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NFB 101

An Introduction To Negative Feedback For Tube Users

by Ian ThompsonBell

Version 0.4 Copyright Ian ThompsonBell 2008

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Introduction

This article was written in response to a request on the rec.audio.tubes newsgroup for a discussion of negative feedback (NFB) as it applies to tube based designs. It aims to cover the general theory of negative feedback and its practical application to tube circuits. Unfortunately any understanding of NFB requires some mathematical ability, a basic understanding of how tubes work and of general electrical theory. If you are not familiar with terms such as plate resistance, transconductance, Thevenin/Norton equivalents and Kirchoff's laws then you need to do some further reading before tackling NFB. The 'Bible' for tube circuit designers is the Radio Designers Handbook (4th Edition) by Radiotronics (RDH4). Hard copies can still be found but there are plenty of electronic copies on the internet. A complete copy can be found here. NFB is covered in Chapter 7 of RDH4. If you have read and understood the previous six chapters you will have all the tools necessary to tackle NFB.

It is unfortunate that a discussion of NFB inevitably involves large quantities of equations. To avoid pages and pages of equations between important results this article has been written in two separate but complementary parts. The first part covers the basics, the properties of feedback and its application to tube circuits. Important mathematical results are simply quoted where their derivation would consume a lot of space. This means the first part can be used as a reference for design work. The second part (the appendices) consists of the full mathematical derivation of all the results quoted in the first section so if you are interested in the details of how the results are obtained you can see the maths in all its gory detail.

Further Reading

There is a very large number of books that cover NFB and quite a few that cover its application to tubes at least in part. Most are still available second hand. In writing this article I have leant heavily on the following books:

1. Feedback Circuit Analysis by S.S. Hakim (1966 published by Iliffe books). The is the NFB bible for tubes. If you are a math head then this is the book for you.

2. Electronics Engineering by C.L. Alley and K.W. Atwood (1966 published by John Wiley & Sons). This was my second year university text book and covers tubes in almost equal proportion to transistors.

3. Operational Amplifiers – Design and Applications by Burr Brown (1971 published by McGrawHill). This is the bible for the semiconductor people but its chapters on NFB and compensation techniques are particularly well written.

4. Electronic Circuits – Discrete and Integrated by D.L. Schilling and C. Belove (1972 published by McGrawHill). Another book that covers both tubes and transistors with good explanations of feedback stability.

5. Feedback Amplifier Principles by Sol Rosenstark (1986 published by MacMillan). This is perhaps the most up to date book on NFB in amplifiers although it does not cover tubes.

6. Integrated Circuits and Semiconductor Devices: Theory and Application by Beboo and Burrous (2nd edition 1977 published by McGrawHill). Very good sections on non ideal amplifiers.


http://www.pmillett.com/

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Fundamentals

Suppose we have an amplifier which has a pair of input terminals across which we apply an input signal voltage Vs and a pair of output terminals across which we measure an output signal voltage Vo . The voltage gain of the amplifier Ao is, by definition:

Ao =VoVs

or Vo = Vs⋅Ao ..........(1)

Now suppose we take a fraction ß of the voltage from the amplifier's output terminals and subtract it from the input voltage, for example by connecting the output of a feedback network in series with the input terminals as in Figure 2:

The input to the amplifier is now Vs−β⋅Vo but we also know that Vo is just this input voltage times Ao (equation1) so:

Vs−β⋅Vo⋅Ao = Vo

Expanding gives: Vo = Vs⋅Ao−ß⋅Vo⋅Ao


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Rearranging gives:

Voß⋅Vo⋅Ao = Vs⋅Ao or Vo 1ß⋅Ao = Vs⋅Ao

So the gain of the new amplifier (An) is:

An =VoVs

Ao

1ß⋅Ao........ 2

This is the basic feedback equation, where:

Ao is the gain before feedback is applied often called open loop gain.

β is the feedback fraction (the proportion of the output fed back).

An is the gain after feedback is applied often called the closed loop gain.

β⋅Ao is called the loop gain often abbreviated to the letter T.

The basic feedback equation is always true.

The change in gain caused by applying feedback is simply:

AnAo

=Ao

1β⋅Ao⋅Ao=

11β⋅Ao

..............(3)

The factor 1β⋅Ao is called the feedback factor F. Expressed in dB it is simply the amount by which the gain has been reduced by applying feedback.

20log An = 20 logAo–20 logF

So if the open loop gain was 60dB and feedback reduced this to 40dB the feedback factor would be 20dB. The feedback factor is commonly referred to simply as the amount of feedback. We will see how important the feedback factor is when we come to examine the effects of NFB on the amplifiers parameters. The feedback factor is commonly designated by the letter F.

