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 methode 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 modelled 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 analyse 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 7.
This tries to explain what happens instaneously and without 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 = R1 / ( R1 + R2 ) and is called
ß, the 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 trianglular 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 else where...
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
fundemental.
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 8.
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 stabilising 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,
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 preceeding 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 8,
an equivalent model of the whole amplifier can be drawn up, and I have
to say it is a dreadfully tedious exercize
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 9, fig 10, and
amplifier model
in Fig 11.
Fig 9.
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
imaginery 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
peices of equipment such as
human imagination and persistance.
Here is another model for a single 1/2 of a 6SN7, 6CG7, or 6J5...
Fig 10.
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
pronciples and Ohm's Law.
Here is the whole Fig 7 triode amplifier modelled more fully......
( Even grandpa had trouble figuring it all out ! )
Fig 11.
Anyone can see this isn't any easier to
understand than the actual amp schematic in Fig 2.
But its not that hard to read the above Fig 10 and Fig 7 together to
know what the
relaltionship 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
amplifer 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 passband 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 satge 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 3.
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 realise 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 peices 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 amout 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 transfromer 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 wil 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 dependant 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 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 13.4vrms into 6 ohms we only need 0.34 vrms 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
wth 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 stablity.
The Fig 2 shows all these shunt and input capacitances alonf the path
in the amp.
Rather than waste time publically 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 1 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 , C4 and R7. 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 Fig1 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
definately has fallen to below unity.
The more NFB used, the more difficult it is to stabilise 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 1 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 12.
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 cpacitances 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
13.....
Fig 13.
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 Amplifer 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.
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