B2 Spice output transformer models?
Danny T
it's almost done except for the output transformer (Lundahl LL1623).
Does anyone have models for a PP OPT ?
Or is it possible to make one with the transformer models available in B2 spice?
Max
> I've drawn the circuit of my PP KT88 amplifier in B2 Spice,
> it's almost done except for the output transformer (Lundahl LL1623).
> Does anyone have models for a PP OPT ?
I do not know if it something that you can use but on
http://www.duncanamps.com/spicemodels.html
is some transformer models.
--
Best Regards
Max
The Fox from Denmark
Robert McLean
possible combinations Single Ended vs Push Pull , and with UltraLinear
screen grid taps, or without. I had tried various modelling approaches :
multiple mutal inductors, dependant current sources and so on, all of which
seemed as if they theoretically should work, but when actually run in Spice
gave weird results, or would only work for a single ended transformer.
What I finally found would work reliably, and that could be extended to
multiple winding primaries, was to use an "ideal" transformer consisting of
mutually coupled inductors, with an equivalent circuit of inductors and
resistors connected to the primary. The equivalent circuit did 2 things.
a) produce the desired frequency response, and b) in the case of multiple
winding primaries (ie any PP type, or a SE type with ultralinear taps )
divide up the voltages appropriately.
Note that I am basing my model solely on measured or published specs for
frequency response and DC resistance. I am not basing it on a physical
model, ie with leakage and core losses and saturation and all that stuff.
The equations for the equivalent circuit portion can be found in many
textbooks, I use Clarke-Hess "Communication Circuits : Analysis and Design"
, 1971, section 2.1 since that is what I have handy.
The first model listed gives the formulas to convert frequency response to
the model parameters. Use the same basic equations for each model. Note
that I use a PSpice compatible program, which can pass parameters to
subcircuits. If yours does not, then you will have to replace {xxx} with
the value of xxx for your particular transformer, and use a separate
subcircuit for each transformer you want to model.
Also note check that whatever tube models you use do not mistakenly permit
negative plate currents to come about due to negative plate voltages,
otherwise when used with any model with inductors you can get very strange
results such as huge voltages like gigavolts on the plate, and huge
currents, and other things like that.
*************************************
*
* MyTransformers.lib
*
* R. McLean
* 29 December 2000
*
**************************************
**************************************
*
* Center Tap Audio Transformer,
* for push-pull tube operation,
* with screen grid taps
*
* TransPPSGT
*
* Primary Terminals
* P1, P2 plate connections
* Sg1, Sg2 screen grid connections
* B primary center tap (+Vpp)
* Secondary terminals
* Sp1, Sp2 speaker connections
*
*
* Parameters :
* RP one half primary winding resistance
* LA one half primary winding inductance
* LP one half primary inductance, plate to screen tap
* LS one half primary inductance, screen tap to center tap
* RA impedance ratio, primary plate-plate to secondary
* RS secondary winding resistance
*
* the parameters are related to frequency response as follows :
* let Rs = source resistance ( ie the load at the seen by the plate )
* let RL = load resistance ( ie the speaker )
* then RA = Rs/RL =n^2, where n is turns ratio, Ts/Tp
* let r1 = total primary winding resistance
* let r2 = total secondary winding resistance
* let F1 = upper 3dB frequency
* let F2 = lower 3dB frequency
* if 3dB points are not known, but 1 dB points are, then
* let F1' = upper 1dB frequency,
* let F2' = lower 1dB frequency
* then F1 = F1' x 2
* F2 = F2' / 2
*
* then calculate total series inductance La
* La = (Rs + r1 + (n^2)( RL + r2))/( 2 x pi x F1 )
*
* then calculate total parallel inductance Lb
* Lb = (Rs +r1)(RL + r2) (n^2)/((Rs + r1 +(n^2)(RL +r2))x 2 x pi x F2)
*
* then divide up La and Lb into the multiple windings required
* let t be the screen grid tap percentage, ie for a 40% screen grid tap, t =
0.40
* then the parameters are
* RP = r1 / 2
* RS = r2 (NOTE : RS is NOT Rs !!)