The most common simplification of the basic feedback equation is to assume β⋅Ao ≫1 in which case the equation reduces to:

An =Aoβ

∗Ao =1β

The gain is thus determined entirely by the components of the feedback network that comprise the term β . If β=0.1 we get a gain of 10. Notice that if the feedback factor β is independent of


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frequency then the response of the new amplifier must also be independent of frequency i.e. it has a flat response. The important thing is that this is only true when β⋅Ao ≫1 .

In the real tube world we are never quite sure of the exact value of Ao as it varies with tube parameters which vary from tube to tube and over time. We would prefer our amplifier's characteristics to be determined entirely by passive components whose values and tolerance we know and can control. To do this we want to be able to fix β to determine our amplifier's gain so the only way we have to make β⋅Ao ≫1 is to have a large Ao . In the world of op amps this is not a big problem ( Ao of 120dB at dc is not uncommon). In the tube world such gains are unheard of. Since the properties of NFB depend largely on the feedback factor F=1β⋅Ao and hence on Ao we will often need to use the basic feedback equation rather than the simplified one.

In the tube world, small values of feedback (less than 20dB) are common. So when people talk about 20dB of feedback in tube design that means F=1β⋅Ao is just 10 or β⋅Ao is 9 and thus definitely not >> 1.

That will do for starters. Next we need to deal with how feedback is derived and applied and how that affects the properties of NFB. After that we might just get onto the effect of the frequency responses of the various elements of the system and hence to stability.

Derivation and Application of NFB

There are two principal ways we can derive the feedback from the output of our amplifier and two similar ways we can apply it to the input.

If we simply connect the feedback network across the output of the amplifier (as in Figure 2 above) we shunt the output with the feedback network and derive the feedback signal from the output voltage of the amplifier. This type of feedback is therefore called shunt or voltage derived.

If we connect a resistor in series with the load and connect our feedback network across it, the feedback is derived from the output current. This type of feedback is therefore known as series or current derived.

Similarly if we apply the output of the feedback network in parallel with the input of the amplifier this is known as shunt feedback. In this case, the input and feedback voltages cannot add as they are connected in parallel. Instead the feedback input currents add, so shunt applied feedback is also called current applied feedback. If we apply feedback in series with the input (to an unbypassed cathode resistor for example) it is known as series applied feedback. In this case the input and feedback voltages add so series applied feedback is also called voltage applied feedback. (The feedback in Figure 2 is series applied, shunt derived).

How the feedback is derived and applied determine key properties of the amplifier with feedback. The maths behind these effects are well documented so I do not intend to repeat all of them here. If you really want the detailed maths check the reading list detailed earlier. At this point we will just summarise the results and look specific tube examples.


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Input and Output Resistance

All practical amplifiers have an input and output resistance. Both are affected by the application of negative feedback.

The method by which feedback is derived affects the output resistance and the way it is applied affects the input resistance.

Shunt derived feedback lowers output resistance and series derived feedback raises output resistance. Shunt applied feedback lowers input resistance and series applied feedback raises input resistance.

Put another way, shunt feedback lowers resistances and series feedback raises them. Figure 1 has shunt derived feedback so we expect the output resistance to be lowered. It has series applied feedback so we expect the input impedance to be raised. The amount these resistances are altered depends on the feedback factor.

Shunt derived feedback lowers output resistance by the feed back factor F=1β⋅Ao . So if an amplifier has 20dB of feedback its output resistance will be lowered 10 times by shunt derived feedback. Series derived feedback raises output resistance by the same factor.

Similarly, series applied feedback raises input resistance by the feedback factor and shunt applied feedback lowers it by the same factor.

Figure 3 shows a real amplifier with input resistance ri and output resistance ro.

If we apply a voltage Vs to the input, a current Ii flows into the input of the amplifier. The input resistance is defined as:

Ri=VsIi

Appendix 1. shows that this input resistance is given by:


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Ro=ri1β⋅Ao =ri⋅F ,that is the input resistance is raised by the feedback factor F

Similarly, if we connect the input to ground and apply a voltage Vo to the output, a current Io will flow in the output. The output resistance is defined as:

Ro=VoIo

Appendix 1. shows that this output resistance is given by:

Ro= ro1β⋅Ao

=roF

so the output resistance has been reduced by the feedback factor F

NFB also changes several other properties of an amplifier. Its bandwidth is (generally) increased, its distortion reduced and its sensitivity to changes in component values and interfering signals (noise) from power supplies for example is reduced. The one thing NFB cannot do is improve signal to noise ratio.

