I have just completed building a class-C AM transmitter with an 829B (two
tetrodes in parallel) driven by a T-368 VFO.

The 829B bias is a fixed -33V (keeping the tube safe against lack of drive) plus
the bias which develops across the grid resistor. It works fine, but I still
have to adjust the grid resistor value for optimum performance

In the 829B data sheet I read a value of about 7 kohm. In my Geloso AM
transmitter (single 6146) they use 12 kohm. In the T-368 (single 4-400A) they
use about 23 kohm.

My practical experience is that lowering the grid resistor I always get more
output power FOR THE SAME GRID CURRENT. In other words, every time I try a
different resistor value I re-adjust the drive power so as to make the grid
current equal to the allowable limit of 15 mA.

At the moment I ended up using just 470 ohm but, despite that value gives me
good output power, I suspect that it may cause some drawbacks that I cannot
presently figure out.

Does anyone have a CLEAR understanding of the trade-offs involved in selecting
the grid bias resistor value? Or where I can find a practical and coincise
discussion of the issue?

Thanks & 73

Tony I0JX
Rome, Italy

On Mon, 1 Feb 2010 21:52:11 +0100, "Antonio Vernucci"
wrote:

I have just completed building a class-C AM transmitter with an 829B (two
tetrodes in parallel) driven by a T-368 VFO.

The 829B bias is a fixed -33V (keeping the tube safe against lack of drive) plus
the bias which develops across the grid resistor. It works fine, but I still
have to adjust the grid resistor value for optimum performance

In the 829B data sheet I read a value of about 7 kohm. In my Geloso AM
transmitter (single 6146) they use 12 kohm. In the T-368 (single 4-400A) they
use about 23 kohm.

My practical experience is that lowering the grid resistor I always get more
output power FOR THE SAME GRID CURRENT. In other words, every time I try a
different resistor value I re-adjust the drive power so as to make the grid
current equal to the allowable limit of 15 mA.

At the moment I ended up using just 470 ohm but, despite that value gives me
good output power, I suspect that it may cause some drawbacks that I cannot
presently figure out.

Does anyone have a CLEAR understanding of the trade-offs involved in selecting
the grid bias resistor value? Or where I can find a practical and coincise
discussion of the issue?

Thanks & 73

Tony I0JX
Rome, Italy


I cannot give you actual values for your set up, but basically, if you
have already provided optimum bias with your fixed bias voltage, then
you will not need to develop any more bias across a grid resistor.

peter

I cannot give you actual values for your set up, but basically, if you
have already provided optimum bias with your fixed bias voltage, then
you will not need to develop any more bias across a grid resistor.

peter

My fixed bias is set for an idling plate current of 10 mA. The only criterion
behind that bias setting was to keep the final tube safe. So, it may not be
optimum with regard to the final stage efficiency.

On the contrary, it is surely not optimum because, without extra bias, the tube
operates class B. More bias is needed (produced by the grid resistor) to have
the tube operating class C.

73

Tony I0JX

On Feb 2, 6:29*pm, "Antonio Vernucci" wrote:
I cannot give you actual values for your set up, but basically, if you
have already provided optimum bias with your fixed bias voltage, then
you will not need to develop any more bias across a grid resistor.

peter

My fixed bias is set for an idling plate current of 10 mA. The only criterion
behind that bias setting was to keep the final tube safe. So, it may not be
optimum with regard to the final stage efficiency.

On the contrary, it is surely not optimum because, without extra bias, the tube
operates class B. More bias is needed (produced by the grid resistor) to have
the tube operating class C.

73

Tony I0JX

With a fixed bias supply , do you really need a grid resistor as
such ? how is the -Ve bias reaching the grid ? ....with the
cathode earthed .. the fixed bias is 'required' .. I would think for
a given drive level ... increasing the -Ve grid bias would reduce
the conduction angle and slide the stage from class b to c as
the peak drive level would need to overcome the bias to allow
the grid to conduct .. thus the conduction angle is reduced ?

