1.32 V then:

The actual values of R_{1} and R_{2} are not so important just their ratio. However, the divider ratio we choose will be correct for only one set of conditions of power supply voltage, transistor threshold voltage and transconductance, and temperature. Actual designs often use more involved bias schemes.

Figure 9.2.2 Voltage divider bias

For the NPN case the calculation is somewhat more involved. We know we want I_{B} to be equal to 50uA. The current that flows in R_{1} is the sum of the current in R_{2} and I_{B} which puts an upper bound on R_{1} when R_{2} is infinite and no current flows in R_{2}. If we assume a nominal V_{BE} of 0.65 volts then R_{1} must be no larger than 7.35V/50uA or 147KΩ. The purpose of the voltage divider is to attenuate the variations in V+ and thus make the DC operating point of the transistor less sensitive to V+. To that end we need to make the current in R_{2} many times larger than I_{B}. If we, for example, choose to make I_{R2} 9 times I_{B} then the current in R_{1} will be 10*I_{B} or 500uA. R_{1} will be 1/10 what we just calculated as the upper bound or 14.7KΩ. R_{2} will be V_{BE} divided by 450uA or 1.444KΩ which is a divider ratio of 0.8921. If we had simply used 8V-V_{BE}/8V as the ratio (assume V_{BE} = 0.65V) the divider ratio would have been 0.8125. Taking I_{B} into account shifted the required ratio. These values would need to be adjusted slightly if the actual V_{BE} was not the 0.65 volts (or β was not 200) we used in this calculation. This points out a major limitation of this bias scheme as we pointed out in the MOS example above. That is the sensitivity to device specific characteristics like V_{BE} and β as well as supply voltage and temperature.

A consequence of including this bias scheme is a lowering of the input impedance. The input now includes the parallel combination of R_{1} and R_{2} across the input. For the MOS case this now sets the input resistance. For the BJT case we now have R_{1}||R_{2}||r_{π} as the effective input resistance.

There is another minor inconvenient problem with this bias scheme when it is connected to a prior stage in the signal path. This bias configuration places the AC input signal source directly in parallel with R_{2} of the voltage divider. This may not be acceptable, as the input source may tend to add or subtract from the DC voltage dropped across R_{2}.

One way to make this scheme work, although it may not be obvious why it will work, is to place a coupling capacitor between the input voltage source and the voltage divider as in figure 9.2.3 below.

Figure 9.2.3 Coupling capacitor C_{C} prevents voltage divider bias current from flowing into the input signal source.

The capacitor forms a high-pass filter between the input source and the DC voltage divider, passing almost the entire AC portion of the input signal on to the transistor while blocking all the DC bias voltage from being shorted through the input signal source. This makes much more sense if you understand the superposition theorem and how it works. According to superposition, any linear, bilateral circuit can be analyzed in a piecemeal fashion by only considering one power source at a time, then algebraically adding the effects of all power sources to find the final result. If we were to separate the capacitor and the R_{1}/R_{2}voltage divider circuit from the rest of the amplifier, it might be easier to understand how this superposition of AC and DC would work.

With only the AC signal source in effect, and a capacitor with an arbitrarily low impedance at the input signal frequency, almost all the AC voltage appears across R_{2}.

To calculate the small signal voltage gain of the common emitter or source amplifier we need to insert a small signal model of the transistor into the circuit. The small signal models of the BJT and MOS FET are actually very similar so the gain calculation for either version is much the same. The small signal hybrid-π models for the BJT and MOS amplifiers are shown in figure 9.2.4.

Figure 9.2.4 Common emitter or source small signal models.

The following are some of the key model equations we will need to calculate the amplifier stage voltage gain. These equations are used for the other amplifier configurations that we will discuss in following sections as well.

The small signal voltage gain A_{v} is the ratio of the input voltage to the output voltage:

The input voltage V_{in} (v_{be} for the BJT and v_{gs} for the MOS) times the transconductance g_{m} is equal to the small signal output current, i_{o} in the collector or drain. V_{out} will be simply this current times the load resistance R_{L,}neglecting the small signal output resistance r_{o} for the moment. Notice the minus sign because of the direction of the current i_{o}.

