TUBE OPERATION 3.

POWER AMP ANALYSIS 

Content of this page :-
How negative feedback works.
Fig 7. Schematic for Basic NFB around an amplifier.
Explanations and formula for NFB gain reductions and effects of NFB.
Fig 8. Schematic for Turner Audio 35 watt class AB triode amp using KT90.
General notes about this amp which has the same overall gain and NFB
as the basic example in Fig1.
Calculation method for output resistance with NFB.
The Model of the tube gain stage as a voltage generator.
Fig 9. Schematic of a power tube gain stage modeled as a generator with resistor
to indicate Ra.
Explanations about the generator model.
Fig 10. Schematic of a tube gain stage using 6SN7.
Fig 11. Schematic of a tube amp drawn with each stage as a generator
with loads and positions of shunt C to analyze the HF response and graph
all the attenuation profiles.

A whole lot more about NFB.

Calculation of output resistance.
A simple formula for calculating output resistance of a real amp after
taking 2 simple voltage measurements.
More on stability of amplifiers with NFB and the use of RC networks
to tailor open loop gain.
Fig 12
. Graph of tube amp frequency response without global NFB
and with global NFB, with no attempts to tailor open loop gain or phase shift.
Fig 13. Graph of tube amp frequency response without and with NFB but
with RC gain and phase shift tailoring networks in place.

More about stability and NFB.

Critical damping methods for tube amps with NFB.
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To have any idea about NFB, we must reduce the complexity of what makes
up an apparently simple triode amplifier to the following basic circuit :-
Fig 8.
  basic-NFB-example1.gif
This tries to explain what happens instantaneously and without significant
time delays at 1 kHz.
 
Without any NFB applied, ie, with R1 shunted to 0V, the amp may produce
3% THD at 30 watts, just before the onset of clipping.

With the NFB network of R1 and R2 is connected, the 3% distortion tries
to appear but instantly the voltage fed back contains a fraction of the
distortion which is then amplified to be the opposite phase of the distortion.

The distortion is amplified because there is no distortion in the signal input,
but there is distortion within the signal fed back to the inverting input.
The amplifier simply amplifies the difference between the input signal and
the feed back signal which is of the same phase, but also amplifies
the distortion signal. The voltage applied as a negative voltage feedback
signal to the inverting input is a fraction of the output. The fraction of output
fed back, ß = R1 / ( R1 + R2 ). ß is a Greek letter.
 
There are two inputs to every amplifier, and in the case of a tube amp with
a single triode one input is the very high impedance grid input, and the other
is the very low impedance cathode input. The real amp input impedances
are different to the model but the effect of voltages regardless of local tube
currents is the same as the model. The V1 single input triode acts as a
differential voltage amplifier. The signal voltage fed back is 1Vrms in this
case, and the same phase as the input signal. Thus the input signal must
be greater in amplitude to the fed back signal. In effect, the signal voltage
between the two inputs must be simply added to the FB voltage to establish
the input level when NFB is used. This type of many different types of
feedback is called 'series voltage negative feedback' sometimes also
called "inverse" voltage NFB, but this refers to the amplification of the
distortion signal by the gain stages to become an output signal which is an
opposite or "inverse" phase of the naturally generated distortion signal that
exists when no FB is used. The word "negative" also means an opposite
phase of distortion signal is produced at the output. Positive voltage FB,
PFB, causes distortion to increase, bandwidth to be reduced, and worsening
phase shift and increased risk of instability. There are several varieties of
NFB and PFB, either series or shunt, or voltage or current types, and a
reader here should refer to the 1955 Radiotron Designer's Handbook,
4th Ed, for a heck of a lot more about NFB than I care to re-produce here.

Note the use of the triangle to represent the whole amplifier. It is engineering
convention to always assume there is a positive phase of output signal
at the "point" of the triangle pointing to the right. The output impedance of the
triangular amp is always regarded as much lower than RL even when NFB
is applied. The model does not include output resistance considerations;
to include them means a more complex model which I deal with elsewhere.
The two inputs each have the opposite effect, ie, one will cause the same
phase of output voltage and is known as the 'non-inverting' input,
marked ( + ), the other is the inverting input marked ( - ).
For the model to work, each input is regarded as a very high input resistance
so we need only consider the external connections without all the complexity
of dc biasing and V1 tube set up in a real amp.