G ..

On Feb 2, 7:05*pm, Graham wrote:
On Feb 2, 6:29*pm, "Antonio Vernucci" wrote:





I cannot give you actual values for your set up, but basically, if you
have already provided optimum bias with your fixed bias voltage, then
you will not need to develop any more bias across a grid resistor.

peter

My fixed bias is set for an idling plate current of 10 mA. The only criterion
behind that bias setting was to keep the final tube safe. So, it may not be
optimum with regard to the final stage efficiency.

On the contrary, it is surely not optimum because, without extra bias, the tube
operates class B. More bias is needed (produced by the grid resistor) to have
the tube operating class C.

73

Tony I0JX

With a fixed bias supply *, do *you *really *need a *grid *resistor as
such ? how is the *-Ve *bias reaching the *grid ? ....with the
cathode earthed .. the fixed bias is *'required' *.. I would think for
a *given drive *level ... increasing the *-Ve *grid bias would *reduce
the *conduction angle and slide the *stage *from class b to *c *as
the *peak drive level *would need *to *overcome the *bias to allow
the *grid to *conduct .. thus the *conduction angle is reduced ?

G ..- Hide quoted text -

- Show quoted text -

The original data has guide lines on the vlaves use in class C
telephony .....

http://www.r-type.org/pdfs/829b.pdf

G ..

With a fixed bias supply , do you really need a grid resistor as
such ?

Yes, because the fixed bias I have chosen causes the tube to operate in class B,
whilst I wish it to operate in class C.

how is the -Ve bias reaching the grid ? ....with the
cathode earthed .. the fixed bias is 'required' .. I would think for
a given drive level ... increasing the -Ve grid bias would reduce
the conduction angle and slide the stage from class b to c as
the peak drive level would need to overcome the bias to allow
the grid to conduct .. thus the conduction angle is reduced ?

G ..- Hide quoted text -

- Show quoted text -

I agree with your statements but they do not help me much with regard to my
original doubts

The original data has guide lines on the vlaves use in class C
telephony .....

http://www.r-type.org/pdfs/829b.pdf

I have several data sheets for the 828B of various manufacturers but with the
grid resistor value they specify I obtain less output power than with a lower
resistance value

73

Tony I0JX

G ..

On Mon, 01 Feb 2010 21:52:11 +0100, Antonio Vernucci wrote:

I have just completed building a class-C AM transmitter with an 829B
(two tetrodes in parallel) driven by a T-368 VFO.

The 829B bias is a fixed -33V (keeping the tube safe against lack of
drive) plus the bias which develops across the grid resistor. It works
fine, but I still have to adjust the grid resistor value for optimum
performance

In the 829B data sheet I read a value of about 7 kohm. In my Geloso AM
transmitter (single 6146) they use 12 kohm. In the T-368 (single 4-400A)
they use about 23 kohm.

My practical experience is that lowering the grid resistor I always get
more output power FOR THE SAME GRID CURRENT. In other words, every time
I try a different resistor value I re-adjust the drive power so as to
make the grid current equal to the allowable limit of 15 mA.

At the moment I ended up using just 470 ohm but, despite that value
gives me good output power, I suspect that it may cause some drawbacks
that I cannot presently figure out.

Does anyone have a CLEAR understanding of the trade-offs involved in
selecting the grid bias resistor value? Or where I can find a practical
and coincise discussion of the issue?

Thanks & 73

Tony I0JX
Rome, Italy

The higher the grid resistor, the higher the bias voltage that must be
overcome by the drive. Hence, higher drive, more power lost in the grid
resistor, and lower conduction angle.

So, too high a grid resistor and you'll need to beef up your drive
stage. Plus (as mentioned), your conduction angle decreases, and your
final-stage efficiency may suffer.

Get the grid resistor too low, your conduction angle will increase, and
your final-stage efficiency may suffer.

Note that I say "may" -- there's an optimum conduction angle. There's
handbook values for it (which I can't remember!) but I'll bet that no one
amplifier works best right at the handbook value.

