Hi,
I am planning to build a Point-to-point K12-G from scratch. I just happen to have all of the parts on hand and had such good results with the K-502 kit. The K12-G uses the 10GV8 tube which is a little different from the tube used in the 12-M version of the kit. The 12-M uses a 300R cathode resistor to set the idle current around 82mA. On the 12-G it is a 150R resistor. Can anybody help me with the bias point on the idle current of the 10GV8? What is the currnet set at with the 150R? Of course I could do a little math myself but being a newbie I want to make sure I get this right (wouldn't want to burn up an expensive $2 10GV8!) I could of course just use a 150R on the tube like the original schematic calls for and move on but, I am considering a CSS using an LM317HV in place of the resistor.
Thanks for your help!
Jeff
I am planning to build a Point-to-point K12-G from scratch. I just happen to have all of the parts on hand and had such good results with the K-502 kit. The K12-G uses the 10GV8 tube which is a little different from the tube used in the 12-M version of the kit. The 12-M uses a 300R cathode resistor to set the idle current around 82mA. On the 12-G it is a 150R resistor. Can anybody help me with the bias point on the idle current of the 10GV8? What is the currnet set at with the 150R? Of course I could do a little math myself but being a newbie I want to make sure I get this right (wouldn't want to burn up an expensive $2 10GV8!) I could of course just use a 150R on the tube like the original schematic calls for and move on but, I am considering a CSS using an LM317HV in place of the resistor.
Thanks for your help!
Jeff
The 10GV8 is pretty much an ECL85. You'll find links to datasheets here:
http://tdsl.duncanamps.com/show.php?des=6GV8
Looking at one set of typical operating conditions, with a 170 volts across the tube you might run it at 43.7 mA (A + G2) and -15 at the cathode. To drop 15 volts at a current of 43.7 mA, you'd need R = V/I = 343 ohms. If you're building a push/pull amp and the cathodes are tied together with a common cathode resistor, use half that value. Of course, this only applies for this specific operating point (specified tube voltage and current).
http://tdsl.duncanamps.com/show.php?des=6GV8
Looking at one set of typical operating conditions, with a 170 volts across the tube you might run it at 43.7 mA (A + G2) and -15 at the cathode. To drop 15 volts at a current of 43.7 mA, you'd need R = V/I = 343 ohms. If you're building a push/pull amp and the cathodes are tied together with a common cathode resistor, use half that value. Of course, this only applies for this specific operating point (specified tube voltage and current).
I wasn't really looking for critisism, just help. So then, why don't you give me the answer? Got anything useful to say?
ummmm, not a decimal issue. If you are using hte "typical" settings that ty directed you to, you are wanting to bias the tube at -15v while running it at 170V / 41ma anode load, 2.7ma g2 load.
You now have 43.7ma through the cathode and hence the cathoide resistor. If you want to drop 15 volts across that resistance, the math is 15/0.043 = 349 ohms. As ty points out, if you have the two pp outputs tied at the cathode, half the resistor value and double its power handling viz, 175 ohms, say... 5w power rating.
If you want to run it at 210V you will need to look at the curves and deduce the operating point...
also, check hte information in this thread http://diyaudioprojects.com/Forum/viewtopic.php?f=9&t=515
You now have 43.7ma through the cathode and hence the cathoide resistor. If you want to drop 15 volts across that resistance, the math is 15/0.043 = 349 ohms. As ty points out, if you have the two pp outputs tied at the cathode, half the resistor value and double its power handling viz, 175 ohms, say... 5w power rating.
If you want to run it at 210V you will need to look at the curves and deduce the operating point...
also, check hte information in this thread http://diyaudioprojects.com/Forum/viewtopic.php?f=9&t=515
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Are you going to run this as a pentode, or UL? Will you put 210V on G2? What's the primary impedance of your output transformer?
