Take one triode section of a 12AX7, with a 100K plate resistor. If I
tie the section of the tube to the same plate resistor, am I doubling
the current through the resistor? I'm trying to understand the gain
effects of running parallel preamp stages.
Thanks.

wrote

Take one triode section of a 12AX7, with a 100K plate resistor. If
I
tie the section of the tube to the same plate resistor, am I
doubling
the current through the resistor? I'm trying to understand the
gain
effects of running parallel preamp stages.
Thanks.

Assuming much about what you haven't said, er, no.

If you parallel them and use a 50k anode resistor, then you will
double the total current. The same is true of the cathode resistor
and grid stopper and grid resistor, but you should double the value
of the bypass and coupling caps.

But it is obvious if you think like this. Take two identical and
independent circuits, each with an ECC83 and a 100k anode resistor.
The voltage at each electrode will be the same for both valves. So
if you connect corresponding pins it won't make any difference. The
resistors halve when they are parallel, the caps double.

cheers, Ian

wrote:

Take one triode section of a 12AX7, with a 100K plate resistor. If I
tie the section of the tube to the same plate resistor, am I doubling
the current through the resistor? I'm trying to understand the gain
effects of running parallel preamp stages.
Thanks.

The gain of a 1/2 section of a 12AX7 with 100k RL, and fully
bypassed cathode resistor is as follows :-

Gain for any tube, A, = U x RL / ( RL + Ra,

where A = Gain, which isn this case is truly a neagtive number because
a positive going grid voltage causes a negative going plate voltage.
But let us not worry about the output signal polarity right now.
U is a amplification factor of the tube from the data sheets,
RL is the plate load resistance, ie the plate DC supply resistance,
in parallel with the following cap coupled grid bias resistor in the
following stage,
or any other resistor, if there is one.
Ra is the dynamic plate resistance from the tube data sheets.

So for a 1/2 12AX7, where RL = 100kohms, Ra = 65 kohms, and U = 100,
A = 100 x 100,000 / ( 100,000 + 65,000 )
= 60.6.

So for -1 volt change at the grid, you should measure + 60.6v at the
plate.
Slight variations from the data will not enable the theoretical gain to
occur, but
it should be within 7%, if you have Ia, the plate current at idle = about
0.7 mA,
and a supply voltage of about +210V, so Ea, the DC voltage between plate
and cathode is about
140V, and the voltage across the 100k is 70V.

When you connect the two 1/2 section of the 12AX7 together,
you make a new triode whose U will remain at 100,
but the Ra will be 1/2 the 65kohms of one 1/2 of the tube.
If you set have the same +210V supply, and you have 100k for RL,
then you won't get Ea = 140V, since the load current will make the
voltage across RL
about 100V, so 110V will be across the tube.
So to get the two parallel sections to have the same Ia and Ea as in the
case where
one section was at 140V and 0.7 mA, the
supply voltage has to be raised to +280V, so that you have Ea = 140V,
ERL = 140V, and Ia for both sections = 1.4 mA, and a cathode bias
resistor
used for one triode has also to be used on the second triode.
Or put another way, if the two 1/2 sections are in parallel, the cathodes

are connected together, and the bias resistor for both tubes is half the
value
used for one triode section.
..

The gain for parallel tubes will be slightly higher than for one 1/2
section of the tube,
and you can calculate it out for yourself,
and confirm the calculation with measurement.

If you reduce the RL from 100k to 50k, and parallel the two 1/2 sections,

the gain will be the same as for one triode section with RL = 100k.

The effect of connecting two triodes in parallel makes the output
resistance of the
stage, ie, the output resistance when measured at the plates measure
lower than
with one triode, and this may be useful where wide bandwidth is wanted.

The Rout, ie, output resistance = Ra in parallel with RL,
and for 65k and 100k, Rout = 39k for one triode.
For two triodes with RL = 100k for both, Rout = 24k, a big drop.

Patrick Turner.

Take one triode section of a 12AX7, with a 100K plate resistor. If I
tie the (other) section of the tube to the same plate resistor, am I
doubling
the current through the resistor? I'm trying to understand the gain
effects of running parallel preamp stages.