If you _really_ want to be scientific about it then for each grid
resistance value monitor your final stage input power, the amplifier
output power, and calculate the grid resistance dissipation. If nothing
else, that'll help you make an informed choice.

Otherwise, if it's given, calculate the grid resistor value to get you
both the desired current and the RF p-p voltage, or the rated bias
voltage, whichever is listed for your tube in that service.

--
www.wescottdesign.com

The higher the grid resistor, the higher the bias voltage that must be
overcome by the drive. Hence, higher drive, more power lost in the grid
resistor, and lower conduction angle.

So, too high a grid resistor and you'll need to beef up your drive
stage. Plus (as mentioned), your conduction angle decreases, and your
final-stage efficiency may suffer.

Get the grid resistor too low, your conduction angle will increase, and
your final-stage efficiency may suffer.

Note that I say "may" -- there's an optimum conduction angle. There's
handbook values for it (which I can't remember!) but I'll bet that no one
amplifier works best right at the handbook value.

If you _really_ want to be scientific about it then for each grid
resistance value monitor your final stage input power, the amplifier
output power, and calculate the grid resistance dissipation. If nothing
else, that'll help you make an informed choice.

Otherwise, if it's given, calculate the grid resistor value to get you
both the desired current and the RF p-p voltage, or the rated bias
voltage, whichever is listed for your tube in that service.

--
www.wescottdesign.com

Thanks for your comments. I agree that there should be an optimum grid
resistance value (even if rather dull), but in my case the optimum occurs at
zero grid resistance.

Let me report you some tests I have made, increasing the grid resistance in
steps (starting from R=0) and then re-adjusting the drive power each time (and
also re-optimizing the Pi-network controls):

- increasing the grid resistance and then adjusting the drive power so as to
keep the GRID current constant, the plate current - and hence the output power -
decreases. Therefore, to obtain maximum output power, the grid resistance must
be zero

- conversely, increasing the grid resistance and then adjusting the drive power
so as to keep the PLATE current constant, the output power remains about the
same for a quite wide range of grid resistance values (except when resistance
becomes very high). It should be noted that, increasing the grid resistance at
constant plate current, the grid current increases significantly, to the extent
that, for fairly high grid resistance values, the grid current gets beyond the
allowable limit.

In conclusion, it looks like the final stage operates best at zero grid
resistance:

- no efficiency loss
- minimum grid current for a given ouptut power.

In such conditions, the tube operates in class B (the fixed -33V bias causes an
idling plate current of about 10 mA), with a circulation angle of more than 180
degrees. Increasing the grid resistor causes a reduction of the circulation
angle, with no practical benefit and some drawbacks.

Where has the class-C efficiency advantage gone?

73

Tony I0JX

Antonio Vernucci wrote:

The higher the grid resistor, the higher the bias voltage that must be
overcome by the drive. Hence, higher drive, more power lost in the grid
resistor, and lower conduction angle.

So, too high a grid resistor and you'll need to beef up your drive
stage. Plus (as mentioned), your conduction angle decreases, and your
final-stage efficiency may suffer.

Get the grid resistor too low, your conduction angle will increase, and
your final-stage efficiency may suffer.

Note that I say "may" -- there's an optimum conduction angle. There's
handbook values for it (which I can't remember!) but I'll bet that no one
amplifier works best right at the handbook value.

If you _really_ want to be scientific about it then for each grid
resistance value monitor your final stage input power, the amplifier
output power, and calculate the grid resistance dissipation. If nothing
else, that'll help you make an informed choice.

Otherwise, if it's given, calculate the grid resistor value to get you
both the desired current and the RF p-p voltage, or the rated bias
voltage, whichever is listed for your tube in that service.

--
www.wescottdesign.com

Thanks for your comments. I agree that there should be an optimum grid
resistance value (even if rather dull), but in my case the optimum
occurs at zero grid resistance.