Rearranging for the gain we get:

Substituting the BJT and MOS g_{m} equations we get:

Comparing these two gain equations we see that they both depend on the DC collector or drain currents. The BJT gain is inversely proportional to V_{T} (the Thermal Voltage) which is approximately 26mV at room temperature. The Thermal Voltage, V_{T} increases with increasing temperature so from the equation we see that the gain will actually decrease with increasing temperature. The MOS gain is inversely proportional to the over drive voltage, V_{ov} (V_{GS} – V_{th}) which is often much larger than V_{T} at similar drain currents leading to the lower gain for the MOS stage vs. the BJT stage for approximately equal bias currents.

If R_{L} is relatively large when compared to the small signal output resistance then the gain will be reduced because the actual output load is the parallel combination of R_{L} and r_{o}. In fact r_{o} puts an upper bound on the possible gain that can be achieved with a single transistor amplifier stage.

Again looking at the small signal models in figure 9.2.4 we see that for the BJT case the input V_{in} will see r_{π} as a load. For the MOS case V_{in} will see basically an open circuit (for low frequencies anyway). This will of course be the case absent any Gate or Base bias circuitry.

Again looking at the small signal models in figure 9.2.4 we see that for both the BJT case and the MOS case the output impedance is the parallel combination of R_{L} and r_{o}. For most practical applications we can ignore r_{o} because it is very often much larger than R_{L}. Below are the BJT and MOS r_{o} equations.

The Current Follower or Common base/gate amplifier has a high voltage gain, relatively low input impedance and high output impedance compared to the voltage follower or common collector/drain amplifier. The BJT and MOS versions are shown in figure 9.3

Figure 9.3: Basic n-type current follower or common base/gate circuit (neglecting biasing details)

In applications where only a positive power supply voltage is provided some means of providing the necessary DC voltage level for the common gate or base terminal is required. This might be as simple as a voltage divider between ground and the supply. In applications where both positive and negative supply voltages are available, ground is a convenient node to use for the common gate or base terminal.

The common gate or base stage is most often used in combination with the common emitter or source amplifier in what is known as the cascode configuration. The cascode will be covered in the next chapter on multi stage amplifiers in greater detail.

To calculate the small signal voltage gain of the common base or gate amplifier we insert the small signal model of the transistor into the circuit. The small signal models for the BJT and MOS amplifiers are shown in figure 9.3.1.

Figure 9.3.1 Current follower or Common base/gate small signal models.

Much like in the common emitter/source amplifier stage the small signal input voltage, V_{in} (v_{be} for the BJT and v_{gs} for the MOS) times the transconductance g_{m} is equal to the small signal output current, i_{o} in the collector or drain. V_{out} will be simply this current times the load resistance R_{L,}neglecting the small signal output resistance r_{o} for the moment.

It is perhaps more useful to consider the current gain of the current follower stage rather than its voltage gain. In the case of the MOS version we know that I_{S} = I_{D}because I_{G}= 0. Thus the MOS stage current gain is exactly 1. In the case of the BJT version we know that the ratio of I_{C} to I_{E}is equal to α and thus will be slightly less than 1.

Again looking at the small signal models in figure 9.3.1 we see that for the BJT case the input V_{in} will see r_{π}in parallel with the series combination of g_{m} and R_{L} as a load. For the MOS case V_{in} will see basically just the series combination of g_{m} and R_{L}. The equation below (from the BJT small signal T model) relates g_{m} and the resistance seen at the emitter r_{E}. We can also use this relationship to give us the resistance seen at the source r_{S}.

It is also important to note here that 100% (neglecting I_{B} in the BJT case) of the current from the input source flows through the transistor and becomes the output current. Thus the name current follower.

Again looking at the small signal models in figure 9.3.1 we see that for both the BJT case and the MOS case the output impedance is the parallel combination of R_{L} and r_{o}. We can generally assume this is true if we consider that V_{in} is driven from a low impedance (nearly ideal) voltage source. If this is not the case then the finite output impedance must be added in series with r_{o}. If the input of the current follower is driven by the relatively high output impedance of a transconductance amplifier such as the common emitter or source amplifier from earlier then the output impedance for the combined amplifier can be very high. For most practical applications we can ignore r_{o} because it is very often much larger than R_{L}.