In this case, it is as if 1.34Vrms applied to the + input creates +52.26Vrms
output and the 1Vrms NFB creates -39Vrms, leaving the difference of
about 13.4Vrms we  see above. But in fact the amplifier only "sees" the
difference between the FB voltage and input voltage, and since the
difference in this case is 0.34Vrms, the open loop gain of 39 multiplies
this 0.34vrms to 13.4Vrms.

The amount of feedback reduces it to a state of equilibrium where there
is 0.75% at the output. The steady state quantity of THD is applied as
indicated in Fig 7 so that the closed loop amplifier has 1/4 of the THD
when the open loop gain is reduced by 12 dB of global NFB.

The voltage gain, A', is the gain with NFB applied, = A / ( 1 + [ A x ß ] )

where A is the open loop gain, ie, gain with no NFB connected, ie, with
R1 shunted to 0V, A' is the closed loop, gain, ie gain with NFB applied,
ie, with R1 and R2 connected as shown in Fig1, ß is the fraction of the
output voltage fed back to the second of two inputs at the front end of
the amp, and 1 is a constant required for all equations to work properly.

In this case A = 39, ß =  50 / ( 50 + 600 ) = 0.077.
So A' with FB = 39 / ( 1 + [ 39 x  0.077 ] ) = 9.75.

The model shows that we have gain with  FB = 13.4 / 1.34 = 10, which
is close to what is calculated. Applied NFB in dB = 20 log ( gain reduction )
=  20 log ( 39 / 9.75 ) = -12dB. Distortion is also reduced -12 dB.

However, this isn't the whole story.
There are complexities below any surface...
Because of intermodulation effects, the THD fed back will produce other
harmonic products not present in the spectra before NFB was applied.
So if there was mainly 2H in the THD spectra, some 3H would be
generated because the IMD products formed are the sum and difference
between the main input signal frequency and the harmonic. In this case we
can call the input signal 1H, and 2H + 1H = 3H, and 1H - 2H = -1H. The
math required to calculate the IMD is somewhat beyond the scope of this
website but the result is that the IMD production in a triode slightly attenuates
the input signal but creates 3H that may not have been present. This whole
IMD effect is rather minimal when THD is at low levels below 1% and fairly
sonically benign if the THD is mainly 2H. Of course the 3H intermodulates
with 1H to form 2H and 4H, and so on to infinity, but usually the noise levels
of the amp cover up what infinite effects occur. The other THD harmonics in
an average triode will swamp the IMD products formed between various
harmonics and the fundamental. In this case where the triode produces 3%
open loop THD of mainly 3H, the 2H may be reduced by about 12dB but
the 3H would maybe only reduced by 6dB, and other products such as
4H, 5H, 6H, 7H, 8H, 9H etc created and will alter the levels of existing
THD products without NFB. The thing to remember is to set the tubes up
to operate with low THD and there won't be any worries with IMD. When the
open loop THD is low, less than 3%, then these other products are also low
and in fact negligible when open loop THD < 1% at low levels at which we
listen to music. 12dB of NFB is most effective at lower output levels when
the there is no class AB action and the amp is in pure class A.

The reductions of THD spectral content according to the formula applied
depends on the open loop phase accuracy and gain at HF. Where the gain
is attenuated and phase shift is large outside the bandwidth of 20Hz to 20kHz,
the amount of effectively applied NFB is low. But this is what is wanted with
any amplifier; we have zero need for electronic heroics and no need for the
same amount of NFB to be applied at 5Hz or 100kHz as between 20Hz and
20kHz. If there is too much open loop gain and phase shift reduction within
the 20-20 pass band we will get more THD and IMD products appearing at
the output and a higher output resistance than we would want. Nearly all
tube amps have higher THD, IMD, and Rout at 20Hz and 20kHz than at
between say 100Hz and 5kHz. The better the OPT is, the broader the
bandwidth possible where the THD, IMD and Rout and phase shift are all
kept low; ie, the NFB is maximally effective. To get the best from a tube amp
with NFB the OPT must be of top quality, rather than have a design where
the correction of phase shift effects of the OPT as well as all other types
of distortions relies solely on the NFB application.
Fig 9.
Schematic for 35w
        triode PP class AB1 amp

