Push Pull P-P Impedance

[Ed]

Hi;

If the typical plate to plate load impedance required of a pair of
power output tubes(in a push pull configuration) is z= 4000 ohms, each
tube each tube sees z=1000 ohms, am I correct?

In order to double the wattage of the amplifier design, I can connect
two power tubes in parallel on each side and use a transformer that
will cause the output tubes to see a p-p impedance of 2000 ohms.

Correct?

Sorry if these seem like dumb questions.

Thanks

Ed

[Patrick Turner]

Ed wrote:

> Hi;
>
> If the typical plate to plate load impedance required of a pair of
> power output tubes(in a push pull configuration) is z= 4000 ohms, each
> tube each tube sees z=1000 ohms, am I correct?

Only if the amp is a class B amp.

If the amp is a class A amp, then the load each tube sees is 2k.

IN a class AB amp, the load starts off being 2k for each tube,
but then reduces to 1k for the class B part of the cycle, after one tube
is cut off,

>
> In order to double the wattage of the amplifier design, I can connect
> two power tubes in parallel on each side and use a transformer that
> will cause the output tubes to see a p-p impedance of 2000 ohms.

Yes.
The operating conditions of a tube will be the same for
2 tubes and 4k load.

But the losses as heat in the output tranny will be double
for the 4 tube amp, if the same OPT is used for either case,
so one never gets exactly twice the power from the quad of tubes.

>
>
> Correct?
>
> Sorry if these seem like dumb questions.

Not dumb. Many have trouble visualizing what goes on in
amplifiers.

Patrick Turner.

>
>
> Thanks
>
> Ed

[Ed]

On Mon, 01 Dec 2003 16:15:33 +1100, Patrick Turner
> wrote:


>Not dumb. Many have trouble visualizing what goes on in
>amplifiers.
>
>Patrick Turner.

How is the plate to plate load resistance of an OPT calculated? I
would assume that the reflected load resistance is equal to the


Zr * RL

impedance ratio Zr
Load Resistance on secondary RL

So a 25 to 1 OPT's reflected impedance at an 8 ohm secondary load
would be equal to (25/1) ² * 8 or 5000 ohms

Since the reflected impedance per center tap is 1/4 of the total
impedance, the reflected impedance to each half of the transformer is
1250 ohms.

What am I missing in my above calculations?

Thanks

Ed













>
>>
>>
>> Thanks
>>
>> Ed

[Phil Allison]

"Ed" > wrote in message
...

> How is the plate to plate load resistance of an OPT calculated? I
> would assume that the reflected load resistance is equal to the
>
>
> Zr * RL
>
> impedance ratio Zr
> Load Resistance on secondary RL
>
> So a 25 to 1 OPT's reflected impedance at an 8 ohm secondary load
> would be equal to (25/1) ² * 8 or 5000 ohms
>
> Since the reflected impedance per center tap is 1/4 of the total
> impedance, the reflected impedance to each half of the transformer is
> 1250 ohms.
>
> What am I missing in my above calculations?


** Nothing - but you are describing the transformer alone.

When two or more tubes are driving it one can "see" the other/s tubes
helping.




............ Phil

[Patrick Turner]

Ed wrote:

> On Mon, 01 Dec 2003 16:15:33 +1100, Patrick Turner
> > wrote:
>
> >Not dumb. Many have trouble visualizing what goes on in
> >amplifiers.
> >
> >Patrick Turner.
>
> How is the plate to plate load resistance of an OPT calculated? I
> would assume that the reflected load resistance is equal to the
>
> Zr * RL
>
> impedance ratio Zr
> Load Resistance on secondary RL
>
> So a 25 to 1 OPT's reflected impedance at an 8 ohm secondary load
> would be equal to (25/1) ² * 8 or 5000 ohms
>
> Since the reflected impedance per center tap is 1/4 of the total
> impedance, the reflected impedance to each half of the transformer is
> 1250 ohms.

The load each tube sees in a two tube PP amp is not
always 1/4 of the anode to anode load of 5 k that you have correctly
calculated.

Imagine you remove one tube.
The transformer then has only half its P winding connected,
with the other half with no current, so effectively,
the transformer ratio becomes 1,25k to 8 ohms.

Now in a class B amp, when one tube has a negative grid voltage applied,
its current is cut off, and its as if someone plucked the tube out of the
amp.
The other tube gets a +ve grid voltage and the load it sees is 1.25k.
Each tube takes its turn to amplify only the +ve or -ve parts of an
applied
signal wave.
But the voltage applied to the load still appears at both sides of the
primary,
so if we measure the impedance ratio between the whole P and S, its still

5k to 8 ohms.