The Emitter or Source follower is often called a common Collector or Drain amplifier because the collector or drain is common to both the input and the output. This amplifier configuration, figure 9.4, has its output taken from the emitter/source resistor and is useful as an impedance matching device since its input impedance is much higher than its output impedance. The voltage follower is also termed a “buffer” for this reason.

Figure 9.4:Basic n-type Voltage follower or common collector/drain circuit (neglecting biasing details)

The gain of the voltage follower is always less than one since r_{E}and R_{L}or r_{S} and R_{L} form a voltage divider. The input to output offset is set by the V_{BE} drop of about 0.65 volts below the base for the BJT and V_{GS} below the gate for the MOS. This configuration’s function is not voltage gain but current or power gain and impedance matching. The input impedance is much higher than its output impedance so that a signal source does not have to supply as much power to the input. This can be seen from the fact that the base current is on the order of 100 times (β) less than the emitter current. The low output impedance of the emitter follower matches a low impedance load and buffers the signal source from that low impedance.

The collector/source current is basically determined by the emitter/source resistor so the main design variables in this case is simply R_{L} and the power supply voltage.

To calculate the small signal voltage gain of the voltage follower configuration we insert the small signal model of the transistor into the circuit. The small signal models for the BJT and MOS amplifiers are shown in figure 9.4.1.

Figure 9.4.1 Voltage Follower small signal models.

For the circuit in figure 9.4.2 calculate the voltage gain A_{V} = V_{out}/V_{in}.

Figure 9.4.2 BJT Voltage gain example

To use the voltage gain formula we just obtained using the small signal models we need to first calculate r_{E}. From section 9.3.3 we are given the equation for r_{E}:

To use this formula we need to know I_{E}. We know that the voltage across R_{L} is V_{out}. We also know that V_{out} = V_{in} - V_{BE}. If we use an estimate of V_{BE} to be 0.6 volts, we get V_{out} = 5.6 - 0.6 or 5 volts. If R_{L} is 1KΩ then I_{E} is 5mA. Using a room temperature value for V_{T} = 25mV, we get r_{E} is equal to 5Ω. Substituting these values into our gain equation we get:

The output impedance is simple the parallel combination of the Emitter (Source) resistor R_{L} and the small signal emitter (source) resistance of the transistor r_{E}. Again from section 9.3.3, the equation for r_{E} is as follows:

Similarly, the small signal source resistance, r_{S}, for a MOS FET is 1/g_{m}.

Referring back to our gain example in figure 9.4.2, we can also calculate the output resistance, which will be the parallel combination of the 1KΩ R_{L} and the 3Ω r_{E} or 2.99Ω.

Common emitter/source amplifiers give the amplifier an inverted output and can have a very high gain and can vary widely from one transistor to the next. The gain is a strong function of both temperature and bias current, and so the actual gain is somewhat unpredictable. Stability is another problem associated with such high gain circuits due to any unintentional positive feedback that may be present. Other problems associated with the circuit are the low input dynamic range imposed by the small-signal limit; there is high distortion if this limit is exceeded and the transistor ceases to behave like its small-signal model. When negative feedback is introduced, many of these problems are reduced, resulting in improved performance. There are several ways to introduce feedback in this simple amplifier stage, the easiest and most reliable of which is accomplished by introducing a small value resistor in the emitter circuit (R_{E}). This is also referred to as series feedback. The amount of feedback is dependent on the relative signal level dropped across this resistor. The signal seen across R_{E} is out of phase with the signal seen at V_{out} and thus subtracts from V_{out} reducing its amplitude. When the emitter resistor value approaches that of the collector load resistor (R_{L}), the gain will approach unity (A_{v} ~ 1).

Figure 9.5: Adding an emitter/source resistor decreases gain. However, with increased linearity and stability

It is much less common to include a degeneration resistor in MOS designs. This is because, in microelectronic integrated circuits, the gain (g_{m}) of the device can be adjusted by changing the W/L ratio. This degree of design freedom is not generally available in Bipolar (BJT) processes.