* LP = (1-t)Lb/2
* LS = t x Lb/2
* LA = La/2
*
* the default parameter values given are for an 8K to 8ohm transformer,
* with 40% screen grid taps,
* and 30 to 30KHz frequency response (15 to 60KHz 3dB response)
*
.SUBCKT TransPPSGT P1 Sg1 B Sg2 P2 Sp1 Sp2
+PARAMS: RP=25 LA=0.02234801 LP=13.3822388 LS=8.921493 RA=1000 RS=0.8
*
* primary
*
RP1 P1 1 {RP}
La1 1 2 {LA}
LP1 2 3 {LP}
LS1 4 5 {LS}
LS2 6 7 {LS}
LP2 8 9 {LP}
La2 9 10 {LA}
RP2 10 P2 {RP}
RP3 3 Sg1 1
RP4 Sg1 4 1
RP5 5 B 1
RP6 B 6 1
RP7 7 Sg2 1
RP8 Sg2 8 1
*
* secondary
*
Rs Sp1 11 {RS}
LSA 11 Sp2 {8*({LP} +{LS})/{RA}}
*
* coupling
*
Kcore1 LP1 LS1 .999999
Kcore2 LP1 LS2 .999999
Kcore3 LP1 LP2 .999999
Kcore4 LP1 LSA .999999
Kcore5 LS1 LS2 .999999
Kcore6 LS1 LP2 .999999
Kcore7 LS1 LSA .999999
Kcore8 LS2 LP2 .999999
Kcore9 LS2 LSA .999999
Kcore10 LP2 LSA .99999
.ENDS TransPPSGT
**************************************
*
* Center Tap Audio Transformer,
* for push-pull tube operation,
* without screen grid taps
*
* TransPP
*
* Primary Terminals
* P1, P2 plate connections
* B primary center tap (+Vpp)
* Secondary terminals
* Sp1, Sp2 speaker connections
*
*
* Parameters :
* RP one half primary winding resistance
* LA one half primary winding inductance (series)
* LB one half primary inductance, (parallel)
* RA impedance ratio, primary plate-plate to secondary
* RS secondary winding resistance
*
*
* the default parameter values given are for an 8K to 8ohm transformer,
* and 30 to 30KHz 1 dB frequency response (15 to 60KHz 3dB response)
*
.SUBCKT TransPP P1 B P2 Sp1 Sp2
+PARAMS: RP=25 LA=0.02234801 LB=22.3037318 RA=1000 RS=0.8
*
* primary
*
RP1 P1 1 {RP}
La1 1 2 {LA}
Lb1 2 B {LB}
Lb2 B 5 {LB}
La2 5 6 {LA}
RP2 6 P2 {RP}
LPA 3 4 {RA}
R1 2 3 1u
R2 5 4 1u
*
* secondary
*
Rs Sp1 9 {RS}
LSA 9 Sp2 1
*
* coupling
*
Kcore LPA LSA .999999
.ENDS TransPP
**************************************
*
* Audio Transformer,
* for single ended tube operation,
* with screen grid taps
*
* TransSESGT
*
* Primary Terminals
* P1 plate connection
* Sg1 screen grid connection
* B +Vpp connection
* Secondary terminals
* Sp1, Sp2 speaker connections
*
*
* Parameters :
* RP primary winding resistance
* LA primary winding inductance
* LP primary inductance, plate to screen tap
* LS primary inductance, screen tap to +Vpp tap
* RA impedance ratio, primary to secondary
* RS secondary winding resistance
*
* the default parameter values given are for an 8K to 8ohm transformer,
* with 40% screen grid tap,
* and 30 to 30KHz frequency response (15 to 60KHz 3dB response)
*
.SUBCKT TransSESGT P1 Sg1 B Sp1 Sp2
+PARAMS: RP=50 LA=0.04469602 LP=26.7644776 LS=17.842986 RA=1000 RS=0.8
*
* primary
*
RP1 P1 1 {RP}
La1 1 2 {LA}
LP1 2 7 {LP}
LS1 7 B {LS}
R3 7 Sg1 1u
LPA 3 4 {RA}
R1 2 3 1u
R2 B 4 1u
*
* secondary
*
Rs Sp1 9 {RS}
LSA 9 Sp2 1
*
* coupling
*
Kcore LPA LSA .999999
.ENDS TransSESGT
**************************************
*
* Audio Transformer,
* for single ended tube operation,
* without screen grid taps
*
* TransSE
*
* Primary Terminals
* P1 plate connections
* B +Vpp connection
* Secondary terminals
* Sp1, Sp2 speaker connections