The table below summarises these basic properties of NFB.

Property Shunt NFB Series NFB

Derived from: OP Voltage OP Current

Applied as: IP Current IP Voltage

OPZ Reduced Reduced

IPZ Increased Increased

Bandwidth Increased Increased

Sensitivity Reduced Reduced

Let's have a look at a couple of tube circuit examples to see how this is applied. First the cathode follower.


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The Cathode Follower

Figure 4a shows a typical cathode follower circuit.

In this example the bias resistor is bypassed to simplify the maths. Figure 4b shows the ac small signal equivalent circuit of Figure 4a. The tube is represented by the voltage generator with an output equal to μ⋅Egk where μ is the mu of the tube and Egk is the voltage between the grid and cathode. For ac, the tube anode resistance ra is connected to ground via the PSU.

We can see that the whole of the output voltage is in series with the input so we have 100% feedback and β=1 . This is shunt derived, series applied NFB.

Appendix 2. shows that the closed loop gain An of the cathode follower is:

An =μ

μ1 raRk

................(9)

Obviously An 1 for all positive values of Ra and Rk .


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Cathode Follower Input Resistance

When we apply a voltage Vi to the input of the cathode follower a current Ii flows into its input. The input resistance is defined simply as:

Ri=ViIi

Appendix 3. shows that the input resistance is:

Ri=Rg1β⋅Ao =Rg⋅F

So, as expected, the input resistance of the amplifier with series applied feedback is the input resistance of the original amplifier multiplied by the feedback factor.

Cathode Follower Output Resistance

The output resistance of the tube before feedback is just ra and in our example the load is Rkso we expect the output resistance to be just:

Ro= ra1β⋅Ao

=raF

Appendix 4. shows that if Rk≫ra this is simply:

Ro =ra1μ

≈1gm


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Common Cathode Unbypassed

Next we will look at a common cathode stage with an unbypassed cathode resistor. Although most common cathode stages use a bypassed cathode resistor (with good reason) we will look at the unbypassed case because the cathode of such a stage is a very common point at which to apply global feedback. Figures 5a shows a typical common cathode stage with unbypassed cathode resistor and 5b shows its small signal equivalent circuit.

Here, the current that flows through Rl also flows through Rk , so our feedback is series (current) derived (because Rk is in series with Rl ). If i is the common signal current flowing in Rl and Rk then the feedback voltage we create from this common current is simply:

Vf = i⋅Rk and the output voltage is simply:

Vo = i⋅Rl so our feed back fraction β is:

β =VfVo

=i⋅Rki⋅Rl

=RkRl

Notice that in the circuit shown β =1

100 and since the open loop gain can never be greater

than μ which for the 12AX7 is 100 then then β⋅Ao 1

100⋅100 1 so we can say that the

common approximation used when β⋅Ao≫1 is emphatically not true.

Appendix 5. shows that the closed loop (stage) gain is:


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An= µ⋅Rlµ1RkraRl

Notice that if Rk was bypassed then for ac signals the µ1Rk term disappears and the stage gain becomes:

An =µ⋅RlraRl

which is the standard expression for a bypassed common cathode stage which

is also the open loop gain of the unbypassed stage Ao .

By not bypassing Rk we have effectively increased Ra by µ1Rk

As an aside, notice that even when bypassed, at dc the unbypassed equation still applies. This gives us an easy way to determine Rk if we know Ra and µ at the operating point. Suppose we pick an operating current Ia and select a plate resistor such that when Ia flows it drops the supply to the required plate voltage Vp. Then we know that:

VpIa

= µ1Rkra and hence:

Rk =

VpIa

−ra

µ1


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NFB and Stability

So far we have studiously avoided any mention of frequency response. In practice Ao is frequency dependant and sometimes β is deliberated made so. So far we have assumed the feedback is always perfectly negative i.e. the applied feedback is exactly 180 degrees out of phase with the input signal. However, if Ao and/or β are frequency dependant they will also introduce a phase shift. If the total phase shift approaches 180 degrees the feedback is now applied in phase with the input signal and we have positive feedback rather than negative. So if the loop gain β⋅Ao ever becomes 1 then then the feedback factor F=1β⋅Ao becomes zero so the closed loop gain becomes infinite and we have created an oscillator.