**DC Biasing example with emitter degeneration**

There are some BJT biasing rules of thumb:

1. Set I_{E} not I_{B} or V_{BE} : less dependence on β and temperature (V_{T})

2. Allow 1/3V_{CC} across R_{C}, V_{CE} and R_{B2}

3. Save power by allowing only 10% of I_{E} in R_{B}

We are given the following for circuit in figure 9.5.1, V_{CC} = 20V ; I_{E} = 2mA ; β = 100. From our rules of thumb we set V_{B} = 1/3*V_{CC} = 6.7 V.

Figure 9.5.1 DC Biasing example

V_{B} = (R_{B2}/(R_{B1}+R_{B2}))*V_{CC} ⇒ 6.7V = (R_{B2}/(R_{B1}+R_{B2}))*20 (1)

V_{CC} /(R_{B1} + R_{B2} ) = 0.1*I_{E} ⇒ 20/(R_{B1} + R_{B2}) = 200 μA (2)

Solving equations (1) and (2) we get:

R_{B1}=2R_{B2} then from (2)

3R_{B2} = 20/200 μA = 100kΩ

So, R_{B2} = 33kΩ and R_{B1} = 66kΩ

Now we have V_{E} = V_{B} – V_{BE} = 6.7 – 0.7 = 6 V and I_{E} is 2 mA: R_{E} = V_{E}/I_{E} = 6/2mA = 3kΩ.

I_{C} = (β/(β+1))*I_{E} = (100/101)*2mA = 1.98 mA and I_{B} = I_{C}/β = 1.98mA/100 = 19.8μA.

From our rules of thumb we know that V_{C} = 2/3*20V = 13.3 V

So to find R_{L} we have: R_{L} = (V_{CC} – V_{C})/I_{C} = (20 – 13.3)/1.98mA = 3.4kΩ

To calculate the small signal voltage gain of the common emitter/source amplifier with the addition of emitter/source degeneration we again insert the small signal model of the transistor into the circuit. The small signal models for the BJT and MOS amplifiers are shown in figure 9.5.1.

Figure 9.5.1 Common emitter/source with degeneration

The impedance R_{E} reduces the overall transconductance g_{m} of the circuit by a factor of g_{m}R_{E} + 1, which makes the voltage gain:

So the voltage gain depends almost exclusively on the ratio of the resistors R_{L} / R_{E} rather than the transistor’s intrinsic and unpredictable characteristics. The distortion and stability characteristics of the circuit are thus improved at the expense of a reduction in gain.

Going back to our earlier biasing example, figure 9.5.1, values for I_{C} = 2mA, R_{L} = 3.4KΩ and R_{E} = 3KΩ to calculate the small signal gain we first find g_{m} = I_{C}/V_{T} = 2mA/25mV = 0.08. Using our formula for A_{V}:

Again looking at the small signal models in figure 9.4.1 we see that for the BJT case the input V_{in} see r_{}in series with degeneration resistor R_{E} as a load. For the MOS case V_{in} see basically an open circuit.

Again looking at the small signal models in figure 9.5.1 we see that for both the BJT case and the MOS case, much like in the earlier common emitter/source stage, the output impedance is the parallel combination of R_{L} and r_{o} but now degeneration resistor R_{E} is in series with r_{o}. For most practical applications we can ignore r_{o} because it is very often much larger than R_{L}.

Basically the same techniques used in the simple common emitter/source amplifier stage, which were discussed in section 9.2.1, can be used when the emitter degeneration resistor is added. The added voltage across the R_{E} (R_{E}*I_{E}) must be added to the bias level. This added voltage drop actually make the operating point (I_{C}) much less sensitive to the bias level.

The small signal voltage gain of the common emitter amplifier with the emitter resistance is approximately R_{L} / R_{E}. For cases when a gain larger than 5-10 is needed, R_{E} may be become so small that the necessary good biasing condition, V_{E} = R_{E}*I_{E} > 10* V_{T} cannot be achieved. A way to restore the small signal voltage gain while maintaining the desired DC operating bias is to use a by-pass capacitor as is figure 9.5.4. The small AC signal sees an emitter resistance of just R_{E1} while for DC bias the emitter resistance is the series combination of R_{E} = R_{E1}+R_{E2}. Calculations for the common emitter amplifier with emitter degeneration can be applied here by replacing R_{E} with R_{E1} when deriving the amplifier gain, and input and output impedances, because a sufficiently large bypass capacitor in effects shorts R_{E2}and is effectively removed from the circuit for sufficiently high frequency inputs.