I don't need to comment too much on the circuit operation except to say that it
will work well and sound excellent providing you can obtain a nice OPT.
You could also run the output tube screens to 50% Ultralinear taps if your OPT
has such taps. I recommend an 8k : 6 ohm  OPT, but a 5k : 4 would be fine if it
is rated for 50 watts+. If the power supply you build does not have exactly the
same  B+ and bias voltages, then be prepared work out the amended values
for R3, R4, R9, R10, R14, R15, R18, and take note that unless you get things
right there *will* be a price to pay. Nevertheless, check that the output tube bias
is not more than 65mA for each KT90 when the 10k pot is turned to balance the
bias current for equality after warm up. There is no actual bias adjustment for
the fixed bias and there is only a balancing pot. Should anyone feel the need
for overall bias adjustment, then use a 10k wire wound pot in series with a 12k
resistance to replace R18. R14, R15, R18 can be adjusted to higher values for
a bias supply > -87V, or to lesser values if bias supply is at the minimum of -75V,
so that when balanced, you get 65mA per tube, and the there is a good range
of bias balance adjustment with the 10k pot which should be a wire wound 3
Watt type because it needs to be reliable. Never use tiny little low power rated
trim pots for bias adjustments, and only wire wound or cermet pots.

There are notes on the schematic about HF stabilizing zobel networks to tailor
the open loop phase shift and gain and still try to get a response with a 6 ohm
resistance load out to 65kHz, but also with unconditional stability which simply
means that you can place a 0.22 uF across the open output and it will not
oscillate at low RF and a 5kHz square wave at low power with a 0.22 uF will
not produce more than 6dB of over shoot and only a few cycles of ring
frequency.

The schematic is shown with 12dB of global NFB applied, and the circuit has
virtually the same open loop and closed loop gain character as in the simplified
NFB example in Fig1.


The output resistance of a tube amp can also be calculated if you know
1, voltage gains of the input and driver stages,
2, µ of the output tubes,
3, turn ratio of primary to secondary which is the unloaded voltage ration
between P and S windings,
4, the anode resistance, Ra of one output tube at the Q point.

Rout of the push-pull amp with FB applied  =          Ra-a   +  Rw total                  
                                                                         ZR x ( 1 + [ A" x { µ/TR } x ß ] ) 

Where Ra-a = twice the Ra of one output tube where two tubes are used
in a PP output stage, Rw total is the sum of OPT primary winding wire
resistance and ZR x secondary winding wire resistance. TR is the turn ratio
of the OPT, or unloaded P to S signal voltage ratio at 1kHz, or square root
of the exact known ZR. ZR is the output transformer impedance ratio which
is the turn ratio squared, A" is the gain of the stages preceding the output
tubes, ie, Vgrid to grid of the output tubes / Vg-k of the input tube, µ is the
amplification factor of an output tube, ß is the fraction of OPT secondary
voltage fed back to be "in series" with the input voltage to V1.

For example in this case we have :-
Ra-a = 2,000 ohms,
Rw = 10% of primary RL of 8k = 800 ohms,
TR =  36.5 : 1
ZR = 8,000 ohms : 6 ohms = 1,333 : 1,
A" = 16.17 x 15.64 = 253.
µ = 7.0,
ß = 50 / ( 50 + 600 ) = 0.077.
Rout' = closed loop output resistance at the secondary output terminals

In this case Rout'  for the above PP triode amp,
Rout  =                 2,000 + 800                               =   0.443 ohms
           1,333 x ( 1 + [ 253 x { 7 / 36.5 } x 0.077 ] )

The Model of the tube gain stage as a voltage generator.
To understand some of what is happening in the "simple PP triode amp"
in Fig 9, an equivalent model of the whole amplifier can be drawn up, and
I have to say it is a dreadfully tedious exercise which nobody I know has
attempted since 1955 when the boffins who wrote the text books decided to
confuse their pupils at university electronic engineering courses.
Nevertheless, you know very little about tube amps if you do not comprehend
the following discussion and modelling of the basic tube in Fig 10, fig 11, and
amplifier model in Fig 12.
Fig 10.
Schematic of triode
        as generator model.
If you had a magic box with unknown contents, but with 3 terminals labelled
anode, cathode and grid, and you carefully explored all the possible ways of
using the mystery box, then sooner or later you would figure out you had the
contents in Fig 9, and you would have a model for the power triode as shown.
Only the 6550 ac signal operation is shown with no regard for the bias or dc
supply considerations although the ac operation is that gained by using one
6550 triode strapped with Ea = 420V and Ia = 73 mA.