Unfortunately, as we shall see later, if you have a multistage amplifier, it is unavoidable for it to have 180 degrees of phase shift at one or more frequencies. To avoid feedback around this amplifier producing an oscillator we have to ensure that when the phase shift becomes 180 degrees the loop gain T=β⋅Ao significantly less than 1. How much below 1 is called the gain margin of the amplifier and is usually expressed in dB, so an amplifier that has a loop gain of 13dB when the phase shift is 180 degrees is said to have a gain margin of 13dB. Similarly we also need to ensure that when the loop gain β⋅Ao does equal one, that the phase shift is as far away from 180 degrees as possible. How far away from 180 degrees is called the phase margin of the amplifier. So if an amplifier has a loop phase shift of 150 degrees when the loop gain is 1, it has a phase margin of 30 degrees.

Clearly the frequency and phase response of the loop gain β⋅Ao is crucial to ensuring the stability of an amplifier employing NFB so we need to examine this in detail. From now on we will refer to the loop gain β⋅Ao by its abbreviation T.

All amplifier circuits introduce phase shift so the design of negative feedback needs to take this into account. There are generally three frequency ranges we need to consider, the pass band, frequencies below the pass band and frequencies above the pass band. The pass band is simply the range of frequencies we want to amplify, say 20Hz to 20KHz in audio circuits. As a rule, we will design our amplifier so that its open loop gain in the pass band produces relatively little phase shift and with tubes this is relatively easy to achieve except perhaps in the case of power amplifiers employing transformers. Below the pass band, any coupling capacitors we have in the circuit will cause the gain to drop and phase to change as the frequency lowers. Above the pass band, stray capacitance in parallel with the tube electrodes and other parts of the circuit will cause the gain to drop and the phase to change as the frequency increases. So it is the areas outside the pass band that deserve our principal attention. As the frequencies above and below the pass band are far apart their responses have no effect on each other. We can therefore consider each separately and then use then principle of superposition to combine them. First we need to understand the basics of how circuit elements affect gain and phase.


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Poles, Zeros and Laplace Transforms and Frequency Response

We are not going to use Laplace transforms for their intended purpose but merely as a mathematical

shorthand. We are all familiar with the impedance of a capacitor C being 1

j⋅ω⋅C and the

impedance of an inductor L being j⋅ω⋅L where j is the square root of 1 and ω is the frequency in radians/second (and ω=2⋅π⋅f where f is the frequency). For our purposes we will simply use the Laplace transform s to replace the j⋅ω terms in the above definitions. So the impedance of a

capacitor becomes just 1

s⋅C and that of an inductor becomes s⋅L .

The beauty of Laplace transforms is that they describe both the steady state and the transient behaviour of a circuit in a single equation. To obtain the steady state response of a circuit you just substitute j⋅ω for s in the Laplace transform of the circuit. To obtain its transient response you just multiply the the Laplace transform of the circuit by the Laplace transform of the input signal. Although, as we shall see later, we are very interested in the transient response of circuits employing NFB, we will not be using Laplace transforms because they are far too complex for an introductory article. However, they are a very useful shorthand so just remember s is short for j⋅ω .

When looking at the equations for networks consisting of resistors, capacitors and inductors we will generally try to arrange our equation to be in the following form:

sa⋅sb sc⋅sd

On the top line (numerator), if s = a then the term sa is zero and the whole function is zero no matter what the values of the other terms. For this reason, any term in the numerator of the formsa is called a zero. Similarly, on the bottom line (denominator), if s = c then the term sc is zero and the function becomes infinite no matter what the value of the other terms. Any

term in the denominator of the form sc is called a pole.

Now let's apply this to some simple R C networks. Consider a simple low pass R C network consisting of a series resistor and a parallel capacitor as shown in the figure below:


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This shows a simple low pass RC network and the frequency and phase response it produces. The ratio of the output voltage Vo to the input voltage Vi is:

VoVi

=

1s⋅C1

R11

s⋅C1

=1

R1⋅C1⋅s1

To get this in the required form we divide top and bottom by R1⋅C1 and get:

VoVi

=

1R1⋅C1

s1

R1⋅C1

R1⋅C1 is the time constant of the RC network so 1

R1⋅C1 represents a frequency that we will

call ω1. Substituting ω1 for 1

R1⋅C1 gives:

VoVin

=1ω

s 1ω which represents a pole at a frequency of ω1.


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Notice that as s tends to zero the output tends to 1ω1ω=1

and as s tends to infinity the output tends to zero.

From the frequency response graph is is clear that:

● At frequencies well below ω1 there is little phase shift or signal loss.

● At ω1, the response drops by 3dB and the phase shift is 45 degrees. From the graph you can see that this occurs at about 16KHz, as expected.

● From ω1 upwards, the response falls at a rate of about 6dB/octave or 20dB per decade and the phase shifts tends to 90 degrees.

In broad terms, the effect of a pole is to drop the output by 6dB/octave above ω1 and to introduce a 90 degree phase shift (lag).