Figure 9.5.4 addition of emitter by-pass capacitor

Using our earlier biasing exercise in figure 9.5.1 as an example but splitting the 3KΩ R_{E} into two resistors as in figure 9.5.4 with R_{E1}= 1KΩ and R_{E2} = 2KΩ with C_{1} = 1uF we can recalculate the small signal gain for high frequencies, where C_{1} effectively shorts out R_{E2}, to be:

The addition of by-pass capacitor C_{1}, however, modifies the low frequency response of the circuit. We know from our two gain calculations that the DC gain of the circuit is -1.13 and the gain increases to -3.36 for high frequencies. We can therefore assume that the frequency response consists of a relatively low frequency zero followed by a somewhat higher frequency pole. The formulas for the zero and pole are as follows:

where R’_{E}= R_{E2} || (R_{E1} + r_{e})

For our example problem with R_{E1} = 1K , R_{E2} = 2K and C_{1} = 1uF we get the frequency for the zero equal to 80 Hz and the frequency for the pole equal to 237 Hz. The simulated frequency response from 1 Hz to 100 KHz for the example circuit is shown in figure 9.5.5.

Figure 9.5.5 simulated frequency response

1. Find DC operating point.

2. Calculate small-signal parameters: g_{m}, r_{}, r_{e} etc.

3. Replace DC voltage sources with AC grounds and DC current sources with open circuits.

4. Replace transistor with small-signal model (hybrid-π model or T model)

At this point we are going to take a diversion to discuss Miller’s Theorem. While the methods we have been using up to this point are completely general, there are certain configurations that lend themselves to be analyzed more simply by Miller’s Theorem. Miller’s theorem states that in a linear circuit, if there is a branch where an impedance Z, connects two nodes with node voltages V_{1}and V_{2}, this branch can be replaced by two other branches connecting the corresponding nodes to ground by impedances respectively Z / (1-K) and KZ / (K-1), where the gain from node 1 to node 2 is K = V_{2} / V_{1}.

Figure 9.6.1 Miller’s Theorem

At this point we will go through the steps that show how the Miller impedances are arrived at. We can use the equivalent two-port network technique to replace the two-port represented in figure 9.6.1(a) to its equivalent in figure 9.6.2.

Figure 9.6.2

Replacing the voltage sources in figure 9.6.2 with their Norton equivalent current sources we get figure 9.6.3.

Figure 9.6.3

Using the source absorption theorem (see the Appendix at the end of this chapter), we get figure 9.6.4.

Figure 9.6.4

Which gives us figure 9.6.5 (which is figure 9.6.1(b) ) when we parallel combine the two impedances.

Figure 9.6.5

Another biasing technique for the common emitter or source amplifier, called shunt feedback, is accomplished by the introduction of some fraction of the collector or drain signal back to the input at the base or gate. This is done via the biasing resistor (R_{F}), as shown in figure 9.7.1. Resistor R_{F} connects between two nodes that have gain, A_{V} (K), between them and thus the application of Miller’s theorem is the best way analyze the small signal characteristics of this circuit.

Figure 9.7.1 Drain-to-Gate (a) and Collector-to-Base (b) shunt feedback

Figure 9.7.1(a) shows a common source NMOS amplifier using drain feedback biasing. This type of biasing is often used with enhancement mode MOSFETS and can be useful when operating with a low voltage power supply (V_{+}). If Vin is AC coupled, the voltage on the gate is equal to the voltage on the drain (V_{GS} = V_{DS}) since no gate current flows through R_{F}. If Vin is DC coupled then a voltage divider is formed by R_{F} and R_{S} and V_{GS} will be less than V_{DS}. It is useful to note that the transistor is always in saturation when V_{GS} = V_{DS}. If the drain current increases for some reason, such as a change in V_{+}, the gate voltage drops. The decreased gate voltage in turn causes the drain current to decreases which causes the gate voltage to increase. The negative feedback loop reaches an equilibrium that is the bias point for the circuit.