If you wanted to measure the generator output voltage you wouldn't find it
anywhere since it is only something which is part of a useful mental model
of the triode. The same model could be used for a beam tetrode where
µ = 190, and Ra = 17,000 ohms so that for the same 174vrms into the 3.5k
load, there is the same 50mA of signal current, so there would be 850Vrms
across the Ra = 17,000 ohms so the gene would have an imaginary
1,024Vrms output. Since the gene produces Vout = µ x Vg, since µ = 190,
Vg would be 5.38Vrms. If there was a tetrode in the mystery box, you would
swear you had a voltage generator which amplified the Vg-k signal to 1,024V,
and due to the internal resistance of Ra = 17k, the anode produced Ra,
there was 174V into the externally connected 3.5k load between the anode
and cathode terminals.

Do you not think this is an easy to understand model for any tube as long
as we know Ra and µ? Grandfather's generation knew about all sorts of
useful tricks, and here is one of them. One can add loops of NFB and
other connections such as used for differential amps and SRPP and µ-follower
and cathode follower topologies and by using the model the gains of the
tube can all be worked out using Ohm's Law and those vital pieces of
equipment such as human imagination and persistence.

Here is another model for a single 1/2 of a 6SN7, 6CG7, or 6J5...
Fig 11.
Schematic for basic
        6SN7 triode model.

If you want to set out exactly what gains and voltages you will have with
feedback and cathode resistors et all, just use the generator model to make
your calculations based on first principles and Ohm's Law.

Here is the whole Fig 8 triode amplifier modeled more fully......
( Even grandpa had trouble figuring it all out ! )
Fig 12.
block diagram 3
        satge PP triode amp.
Anyone can see this isn't any easier to understand than the actual amp
schematic in Fig 3 ( basics 1 ).
But its not that hard to read the above Fig 11 and Fig 8 together to know
what the relationship is between gain and circuit resistances. We are
concerned at the outset to understand the way the amp works at the mid
frequency, and once we learn that we can then move to understand why
there is attenuation at HF and LF and that in fact the amplifier we have
is not only an amplifier but also a band pass filter with at least two
specific specific -3dB poles and with additional poles further outside the
pass-band determined by these first two poles.

So to start, just ignore all the capacitances. They have virtually zero effect
at 1kHz which is the mid-frequency which concerns us first.

Each amplifier stage is drawn as a triangle, the universally accepted
manner in which amplifiers are simply drawn. We are not concerned
with biasing and B+ supplies; we just want the two ac signal inputs and
the output of each stage, and all the voltage inputs and outputs are
in Vrms, or exactly what you will measure with an ac voltmeter.

In this case each I have drawn each stage as a perfect low impedance
voltage generator triangle which has an output of µ x Vg-k.  For example,
for V1, the grid accepts the signal input from an external signal generator,
and the cathode accepts the NFB voltage. The difference between the
two in this case = 0.34Vrms. The 6CG7 input tube µ = 19.5 so the gene
output which is the "point" of the triangle is 6.6Vrms. Now the model
of the tube as a generator with Ra in series is only a model. if you went
looking to measure 6.6vrms you won't ever find it. The model is a
convenient way to describe the ac action of the tube; it is as if there was
a gene inside the tube. This is outlined in Fig 4. The Ra of the tube which
is a paralleled 6CG7 = 7k and is shown in series with the grid of the next
stage. The output end of the 7k is the anode of V1. V1 anode
has a dc load of 39k, and an ac load of 220k, so total RL = 33k which is
shown taken to 0V, a convenient fixed point for our model where we are
not interested in the dc operation. See R3 and R7 on the amp schematic.
Oh, and you do have to realize the caps C3 and C4 are such a high value
that at 1 kHz they are such low impedances that they are virtual short
circuits and their impedance can be ignored.  Also as you can see, there
is R1 and C1 and Csh, but remember, these C values are in pF, and will
only have significant impedance when the frequency is above 20kHz.