** Where several tubes are paralleled the plate current increases in direct
proportion to the number of tubes. For example: operating multiple tubes in
parallel in an output stage increases the output current for the same supply
voltage and drive voltage conditions. This means the load impedance can be
made smaller by the same ratio to take advantage of the increased current
available.




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

"Patrick Turner" wrote in message
...


wrote:

Take one triode section of a 12AX7, with a 100K plate resistor. If I
tie the section of the tube to the same plate resistor, am I doubling
the current through the resistor? I'm trying to understand the gain
effects of running parallel preamp stages.
Thanks.

The gain of a 1/2 section of a 12AX7 with 100k RL, and fully
bypassed cathode resistor is as follows :-

Gain for any tube, A, = U x RL / ( RL + Ra,

where A = Gain, which isn this case is truly a neagtive number because
a positive going grid voltage causes a negative going plate voltage.
But let us not worry about the output signal polarity right now.
U is a amplification factor of the tube from the data sheets,
RL is the plate load resistance, ie the plate DC supply resistance,
in parallel with the following cap coupled grid bias resistor in the
following stage,
or any other resistor, if there is one.
Ra is the dynamic plate resistance from the tube data sheets.

So for a 1/2 12AX7, where RL = 100kohms, Ra = 65 kohms, and U = 100,
A = 100 x 100,000 / ( 100,000 + 65,000 )
= 60.6.

So for -1 volt change at the grid, you should measure + 60.6v at the
plate.
Slight variations from the data will not enable the theoretical gain to
occur, but
it should be within 7%, if you have Ia, the plate current at idle = about
0.7 mA,
and a supply voltage of about +210V, so Ea, the DC voltage between plate
and cathode is about
140V, and the voltage across the 100k is 70V.

When you connect the two 1/2 section of the 12AX7 together,
you make a new triode whose U will remain at 100,
but the Ra will be 1/2 the 65kohms of one 1/2 of the tube.
If you set have the same +210V supply, and you have 100k for RL,
then you won't get Ea = 140V, since the load current will make the
voltage across RL
about 100V, so 110V will be across the tube.
So to get the two parallel sections to have the same Ia and Ea as in the
case where
one section was at 140V and 0.7 mA, the
supply voltage has to be raised to +280V, so that you have Ea = 140V,
ERL = 140V, and Ia for both sections = 1.4 mA, and a cathode bias
resistor
used for one triode has also to be used on the second triode.
Or put another way, if the two 1/2 sections are in parallel, the cathodes

are connected together, and the bias resistor for both tubes is half the
value
used for one triode section.
.

The gain for parallel tubes will be slightly higher than for one 1/2
section of the tube,
and you can calculate it out for yourself,
and confirm the calculation with measurement.

If you reduce the RL from 100k to 50k, and parallel the two 1/2 sections,

the gain will be the same as for one triode section with RL = 100k.

The effect of connecting two triodes in parallel makes the output
resistance of the
stage, ie, the output resistance when measured at the plates measure
lower than
with one triode, and this may be useful where wide bandwidth is wanted.

The Rout, ie, output resistance = Ra in parallel with RL,
and for 65k and 100k, Rout = 39k for one triode.
For two triodes with RL = 100k for both, Rout = 24k, a big drop.

Patrick Turner.


Hi Patrick -
Hey, a nice complete answer! I've got a question about paralleling toobs,
though - the reduction of noise through paralleling devices - here's the
theory, as I see it:
In my simplified model, a toob circuit produces X gain on signal A,
(download Tom Wait's "on the road" - just came on & made me feel better) and
Y *noise*. I'm ignoring the distortion here - just random Brownian noise,
and other "random" noise. This noise (to be redundant) is completely random
in nature, while the *signal* is not.
It seems to follow that when you add 2 sections of a tube together, the
signal''s amplification would be mostly additive, while the noise would *not
be*. Again, repeating myself, we're not talking about common-mode stuff
like distortion - just the noise.
Dumming it down further, mainly for myself, imagine 2 jars, 1 labeled
signal, other labeled "noise". With each iteration of a parallel device,
you flip a coin, and place a coin in the "signal" can, and, *if the coin
lands on "heads"*, you put a coin in the "noise" can. With this model, it's
clear that S/N ratio improves with every iteration of this process, though
with progressively diminishing returns.
The math is simple, and i know that many SS devices use this method (IC's
containing hundreds of paralleled devices) to improve S/N ratio, but...
I am ignoring a bunch of variables - can you list the ones you can think
of? Of course the input C increases, interelement C increases, input R
decreases, but what other factors are involved?
As you may have guessed, I'm trying to figure out what to use for the
first stage of a phono preamp. I think I could afford the increased C at
the input - while Morgan Jones is thinking in terms ofstandard "stereo
grade" RCA cable, there's an abundance of really nice microwave cable, with
really low losses / C per foot. When you get into GHz area, cable losses
are grotesquely high, and manufacturers jump through hoops. The stuff's in
surplus stores now. Most designed for small signal (~200mW), better than
'scope probe cable.
Any thoughts?
-dim