*
*
* Parameters :
* RP primary winding resistance
* LA primary winding inductance, series
* LB primary inductance, parallel
* RA impedance ratio, primary to secondary
* RS secondary winding resistance
*
*
* the default parameter values given are for an 8K to 8ohm transformer,
* and 30 to 30KHz frequency response (15 to 60KHz 3dB response)
*
.SUBCKT TransSE P1 B Sp1 Sp2
+PARAMS: RP=50 LA=0.04469602 LB=44.6074636 RA=1000 RS=0.8
*
* primary
*
RP1 P1 1 {RP}
La1 1 2 {LA}
Lb1 2 B {LB}
LPA 3 4 {RA}
R1 2 3 1u
R2 B 4 1u
*
* secondary
*
Rs Sp1 9 {RS}
LSA 9 Sp2 1
*
* coupling
*
Kcore LPA LSA .999999
.ENDS TransSE
"Danny T" <t.d...@excite.com> wrote in message
news:aaa3b9fb.02040...@posting.google.com...
Danny T
that's just what I was looking for.
Patrick Turner
triodes work,
and whether or not there was or wasn't negative
feedback acting within them.
From reading from Terman's 1937 book, and
from what John Stewart posted in an article from
Wireless Engineer,
April 1953, by Stockman, I was left still is some
doubt
about a basic extended idea about triodes.
I had always thought of them as an equivalent of a
voltage generator of
mu x eg, with some resistance in series = Ra.
This basic model of any tube works well for
practical purposes.
But let's say someone gives us a tube and all we
know is mu = 20,
and gm = 2 ma / V.
Straightaway, since we know Gm = mu / Ra,
we could calculate Ra.
But the gm given was for the change of Ia / change
of eg, when the anode is
connected to a fixed B+, which it never is, because
a load resistance is
used between anode and B+.
So the Ia change per volt of grid voltage change is
never as high
as 2 ma / volt. Why?
The anode voltage, Va, has an effect on its own
current, Ip.
In fact, the plate and grid are at different
distances
from the cloud of electrons around the cathode,
and the distances chosen by the makers results in a
given mu for the tube.
mu takes into account both grid distance, and anode
distance.
And we can say that the anode voltage change will
cause an Ip change of Va x gm / u .
So in effect, "Plate gm" is gm / u.
when we test the tube, using a fixed grid voltage,
we find
changing plate voltage results in a current of Va /
Ra.
Since Ra = mu / gm, then Ia = Va x gm / mu.
So let's get back to the tube given to us.
We have identified it as a 6J5.
WE connect it up with a grounded grid, and a fixed
bias grid,
and a 100 k resistor from a B+ supply.
So we apply +1 volt to the grid.
since gm is 2 ma / volt, you might expect 2 ma
change in Ia,
and a swing of - 200 v in the 100 k load.
Why don't we get it?
Because the falling anode voltage attracts less
electrons,
and opposes the increase of current dictated by the
grid.
Suppose we arranged things so the anode voltage
wanted was 50 vrms, caused by some yet undertermined
grid voltage change.
Then Ia = 50 / 100 k = 0.5 ma.
If no grid voltage change happened at the grid,
The effect on plate current would be Va x gm / mu
= 50 x 2 / 20 = 5 ma
But we only see 0.5 ma of Ia change.
We can say then that the grid must have caused a
current change
of 5 ma + 0.5 ma, or 5.5 ma.