So we can say that the Ra of V1 in parallel with the total anode RL = 5.8k,
and that is the output resistance of the stage with all its bits and pieces at
1kHz. Its actual gain = anode voltage / Vg-k = 5.5 / 0.34 = 16.2.

The V1 stage will not change its gain regardless of the load connected
at the output of the amp; it has fixed gain.

I have combined the differential amplifier formed by two 6CG7 in V2 and V3.
The amp schematic shows one of the two grids grounded, one driven so
I have the same set up in the above block diagram. However rather than
show two oppositely phased outputs from two anodes, I have shown just
one output with the summed phases, and an amount of Ra-a equal to the
sum or Ra of each 6CG7. This all a valid way to keep it simple. So we have
the two tubes working as a voltage gene with µ x Vg-g  as the output, with
Ra-a = 14k, thus giving 86Vrms which is the grid to grid driving voltage to
the output stage. The overall gain of the differential pair = 86 / 5.5 = 15.63,
and this gain is also fixed, and not affected by the output stage.

The output stage with two KT90 in triode with µ = 7 can also be reduced
to a single generator of µ x Vg-g output and with the Ra-a shown as the
summed Ra of both triodes and in series with whole push-pull primary load.
So 86V is the generator input voltage and output voltage of the gene = 602Vrms.
The Ra-a of the two KT90 is about 2k, and then the 8k primary load is
connected to the output end of the 2k. There is leakage inductance LL and
shunt C of the OPT forming a 3rd order low pass filter at the OPT. But the
values of L and C will have no effect on the performance below 10kHz.

The gain of the output stage = voltage at the input of the 8k load / Vg-g
= 488 / 86 = 5.67.

The output transformer has a ZR of 8k : 6, so the turn ratio which is the
voltage ratio or current ratio = 36.5 : 1. Therefore the 488Vrms get
transformed to 13.4Vrms at the 6 ohm load, thus giving us 30 watts.

The gain of the output stage varies somewhat with load. You can try
different output loads and calculate the outcome using Ohm's Law,
or just calculate the output gain = µ x RL /  ( RL + Ra ).

The amount of applied NFB is only valid for a specific load value.
The amount of applied global NFB is proportional to A / ( 1 + [ A x ß ] )
where A is the output tube gain x input & driver stage gain, and ß is the
fraction of the output voltage fed back to the feedback input port.

Rout of any amp with an output is considered  without having a load
connected, ie, the Ra of the output tube/s. Voltage amplifiers may have
a dc load resistance between anode and B+ and the Ra is in parallel with
the dc RL. Power amps and some transformer coupled preamps will have a
very high and negligible load from the anodes, so the Ra is considered
to be the Rout. At the secondary of the OPT and with no NFB Rout is
purely dependent on the Ra & µ of the output tubes and the OPT winding
ratio and winding resistances.
With NFB, Rout depends on the basic formula,
Rout' =  Ra / ( 1 + [ A" x µ x ß ] ) all in parallel with any dc supply R.
where A" is the driver stage gain and µ is the output tube amplification factor.
The formula is basic, and applies to all voltage amplifiers where the output
is taken from the anode outputs and also from where the FB signal is fed back
to the input. For Rout at the secondary of an OPT there is the full formula for
Rout of the PP amp under analysis above.

The point I make here is that NFB reduces Rout more than it reduces voltage
gain and distortions. A little NFB goes a long way when it comes to reducing
Rout thus getting a good damping factor < 10 to stop bass becoming boomy
and loose. DF = Speaker nominal Z / Amplifier Rout.