Some data sheets for enhancement MOSFETS give a value for I_{D}(on), where V_{GS} = V_{DS} lf I_{D}(on) is known, the circuit component can be easily calculated as shown in Example 9.3. The input impedance of a circuit using drain feedback biasing is equal to the value of R_{F} divided by the voltage gain plus one.

This configuration employs negative feedback to stabilize the operating point. In this form of biasing, the base feedback resistor R_{F} is connected to the collector instead of connecting it to the DC source V_{+}. So any large increase in the collector current will induce a voltage drop across the R_{L} resistor that will in turn reduce the transistor’s base current.

If we assume that the input source Vin is AC coupled and no DC bias current flows in R_{S}, from Kirchhoff’s voltage law, the voltage V_{RF}across the base resistor R_{F} is:

By the Ebers–Moll model, I_{c} = βI_{b}, and so:

From Ohm’s law, the base current I_{b}=V_{RF}/R_{F}, and so:

Hence, the base current I_{b} is:

If V_{BE} is held constant and temperature increases, then the collector current I_{c} increases. However, a larger I_{c} causes the voltage drop across resistor R_{L} to increase, which in turn reduces the voltage V_{RF} across the base resistor R_{F}. A lower base-resistor voltage drop reduces the base current I_{b}, which results in less collector current I_{c}. Because an increase in collector current with temperature is opposed, the operating point is kept more stable.

**Pros:**

- Circuit stabilizes the operating point against variations in temperature and β (ie. Transistor process variations)

**Cons:**

- In this circuit, to keep I
_{c}independent of β, the following condition must be met:

which is the case when:

- As β is fixed (and generally not known precisely) for a given transistor, this relation can be satisfied either by keeping R
_{L}fairly large or making R_{F}very low.

- If R
_{L}is large, a high V_{+}is necessary, which increases cost as well as precautions necessary while handling.

- If R
_{F}is low, the reverse bias of the collector–base region is small, which limits the range of collector voltage swing that leaves the transistor in active mode.

- The resistor R
_{F}causes an AC feedback, reducing the voltage gain of the amplifier. This undesirable effect is a trade-off for greater quiescent operating point stability.

**Usage:** The feedback also decreases the input impedance of the amplifier as seen from the base, which can be advantageous. Due to the gain reduction from feedback, this biasing form is used only when the trade-off for stability is warranted.

For the amplifier shown in figure 9.7.2(a) with a DC coupled input source V_{in} calculate the input and output resistance and voltage gain A_{V}. We first need to start with some preliminary DC analysis to determine the operating point of Q_{1}. For this we set V_{in} to zero volts, i.e. short it out. If we assume a V_{BE} of 0.65 volts we will have 65 uA flowing in the 10K resistor R_{S}. Given that V_{+} is 10V, we would like V_{out} to be 5 volts. The current in R_{L} is equal to 500uA and will split between the collector of Q_{1} and the feedback resistor R_{F}. The voltage across the 62.7KΩ feedback resistor is 5-0.65 or 4.35 volts. The current in R_{F} splits between the current in R_{S} and I_{B}. The base current I_{B} is equal to 4.35/62.7KΩ – 65uA or 4.3 uA. We should get a collector current of 500uA - 69.3uA or 430.3uA with a β of about 100.

If we use Miller’s theorem to replace the feedback resistor R_{F} with its two equivalent impedances we get figure 9.7.2(b). Assuming that the voltage gain from base to collector A_{V} is significantly greater than 1 we can make the simplification that A_{V}/(A_{V}-1) is close to 1. The effective load resistance, R_{Leq} we will use to calculate the gain will be 10KΩ||62.7KΩ or 8.62KΩ. Now we can use the same common emitter or source small signal gain equations we used in section 9.2.2. The 430uA collector currents gives us a g_{m} of 430uA/25mV or 0.0172. We know that A_{V} = -g_{m}R_{Leq} or A_{V} = -0.0172*8.62K = -148 which is » 1. The input resistance seen at the base of Q_{1} will be the r_{π}of Q_{1}, which is equal to β/g_{m} or 100/0.0172 = 5.814KΩ, in parallel with the Miller resistance 62.7KΩ/149 = 421Ω thus the effective input resistance, R_{base} will be about 392.5Ω.