Next we examine a high pass network consisting of a series capacitor and a parallel resistor as shown in the figure below:

The ratio of the output voltage Vo to the input voltage Vi is:


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VoVi

=R1

R11

s⋅C1

=R1⋅C1⋅s

R1⋅C1⋅s1=

s

s1

R1⋅C1

If again we substitute ω1 for 1

R1⋅C1 we get:

VoVi

=s

s 1ω =

s0s 1ω

We can immediately see that once again we have a pole at ω1, but what is going on with this lone s term on the top line which we have written as s + 0. What this means is that the function is zero when s is zero. In other words we have a zero at zero frequency.

We can see that as s tends to zero the output tends to zero.

As s tends to infinity the output tends to ss=1 .

Looking at the frequency response we can see the effect of this zero is for the output to rise at 6dB/octave from zero Hz. When we approach ω1, the pole begins to act. Remember this drops the response by 6dB/octave, the exact opposite of what the zero is trying to do so the overall effect is that above ω1 the response flattens out.

At zero Hz, the zero adds a +90 degree phase shift(lead). The pole wants to add a 90 degree phase shift. The result is that as we begin to approach ω1the phase starts to shift and at ω1 the pole has managed to change the phase shift to 45 degrees. At frequencies well above ω1 the phase shift tends to zero.

We can carry out the same exercise for networks consisting of resistors and inductors and again we find either single poles or a pole zero combination as we did for the simple RC networks. As a rule a pole can exist on its own but every zero has an associated pole. For this reason there are always more poles than zeros in a circuit.

The above RC networks are important because they are representative of the most common frequency dependant parts of tube circuits, namely RC coupling and interelectrode capacitance. The figure below is a simple equivalent circuit of the coupling between two tube stages:


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RL represents the parallel combination of the plate resistance ra and the anode load resistor, C1 is the coupling capacitor and Rg is the grid resistor of the following stage. So:

VgVa

=Rg

RLRg1

S⋅C1

=Rg⋅C1⋅s

raRg⋅C1⋅S1=

RgraRg

⋅s

s1

raRg⋅C1

If as before we let the time constant 1

raRg⋅C1= 1ω we get:

VgVa

=Rg

raRg⋅

s s 1ω

So we can see that this is the very similar to one of the RC networks discussed earlier:

● It has a zero at zero Hz

● It has a pole at ω1 so below that frequency the output falls at 6dB/octave

● As s tends to infinity the output tends to Rg

raRg

● The time constant is simply the sum of the two resistors multiplied by the capacitor

This circuit affects only the low frequency response of the circuit i.e. below the pass band. Next we


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will examine the high frequency response. The figure below is a simple equivalent circuit of the high frequency coupling between two tube stages.

Once again RL represents the parallel combination of the plate resistance ra and the anode load resistor. Cak represents the anode to cathode capacitance plus any stray capacitance between the anode and ground. Appendix 6 shows that:

VgVa

=K⋅ 1ωs 1ω

where:

K =Rg

RgRL and

1ω =RL⋅Rg

RLRg⋅Cak

We note that:

● There is a pole at 1 so above 1 the response falls at 6dB/octaveω ω

● As s tends to zero, the response tends to K

● As s tends to infinity the response tends to zero

So this is just like the simple RC network with a time constant determined by the parallel combination of Rg and Rl with Cak and at low frequencies the response is defined by the pot


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divider formed by Rg and Rl rather than being unity.

Now we now what poles and zeroes are and the effect the have on amplitude and phase, we are in a position to consider the stability of circuits incorporating NFB.

Stability

We have seen that RC coupling between tube stages introduces a low frequency pole and a zero and at most a 90 degree phase shift. We have also seen that the stray capacitances from the anode introduce a high frequency pole and at most a 90 degree phase shift. So it is clear that a single tube stage is with NFB is unconditionally stable because the phase shift is at most 90 degrees. However, as soon as we connect two of them together we have the possibility of 180 degrees of phase shift and some serious instability problems. Let's have a look at a simple teo stage RC coupled circuit.

The above figure shows a two stage RC couple amplifier using a single 12AX7. The cathode resistor of the first stage is unbypassed because later we are going to use if to apply NFB. Both halves are run at an plate current of about 1mA and a plate voltage of about 170V. At this operating point both triodes have an ra of about 60K. The specified plate to cathode capacitance is about 0.5pF , the grid to cathode capacitance is 1.6pF and the plate to grid capacitance is 1.7pF.

Since the first triode has an unbypassed cathode this increases its effective ra to about 180K so at high frequencies would would expect to see a pole at about 1.2MHz due to the 2.1pF of plate to cathode and grid to cathode capacitance. Similarly the second stage should have a pole at about 2.0MHz.

