EE3300/EE5300 Electronics Applications

Week 2 Self-Study Notes

Recap of transfer functions

Finding poles by inspection

It’s time to learn an engineer “superpower”: to be able to look at a circuit and predict its frequency response.

The frequency response of a circuit is determined by the poles and zeros of its transfer function. Poles create low-pass characteristics (i.e. dB/decade). If the goal is to create a faster circuit, then low-pass characteristics will limit its behaviour.

Recognising poles in a circuit design allows you to intuitively see which aspect of the circuit will limit its bandwidth. With practice, you will be able to look at many circuit schematics and understand their frequency response with only very minimal calculations.

The rule is simple: Poles are contributed by every node that has a capacitance and a resistance in parallel to AC ground. In more detail: any node may contribute a pole. To check, scan the signal path through the circuit from input through to output, with independent sources turned off (i.e. voltages source set to 0 V, current sources set to 0 A). If a node has a capacitance and a resistance to AC ground then it contributes a pole at .

Some examples will illustrate this technique.

Example 1

Consider the common source amplifier shown in Figure 1, where is an ideal voltage source, , μ , , , , and . Neglect channel-length modulation.

Write down the transfer function and sketch the Bode plot. Which capacitor limits the bandwidth of the circuit?

Solution

Tracing the signal path from left to right, we observe:

• is not between a node and ground, so we cannot use the inspection method we have studied so far. However, has a large value and therefore will have a small impedance. As an approximation, we neglect the contribution of at the moment and consider a small signal model where is replaced by a short circuit. Therefore, we imagine connecting directly to .
• is an independent source so it’s turned off (i.e. set to 0 V). Since is shorted to , we do not observe a pole at . (To elaborate, the voltage on is constrained by and therefore the capacitor is not acting as an independent energy storage element.)
• At node , we see a pole! We have a capacitance and resistance in parallel to AC ground. Specifically,

Therefore, the circuit has one pole, which will give it a low-pass characteristic. Its -3dB cutoff frequency is controlled by the lowest-frequency pole, which in this case is due to the only pole at .

To sketch the magnitude plot, we need to know the DC gain. We recognise this circuit as a degenerated common source amplifier and recall that the gain is

In decibels, this is equivalent to .

Using these results we can sketch the magnitude response as shown in Figure 2.

Comparison to accurate simulation

Is it really that easy to find transfer functions by inspection?

To verify the result, we can simulate the circuit to obtain a detailed Bode plot.

Example 2

How would Example 1 change if was modelled as a realistic voltage source having an output impedance of ?

Solution

Draw the circuit with the addition of the new source resistance, as shown in Figure 4.

The voltage on is now independent of , and therefore the node can contribute a pole.

Recall that when we identify poles by inspection, we set all independent sources to zero. Hence is an AC ground. Treating as a short, the node sees to AC ground. Since the latter resistors are in the kiloohm range, we can approximate the resistance to ground as simply 10 Ω.

Therefore the new pole is found at

This is considerably lower than the pole previously we found at the output. We would say that the pole at is the “dominant pole” in this circuit because it limits the bandwidth.

Based on this analysis, we can sketch the new magnitude plot as shown in Figure 5.

Comparison to accurate simulation

The circuit simulation in Figure 6 confirms the hand sketch.

Discussion

To illustrate the power of this technique, suppose that we have the opportunity to modify the previous circuit stage to change its output resistance. Would a higher or lower output resistance improve the bandwidth?

The pole location is and therefore to increase the pole frequency, we want a lower . Therefore, to increase the bandwidth of the circuit, lower the output resistance of the previous stage.

Pole locations can be predicted simply by looking at the circuit! You can mentally consider how changes to the design of a circuit will affect its frequency response.

Example 3

Pushing poles to the right on the Bode plot is not always desirable. Sometimes a pole is placed in a circuit in order to attenuate noise at a given frequency.

Consider the circuit shown in Figure 7. This circuit uses a Zener diode to create a reference voltage . Recall that a Zener is a diode with a well-defined reverse breakdown voltage. The capacitor is intended to filter out ripple in the unregulated power supply . Resistor biases the Zener; it will be in the kiloohm range.