So you can see the grid says +5.5, the anode says
-5, and
only the balance of +0.5 is actually produced.
To make its bid of 5.5 ma on the size of plate
current,
and since we know Gm is 2 ma /V,
then the grid voltage change is 5.5 / 2 = 2.75
volts.
So we can say the tube gain, A = 50 / 2.75 = 18.18.
So we have been able to calculate tube gain without
the simple formula , A = mu x RL / ( Ra + RL ).
This would give 18.18 for the 6J5, with a 100 k RL.
Now suppose the 6J5 wasn't very linear, and some
distortion
voltage appeared at the anode, in the process of the
mutual interaction
of the anode and grid voltages on anode current.
We can see that this would occur by perusing the
plate voltage/ plate current curves,
which are not straight lines, and have differences
between themselves,
and slight differences between the spacings of Ip
for various Eg1 values.
Let's say the distortion anode voltage
which is produced by the distorted anode current
that would be produced by the grid voltage change
and was -dv.
Remember, no such distortion voltage exists at the
grid.
Then Ia would be reduced by -dv x gm / mu
so the distortion voltage which would try to appear
at the anode
were it not for the action of the anode on its own
current is much reduced.
As fast as some THD voltage tries to appear at the
anode, the effect
of this voltage turns itself off.
Not all of it does, of course, and what we see is
the distortion we get
with triodes.
Imperfections in the arrangements of wires and plate
and grid coils,
all lead to distortion, as well as the fact that
triode plate current
= a constant x square root of Ep cubed, for fixed
values of grid voltage,
and this leads to the curved lines on the data
graphs.
The most linear way we can use a triode is when RL =
very high ohms.
The fact is also that we cannot change what
has been arranged to happen inside the glassware by
the GOT.
So simple models of gene plus Ra will do fine for
everyone.
Patrick Turner.
Fred Nachbaur
that both viewpoints are valid, but that simplicity is the most practical.
Reminds of the story of the three blind men and the elephant; one
described it as a rope, another as a snake, and the third as a tree
trunk. They can each get a better understanding by walking around the
elephant and feeling what the other two feel, but real understanding
isn't achieved until they have a ride on it.
Fred N
Patrick Turner wrote:
--
+---------------------------------------+
| Music: http://www.netidea.com/~fredn/ |
| Vacuum Tube projects & other stuff: |
| http://www.fna.muohio.edu/dogstar/ |
+---------------------------------------+
Patrick Turner
Fred Nachbaur wrote:
> Terrific description, Patrick. You've travelled the full loop, showing
> that both viewpoints are valid, but that simplicity is the most practical.
Well, I feel a bit loopy.
>
> Reminds of the story of the three blind men and the elephant; one
> described it as a rope, another as a snake, and the third as a tree
> trunk. They can each get a better understanding by walking around the
> elephant and feeling what the other two feel, but real understanding
> isn't achieved until they have a ride on it.
Just as well none of them lingered at the rear end,
mighta gotta distorted opinion about nature,
in the form of a dark awakening from above.
Patrick Turner
BFoelsch
What happens if we don't let the plate voltage change? Perhaps the load is
an arbitrarily large capacitance, connected from plate to ground, with a
resistance from B+ to plate to supply current. Work is done my moving charge
in and out of the capacitor, but the voltage will change only a tiny bit if
the capacitor is large enough. A huge electrostatic speaker, perhaps?
I believe it is correct to say that this circuit would provide power gain
but no voltage gain.
Would the "inherent triode feedback" exist in this case or not? If it
doesn't exist in this case, is it truly inherent?
Perhaps by inherent we really mean "using the standard assumptions." There
is nothing wrong with that as long as we all agree on the assumptions.
No flame intended. Just food for thought.
"Patrick Turner" <in...@turneraudio.com.au> wrote in message
news:3CB19EF5...@turneraudio.com.au...
Fred Nachbaur
BFoelsch wrote:
> Just for giggles, consider the following...