The RC coupling between the two stages will produce a zero a 0Hz and a pole at about 6Hz and the output coupling circuit should also produce a zero at 0Hz and a pole at about 5Hz.


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So overall we expect the response to rise at 12dB/octave from 0Hz initially with a phase shift of 180 degrees then flattening out around 5Hz with a 5dB drop and 90 degrees phase shift. The mid band should be flat with zero phase shift then we expect to see a 3dB drop and 45 degree phase shift at 5MHz, the response dropping at 6dB/octave thereafter and phase increasing to until we reach 8MHz when this is repeated ending up with a 12dB octave drop and a final phase shift approaching 180 degrees.

The figure below shows what we get in practice.

The low frequency response looks pretty much as expected. The phase shift is 90 degrees at about 6Hz where the gain has dropped by just under 6dB. The ultimate slope is 12dB/octave.

The high frequency end looks completely wrong. The first 45 degree phase shift occurs at about 29KHz rather than about 1.2MHz as expected. We have just rediscovered the Miller Effect which is due to shunt derived, shunt applied NFB caused by the plate to grid capacitance around the second triode. Remember that shunt applied NFB reduces input impedance and it can be shown that it

reduces it by approximatelyRfAo

where Rf is the feedback resistance and Ao is the open loop

gain. In our case Rf is in fact a pure capacitance, the 1.7pF of plate to grid capacitance. Since the impedance of this 1.7pF capacitor is reduced by Ao times it is exactly the same as placing an extra capacitor Ao times as big from grid to ground. The second triode has a anode resistor of 120K and a load resistor of 120K so its total load is 60K. Since its ra is also 60K and μ is 100 its open loop gain is:

Ao =µ⋅RlraRl

=100⋅60K60K60K

=1002

= 50


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The Miller effect therefore adds 50 x 1.7pF = 85pF across the grid of the second stage. If we recalculate the high frequency pole for the output of the first triode using this 85pF plus the original 2.1pF we get a pole at 29KHz. So the first thing we learn is that, with triodes at least, the high frequency open loop response is dominated by poles produced by the Miller effect. In practice the other poles are so far up the frequency spectrum we safely ignore them. A pole, such as a Miller pole, which dominates part of the open loop response is called a dominant pole.

Now we understand why the open loop response looks the way it does, what does it means for stability in a closed loop system? First let's assume we are going to aim for a phase margin of 45 degrees (we'll discuss appropriate values for gain and phase margin later). This means the maximum phase shift we can allow is 180 45 = 135 degrees. Recall that phase margin is how far away from 180 degrees phase shift we are when the loop gain T=β⋅Ao=1. Expressed in dBs this is:

β dB + OpenLoopGain(dB) = 0dB.

Next we note that the mid band open loop gain is about 64dB.

If we look at the high frequency response we see that when the phase shift is 135 degrees, the gain has dropped to just under 10dB and this happens at about 16MHz. This means if we set β to 10dB, the loop gain T would be 0dB so it meets the phase margin requirement. At any frequency above this T will be less than 1 so the phase shift does not matter. At any frequency below this down to the mid band, the phase shift is less so we always do better than the specified phase margin. So as long as we set beta to 10dB or less we will meet the phase margin requirement. The figure below shows this in a graphical format on a typical frequency response plot.


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This just shows the high frequency part of the plot. On the right hand yaxis, the mid band open loop gain is labelled Ao. The dotted line labelled 135° shows where the selected maximum allowed phase shift intersects the falling gain plot and the horizontal 0dB line shows where the gain falls to zero.

The two vertical arrowed lines show, at the bottom, the minimum closed loop gain necessary to meet the phase margin. The closed loop gain can be anything greater than this up to the open loop gain. The top arrowed line shows the maximum allowed feedback F. It can be any value less than this and the phase margin will be met. This diagram makes it very simple to see the range of closed loop gains over which the amplifier will be stable when NFB is applied. In our example two stage triode circuit, at high frequencies we can see it will meet the 45 degree phase margin for closed loop gains from 10dB to 64dB.

So in this case, at high frequencieas, the Miller effect actually helps use achive stability. Its dominant pole means the open loop gain has been considerably reduced, and the phase shift limited to 90 degrees before the othe poles start to act and increase the phase shift towards the unstable 180 degrees. With the Miller pole at about 30KHz and the next at 1.2MHz there are five octaves separatingng the two poles so the response will have dropped by 30dB before the highr pole begins to act.