Figure 9.7.2 Example using Miller’s theorem

The input source resistance R_{S} and the equivalent resistance at the base, R_{base} form a voltage divider. To calculate the overall voltage gain from voltage source V_{in} to V_{out} we multiply this divider ratio times the base to collector gain, A_{V} we just calculated.

From our investigation of the inverting op amp configuration in Chapter 3 we learned that for amplifiers with less than infinite gain the actual gain will be less than the ideal gain equation, Gain = -R_{F}/R_{S} predicts. If our single transistor amplifier had infinite gain the gain from V_{in} to V_{out} would be 62.7KΩ/10KΩ or 6.27. In Chapter 3 we got an estimation of the percentage error, ε, due to finite gain A_{V} (remember β in this equation is the feedback factor not the current gain of the transistor):

The actual gain of 5.6 is about 10% smaller than the ideal gain of 6.27.

Part 1 DC operating point:

For the circuit in figure 9.7.3 calculate the required R_{F} to bias the DC operating point such that V_{out} is equal to ½ the supply voltage or +5V when Vin = 0. Assume V_{BE} = 0.65V and β = 200.

Figure 9.7.3

Part 2 Small signal gain and impedance:

Given the value for R_{F} calculated in part 1 calculate the voltage gain A_{V}, the input resistance R_{base} and the output resistance R_{out}. Also calculate the overall voltage gain V_{out}/V_{in} and explain why this is different than the ideal value of –R_{F}/R_{S}.

The Miller effect is key to predicting the frequency response of an inverting amplifier stage where capacitive feedback is included. Typically there’s a low-pass pole in the voltage gain stage created by R_{S} of the signal source and a feedback capacitor C_{C}. But, the low pass cutoff is not simply determined by R_{S} and C_{C}. The Miller effect creates an effective capacitance at the base/gate of the transistor that appears as C_{C} scaled by the amplifier’s voltage gain.

Figure 9.7.3 Miller feedback capacitor

The Miller effect is especially useful when you’re trying to produce a low-pass filter on an IC op amp with a relatively low frequency cut-off. The difficulty is that large capacitors are difficult to make because they take up so much space on the IC. The solution is to make a small capacitor and then scale up its behavior using the Miller effect.

Equivalent Circuit

Here’s a simplified version of the circuit above.

Figure 9.7.4 Miller feedback equivalent circuit

Miller said that you can approximate the input capacitance by replacing C_{C} with a different capacitance C_{M} across the R_{IN}. How much bigger is C_{M}? C_{C} is multiplied by the voltage gain (A_{V} = g_{m}R_{L}) of the amplifier. Miller’s theorem also states there will be a capacitor C’_{C} across R_{L} that is equal to C_{C} times (A_{V}+1)/A_{V} which for large values of A_{V} we assume to be 1.

How does this work? Well, we know that forcing a voltage across a capacitor causes a current to flow. How much current depends on the capacitance: I = C_{C} · ΔV/Δt. However, in this circuit, the voltage gain at R_{L} causes a much larger ΔV across C_{C} - causing an even larger current to flow through C_{C}. Therefore, it looks like a much larger capacitance from the point of view of V_{IN}.

In this example we will use the circuit shown in figure 9.7.5 to illustrate the Miller multiplication of the feedback capacitor C_{C}. Bias resistors R_{1} and R_{S} are chosen to set the DC operating point such that V_{out} is at a DC value of approximately V+/2 or 5V. With the given R_{L} of 10KΩ the low frequency small signal voltage gain A_{V} is approximately 80.