The gain is of great importance with beam tetrodes or pentodes in the output
stage. The gain with a 6550 in beam tetrode with RL 3.5k = 32. With no load
the gain = µ = 190, which is 5.9 times higher, or +15.4dB. If we have a class
A 6550 tetrode amp with 15dB of global NFB with RLa-a = 7k, then with no
load the output stage gain increases by 15.4dB so the applied NFB also
increases by 15.4dB so you have 30.4dB of applied NFB when no load is
connected. It is no wonder then that beam tetrode amplifiers are renowned
for oscillating at some low RF if their open loop gain has not been curtailed
at HF with suitable zobel networks, because the greater the NFB applied,
the more likely you will get instability, and 30dB of global NFB for any tube
amp is a huge amount of NFB.

The Rout of amp amp is reduced more by a given value of ß than is the
reduction of gain and distortion with load connected.

Consider again the global NFB. Suppose we disconnect it for awhile.
Then to have 13.4 Vrms into 6 ohms we only need 0.34Vrms input. If someone
connects a 3 ohm load, the signals all the way through the amp are exactly the
same right to the output of the output stage µVg-g generator, and there will be
the same 602Vrms at the model gene output. The load however has halved
from 6 to 3 ohms, so load at the primary falls from 8k to 4k. We have 602V
applied from the gene through Raa of 2k to the load of 3k, so the output at
the load = 602 x 3 / ( 2 + 3 ) = 361Vrms, and power output should be 43 Watts.
But the model here does not include about 800 ohms of the OPT total winding
resistance based on 10% of 8ka-a RL. Then you have increasing winding
resistance % loss as loads less than 6 ohms are connected; if there are 10%
losses with 6 ohms, there will be 20% winding losses with 3 ohms. Then there
is the inability of the output tubes to actually produce such power as seen by
a look at the loadlines and curves. The model is limited. Feel free to insert the
winding resistances into the model of the OPT. But what we can say is that the
overall open loop gain which is the gain without global FB = Vout / Vin and with
6 ohms it is 13.4 / 0.34 =  39.4. If the load was removed, then the 602V would
appear at the primary since no load current flows and the output voltage at the
sec = 602 / 36.5 = 16.5Vrms. Any amp can have its Rout measured easily by
simply setting the amp up without a load and recording the output voltage,
and then connecting a load which causes little distortion and measuring
the voltage drop.

Rout = ( Vout with no load - Vout with a load ) / Load current.

In this case Rout = ( 16.5V - 13.4V ) / 2.23A  = 1.4 ohms.  This method
always includes the winding resistances, so when calculating Rout , never
expect your measured Rout to be as low as the calculations unless you
have skillfully measured all the winding resistances. Many OPTs have
primary winding R = 4% of the primary rated load value, so for 4ka-a
RwP = 160 ohms, but often they have 10% in the secondary, so that
RwS = 0.8 ohms. In such an OPT, the 0.8 ohms is transformed by the
impedance ratio of the OPT when looking into the primary, and because
ZR = 500, we get the secondary winding resistance appearing effectively
in series with the primary winding and its resistance, so RwS at the
primary = 500 x 0.8 ohms = 400 ohms, The total Rw "referred to the primary"
= 160 + 400 = 560 ohms = 14% of the primary load. There are power
losses in transformer windings and an amplifier with 14% winding losses
would have to produce 100 Watts at the anodes to make 86 watts at the
speaker terminals. The total winding resistance losses in a Quad-II
amplifier when the 8 ohm secondary configuration is chosen = 17%,
quite poor considering what could have been achieved had the amps
been 10% larger to contain the more adequate amount of OPT iron
and wire. Many old amps have high Rw, but I like to keep all Rw
below 5%.

When NFB is applied as shown in the schematic, there will be be some
distortion appearing at the output even though it has been reduced by
the amount of gain reduction. We still have to consider stability.
The Fig 12 shows all these shunt and input capacitances along the
signal path in the amp.

Rather than waste time publicly calculating all the accumulative effects
of all the RC filter poles due to Miller capacitances and the LCR poles of
the OPT, allow me to say that the tube line up shown in Fig 9 will provide
very adequate bandwidth compared to amplifiers using pentode input tubes
such as EF86 and a 12AX7 differential driver amp such as the Mullard 520.
The 520 and others were often built by many enthusiasts in the 1950s and
1960s using very poor OPT with high LL and Cshunt and with low primary
inductance so that some samples could not be turned on without a load
connected and they would not drive electrostatic speakers. While Mullard
may have liked us all to use high Ra pentodes and triodes for the driver tubes
I would always prefer the low Ra low µ triodes for signal stages which are
less affected by stray C effects and Miller capacitances. Many mass produced
amplifiers I have tested will oscillate at LF without a load, and some with a
load, hence my use of the network after V1 anode with R6, R7, C4.
This will kill LF oscillations in any amp without such a network.