By contrast, if we look at the low frequency response we see that when the phase shift is 135 degrees the open loop gain is a whopping 50dB. Applying the same criteria as for the high frequencies we can say this is meets the phase margin only for closed loop gains from 50dB to 64dB, in other words we can apply no more than 14dB of NFB before the phase margin is not met at low frequencies. The reason for this is that we have two dominant poles at pretty much the same frequency (~5Hz) so the phase change is rapid (it is already 90 degrees at the common pole frequency). To improve matters we need to separate these poles so we have a single dominant pole as in the high frequency case. We can either move one down significantly in frequency or move one up significantly or move one up and one down. If we lower the pole of the interstage coupling by a decade (10 times) by increasing the coupling capacitor from 0.05uF to 0.5uF we find that for 135 degrees of phase shift the open loop gain has now dropped to 39dB, an improvement of 11dB. If we lower it by another decade by increasing this capacitor to 5uF we find the closed loop gain drops to just over 18dB, a further 21dB improvement. We can now have as much as 46dB of feedback and still meet our phase margin.

This method clearly works but a 5uF interstage coupling capacitor is hardly practical. If we needed to make similar changes at high frequencies the component values would be quite small and readily obtained but solving stability problems at low frequencies seems to require large component values. This is a serious problem for RC coupled tube amplifiers with NFB. Somehow we do need to separate the poles by a couple of decades for the low frequency stability to be close to that of the high frequency. The most practical method is to move both poles. The common pole of the original circuit is at 6Hz so we can probably afford to move this up to 20 Hz by reducing the output coupling capacitor to 0.05uF without seriously compromising the amount of feedback available at the low end of the pass band. If we change the interstage capacitor to 0.5uF and increase the second stage grid resistor to 1M this lowers the other pole to about 0.4Hz. The two poles are now over five octaves apart. The result of these changes is shown on the open loop response below.


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Now the open loop gin has dropped to 23dB when the phase reaches 135 degrees. So we can achieve stability for closed loop gains above this values i.e. we can have as much as 41dB of feedback.

So far, so good, but remember we are at present only considering a two stage amplifier with two sets of RC coupling. A three stage amplifier with three RC coupling networks would just make the problem worse and we would need to move two low frequency poles downward to ensure stability. Transistor circuits, and op amps in particular, do not have this low frequency problem simply because they are dc coupled instead of RC coupled. Clearly dc coupling of tubes would be advantageous.

Lastly in this section we should mention transformers because they will be important when we come to consider power amplifiers. Clearly a transformer has zero gain at zero Hz so we expect them to have a zero at zero Hz and a pole at some low frequency just like an RC coupling network. We therefore expect them to have the same issues at low frequencies as do RC coupling networks.


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Appendices

1. Shunt derived, series applied feedback, effect on input and output resistance.

Figure 3 shows a real amplifier with input resistance ri and output resistance ro.

If we apply a voltage Vs to the input, a current Ii flows into the input of the amplifier. The input resistance is

defined as:

Ri=VsIi

The voltage appearing across the amplifier input terminals is just Ii⋅ri so the output voltage is :

Vo=Ao⋅Ii⋅ri

But, we know that because of the feedback network that:

Vs=Ii⋅riβ⋅Vo

Substituting for Vo gives:

Vs = Ii⋅ri β⋅Ao⋅Ii⋅ri = Ii⋅ri1β⋅Ao

Hence Ro=VsIi

=ri1β⋅Ao ,that is the input resistance is raised by the feedback factor

1β⋅Ao

Similarly, if we connect the input to ground and apply a voltage Vo to the output, a current Io will flow in the output. The output resistance is defined as:


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Ro=VoIo

Because of the feedback network and Vs being zero, the voltage at the amplifier input will be −Vo⋅β and so the voltage on the left hand side of the amplifier output resistance ro will be:

– β⋅Ao⋅Vo

Since the voltage on the right hand side of ro is Vo the total voltage across ro is:

Vo – −Ao⋅Vo=Vo 1β⋅Ao

So the current Io through ro is:

Io=Vo 1β⋅Aoro

and thus the output impedance Ro is:

Ro =VoIo

=ro

1β⋅Ao

So the output resistance has been reduced by the feedback factor 1β⋅Ao


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2. Cathode Follower Closed Loop Gain

Figure 4a shows a typical cathode follower circuit.

In this example the bias resistor is bypassed to simplify the maths. Figure 4b shows the ac small signal equivalent circuit of Figure 4a. The tube is represented by the voltage generator with an output equal to μ⋅Egk where μ is the mu of the tube and Egk is the voltage between the grid and cathode. For ac, the tube anode resistance ra is connected to ground via the PSU.

