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.
-----------------------------------------------------------------------------------------
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.
_files/basic-NFB-example1.gif)
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.
_files/schem-35w-triode-pp-ab1-april-2006.gif)
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.
_files/schem-triode-generator-model.gif)
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.
_files/schem-6SN7-generator-model.gif)
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.
_files/basic-triode-model-3stage-amp.gif)
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.
_files/graph-typical-response-amp2.gif)
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.
_files/graph-typical-response-amp3.gif)
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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