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16th August 2008, 02:23 AM  #1 
diyAudio Member
Join Date: Jul 2008

6ak5/5654 plate resistance
I am a noob trying to put synthesize various things that I have read concerning the 6ak5/5654 pentode and DIY amps. I remember (1) reading that 6ak5s make a nice sounding triode ; (2) an article on "Screwing Around with Load Lines" by someone in SE Asia; and (3) an article in Audioxpress about using a 6j10 compactron as an output tube.
What I want to do is use the 6ak5 as the driver for the audio part of a 6AL11 compactron (which I think is identical to the 6J10). The "Load Lines" article says that the triode must be loaded several times the plate load of the tube. As a pentode, the 6ak5 has a plate resistance of .34 megohms. When it is triode strapped, does the plate load drop precipitously? Any ideas on how to use a triode strapped 6ak5 as a driver tube? Thanks in advance for any assistance you can offer. 
16th August 2008, 03:58 AM  #2  
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Join Date: Dec 2004
Location: Cincinnati, Ohio

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16th August 2008, 04:06 AM  #3 
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Join Date: Jun 2004
Location: Pittsburgh, crumbling wasteland

The last page of this datasheet gives triode curves
http://www.mif.pg.gda.pl/homepages/f...138/5/5654.pdf 
17th August 2008, 06:03 AM  #4 
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Join Date: Jul 2008

I have seen the curves linked above, but I don't know how to use them to define a load for the plate. This is the link to the article that I am trying to follow
http://www.diyparadise.com/tubeloadl...loadlines.html According to this article, you start with the power supply on the x axis and then find a point on the y axis that is calculated, in part, by using the plate load. Thanks for those who responded. I appreciate you sharing your expertise. 
17th August 2008, 12:11 PM  #5 
diyAudio Member
Join Date: Feb 2005
Location: Zagreb

