"Kenneth Scharf" ha scritto nel messaggio
...
COLIN LAMB wrote:
Hi Tony:

My 1936 Radio Handbook gives the following information:

"Grid-leak bias is quite flexible and more or less automatically adjusts
itself with any variation in RF excitation. The value of grid-leak resistor
is not particularly critical because the DC grid current usually decreases as
the grid-leak resistance increases, theeby keeping the product of the two
more or less constant for a given amount of RF excitation. Hence, the value
of the grid-leak resistance can vary from one-half to two times the optimum
value, a ration of four to one, without materially affecting the negative DC
bias voltages actually applied to the grid of the amplifier tube.

One of the disadvantages of grid-leak bias is that the bias voltage is
proportioonal to the RF excitation, thus precluding the use in grid modulated
or linear amlifiers, whose bias must be supplied from a well-regulated
voltage source so that the bias voltage is independent of grid current."

So, I guess the answer is "use whatever value that makes the tube happy".

73, Colin K7FM
Actually the value of the grid leak bias resistor used in a class C amplifier
is going to be a function of the available peak RF voltage being supplied by
the driver stage, and the required grid drive / bias voltage of the final.
The maximum allowed grid current rating of the final tube must not be exceeded
either. If you look at some classical ham transmitter circuits the grid
resistor value varied quite a bit. For the 807 tube a typical value was 15k
(as recommended by RCA) but where drive current was limited (such as on ten
meters driven by a quadrupler from 40) a 22k or higher value was common.

Thanks both for the quote and the comments.

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.

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.

I cannot quite follow your reasoning. The tube temperature is bound to the
dissipated power. And the dissipated power simply is the difference between the
average DC power and the RF output power (neglecting losses in the Pi network).
So, if varying the signal duty cycles and tube conduction angle, I anyway read
the same output power and the same DC power, the stage efficiency 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'.

That's OK.

73

Tony I0JX

Im still confused as to the role the grid resistor is playing in
the amplifier, if the bias is generated by a power-supply , then any
change to the psu output voltage would be resisted by the psu ...
assuming the feed is via a choke ?

Yes, the bias supply is a zener-stabilized -33V supply. I have verified that the
supply voltage stays constant independently of grid current.

where is the grid resistor .. from the grid to real earth ..or
in series with the power supply ??

The grid resistor (and a low-resistance RF choke) are both in series with the
bias supply. So, for zero grid current (no excitation), the grid bias is
just -33V (so keeping plate current at about 10 mA). When I apply RF drive, the
bias grows up (sum of -33V and the voltage that develops across the grid
resistor), so pushing the tube into class-C operation

Convention was to feed via a low value resistor / rfc or both
with a medium value bleed to earth in the 1 K range

To obtain a total of -50V @ 12 mA grid current (class-C operating conditions
recommended by the tube manufacturer), a grid resistance of about 1,400 ohm is
needed (33 + 0.012 * 1400).

Liner service then requiring the psu to resist voltage
fluctuation under drive variation and as such required a low Z path
to the valve and the ability to dissipate voltage .ie gas
discharge tube or dc feed back loop ... zener diode

I have the feeling this is a driver / pa grid matching problem ?

As I mentioned above the bias supply resists to grid current and its voltage is
perfectly stable.

73

Tony I0JX

Grid leak resistance isn't only some discrete resistor, there's
also your bias supply to be considered. Is your bias supply able to
sink current, or does its output voltage rise with grid current?

The bias supply is a zener-stabilized -33V supply. I have verified that the
supply voltage stays constant independently of grid current. So the effective
grid resistance coincides with the grid resistor.


[...]
Where has the class-C efficiency advantage gone?

There was a rule of thumb for optimum anode efficiency, which went
along the lines of "when the anode voltage at its lowest swing equals
the control grid voltage" IIRC.

Thanks for tip, though not immediate to measure.

73

Tony I0JX

Antonio Vernucci wrote:
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.

I cannot quite follow your reasoning. The tube temperature is bound to
the dissipated power. And the dissipated power simply is the difference
between the average DC power and the RF output power (neglecting losses
in the Pi network). So, if varying the signal duty cycles and tube
conduction angle, I anyway read the same output power and the same DC
power, the stage efficiency is the same.

Tubes don't self destruct in an instant when they are asked to dissipate
more than their max rated power. So long as the AVERAGE dissipated
power over time does not exceed the max rating things are safe. The
duty cycle will change the average power dissipation. Also the
temperature isn't bound instantly to the instantaneous power dissipation
due the the tubes' thermal mass.


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'.

That's OK.

73

Tony I0JX

In this case the grid resistor, connected between the rf-bypassed
negative port of the bias supply and the cold end of the grid choke has
to provide the difference between the protective bias (above mentioned
-33V) and the class C bias specified in the valve data sheet. Simple
Ohm's law can be applied. If the grid current is say 2mA and the desired
grid bias is -63V, i.e. a difference of 30V, the control grid resistor
needs to have 15kOhm.

73, Eddi ._._.

Yes, Ohm's law is OK, but the my issue was that whatever "desired grid bias" I
take, the final stage efficiency does not change, So, changing the grid resistor
makes almost no difference, whilst I would have expected that biasing the tube
in the class-C region would yield more RF power that when it operates in class B
(i.e. with grid resistor = 0 ohm)

73

Tony I0JX

On Feb 6, 12:12*am, "Antonio Vernucci" wrote:
* In this case the grid resistor, connected between the rf-bypassed
negative port of the bias supply and the cold end of the grid choke has
to provide the difference between the protective bias (above mentioned
-33V) and the class C bias specified in the valve data sheet. Simple
Ohm's law can be applied. If the grid current is say 2mA and the desired
grid bias is -63V, i.e. a difference of 30V, the control grid resistor
needs to have 15kOhm.

* 73, Eddi ._._.

Yes, Ohm's law is OK, but the my issue was that whatever "desired grid bias" I
take, the final stage efficiency does not change, So, changing the grid resistor
makes almost no difference, whilst I would have expected that biasing the tube
in the class-C region would yield more RF power that when it operates in class B
(i.e. with grid resistor = 0 ohm)

73

Tony I0J

My practical experience is that lowering the grid resistor I always
get more
output power FOR THE SAME GRID CURRENT.

So ... how do you know its the same ...if you use a meter ..
then the shape factor of the grid pulse will change the meter
reading .. wider pulse lower peak will still give the same
reading ?

need to check the waveform .. not easey .. simple way is to
forget the resistor and provide a variable bias psu . and inject
via a low ohm rf-choke ... and add some link to the bias voltage
and the pa plate supply ..no bias no supply .. simple relay ?

My practical experience is that lowering the grid resistor I always
get more
output power FOR THE SAME GRID CURRENT.

Same experience here.


So ... how do you know its the same ...if you use a meter ..
then the shape factor of the grid pulse will change the meter
reading .. wider pulse lower peak will still give the same
reading ?

need to check the waveform .. not easey .. simple way is to
forget the resistor and provide a variable bias psu . and inject
via a low ohm rf-choke ... and add some link to the bias voltage
and the pa plate supply ..no bias no supply .. simple relay ?


What is commonly called "grid current" is not the instantaneous grid current,
but its average value (i,e, its DC component). So, when one measures the grid
current using a DC meter, he has not to worry about the actual current waveform.

73

Tony I0JX