If you've got 210V on G2, according to the charts you'll need to keep about -21V on the grid to properly bias the tube and keep plate dissipation under the tube's limit (8W). You'll be running at about 35mA at that point. I'd expect a 620 ohm cathode resistor would be about right for a single tube. Make it a little higher (680 ohm) if you want to err on the side of caution.
Ok, I get it now! I see where you got the values from the specs on the tube.
As for the 210V, that is where I believe the B+ sits for the K-12G, I could be wrong (as I am frequently).
What I am trying to figure out here is if it would be possible to use a LM317 as a CCS in this circuit. Could it be done, and how? Would you just set the resistor to yield the current needed> What about the -15 volt drop, how does that come into play with a CCS?
Thanks,
Jeff
As for the 210V, that is where I believe the B+ sits for the K-12G, I could be wrong (as I am frequently).
What I am trying to figure out here is if it would be possible to use a LM317 as a CCS in this circuit. Could it be done, and how? Would you just set the resistor to yield the current needed> What about the -15 volt drop, how does that come into play with a CCS?
Thanks,
Jeff
Here's an example of a circuit using the LM317 as a constant current source for the finals. Do you have the schematic for the K-12G handy? I suspect it uses a completely different arrangement for its phase splitter.
DIY ECC802S SRPP / EL84 (6BQ5) Push-Pull Tube Amplifier
Edit: I found the schematic for the K-12G. It has a AF gain stage DC coupled to a cathodyne phase inverter. Pretty normal stuff.
http://i69.photobucket.com/albums/i43/Ty_Bower/Tubes/S-5-Electronics-K-12G-Tube-Amp-Kit-.png
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Here's an example of what you're probably trying to do... different tubes and a different front end topology, but the CCS on the cathodes of the finals is what you're after.
diytube.com :: View topic - for Shannon: a CCS for ST35?
diytube.com :: View topic - for Shannon: a CCS for ST35?
Does anybody know where I can find the curves for the 10GV8 (or 6VG8,18GV8)?
So, I have another question....if I understand correctly, when you use a resistor on the cathode to set its operating point....lets say at 43.7mA like the datasheet example, the resistor value is selected to bring the cathode to -15V relative to the control grid. This in effect SETS the current at 43.7mA. So, when using a CCS....do you not have to worry about the -15V drop and just set the current at 43.7mA with the CCS? Doest setting the current at 43.7mA force the drop to -15V? This is where I get confused with the use of a CCS.
My understanding is at this point to select the cathode resistor you look at the curve, find your operating voltage, find your current above it on the graph, and at that point the cathode relative to control grid voltage is found.
Second question....what is the benefit of a CCS over a cathode resistor with a cathode bypass cap?
Still trying to figure it all out,
Jeff
So, I have another question....if I understand correctly, when you use a resistor on the cathode to set its operating point....lets say at 43.7mA like the datasheet example, the resistor value is selected to bring the cathode to -15V relative to the control grid. This in effect SETS the current at 43.7mA. So, when using a CCS....do you not have to worry about the -15V drop and just set the current at 43.7mA with the CCS? Doest setting the current at 43.7mA force the drop to -15V? This is where I get confused with the use of a CCS.
My understanding is at this point to select the cathode resistor you look at the curve, find your operating voltage, find your current above it on the graph, and at that point the cathode relative to control grid voltage is found.
Second question....what is the benefit of a CCS over a cathode resistor with a cathode bypass cap?
Still trying to figure it all out,
Jeff
The 10GV8, as you have discovered, will have the same curves as the more common 6GV8. The other name for the 6GV8 is ECL85. Referring to the Philips datasheets for the ECL85 says "for curves, see PCL85." I'd suggest searching for those.
TDSL Tube data [PCL85]
Yes, if you choose a current sense resistor for your CCS such that the tube's current is 35mA (or whatever), then the cathode voltage will be wherever it falls on the tube's curve given that current and the plate/G2 voltage. You won't need to worry exactly where that cathode voltage will fall, and it is possible that two unmatched tubes will have different cathode voltages.
