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Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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. |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 .. |
Class-C stage grid resistor
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 .. |
Class-C stage grid resistor
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 .. |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
"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 |
Class-C stage grid resistor
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'. |
Class-C stage grid resistor
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. |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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 ? |
Class-C stage grid resistor
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 |
Class-C stage grid resistor
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) Output power will not rise for a given anode voltage and a given excitation (drive). But with a reduced conduction angle the anode input power will be less. On the other hand you'll need more drive to arrive at the same anode currrent. But drive power doesn't enter the anode efficiency calculation, which is rf output power divided by dc anode input power (the quotient obtained thus multiplied by 100 if you want it in percent). I try to explain me better. When I change the grid resistor, I see no chage in the ratio between RF output power and DC input power, this meaning that efficiency does not change, I would have exprrcted that, going class-C, I would have got a higher efficiency (more RF power, or lower DC power, or both). 73 Tony I0JX |
Class-C stage grid resistor
When I change the grid resistor, I see no chage in the ratio between
RF output power and DC input power, this meaning that efficiency does not change, I would have exprrcted that, going class-C, I would have got a higher efficiency (more RF power, or lower DC power, or both). So would I. How do you feed the 829B's screen grid? The screen grid is fed by a dedicated 200V supply, which is very stable (within the screen current excursion). 73 Tony I0JX |
Class-C stage grid resistor
On Feb 6, 9:10*pm, (Edmund H. Ramm) wrote:
In "Antonio Vernucci" writes: [...] When I change the grid resistor, I see no chage in the ratio between RF output power and DC input power, this meaning that efficiency does not change, I would have exprrcted that, going class-C, I would have got a higher efficiency (more RF power, or lower DC power, or both). * *So would I. How do you feed the 829B's screen grid? * *73, Eddi ._._. -- * * * e-mail: dk3uz AT arrl DOT net *| *AMPRNET: * * * If replying to a Usenet article, please use above e-mail address. * * * * * * * *Linux/m68k, the best U**x ever to hit an Atari! Your not supposed to ask questions like that Eddi ! My point with the grid supply wave form , is that the point of conduction is set by the voltage .. so depending on th voltage the sinusodial drive waveform will redue to a pulse train .same peak .. but lower average ... I think there is some form of operation in the 'negative' region .. now if the screen is dropper fed, then its floating and may be subject to secondary emission .. so the working point of the valve is moving ... may be not if the screen is stabllised .. but if so . would than not affect the am mode ? G .. |
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