If has noise (such as ripple arising from rectifying AC to DC), then this ripple will corrupt . Specifically, the bias current flowing into the Zener will vary with time, and therefore will vary.

Given that is derived from rectifying 50 Hz AC and therefore has 100 Hz ripple (assuming a full-bridge rectifier), design a capacitor value such that the ripple is reduced by 20 dB. The Zener has a small-signal resistance of

Solution

First we must understand the meaning of a Zener diode’s small-signal resistance. Consider the current-voltage characteristic sketched in Figure 8a. The Zener diode breaks down at 3.3 V, but the curve has a finite slope. For small perturbations about a given operating point, we can linearise the slope and define a small signal model resistance . This leads to the small signal model shown in Figure 8b.

The question asks us to attenuate the 100 Hz ripple by 20 dB. This means that we need a pole at 10 Hz because one pole results in a dB/decade slope, as shown in Figure 9.

Recognising that , we can approximate the pole location as

μ

This is a very large capacitor, so it would be preferable to design an alternative circuit that can use a smaller value of . We will consider such a design in tutorial questions.

The Miller Effect

At high frequencies, parasitic capacitances can have a significant impact on the circuit’s behaviour. A particularly important case arises when there is a capacitance between the input and output of an amplifier, as shown in Figure 10.

Consider a step change , as shown in the figure. A larger step change appears at the output, therefore the capacitor must charge up “more than expected” compared to the size of Put another way, instead of the circuit having to supply coulombs, instead it must supply coulombs. From the point of view of the rest of the circuit, it is as if there was a larger capacitor there.

Miller’s Theorem

Miller’s Theorem is a equivalent circuit transformation that we can use to analyse the impact of a capacitor connected between input and output of an amplifier. The theorem applies to impedances in general, but we usually apply it to capacitors.

The theorem states that an impedance connected between the input and output of an amplifier can be replaced by two impedances and , as shown in Figure 11.

To derive this equivalent circuit transformation, we recognise that the circuits will be equivalent if the currents and voltages and are the same. From the original circuit (Figure 11 top), we have

where we have substituted .

From the new circuit, we have

and

Equating Eqs. and ,

Equating Eqs. and ,

This equivalent circuit transformation is exact if a frequency-dependent gain is used for , i.e. the full transfer function instead of just the DC gain.

If the DC gain is used, then the result is approximate. Using the DC gain in this manner is called the Miller approximation.

Miller Multiplication

We see an effect called Miller multiplication when:

1. There is a capacitor connected between the input and output of an amplifier, and
2. That amplifier is inverting (i.e. ).

Using the Miller Theorem (Eqs. and ), notice that an inverting amplifier will cause to be a smaller impedance than . A smaller impedance is equivalent to a larger capacitor.

If the impedances in the Miller theorem are capacitances then it reduces to

Notice that is bigger than , especially if the amplifier has a high negative gain. On the other hand, becomes slightly smaller.

The Miller theorem is mathematically true for both inverting and non-inverting amplifiers (i.e. negative and positive gains). For non-inverting amplifiers (), the impedance becomes negative, that is, a capacitor becomes an inductor. In this case, the physical interpretation is less intuitive and you may prefer to resort to a circuit simulation to understand the resultant frequency response. The term Miller effect is usually reserved specifically for the scenario where an inverting amplifier causes a small capacitance to behave as if it were a large capacitance.

Example 4

The MOSFET amplifier in Figure 12 has a parasitic capacitance between the input and output of 10 pF. Use the Miller approximation to estimate the circuit’s bandwidth.

Solution

We recognise this circuit as a common source amplifier, and recall that the DC gain is

The 1 μF coupling capacitors are much larger than the 10 pF parasitic capacitor, so we will neglect their contribution to the frequency response. Similarly, the bias resistors are much larger than the 50 Ω output impedance of the source, so we will also neglect their contribution. These considerations lead to the approximate high frequency model shown in Figure 13.

The capacitor is the Miller-multiplied version of the 10 pF capacitor that appears in the real circuit. We calculate its value to be

Also,

Therefore we identify a pole at the input as

At the output, the pole is

Therefore the frequency response will be limited by the input pole, and the -3 dB cut-off frequency is approximately 20 MHz.