The current through Rk and ra is the same and equal to:

i =μ⋅EgkraRk

............(4)

So the output voltage across Rk is just i⋅Rk or:

Vo =μ⋅Egk⋅RkraRk

...........(5)

Since Egk is the input to the amplifier the open loop gain is just:


27

VoEgk

=μ⋅Rk

raRk ...............(6)

However, the output voltage Vo is in series with the input (series applied) so:

Vi = VoEgk and hence Egk = Vi−Vo ........(7)

Substituting this in equation (5) gives:

Vo =μ⋅Vi−Vo⋅Rk

raRk ...............(8)

Expanding gives:

Vo⋅raRkRk

= μ⋅Vi−μ⋅Vo and rearranging gives:

Vo⋅raRkRk⋅μ

= Vi−Vo and

Vo⋅raRkμ⋅Rkμ⋅Rk

= Vi

So the closed loop gain, An is:

An =VoVi

=μ⋅Rk

raμ1⋅Rk which simplifies to:

An =μ

μ1 raRk

................(9)

That is the classic way of finding the closed loop gain of a cathode follower. We can get there a bit quicker by applying what we know of NFB theory. The open loop gain we derivedin equation (6) as:

Ao =μ⋅RkraRk

Since the output voltage is directly in series with the input we have 100% NFB so ß = 1. We can now write the closed loop gain as:

An =Ao

1β⋅Ao and substituting for Ao and also setting β=1 we get:


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An =

μ⋅RkraRk

1μ⋅Rk

raRk

Multiplying top and bottom by raRk gives:

An =μ⋅Rk

raRkμ∗Rk=

μ⋅Rkμ1Rkra

Dividing both by Rk gives:

An =μ

μ1 raRk

Obviously An 1 for all positive values of Ra and Rk .


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3. Cathode Follower Input Impedance

When we apply a voltage Vi to the input of the cathode follower a current Ii flows into its input. The input resistance is defined simply as:

Ri=ViIi

Ignoring grid current, the input current flows through Rg which, for ac is connected between the grid and cathode, so:

Egk=Ii⋅Rg and hence the output voltage Vo = Egk⋅Ao = Ii⋅Rg⋅Ao

But, we know that Vi = Ii⋅Rgβ⋅Vo = Ii⋅Rgβ⋅Ii⋅Rg⋅Ao so:

Vi = Ii⋅ri1β⋅Ao and hence:

Ri=ViIi=Rg1β⋅Ao

So, as expected, the input resistance of the amplifier with series applied feedback is the input resistance of the original amplifier multiplied by the feedback factor.


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4. Cathode Follower Output Resistance

The output resistance of the tube before feedback is just ra and in our example the load is Rkso we expect the output resistance to be just:

Ro= ra1β⋅Ao

Substituting β=1 and Ao= μ⋅RkraRk

gives:

Ro =ra

1 μ⋅RkraRk

if Rk≫ra then this reduces to Ro =ra1μ

≈1gm


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5. Common Cathode Unbypassed

Figures 5a shows a typical common cathode stage with an unbypassed cathode resistor and 5b shows its small signal equivalent circuit.

A classic analysis of this stage is relatively straight forward. Rk , ra and Rl all pass the same current I which is equal to:

I =µ⋅Egk

RkraRl

So the output voltage is just Vo = I⋅Rl =µ⋅Egk⋅RlRkraRl

And Vi = EgkI⋅Rk = Egk µ⋅Egk⋅RkRkraRl

So the stage gain is An =VoVi

=

µ⋅Egk⋅RlRkraRl

Egk µ⋅Egk⋅RkRkraRl

Cancelling Egk yields An =

µ⋅RlkraRl

1 µ⋅RkRkraRl

Multiplying top and bottom by RkraRl gives:


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An =µ⋅Rl

RkraRlµ⋅Rk=

µ⋅Rlµ1RkraRl

Notice that if Rk was bypassed then for ac signals the µ1Rk term disappears and the stage gain becomes:

An =µ⋅Rl

raRl which is the standard expression for a bypassed common cathode stage.

By not bypassing Rk we have effectively increased Ra by µ1Rk

As an aside, notice that even when bypassed, at dc the unbypassed equation still applies. This gives us an easy way to determine Rk if we know Ra and µ at the operating point. Suppose we pick an operating current Ia and select a plate resistor such that when Ia flows it drops the supply to the required plate voltage Vp. Then we know that:

VpIa

= µ1Rkra and hence:

Rk =

VpIa

−ra

µ1


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Version History

0.1 First draft

0.2 added CF

0.3 Added Introduction, CC, proper symbols, appendices, Laplace stuff, changed Ra to ra and various corrections

0.4


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