Negative Feedback In Tube Amplifiers
This article was written in response to a request on the rec.audio.tubes newsgroup for a discussion of negative feedback as it applies to tube based designs.
Fundamentals
Suppose we have an amplifier with a gain of Ao. If its output voltage is Vo and the signal we feed its input is Vs (Figure 1) then:
Ao = Vo/Vs or Vo = Vs*Ao ..........(1)
Now suppose we take a fraction B of the output voltage and subtract it from the input voltage, for example by connecting the output of a feedback network in series with the input as in Figure 2:
The input to the amplifier is now Vs - B*Vo but we also know that Vo is just this input voltage times Ao, (equation1) so:
(Vs - B*Vo)*Ao = Vo
Dividing by Vs gives:
(1 - B*Vo/Vs)*Ao = Vo/Vs
The gain of the new amplifier An is, by definition, just Vo/Vs so substituting An for Vo/Vs gives:
(1 - B*An)*Ao = An
Expanding gives Ao - B*An*Ao = An, so Ao = An(1 + B*Ao) and hence:
An = Ao/(1 + B*Ao) ...................(2)
This is the basic feedback equation, where:
Ao is the gain before feedback is applied - often called open loop gain.
B 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.
This equation is always true.
The change in gain caused by applying feedback is simply:
An/Ao = Ao/(1 + B*Ao)*Ao = 1 + B*Ao ..............(3)
The factor 1 + B*Ao is called the feedback factor. Expressed in dB it is simply the amount by which the gain has been reduced by applying feedback. 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 most common simplification of the basic feedback equation is to assume B*Ao >> 1 in which case the equation reduces to:
An = Ao/B*Ao = 1/B
The gain is thus determined entirely by the components of the feedback network that comprise the term B. If B = 0.1 we get a gain of 10. Notice that if the feedback factor B is independent of 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 B*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 B to determine our amplifier's gain so the only way we have to make B*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 1 + B*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 1 + B*Ao is just 10 or B*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 voltage 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 voltage 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 or voltage applied feedback. If we apply it in series with the input (to an un-bypassed cathode resistor for example) it is known as series or current applied feedback. The feedback in Figure 2 is series feedback).
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 them here. If you really want the detailed maths, particularly as they apply to tubes, I suggest reading Chapter 7 of the Radiotron Radio Designers' Manual 4th Edition. You can download a copy from here. At this point we will just summarise the results and look specific tube examples.
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 1 + B*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.
Let's have a look at a couple of tube circuit examples to see how this is applied. First the cathode follower.
The Cathode Follower
Figure 3a shows a typical cathode follower circuit. In this example the bias resistor is bypassed to simplify the maths. Figure 3b shows the ac small signal equivalent circuit of Figure 3a. The tube is represented by the voltage generator with an output equal to u*Egk where u 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 = u*Egk/(Ra + Rk) ............(4)
So the output voltage acroos Rk is just i*Rk or:
Vo = u*Egk*Rk/(Ra + Rk) ...........(5)
Since Egk is the input to the amplifier the open loop gain is just:
Vo/Egk =u*Rk/(Ra + Rk) ...............(6)
However, the output voltage Vo is in series with the input (series applied) so:
Vi = Vo + Egk and hence Egk = Vi - Vo ........(7)
Substituting this in equation (5) gives:
Vo = u*(Vi -Vo)*Rk/(Ra + Rk) ...............(8)
Expanding gives:
Vo*(Ra + Rk)/Rk = u*Vi - u*Vo and rearranging gives:
Vo((Ra + Rk)/Rk*u = Vi - Vo and
Vo (Ra + Rk + u*Rk)/u*Rk =Vi
So the closed loop gain, Vo/Vi (An) is:
An = u*Rk/(Ra + Rk + u*Rk) which simplifies to:
An = u/(u + 1 + Ra/Rk) ................(9)
That took quite a bit of work to find the closed loop gain. We can get there a bit quicker by applying what we know of NFB theory. The open loop gain we derived
in equation (6) as:
Ao = u*Rk/(Ra + Rk)
Since the output voltage is directly in series with the input we have 100% NFB so B = 1. We can now write the closed loop gain as:
An = Ao/(1 + B*Ao)
An = (u*Rk/Ra + Rk))/(1 + (u*Rk/(Ra + Rk))
Multiplying top and bottom by Ra + Rk gives:
An = u*Rk/(Ra + Rk + u*Rk)
Dividing both by Rk gives:
An = u/(u+1 + Ra/Rk)