Notice that we obtained the bandwidth of the circuit without detailed calculations. Using simple formulas and intuitive analysis, we were able to identify which components limit the bandwidth.

Comparison to accurate simulation

The circuit simulation in Figure 14 confirms that the bandwidth (which is the point where the amplitude response has dropped by 3 dB) is indeed approximately 20 MHz. Therefore this approximation is useful in the design phase of a project to give intuition. The approximations can be used to understand specifically which parts of the circuit are limiting the bandwidth.

High frequency models of transistors

At high frequencies, parasitic capacitances between terminals of a transistor can have a significant impact on the circuit’s behaviour.

Parasitic capacitances in BJTs

Unfortunately, there are multiple ways to define and measure the capacitances between terminals of a transistor. Theoretical models, popular in textbooks, consider capacitances caused by p-n junctions. The parasitic capacitances for a typical integrated BJT geometry are shown in Figure 15.

You may see various notations for these capacitances. There is a capacitance between base and emitter, a capacitance between base and collector, and a capacitance between the collector and a fixed voltage (AC ground) denoted If the transistor has its emitter junction grounded, then appears between collector and emitter.

If you’re using discrete transistors, then additional capacitive effects will be caused by the transistor’s package. For instance, even though Figure 15 shows no direct capacitance between collector and emitter, such a capacitance may be present in a discrete device.

A further complication is that transistor datasheets will not give capacitor values defined using this simple theoretical model. This is because capacitances depend on the bias conditions of the transistor, so datasheets need to define the precise conditions under which the measurement was made.

A common procedure is to place the transistor into a common-base configuration and measure the capacitance looking into the common-base’s input and output. The common-base configuration is useful for this measurement because the most significant capacitances ( and ) are approximately decoupled from each other. Another measurement choice is what to connect to the output when measuring the input and vice-versa. Common choices are to leave the output (or input) open or shorted. This leads to the notation as shown in Figure 16.

Transistor datasheets most often provide measurements of and . We see from Figure 16 that these are related to the theoretical junction capacitances by

In the case when there is no coupling between collector and emitter (), these equations simplify to

and

Parasitic capacitances in MOSFETs

Focussing on discrete MOSFETs, there are three capacitances that can be defined between each terminal (Figure 17).

Integrated MOSFETs will also show capacitances to AC ground due to the presence of the substrate (Figure 18). Note that the source terminal is usually shorted to the bulk, neutralising .

Datasheets for discrete MOSFETs typically provide measurements of the three capacitances shown in Figure 17. Notice that the measurement circuits are based on common-source amplifier topologies, in contrast to the BJT case.

From the diagram, we notice:

Feedback

Feedback is when the output state is measured and “fed back” to the input. In the most generic sense, a block diagram of feedback is shown in Figure 20. An input signal (e.g. a voltage or current) runs through the network to produce the output . Feedback occurs when the output is connected to a network and then subtracted from the input. The fact that there is a subtraction leads to the name of negative feedback. If the feedback signal was added instead, then the result would be called positive feedback. Note that negative feedback is the “good kind” that we generally use.

The block represents a transfer function and is called the “open loop gain” because it is the gain that would be applied if the feedback path were not connected. and should be understood as referring to Laplace-domain transfer functions, so that the impact of passing a signal through a network is to simply multiply it by the transfer function.

It is worth remarking upon an important feature of well-designed negative feedback systems. The difference between the input and the feedback is called the error signal (Figure 20b). The error signal should be approximately zero. Note that it cannot be exactly zero because otherwise . Therefore it is the small (non-zero) value of the error signal that drives the output. This implies that must be a powerful amplifier in order to turn the very small error signal into a useful output.

Closed loop gain

To determine the transfer function of the closed loop network, we follow the signal path to write as

Equation is so important that each part has been given its own name:

The loop gain is positive for negative feedback (the normal situation) and negative otherwise.

There is an important simplification of Eq. in the common case that ,

If we want our overall closed loop amplifier to have a gain greater than one, then this implies that .