Well, the CCS will force the tube current to a fixed value. Some have suggested this is more useful if your amplifier is going to run in pure class A. Typically, push/pull amps will run class AB and the total current between both tubes is not fixed. During loud passages the pair will draw more current than at idle. I don't think this is possible if you are using a CCS at the cathodes. You may end up limiting your maximum power output.
I think that even with the CCS, you still require the bypass capacitor from cathode to ground. The AC impedance between the cathodes of both tubes in the pair must be close to zero, or evil things will occur. Note that this is not the same thing as the DC resistance between the cathodes, which is clearly much greater than zero (hundreds of ohms in this case).
I believe the biggest advantage of the CCS is that each side of the pair is set to a fixed current, so DC imbalance in the output transformer is reduced or eliminated. It also allows the use of completely mismatched output tubes, and no user adjustment is required to set the tube bias.
TDSL Tube data [PCL85]
Yes, if you choose a current sense resistor for your CCS such that the tube's current is 35mA (or whatever), then the cathode voltage will be wherever it falls on the tube's curve given that current and the plate/G2 voltage. You won't need to worry exactly where that cathode voltage will fall, and it is possible that two unmatched tubes will have different cathode voltages.
Well, the CCS will force the tube current to a fixed value. Some have suggested this is more useful if your amplifier is going to run in pure class A. Typically, push/pull amps will run class AB and the total current between both tubes is not fixed. During loud passages the pair will draw more current than at idle. I don't think this is possible if you are using a CCS at the cathodes. You may end up limiting your maximum power output.
I think that even with the CCS, you still require the bypass capacitor from cathode to ground. The AC impedance between the cathodes of both tubes in the pair must be close to zero, or evil things will occur. Note that this is not the same thing as the DC resistance between the cathodes, which is clearly much greater than zero (hundreds of ohms in this case).
I believe the biggest advantage of the CCS is that each side of the pair is set to a fixed current, so DC imbalance in the output transformer is reduced or eliminated. It also allows the use of completely mismatched output tubes, and no user adjustment is required to set the tube bias.
I did not criticise you; I pointed out the miscalcualtion on your part and that you two (Ty and you) were talking about two completely different things, which resulted in incorrect assumption you based your miscalculation on in the first place. That might be useful to some people, but then again, YMMV
I don't know what K-12G is so I can't help you there. Ty is doing a great job explaining the basics, I'm not a native English speaker so I prefer to stay away when somebody else (fluent in language) speaks up.
I think i got it
Ok, I finally found the curves. It appears that they are running at an idle of 66.5mA with the configuration of a 150R tying both cathodes at 210V. I figured this out this way.....at 210V, a cathode resistor of 300R (150R when tied together), the cathode sits at -20V. Thus giving the following calculation: 20/0.0665=300.7. This calulation does not take into consideration the G2 current which does not graph below -12V but if I extrapolate, it is around 5mA so the actual idle current would be closer to 61.5mA.
Am I getting it now?
So, if I were to set the LM317 up to run at this current do I include the current for g2? (66.5mA) or would I set it up for just 61.5mA (no g2 current accounted for)?
Lets just assume I need to include the g2 in the calculation. The current set by the LM317 is 1.25 divided by the value in ohms for the resistor. Thus 133mA (66.5mAx2) = 1.25/R......re-arrange 1.25/0.133=9.4R.
So, if I use a 9.4R resistor on the LM317 that should turn out a CCS with a current of 66.5mA per tube which, if I am correct, approximates the current set at idle in the standard cathode resistor setup with an 150R resistor tied to both cathodes......how am I doing so far?
Jeff
Ok, I finally found the curves. It appears that they are running at an idle of 66.5mA with the configuration of a 150R tying both cathodes at 210V. I figured this out this way.....at 210V, a cathode resistor of 300R (150R when tied together), the cathode sits at -20V. Thus giving the following calculation: 20/0.0665=300.7. This calulation does not take into consideration the G2 current which does not graph below -12V but if I extrapolate, it is around 5mA so the actual idle current would be closer to 61.5mA.