>
> What happens if we don't let the plate voltage change? Perhaps the load is
> an arbitrarily large capacitance, connected from plate to ground, with a
> resistance from B+ to plate to supply current. Work is done my moving charge
> in and out of the capacitor, but the voltage will change only a tiny bit if
> the capacitor is large enough. A huge electrostatic speaker, perhaps?
>
> I believe it is correct to say that this circuit would provide power gain
> but no voltage gain.
Yes, indeed there would be power gain without voltage gain, just as in
the case of the common-grid (voltage follower) circuit.
> Would the "inherent triode feedback" exist in this case or not? If it
> doesn't exist in this case, is it truly inherent?
I would say not, because the "inherent feedback" relies on the plate
load resistor in order to operate. It's therefore not the triode itself
which has the "inherent feedback", witness the k*Vg^(4/3) plate current
relationship at fixed plate voltage, rather it's the triode *circuit* in
the common cathode configuration which exhibits this linearising
property, which we can describe as a feedback mechanism.
> Perhaps by inherent we really mean "using the standard assumptions." There
> is nothing wrong with that as long as we all agree on the assumptions.
>
> No flame intended. Just food for thought.
Indeed. Thanks for the "afternoon snack". ;-)
Fred N
John Byrns
"BFoelsch" <BFoe...@no.spam.att.net> wrote:
> Just for giggles, consider the following...
>
> What happens if we don't let the plate voltage change? Perhaps the load is
> an arbitrarily large capacitance, connected from plate to ground, with a
> resistance from B+ to plate to supply current. Work is done my moving charge
> in and out of the capacitor, but the voltage will change only a tiny bit if
> the capacitor is large enough. A huge electrostatic speaker, perhaps?
>
> I believe it is correct to say that this circuit would provide power gain
> but no voltage gain.
>
> Would the "inherent triode feedback" exist in this case or not? If it
> doesn't exist in this case, is it truly inherent?
I would say it still exists, you could use the same argument to say that
externally connected feedback doesn't exist either in a similar case. In
either case the negative feedback is doing what it is supposed to do,
jacking up the current as much as it can when driving a short or other
difficult load.
Regards,
John Byrns
Surf my web pages at, http://www.enteract.com/~jbyrns/index.html
Denis
> Just for giggles, consider the following...
>
> What happens if we don't let the plate voltage change? Perhaps the load is
> an arbitrarily large capacitance, connected from plate to ground, with a
> resistance from B+ to plate to supply current. Work is done my moving charge
> in and out of the capacitor, but the voltage will change only a tiny bit if
> the capacitor is large enough. A huge electrostatic speaker, perhaps?
>
> I believe it is correct to say that this circuit would provide power gain
> but no voltage gain.
>
> Would the "inherent triode feedback" exist in this case or not? If it
> doesn't exist in this case, is it truly inherent?
>
In this situation the internal feedback of a triode is close to zero.
Indeed, one can consider the triode feedback "inherent", simply
because the physical mechanism providing this feedback, i.e, the
transfer of some part of the output signal to the input of an
amplification device is always present inside this device, and there
is no need to connect additional components(resistors, transformers,
etc.) to establish a feedback.
> Perhaps by inherent we really mean "using the standard assumptions." There
> is nothing wrong with that as long as we all agree on the assumptions.
>
The internal feedback also works in such circuits as cathode follower
or grounded grid amp. In these cases another feedback loop appears in
addition to the inherent one. The unbypassed cathode resistor adds
some external feedback too.
Regards,
Denis
Patrick Turner
BFoelsch wrote:
> Just for giggles, consider the following...
>
> What happens if we don't let the plate voltage change? Perhaps the load is
> an arbitrarily large capacitance, connected from plate to ground, with a
> resistance from B+ to plate to supply current. Work is done my moving charge
> in and out of the capacitor, but the voltage will change only a tiny bit if
> the capacitor is large enough. A huge electrostatic speaker, perhaps?
>
> I believe it is correct to say that this circuit would provide power gain
> but no voltage gain.
Power = volts x amps.