The block diagram is a rather generic description. An example will help connect these concepts to a familiar setting in electronics.

Example 5

Analyse the op-amp circuit shown in Figure 21 using the concept of feedback.

Solution

The op-amp provides the subtraction operation and an open-loop gain . If this is not obvious, recall that an op-amp produces an output given by . Therefore we can see that the subtraction operation and open-loop gain are internal to the op-amp.

The voltage divider provides the feedback network, with . Therefore the transfer function of this circuit is

To cast this in the familiar form for a non-inverting amplifier, we realise that an ideal op-amp has . Therefore the second term in the denominator is much greater than 1, and we find

This of course is the well-known expression for a non-inverting amplifier.

We see that the feedback perspective allows for the non-ideality in the op-amp’s open loop gain to be easily incorporated in the analysis. As an example, the Texas Instruments OPA656 op-amp has a typical open loop gain of . We could have used this value directly to account for the realistic behaviour of the op-amp.

Gain desensitisation

In many cases, the open-loop gain may be poorly controlled. For example, the MCP6001 op-amp’s datasheet says that its DC open loop gain is a minimum of and is typically . The datasheet does not list a maximum value for the open loop gain. These are impressively high voltage gains but are subject to large variations. If it were not for feedback circuits, such a “wild, untamed” amplifier would be basically useless. Variability with frequency is also a concern, as illustrated by the pictures from the original feedback patent reproduced in the AoE book (p. 118).

We saw above that when the loop gain satisfies then the closed-loop transfer function becomes

This result is independent of the “untamed” open loop gain provided that the condition is met. Therefore we can begin with a high but poorly controlled gain, and use negative feedback to create an amplifier with a lower but more precisely controlled gain. Of course this requires that be controllable, for example, may be formed by the combination of multiple identical resistors such that temperature variations affect them all equally and the voltage divider ratio is unaffected (Figure 22).

The gain-bandwidth product

Let’s consider a open-loop amplifier with transfer function

where is the DC gain and is the -3 dB bandwidth of the amplifier. If the amplifier has additional poles, then Eq. can be considered an approximation where only the dominant pole (the one at the lowest frequency) is included.

If this amplifier is connected in negative feedback with loop gain then the closed loop transfer function is

To read off the gain and bandwidth, we must rearrange this into the standard form for one pole transfer functions. Divide the numerator and denominator by to obtain

Studying this transfer function, we notice the following results.

Therefore, the gain has reduced by a factor of (which can be remembered as “one plus the loop gain”). Meanwhile the bandwidth has risen by the same factor. The gain and bandwidth display a constant product given by . This quantity is called the gain-bandwidth product of the amplifier; it is a property of the amplifier itself rather than the feedback network into which it is placed.

We see that feedback allows us to trade off gain and bandwidth against each other, subject to the constraint that their product remains constant. For off-the-shelf amplifiers, gain-bandwidth products are usually shown on datasheets. For instance, the gain-bandwidth product of the aforementioned MCP6001 amplifier is 1 MHz, while the OPA656 (which is specifically marketed as a “wideband” amplifier) has a gain-bandwidth product of 230 MHz.

Linearity improvement

Feedback also helps lessen the effect of non-linearities in the amplifier’s response. For example, transistor amplifiers have non-linear responses, e.g. for a MOSFET with the quadratic characteristic we see that the response to changes in is non-linear. The small signal model is an approximation, akin to the linear tangent to the curve at the bias point.

The simplest way to see how linearity will improve is by looking at the block diagram of negative feedback (Figure 20). Observe that the feedback system will drive the error signal to nearly zero. Therefore, despite a non-linear response inside , the effect overall is that the system will “self-correct” to keep the error signal small.

Types of feedback

At the input

The feedback connection at the input can be arranged in such a way as to subtract either a voltage or a current.

To add/subtract a voltage, the feedback network must be in series with the input, as shown in Figure 23. This type of connection is called “voltage feedback” or “series feedback”.

A simple way to recognise voltage feedback (series feedback) is to imagine an ant walking along the wire from the source. If you can walk all the way to the amplifier without touching the feedback network, then you have series feedback.