The open loop HF response and phase shift character depends on the sum
of all the attenuations of the RC and RL low pass de-facto filters caused by
Miller leakage inductance. The addition of the series network with C5&R8
in the Fig 9 schematic is a commonly used method to reduce the gain at HF
by placing a "shelf" in the HF response. The amplifier without the zobel
network at V1 may have too much phase shift at say 70kHz, and the addition
of a 0.22 uF at the output will shift the phase additionally to cause the phase
shift to be more than 180 degrees lagging while the open loop gain still
remains above 1.0. In this case the amp will oscillate at the the available
frequency above the audio band, When ESL speakers are connected,
often there is a peaking in the response of 6db at 25kHz with a few dB at
16kHz and the sound is not optimal. The zobel network begins to reduce gain
as F rises above 15kHz, and in some instances at 50kHz gain can be reduced
by 12dB and because the load is mainly resistive at 50kHz, the phase shift can
be reduced by 45 degrees at 50kHz and so the ultimate phase shift is delayed
for above say 150kHz where open gain most definitely has fallen to below unity.
The more NFB used, the more difficult it is to stabilize the amp. The triode output
stage is not immune to oscillations but a zobel is seldom needed across the 1/2
primaries of the OPT with triodes. I always connect them with UL or pentode amps.
The pair of zobels are typically 0.001uF plus 3k9 for an 8k a-a primary load.
These zobels do two things. They provide a resistive load to the output anode
circuit at above 80 kHz, and also provide some R to load the resonant effects
of the leakage L and shunt C of the OPT. I also nearly always at least have a
zobel network across the output, 0.1uF plus 4.7 ohms is typical and it also loads
the OPT resistively at HF and helps prevent oscillations when NFB is used.
The additional stability measure uses the C across the feedback resistor,
see C8 in Fig 9 schematic. This capacitor must be chosen carefully during
setting up an amp as too big a C value will make HF oscillations worse.
The values shown will give an slight advance to the phase of the feedback
signal.

There is no easy way to set up any amp with NFB and make it
unconditionally stable so that the response with NFB gives the following :-
A -3dB pole at 65 kHz with 1/2 the rated value of resistance load.
Less than 6dB over shoot on 10khz square waves with no more than 3 ring cycles
before settlement. Complete tolerance of pure capacitance loads without any C
value causing oscillations. No value of C should result with a rise in sine wave
response of more than 2dB below 20kHz. Peaking in the response due to C
loads should not exceed 6dB above 20kHz. I use old radio tuning caps with a
pot soldered onto the side to establish V1 anode zobel values in conjunction
with the adjustment of the compensation cap across the FB resistance.
The other zobels on the OPT are usually established on a try and see method.
A CRO is used to monitor the square wave response conducted at low levels
where there is less chance of roasting the zobel resistance by large HF signals.

Here are some typical graphs which the audio enthusiast can plot using a sine
wave generator and CRO :-
Fig 13.
Graph of response
        with/without NFB.
The above graph shows the sine wave response a very ordinary tube amp
with average quality OPT without NFB and with NFB, and with no attempt to
tailor the open loop gain or phase shift to prevent the peaks in sine wave
response just outside the audio band. The responses here are with the amp
loaded with a resistance only. The LF peaks are due to LF phase shift so that
between 1Hz and 5Hz there is perhaps more than 110degrees of phase lead
so that the FB which is applied is so much out of phase with the input signal
that little gain reduction occurs due to the 12dN of applied FB. At HF the
situation is similar at 60kHz, and the dips and peaks above 100kHz are due
to output transformer resonances series or parallel L&C combinations of
leakage inductance stray shunt capacitances between windings.