When you shunt the plate to ground via an electrolytic cap,
there will be almost no output plate voltage, and
almost no output work done, hence almost no sound.
If the tube is SET, there can then be a maximum
and minimum current in and out of the cap.
there will be a lot of distortion when clipping in the current wave form occurs,
generally at grid current.
In a PP class AB amp, the low load will cause much higher than
normal average current flow, and plate dissipation, and the tubes will over heat
with
a sine wave taken to current clipping.
You can see what the current swing will be if you plot
the load line so it is vertical, through the operating point, where C is very
large.
When C is not so large, it becomes an eliptical load line,
due to phase shift.
>
> Would the "inherent triode feedback" exist in this case or not? If it
> doesn't exist in this case, is it truly inherent?
It does not exist in this case, and with a shunted output,
the triode acts like a pentode.
AS soon as the plate is free to wander, with a load
that gives the tube gain, the effect of the plate voltage changes is felt
by the cathode electrons.
With the plate shorted, +1 volt grid change causes +0.011 amps current change in
a triode connected EL34.
But when there is a load, the voltage swings down with the increased load
current,
and this changing electric voltage field tells the cathode to stop sending
so many electrons, even though the grid has told the cathode to send a lot.
The load current change is always less than when there is no load, ie,
a short circuit.
In fact, when an infinite resistance, ie a high value load, is placed as a load
on the triode,
the grid says I want ten electrons to flow, and when the plate voltage descends,
the plate says, no, no, we don't want any more electrons today.
and no current increase occurs, just a voltage change at electrodes.
The plate and grid are doing deals all the time.
>
> Perhaps by inherent we really mean "using the standard assumptions." There
> is nothing wrong with that as long as we all agree on the assumptions.
Inherent means "as percieved by the describer to be within
the device" , IMHO.
> No flame intended. Just food for thought.
Yup,
Patrick Turner.
Patrick Turner
Fred Nachbaur wrote:
> BFoelsch wrote:
>
> > Just for giggles, consider the following...
> >
> > What happens if we don't let the plate voltage change? Perhaps the load is
> > an arbitrarily large capacitance, connected from plate to ground, with a
> > resistance from B+ to plate to supply current. Work is done my moving charge
> > in and out of the capacitor, but the voltage will change only a tiny bit if
> > the capacitor is large enough. A huge electrostatic speaker, perhaps?
> >
> > I believe it is correct to say that this circuit would provide power gain
> > but no voltage gain.
>
> Yes, indeed there would be power gain without voltage gain, just as in
> the case of the common-grid (voltage follower) circuit.
>
> > Would the "inherent triode feedback" exist in this case or not? If it
> > doesn't exist in this case, is it truly inherent?
>
> I would say not, because the "inherent feedback" relies on the plate
> load resistor in order to operate. It's therefore not the triode itself
> which has the "inherent feedback", witness the k*Vg^(4/3) plate current
> relationship at fixed plate voltage, rather it's the triode *circuit* in
> the common cathode configuration which exhibits this linearising
> property, which we can describe as a feedback mechanism.
So the triode depends on anode voltage change for its feedback.
But that's all.
With a pentode, there is no such self regulatory process,
so you gotta incude some other loop of feedback,
such as a cathode resistor, which raises Ro, reduces THD,
or some shunt FB, using two resistors, to get voltage feedback,
which reduces THD, and Ro, etc.
Go back to the shunted plate condition.
when + 1 grid volts occurs, you get +1 x gm of current,
which is the most the triode can muster, for a volt of grid change.
But when a load is placed, less current is made for the same 1 volt
at the grid.
Something is telling the triode to ease off with plate current
as we increase the load value.
Its the plate voltage doing it.
Put it another way, say you have a load, and for 1 volt of grid change,
you get 2 ma of load current change.
When you short circuit that load, for the same grid change, you get
11 ma produced.
Something is telling the tube to make more current for the same
grid voltage change, in a vain attempt to maintain output voltage.
This is how all feedback amps behave.
We have found feathers, see webbed footprints, and hear ducks quacking,
maybe there are ducks around.
Patrick Turner.