On the other hand, to add/subtract a current, the feedback network must be in parallel with the input. Current flowing in or out of the forward network can be partially diverted through the feedback network . Such a topology is also called shunt feedback.

You can recognise current feedback (shunt feedback) if an ant walking from the source could touch the feedback network without reaching the amplifier first.

At the output

The sense connections can also be designed to sense a voltage or a current.

To sense a voltage, you place a voltmeter across the output. In other words, to sense a voltage, place the sense part of the feedback network in parallel with the output.

To sense a current, you break the circuit and place an ammeter in series with the output. This means that the sense portion of the feedback network will be in series with the output.

Types of feedback

Given that the input side can add/subtract a voltage or current, and the output side can sense a voltage or current, there are four possible combinations of feedback. These are summarised in the table below.

Self-Check Quiz 1

What type of feedback is shown in Figure 25?

(a)

Voltage-controlled voltage feedback (series-shunt).

(b)

Current-controlled voltage feedback (series-series).

(c)

Voltage-controlled current feedback (shunt-shunt).

(d)

Current-controlled current feedback (shunt-series).

Check Start again

✔️ Correct!

Self-Check Quiz 2

What type of feedback is shown in Figure 26?

(a)

Voltage-controlled voltage feedback (series-shunt).

(b)

Current-controlled voltage feedback (series-series).

(c)

Voltage-controlled current feedback (shunt-shunt).

(d)

Current-controlled current feedback (shunt-series).

Check Start again

✔️ Correct!

Self-Check Quiz 3

What type of feedback is shown in Figure 27?

(a)

Voltage-controlled voltage feedback (series-shunt).

(b)

Current-controlled voltage feedback (series-series).

(c)

Voltage-controlled current feedback (shunt-shunt).

(d)

Current-controlled current feedback (shunt-series).

Check Start again

✔️ Correct!

Self-Check Quiz 4

What type of feedback is shown in Figure 28?

(a)

Voltage-controlled voltage feedback (series-shunt).

(b)

Current-controlled voltage feedback (series-series).

(c)

Voltage-controlled current feedback (shunt-shunt).

(d)

Current-controlled current feedback (shunt-series).

Check Start again

✔️ Correct!

Improvements in input/output impedances

Feedback acts to improve the input and output impedances of an amplifier, as shown in the table below. Recall from Week 1 that the desired input/output impedances depend on whether the signal is a voltage or current.

You can easily remember these results by remembering that the impedance is improved in all cases, meaning it goes up or down depending on whether the input or output signals are voltages or currents.

Detailed proofs of each case are given in Razavi Chapter 12, however, a simple intuitive argument is as follows. Consider firstly the input impedances. In the absence of feedback, the entire input signal appears at the amplifier. However, due to feedback, the amplifier instead sees the error signal, which is smaller. Therefore the input impedance of the amplifier produces less of an impact because it only experiences a smaller signal.

In terms of output impedances, the improvement can be intuitively predicted as follows. The loading of the next circuit affects the output of the amplifier, but this error is partially corrected by the feedback mechanism. Hence, the circuit is less sensitive to loading by the downstream circuit, and therefore behaves as if it had a better output impedance.

Types of amplifiers

Given that inputs and outputs can be voltages or current, there are four possible types of amplifier, as shown in the table below.

Impedance is the quantity in Ohm’s law that has units of , whereas “transimpedance” is the gain of an amplifier that reads in current and produces a voltage. Therefore the gain of the amplifier has units of impedance, i.e. . The gain means how many volts of output are produced per amp of input current. The “trans-” prefix indicates that the voltage-current relationship applies “to the other side” of the amplifier.

“Transconductance” should be familiar from the small signal model of transistors, where it relates input voltage to output current. It is called “-conductance” because the gain has units of .

Calculating the loop gain

We see that the loop gain is an important property of feedback systems, so we need a precise procedure for calculating it.

The method is:

1. Set the input signal (voltage or current) to zero.
2. Disconnect a wire to break the feedback loop at any convenient location. It does not matter where in the feedback loop we make the break, so we can choose it based upon ease of analysis.
3. Apply a test signal (a voltage or current) at the point where the loop was broken. Follow this signal around the loop back to the other side of the break.
4. The loop gain is the negative of the how much the signal was amplified as it passed around the loop. Don’t forget the negative sign!