Let me report you some tests I have made, increasing the grid resistance
in steps (starting from R=0) and then re-adjusting the drive power each
time (and also re-optimizing the Pi-network controls):

- increasing the grid resistance and then adjusting the drive power so
as to keep the GRID current constant, the plate current - and hence the
output power - decreases. Therefore, to obtain maximum output power, the
grid resistance must be zero

- conversely, increasing the grid resistance and then adjusting the
drive power so as to keep the PLATE current constant, the output power
remains about the same for a quite wide range of grid resistance values
(except when resistance becomes very high). It should be noted that,
increasing the grid resistance at constant plate current, the grid
current increases significantly, to the extent that, for fairly high
grid resistance values, the grid current gets beyond the allowable limit.

In conclusion, it looks like the final stage operates best at zero grid
resistance:

- no efficiency loss
- minimum grid current for a given ouptut power.

In such conditions, the tube operates in class B (the fixed -33V bias
causes an idling plate current of about 10 mA), with a circulation angle
of more than 180 degrees. Increasing the grid resistor causes a
reduction of the circulation angle, with no practical benefit and some
drawbacks.

Where has the class-C efficiency advantage gone?

73

Tony I0JX

The advantage of class C isn't necessary greater efficiency. By
reducing the conduction angle the tube is drawing current for a short
period of time and therefor can run cooler. It also means that the tube
can be run at a bit higher power level than it could in class B since
the AVERAGE power dissipated is the same. HOWEVER the duty cycle of
both the time transmitting vs not transmitting and that of the signal
also play a role. In other words a class C CW transmitter in theory
could be run at higher power than a class C FM phone transmitter (even
though both are usually run at the same typical parameters) since the
tube can cool between elements on CW, while FM is key down forever.
Also class B audio has a different duty cycle than a class B RF linear
amplifier running FM (don't need to be linear for FM 'thou).

In the 30's there was an article in QST on how someone ran a 200 watt
tube at a KW CW. It worked because of CW's short duty cycle, but the
editor suspected 'short dashes'.

On 3 Feb, 17:54, Kenneth Scharf wrote:
Antonio Vernucci wrote:

The higher the grid resistor, the higher the bias voltage that must be
overcome by the drive. *Hence, higher drive, more power lost in the grid
resistor, and lower conduction angle.

So, too high a grid resistor and you'll need to beef up your drive
stage. *Plus (as mentioned), your conduction angle decreases, and your
final-stage efficiency may suffer.

Get the grid resistor too low, your conduction angle will increase, and
your final-stage efficiency may suffer.

Note that I say "may" -- there's an optimum conduction angle. *There's
handbook values for it (which I can't remember!) but I'll bet that no one
amplifier works best right at the handbook value.

If you _really_ want to be scientific about it then for each grid
resistance value monitor your final stage input power, the amplifier
output power, and calculate the grid resistance dissipation. *If nothing
else, that'll help you make an informed choice.

Otherwise, if it's given, calculate the grid resistor value to get you
both the desired current and the RF p-p voltage, or the rated bias
voltage, whichever is listed for your tube in that service.

--
www.wescottdesign.com

Thanks for your comments. I agree that there should be an optimum grid
resistance value (even if rather dull), but in my case the optimum
occurs at zero grid resistance.

Let me report you some tests I have made, increasing the grid resistance
in steps (starting from R=0) and then re-adjusting the drive power each
time (and also re-optimizing the Pi-network controls):

- increasing the grid resistance and then adjusting the drive power so
as to keep the GRID current constant, the plate current - and hence the
output power - decreases. Therefore, to obtain maximum output power, the
grid resistance must be zero

- conversely, increasing the grid resistance and then adjusting the
drive power so as to keep the PLATE current constant, the output power
remains about the same for a quite wide range of grid resistance values
(except when resistance becomes very high). It should be noted that,
increasing the grid resistance at constant plate current, the grid
current increases significantly, to the extent that, for fairly high
grid resistance values, the grid current gets beyond the allowable limit.