We can now calculate the -3 dB frequency and unity gain (0dB) frequency for a feedback capacitor, C_{C}, of 0.001 uF. The frequency where the gain from V_{in} to V_{out} falls by -3 dB from its DC values is approximately equal to:

The unity gain frequency is approximately equal to :

Figure 9.7.5 Miller Capacitance Example

The circuit in figure 9.7.5 was simulated and the AC frequency response from 1 Hz to 1 MHz is plotted in figure 9.7.6. The gain from V_{in} to V_{out} in dB is 20Log(A_{V}) or about 38 dB. The -3 dB frequency in this case would be where the gain curve crosses 35 dB (~263 Hz) and the unit gain frequency would be where the gain curve crosses the 0 dB line (~21.7 KHz ). The simulation results are in reasonably close agreement with our approximate hand calculations. For our hand calculations we assumed that R_{1} was sufficiently larger than R_{S} so it could be ignored and likewise the r_{π} of Q_{1} was large enough to not materially affect R_{S}.

Figure 9.7.6 Frequency sweep simulation

- The Common Emitter stage has high gain but low input and high output impedance.

- R
_{E}emitter degeneration improves input impedance and provides negative feedback to stabilize DC operating point but with some loss in gain.

- The Common Base stage has low input, high output impedance but is good at high frequencies. Good current buffer sometimes called the current follower.

- The Common Collector or Emitter follower can be biased with large input impedance, low output impedance but has approximately unity gain. Good voltage buffer.

The source absorption theorem has two dual forms: the voltage source absorption and the current source absorption theorems.

The voltage source absorption theorem states that if, in one branch of the circuit with current I, there is a voltage source controlled by I, the source can be replaced by a simple impedance with value equal to the source controlling factor.

Figure 9A.1

The proof is trivial. An impedance Z where a current I flows has the same voltage drop the I controlled source generates at its terminals.

The current source absorption theorem states that if, in one branch of the circuit there is a current source controlled by a voltage V, the source can be replaced by a simple admittance with value equal to the source controlling factor.

Figure 9A.2

The proof is again trivial. An admittance Y submitted to a voltage V imposes the same current that the source Y V provides.

Figure A9.3 shows the small signal equivalent circuit model of a transistor. Find the resistance Rin looking into the emitter (with base and collector at small signal AC grounds).

Figure 9A.3

Using what we just learned about the source absorption theorem for current sources we know we can replace the controlled source with a resistance equal to 1/g_{m}its transconductance.

Figure AT1.1 Inserting a Diode connected device in the bias divider

Figure AT1.2 Inserting R_{2}increases the input resistance

Depending on your component choices and signal source, the circuit in figure AT2.1 may load the source so that the input signal is noticeably attenuated when connected to the circuit. That is, at signal frequencies, the input impedance of the circuit may be low compared to the output impedance of the signal source, and so dissipation in the signal source causes attenuation of the signal entering the circuit. To ensure the current into the base of the transistor is negligible, the biasing network must have a relatively low equivalent resistance at DC when looking out of the base. However, there is a clever method we can use to raise the impedance of the network at signal frequencies when looking out of the capacitor. By bootstrapping some of the transistor’s output signal back into the input, we can make the input impedance (at signal frequencies) very large (i.e., approximately R_{E}, the input impedance of the transistor). Consider the modified circuit in figure AT2.1.

Instead of Equation (3.3) and Equation (3.4), assume that

(R1||R2) + R_{B} ≪ βR_{E} and β ≈ 100 and C ≥ 1 / 2πf (R1 || R2 + R_{B}). (A.1)

Otherwise, components can be chosen exactly as before. The bootstrapping capacitor C_{B} must be very large so that it looks like a short circuit to signal frequencies. Theoretically, the resistor R_{B} can be chosen arbitrarily. As long as Equation (A.1) can be met, a high choice of R_{B} (e.g., R_{B} > 1 kΩ) is a good idea. The signal at the transistor’s emitter follows the signal at its base. At signal frequencies, C_{B} acts like a short circuit, and so both ends of R_{B} see the same potential. Hence, R_{B} carries no current at signal frequencies. Thus, the R1–R2 divider cannot load the input source because no current from the source makes its way across R_{B} (i.e., R_{B} ≈ ∞ at signal frequencies). The current through R1–R2 that would normally come from the source comes from the output instead. This method is called bootstrapping because we use the circuit’s own output to reduce current required from the input.

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