shiva wrote:

"Patrick Turner" wrote in message
...


wrote:

Take one triode section of a 12AX7, with a 100K plate resistor. If I
tie the section of the tube to the same plate resistor, am I doubling
the current through the resistor? I'm trying to understand the gain
effects of running parallel preamp stages.
Thanks.

The gain of a 1/2 section of a 12AX7 with 100k RL, and fully
bypassed cathode resistor is as follows :-

Gain for any tube, A, = U x RL / ( RL + Ra,

where A = Gain, which isn this case is truly a neagtive number because
a positive going grid voltage causes a negative going plate voltage.
But let us not worry about the output signal polarity right now.
U is a amplification factor of the tube from the data sheets,
RL is the plate load resistance, ie the plate DC supply resistance,
in parallel with the following cap coupled grid bias resistor in the
following stage,
or any other resistor, if there is one.
Ra is the dynamic plate resistance from the tube data sheets.

So for a 1/2 12AX7, where RL = 100kohms, Ra = 65 kohms, and U = 100,
A = 100 x 100,000 / ( 100,000 + 65,000 )
= 60.6.

So for -1 volt change at the grid, you should measure + 60.6v at the
plate.
Slight variations from the data will not enable the theoretical gain to
occur, but
it should be within 7%, if you have Ia, the plate current at idle = about
0.7 mA,
and a supply voltage of about +210V, so Ea, the DC voltage between plate
and cathode is about
140V, and the voltage across the 100k is 70V.

When you connect the two 1/2 section of the 12AX7 together,
you make a new triode whose U will remain at 100,
but the Ra will be 1/2 the 65kohms of one 1/2 of the tube.
If you set have the same +210V supply, and you have 100k for RL,
then you won't get Ea = 140V, since the load current will make the
voltage across RL
about 100V, so 110V will be across the tube.
So to get the two parallel sections to have the same Ia and Ea as in the
case where
one section was at 140V and 0.7 mA, the
supply voltage has to be raised to +280V, so that you have Ea = 140V,
ERL = 140V, and Ia for both sections = 1.4 mA, and a cathode bias
resistor
used for one triode has also to be used on the second triode.
Or put another way, if the two 1/2 sections are in parallel, the cathodes

are connected together, and the bias resistor for both tubes is half the
value
used for one triode section.
.

The gain for parallel tubes will be slightly higher than for one 1/2
section of the tube,
and you can calculate it out for yourself,
and confirm the calculation with measurement.

If you reduce the RL from 100k to 50k, and parallel the two 1/2 sections,

the gain will be the same as for one triode section with RL = 100k.

The effect of connecting two triodes in parallel makes the output
resistance of the
stage, ie, the output resistance when measured at the plates measure
lower than
with one triode, and this may be useful where wide bandwidth is wanted.

The Rout, ie, output resistance = Ra in parallel with RL,
and for 65k and 100k, Rout = 39k for one triode.
For two triodes with RL = 100k for both, Rout = 24k, a big drop.

Patrick Turner.