The action is different for a class A PP amp, where each tube does not
have its current cut off, and BOTH remain connected to the load,
and both act as if they each see 2.5k, like a pair of SE amps, except
that
in a PP amp the anode voltage are opposite phase and applied to the ends
of an OPT,
so the loads of the tube are said to be in series, so 5k is the load
anode to anode,
and current flows all the time, and nothing cuts off.

As one tube turns on harder to make its anode voltage get more -ve,
the other tube is turning off as its anode voltage gets more +ve.
The summed effect is that both tubes contribute nearly equally to the
power at the output, and if you doubt this, then measure
all the currents and voltages involved.

I say that each tubes conribution is nearly equal, but the 2H distortion
means that one tube pulls harder than the other pushes,
and perfect cancellation of 2H in a PP stage does not occur,
but it is still reduced almost completely for most of the power spread in
a PP
class A amp.

Since the distortion currents exist, each tube's share of the load is
affected by the others,
and the load line each tube sees is not a constant value load,
and it is described best by a curved line, not a straight one.

Load line analysis assists us in producing a visual mental
picture, and I suggest you spend some time
reading about and drawing up some load lines across
the data curves for some triodes so you understand load lines,
and what I am trying to tell you. In the process, you will undestand what

each curved line of the data sheets mean, and you need to be clear about
that as well.
RDH4 is an excellent book from which to learn how to
draw load lines for triodes, pentodes, beam tetrodes,
or any other tubes that exist,
and the exerience could be applied to transistor circuits.


In class AB amps, the tube start off sharing the load, with a nearly
constant load value, but when one tube reaches its cut off point, the
other tube
then sees a load which is half that when both tubes are conducting in the
class A amp.

The load line for one tube then is a very much curved line on
any graphical depiction of its load, starting at 2.5k, and swinging to
1.25k.
The triode transition from A to AB is the gentlest, and one which
generates the
least harmonics, just mainly 3H.
The transfer curve for such amps is a line like an S.
Pentode or beam tetrode amps have kinked load line for one of the
tubes taken alone, because of their sudden cut off behaviour,
and the transfer curve for both tubes looks more like a Z,
and the production of odd order harmonics will contain much more 5,7,9H,
as well as the dominant 3H.

>
>
> What am I missing in my above calculations?

The mental concept.

It takes time to accept.
Try measuring the amp, with little 10 ohm cathode and anode resistors,
to enable current measurements.
Use a cro, and watch the distortions in the current waves, compared to
the relatively
undistorted voltage waves at the output.

And read the old books,

Then it gets clearer.

Patrick Turner.

[Ian Iveson]

"Ed" > wrote in message
...
> Hi;
>
> If the typical plate to plate load impedance required of a pair of
> power output tubes(in a push pull configuration) is z= 4000 ohms,
each
> tube each tube sees z=1000 ohms, am I correct?
>
> In order to double the wattage of the amplifier design, I can
connect
> two power tubes in parallel on each side and use a transformer
that
> will cause the output tubes to see a p-p impedance of 2000 ohms.
>
> Correct?
>
> Sorry if these seem like dumb questions.

Ed, each valve sees the (relatively passive and constant) reflected
secondary impedance in parallel with the (varying and active)
reflected impedance of the other valve.

Perhaps you cannot really, sensibly, separate the two valves as you
are trying to do, because they are interdependent as long as both
have bias current.

For a full appreciation, you might try plotting the anode
characteristics of a PP pair. There is a good explanation of this
somewhere. What you get is a much more constant output impedance
than a SE valve, because as one rises the other falls.

Er...Norman Koren or Steve Bench?...found it...

http://members.aol.com/sbench102/composit.html

cheers, Ian

[Paul D. Spiegel]

I can see where Ed is getting the quarter load from. Both the _Radio
Designer's Handbook_ and Terman's _Radio Engineering_ describe the slope
of a composite PP class A triode load line as "...one fourth the
plate-to-plate load resistance" (Ternan) "...corresponding to Rl'=1/4Rl,
where Rl = load resistance plate to plate" (RDH). The graphic examples
in Ternan seem to support the 1/4 primary load.

[Patrick Turner]

"Paul D. Spiegel" wrote:

> I can see where Ed is getting the quarter load from. Both the _Radio
> Designer's Handbook_ and Terman's _Radio Engineering_ describe the slope
> of a composite PP class A triode load line as "...one fourth the
> plate-to-plate load resistance" (Ternan) "...corresponding to Rl'=1/4Rl,
> where Rl = load resistance plate to plate" (RDH). The graphic examples
> in Ternan seem to support the 1/4 primary load.

The load for one tube whilst in class A is 1/2 the RLa-a.
Whilst in class B, the load is 1/4 RLa-a.
The load seen by one tube in a class AB amp varies from 1/2
to 1/4 RLa-a during a full power cycle, and the load line appears as a curve

across the data curves.

Patrick Turner.