Am I getting it now?
So, if I were to set the LM317 up to run at this current do I include the current for g2? (66.5mA) or would I set it up for just 61.5mA (no g2 current accounted for)?
Lets just assume I need to include the g2 in the calculation. The current set by the LM317 is 1.25 divided by the value in ohms for the resistor. Thus 133mA (66.5mAx2) = 1.25/R......re-arrange 1.25/0.133=9.4R.
So, if I use a 9.4R resistor on the LM317 that should turn out a CCS with a current of 66.5mA per tube which, if I am correct, approximates the current set at idle in the standard cathode resistor setup with an 150R resistor tied to both cathodes......how am I doing so far?
Jeff
210V on the plate (anode) and 300R common cathode resistance giving an idle current of 66.5ma (per tube?) I think is what you are trying to say. Its REALLY important here to get the various terms right as it can get very confusing for everyone otherwise
Yep, although for calculating the cathode resistor, the anode voltage is irrelevant - you are only interested in what CURRENT is flowing through the tubes at idle (I guess to that extent you need to know the anode voltage, but hey...) In your example, the cathode is actually sitting at 20 volts ABOVE ground, putting the G1 at 20v BELOW the cathode.
Had a quick look at the curves for the PCL85 - I think you may have either misintrepreted them, or you are running the tube fairly heroically, but its a little hard to be sure given the confusing first paragraph of your last post...
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Is that 20 volts Vk a value you measured on an actual working amp? Are you certain the B+ is 210V?
Well, Ik = Ia + Ig2 (current through the cathode equals plate current plus screen current). If you're thinking you've got Ia = 61.5 and Ig2 = 5, that doesn't quite work out right given the curves for the PCL85. If I go by the chart where Vg2=210, then at Va~190 the plate current is just over 61mA where the Vg1 line equals -18. The lines for Ig2 kinda all flatten out and bunch together, but as you say it'll be something around 5mA. This is close, but we're off by two volts at g1. There is a certain amount of variation between actual tubes and the curves, but I'd like to see this a little closer. I'm also bothered by the fact that if Ia=61mA, that's around 11.5 watts dissipated by the plate. That's too much - the tube is rated for 8 watts dissipation.
Something else to keep in mind when reading these curves. The curves are presented where Vk is at ground (0 volts). That's why all the Vg1 numbers are shown negative - the grid must be at a lower potential than the cathode. Our circuit is cathode biased, and g1 is at ground. Vk is kept at some voltage above ground by means of the drop across the cathode resistor. We still know the potential between k and g1, and can read that straight off the chart. Our actual working anode voltage should really be read as the voltage across the tube (Va - Vk).
I'd much rather see the tube operated below and to the left of the dotted-dashed line (the 8 watt line). With a Vg2=210, a better operating point might be somewhere around Ia=41mA and Ig1=-20V. I just don't see how you can get that with a 150 ohm cathode resistor for the pair of output tubes. I'd imagine it should be at least 210 ohms, preferably more. It seems the design choice made for the K12-M was reasonable, but what they've got on the K12-G should be roasting the poor tubes.
Well, Ik = Ia + Ig2 (current through the cathode equals plate current plus screen current). If you're thinking you've got Ia = 61.5 and Ig2 = 5, that doesn't quite work out right given the curves for the PCL85. If I go by the chart where Vg2=210, then at Va~190 the plate current is just over 61mA where the Vg1 line equals -18. The lines for Ig2 kinda all flatten out and bunch together, but as you say it'll be something around 5mA. This is close, but we're off by two volts at g1. There is a certain amount of variation between actual tubes and the curves, but I'd like to see this a little closer. I'm also bothered by the fact that if Ia=61mA, that's around 11.5 watts dissipated by the plate. That's too much - the tube is rated for 8 watts dissipation.