To be able to measure a tube amp with global FB isn't always easy with a
high amount of NFB applied initially. A basic circuit without any correction RC
networks will often oscillate quite badly as soon as any NFB is connected for
the first time. For any new amp, the FB used should be 6dB then 12dB, and if
it doesn't oscillate, one is very lucky, but usually the response for sine waves
is peaked as in the above graph, and steps must then be taken to decide on
values of the RC compensation networks and zobel networks. When all that
is done properly, a graph of the amp with networks in place will more likely
resemble the graphs in Fig 14.....
Fig 14.
Graph response
        with/without NFB, compensated.
Notice that the open loop gain has been reduced at the extreme ends of the
audio band and the rate of phase shift increase for frequencies where
oscillation may possibly otherwise occur has been reduced. With a very good
OPT, it is possible to apply up to about 40dB of global NFB into a pure
resistance load. A good OPT would give you an amp with 3Hz to 65kHz
of bandwidth without any NFB connected at least and with few resonances
above 80kHz. The bandwidth of the tubes and their couplings must be wider
than the OPT to be able to measure the open loop bandwidth of the OPT.
The output tubes should be triodes because the bandwidth of the output stage
tubes plus OPT is much less when beam tetrodes or pentodes or even UL
are used because Ra is high and the primary inductance, Lp, and shunt
capacitance looking into the primary terminals of the OPT is also high,
and tends to shunt the signal across the rated load resistance.

But many OPT are capable of being used with only 16dB maximum of
global NFB. 40dB is about the limit of NFB application for a tube amp with
an OPT and with any more NFB, usually it will oscillate and there is nothing
we can do to stop it. When the point at which more FB causes unavoidable
oscillations and instability, there is no "margin" left for stability. So if an amplifier
takes a max of 35 dB without oscillations, and this amplifier has its NFB
reduced to say 12dB, then the margin of stability = 35dB - 12dB = 13dB, which
is a good margin of stability. This only applies to resistive loads.
The maximum amount of NFB applicable with a 0.22uF cap is across the output
will usually be less than for the pure resistance load, or resistance plus cap in
parallel. Usually if the margin of stability with R loads is 12dB or better, cap
loads or any other type of load including no load at all should cause no
oscillations even though some peaking in the response will occur.
 
The square wave response at 5kHz should show minimal over shoot
not exceeding 3dB for any value of capacitor load and with only a few
cycles of "ring" frequency. The square wave response into any pure resistor
load should have virtually no over shoot. When these two conditions are
met with R load response at 65kHz, the amplifier has been constructed
correctly. The amplifier then behaves as a critically damped band pass
filter with second order attenuation which is quite OK as long as the F
poles are below 20Hz and above 60 kHz.

For another graph and notes and schematic details for dealing with
problematical amplifiers go my web page on 'Leak Amplifier re-engineering',
where considerable phase tweaking is used to obtain healthy bandwidth into
a resistive load and unconditional stability for a 1954 Leak TL12 Pt1 which
has an awful OPT. I have used all the tricks available to tame the Leak.
There is a Zobel network across the output to 0V, and I tried Zobel networks
from V1 anode to 0V, and across the OPT primaries, but they did little to
reduce response peaking or improve the square wave response so I relied
on the negative current FB operating above 20kHz and the normal shunting
of the global NFB resistance with a small cap. The Leak thus gives a healthy
bandwidth with some peaking in the response but it ended up much more
stable than in the original circuit.

Zobel values should be about as follows :-
V1 anode to 0V, R = 1/10 of total ac and dc RL in parallel, C should have
its reactance in ohms = 1/10 V1 RL at 100kHz. Use a pot and a radio tuning
gang to establish correct values if this Zobel is used; wrong values will worsen
the outcome and cause more peaking and less bandwidth for resistive loads,
thus reducing the margin of stability.
Eg, where  RL = 30k, R = 3.3k, C = 470pF.

Across each half of the OPT primary. R = Class A  RLa-a / 2, C has reactance
= R at 100kHz. Eg, RLa-a = 8k, so R across each 1/2 P winding from B+ to
each anode = 3.9k, and wire wound, with C = 390pF. I find maybe 1,000pF
is better.

Across the output secondary of the OPT, R = Rated load of the amp, C has
reactance = R at 100kHz Class A, Eg , RL = 6 ohms, so R = 6 ohms,
C = 0.27 uF.

Good luck!

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