Hi Patrick -
Hey, a nice complete answer! I've got a question about paralleling toobs,
though - the reduction of noise through paralleling devices - here's the
theory, as I see it:
In my simplified model, a toob circuit produces X gain on signal A,
(download Tom Wait's "on the road" - just came on & made me feel better) and
Y *noise*. I'm ignoring the distortion here - just random Brownian noise,
and other "random" noise. This noise (to be redundant) is completely random
in nature, while the *signal* is not.
It seems to follow that when you add 2 sections of a tube together, the
signal''s amplification would be mostly additive, while the noise would *not
be*. Again, repeating myself, we're not talking about common-mode stuff
like distortion - just the noise.
Dumming it down further, mainly for myself, imagine 2 jars, 1 labeled
signal, other labeled "noise". With each iteration of a parallel device,
you flip a coin, and place a coin in the "signal" can, and, *if the coin
lands on "heads"*, you put a coin in the "noise" can. With this model, it's
clear that S/N ratio improves with every iteration of this process, though
with progressively diminishing returns.
The math is simple, and i know that many SS devices use this method (IC's
containing hundreds of paralleled devices) to improve S/N ratio, but...
I am ignoring a bunch of variables - can you list the ones you can think
of? Of course the input C increases, interelement C increases, input R
decreases, but what other factors are involved?
As you may have guessed, I'm trying to figure out what to use for the
first stage of a phono preamp. I think I could afford the increased C at
the input - while Morgan Jones is thinking in terms ofstandard "stereo
grade" RCA cable, there's an abundance of really nice microwave cable, with
really low losses / C per foot. When you get into GHz area, cable losses
are grotesquely high, and manufacturers jump through hoops. The stuff's in
surplus stores now. Most designed for small signal (~200mW), better than
'scope probe cable.
Any thoughts?
-dim

The use of higher Gm tubes results in a lower
"equivalent noise resistance" and I suggest strongly that you read RHH4
on the subject of tube input noise causes, and search the eb
on noise, rather than expect 20 pages from me on the matter.

In theory, the doubling of gm by parallelling two tubes will reduce the noise
by 1 / sq.rt of devices paralleled.

So 5 12AX7 tubes with 10 triodes all in parallel will have noise
of one triode x 1 / 3.16, or about 1/3 of the noise.
That's in theory.
The input equivalent 'noise resistance' = 2.5 / gm.
So a 6DJ8 with both halves paralleled should have about
1/3 the noise of a 1/2 of a 12AX7.

Unfortunately, this seldom actually occurs in practice.

Theory says noise varies as the square root of resistance, so
reducing NR by 4 times reduces noise by half,
or increasing NR 4 times gives twice the noise.


But tubes still are OK for MM phono inputs, or MC with a step up tranny.
A good 1/2 of a 12AX7 has a calculated NR = 2.5 / 1.6 mA/V, so 1.6kohms,
and that makes about 2 uV of noise, so if gain = 60, expect 120 uV of noise at
the anode,
plus the noise of the 100k load in parallel with ra, so for 40k
the plate noise at the plate is 10 uV, and this has a tiny
effect on the noise already at the anode from the input.

So if there is 2 mV of signal at the input, the
SNR will be approx 2 mV / 2 uV, or 1000:1,
which an snr = 60 dB.

What happens at the input is the chief noise determing factor.

But for direct amplification of an MC cart, a j-fet cascode circuit is by far
the
quietest circuit, since fet eq noise resistance = 0.7 / gm.

With a 2SK147, or 2SK369, Gm at 5 mA = 40 mA/V, so
NR = 0.7 / 0.04 A/V = 17.5 ohms, and the noise
generated is about 10 times less than a 1/2 a 12AX7,
say 0.2 uV.
Then if the signal from an MC is 0.4 mV, and noise is 0.2 uV,
SNR = 0.4mV / 0.2uV = 2000:1 = 66 dB, unweighted.

The RIAA filter also has an effect to reduce noise.

In practice, MM tube stages are barely quiet enough
for loud replay of vinyl, since the amp snr is only just lower
than the noise made by the vinyl, even with an unmodulated groove.
And the nature of tube noise is worse with more flicker noise and hum.
With a fet input the noise seems more highly pitched, and far less noticeable,
and its is a lower level than any normal tube.

Triodes such as the 6C45pi may good enough for MC, since they have
gm up to 35mA/V.

For a quiet MC amp see http://www.vacuumstate.com

Patrick Turner.