If the negative sign seems confusing, think back to the block diagram in Figure 20. We defined the loop gain to be (omitting the negative sign at the summing junction). Therefore, the loop gain is the negative of the gain around the loop.

An example will illustrate this method.

Example 6

Show that the loop gain of the network in Figure 29 is .

Following the loop gain calculation procedure, we set the input to zero, which in this case means zero volts. Then we break the loop at any convenient point, e.g. before the feedback network, and apply a test signal . Tracing this around the loop, we find

Therefore, the loop gain is the negative of how much the signal was amplifier around the loop:

as expected.

Example 7

Find the loop gain of the network shown in Figure 30. Assume is large and is small. Therefore find the closed-loop transfer function.

This is current-controlled current feedback. To see this, notice on the input side that the feedback network (, etc) is in parallel with the amplifier input (). Therefore the input and feedback signals are currents. On the output side, the feedback network is in series with the output, so the output signal is again a current.

For loop gain analysis, we set (i.e. the current source becomes an open circuit). The injected current faces a current divider formed by and . (Recall the impedance looking “up” into the source of a MOSFET is ) Since we assume and , we have

This AC small signal current must pass entirely through and therefore the AC small signal perturbation value of is

Continuing around the loop, is a voltage-controlled current source, but since it is p-type, the drain current points in the opposite direction to . Therefore there is a negative sign.

Hence we have

For the open-loop gain, consider removing the feedback path (remove ) and then . This will cause a current .

Therefore,

The overall small signal transfer function is

Note that this transfer function is valid for small signal perturbations of because it considers the transistor biasing to be constant.

Op-amp departures from ideal

We will often implement feedback using op-amps. However, op-amps are not ideal devices. The most common non-idealities are listed below.

Input offset voltage

Suppose you short both inputs of an op-amp to ground, as shown in Figure 31 (a). You would expect the op-amp to output zero volts (since the difference between the inputs is zero).

What actually happens in reality is that as soon as you power on the op-amp, the output quickly saturates to either the positive or negative supply rail, as shown in Figure 31 (b). You can’t predict which one. Different op-amps from the same manufacturing batch may saturate to different supply rails. The effect is unpredictable.

This circuit illustrates an underlying defect present in all op-amps called the input offset voltage. The two input terminals are never perfectly symmetric. The op-amp output voltage is better thought of as

where is called the input offset voltage. Typical values for are hundreds of microvolts for good devices or a few millivolts for older op-amp designs.

A helpful circuit model is shown in Figure 31 (c). The input offset voltage can be understood as a voltage source in series with one of the inputs (it doesn’t matter which one, although the polarity shown in the figure is the most common convention).

Input bias current

The input bias current is the current that flows into the op-amp’s inputs. Its main impact from a circuit design perspective is to create a voltage drop across resistors connected to the inputs. It can also present a challenge for precision current sensing applications, e.g. when building a transimpedance amplifier.

Slew rate

At the output of an op-amp, the voltage cannot change instantaneously. The rate of change of the output voltage is limited by the slew rate. The slew rate is typically given in volts per microsecond.

Input and output voltage range

Summary

Here are some of the key ideas from this week:

1. Every node that has a resistance and capacitance to AC ground forms a pole. The pole frequency is . By looking for elements in this way, you can quickly spot which part of a circuit will limit its bandwidth.
2. Capacitances connected between input and output of inverting amplifiers act as if their capacitance is larger, an effect called Miller multiplication.
3. Parasitic capacitances in transistors need to be considered when analysing frequency responses of circuits, especially if the circuit topology is such that Miller multiplication will occur.
4. Feedback is a powerful tool for improving the performance of circuits. It improves linearity, stabilises gain, and improves input and output impedances.
5. In one-pole amplifiers, the gain-bandwidth product is constant. This means that if you increase the gain, the bandwidth will decrease, and vice versa.
6. Precision design with op-amps requires consideration of their non-idealities, which affect both the input and output sides of the op-amp.