Something else to keep in mind when reading these curves. The curves are presented where Vk is at ground (0 volts). That's why all the Vg1 numbers are shown negative - the grid must be at a lower potential than the cathode. Our circuit is cathode biased, and g1 is at ground. Vk is kept at some voltage above ground by means of the drop across the cathode resistor. We still know the potential between k and g1, and can read that straight off the chart. Our actual working anode voltage should really be read as the voltage across the tube (Va - Vk).
I'd much rather see the tube operated below and to the left of the dotted-dashed line (the 8 watt line). With a Vg2=210, a better operating point might be somewhere around Ia=41mA and Ig1=-20V. I just don't see how you can get that with a 150 ohm cathode resistor for the pair of output tubes. I'd imagine it should be at least 210 ohms, preferably more. It seems the design choice made for the K12-M was reasonable, but what they've got on the K12-G should be roasting the poor tubes.
I think I actually don't remember how I came to that conclusion now that I look at it again.
Again, I have been mistaken. What I know for a fact now is that the B+ is 250V and the effective cathode resistor value is 300R. The B+ is 250V at the center tap of the OPT which then gives rise to each plate. Does this mean the actual anode voltage is 1/2 of the B+?
Also, I think I assumed the g2 was at -20V, why I did this I don't know at the moment...it all seemed to click when I ran through this at that time. Can you help me through this? the values above are from the schematic and the knowlegde that the unrectified trafo voltage is 180-0 through a bridge rectifier, 180x1.4--->250V.
When I now look at the curve if I pick about 50mA (just below max power curve) above the 125V X-axis, the point falls right at the -15V line. If I calculate this out 15/0.05=300....300R is exactly the value (300/2=150 on the schematic) of the original design. Now I have come up with 300R 2 different ways, obviously one is incorrect (probably both).
I can say one thing for sure, the more I work through this stuff the more I begin to understand it (I think!).
50mA is much more in line with where this tube should operate, from what I can tell.
So, back to the LM317...if I am now correct in my calculation Of 50mA then I would need the CCS to create 100mA. This should be done simply by the equation 1.25/I=R------>1.25/0.1=12.5R. So, a 12.5R resistor should give me what I think I want. I hope.
Comments, observations, critisism...
Thanks,
Jeff
Again, I have been mistaken. What I know for a fact now is that the B+ is 250V and the effective cathode resistor value is 300R. The B+ is 250V at the center tap of the OPT which then gives rise to each plate. Does this mean the actual anode voltage is 1/2 of the B+?
Also, I think I assumed the g2 was at -20V, why I did this I don't know at the moment...it all seemed to click when I ran through this at that time. Can you help me through this? the values above are from the schematic and the knowlegde that the unrectified trafo voltage is 180-0 through a bridge rectifier, 180x1.4--->250V.
When I now look at the curve if I pick about 50mA (just below max power curve) above the 125V X-axis, the point falls right at the -15V line. If I calculate this out 15/0.05=300....300R is exactly the value (300/2=150 on the schematic) of the original design. Now I have come up with 300R 2 different ways, obviously one is incorrect (probably both).
I can say one thing for sure, the more I work through this stuff the more I begin to understand it (I think!).
50mA is much more in line with where this tube should operate, from what I can tell.
So, back to the LM317...if I am now correct in my calculation Of 50mA then I would need the CCS to create 100mA. This should be done simply by the equation 1.25/I=R------>1.25/0.1=12.5R. So, a 12.5R resistor should give me what I think I want. I hope.
Comments, observations, critisism...
Thanks,
Jeff
Yep - know that feeling!
no - it will be almost exactly the B+ applied to the opt centre tap.
Yep again.
You may want to move your quiescent voltage across to the right a bit, more toward 2/3 of the B+ voltage, say 160-ish volts. This will be closer to the centre of the real operating range of the tube. Again, open to criticism or real-world experience with this tube on this one.
Here's another site with a good simple explanation - follow the link to the description of designing a p-p output stage.
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