Well, basically you take the graphed characteristics in trode mode nad play around on them with a ruler.
First you chose the B+ lik ein the article, and mark a point for it on the X axis (plate voltage). You use this point as an 'anchor' around which you can pivot a line (your ruler) at various angles, intersecting the y axis (plate current). Before you do that, it is wise to add a few things to the curve plot: Look up the maximum DC plate current in the data sheet. Mark it off on the y axis and draw a horizontal line all the way through the graph at this current. Your chosen operating point should never be above (and it is recomended it is substantially below) this line, i.e. this is a current limit. Keep in mind that it IS possible fot the load line to intersect the y axis (plate current) at a higher value than that, we shall get to this later on. Also, look up the maximum plate dissipation in Watts, in the datasheet. You may also look up the max dissipation of grid 2 (screen grid). As a rule of thumb, the maximum dissipation in throide mode, is the two added up, however, screen current and plate current may have such proportions in a triode conneced pentode that G2 dissipation can be violated. Unless the data sheet states plate dissipation in triode mode, or shows it as being the two above, added, assume that triode conneced plate dissipation is the same as pentode. In reality it should be higher, but this way you will be sure not to violate any spec, and in all probability run the tube with higher longevity. Now, keep in mind that plate dissipation is plate voltage (in V) times plate current (in A). Moving along the x axis by convenient intervals, plot voltagecurrent points at which voltage x current equals the max plate dissipation. Then, connect the points. This will give you a curve above which the plate dissipation is over the maximum alowable limit, so your chosen operating point should always be below this curve. With all these curves in place, assuming you are using a regular RC coupled amp, or current source loaded plate (but NOT a choke loaded plate) you can use something like a red pen or similar, to draw around the alowable area where the operating point should be. You start at coordinates 0V, 0A, draw left along the x axis up to B+. This is because obviously B+ is the highest voltage in the ciruit so plate voltage will at most be that (in reality it will only approach it). Then, go straight up from B+ volts, 0 plate mA, until you get to the maximum dissipation curve. Then go back towards the left along the maximum dissipation curve until you get to the plate curve which says grid voltage = 0V, or, until you get to the line denoting maximum plate current. If you get to the latter first, then follow along it to the left until you get to the former, the 0V grid voltage plate curve. Finally, follow down this 0V grid plate curve till you gto to coordinates 0,0. Your chosen operating point must be inside the shape you just drew. And now you are ready to start looking for a static good operating point. This would be the plate current and voltage of your tube with no input signal. With an input signal, the actual, or dynamic operating point moves to the left and right of the static operating point along the plate load line. The tricky part about this is that there are actually also two plate load lines, a static and a dynamic one. The reason for this is that for a static signal (no AC), in an RC coupled amplifier, the tube only sees it's plate load, because the output coupling capacitor presents an open circuit for DC. Once you apply a signel, however, the tube sees it's plate load in parallel with the input impedance (which we normally approximate as the input resistance) of the next stage. These two resistances can be qoute different, and the reason why it is important will become apparent in a minute. Drawing a load line on a graph becomes a thing of playing with a ruler and having a good eye. What you are looking for, are intersections of your ruler with the various grid voltage lines on the graph. A good linear operating point (assuming that's what you want) will be on a load line where these intersections are evenly spaced. You will find that there are areas of the graph where the intersections are very much unevenly spaced  usually at low plate currents and/or low plate voltages. Also, it will be impossible to get even intersections along the whole lenght of the load line, their distance normally compresses as you move towards the right. It is possible to gauge how flexible a triode is from a raw graph, just looking at the spacing of the various grid voltage graphs. If they are mostly nicely evenly spaced for an even increment in grid voltage, you have lots of choices in finding a good operating point. In other cases, you may still find one but you'll have to look real hard. Once you have found a promising looking load line, it is time to chose a static operating point on it, within the voltage, current and dissipation limit shape you drew on the graph earlyer. Keep in mind that your input signal is AC, so the static operating point (which is what you are chosing) must preferably not be right on any limit, because the actual oprating point moves along the load line to the left and right of the static one, and the maximum size of this 'movement' must also be enclosed within the limit shape you drew into the graph before. So, how much does it move? Well, given that you apply the input signal to the grid, calculate or look up the maximum peak input voltage swing (keep in mind that input voltages are usually given RMS, to get peak, multiply by the square root of two, approx. 1.4142). Mark intersections of grid voltage lines along the load line. The operating point moves to the left and right on the load line to where the grid voltage changes by the same amount. For example, if you operating point lies on the intersection of the load line and the 3V grid line, and your input voltage is 1V peak, then the dynamic operating point moves along the load line + or  1V from the static point, i.e. to 2V on one side, to 4V on the other. Now three things become apparent: 1) Because grid voltages >0 have repercussions (grid current) they are normally not used for simple RC coupled amps. The statis point chosen, plus the maximum peak input voltage swing, whould never become positive. This is why we used the 0V grid line as one of the limits to where our operating point shhould be. In actual fact it is usually wise to keep well away from this limit, as some tubes exhibit grid current even for small negative grid voltages. Usually 1V away from it for maximum input swing is more than enough. 2) We 'only' have to find a load line where the intersections with grid lines are nice and even for a grid voltage interval equal to twice the maximum peak swing (once for the negative peak, once for the positive). THis makes things easyer but we have to have sensible limits or at least an educated guess for the maximum input voltage swing. 3) Things are considerably easyer to see and construc tif the static operating point is right at a grid voltage line intersection with the load line. Unfortunately, often only a few grid voltage lines are plotted and one has to approximate. Once you find a load line and an operating point on it that fits all of the above, you have your static operating point. Basically, you have to play around with the ruler and have a good eye, or, in some instances do some doodling to approximate grid voltage lines that are not drawn, and see what you get, then if it doesn't look good, repeat the process. Once you are done with that, you are nearly finished. When you chose a static load line, you can calculate the plate resistor (careful, NOT plate resistance!) by dividing the B+ voltage by the current you read off from the intersection of the load line and the y axis. You will notice that because the position of the operating point is limited by the 0V grid line and maximum current, but the load line extends past that and towards higher plate currents, that the intersection with the y axis can happen at a current higher than the maximum alowable. This is fine because the tube will never operate at this current  a positive grid voltage would be required for this. (continued below) 
17th August 2008, 12:13 PM  #6 
diyAudio Member
Join Date: Feb 2005
Location: Zagreb

Next, once you have your static operating point, you use this as a 'pivot point' to check if it suits the dynamic load line. The dynamic load line is always tilted clockwise around the static operating point, with respect to the statis load line imposed by the plate resistor. This is because it represents a lower resistance, which results by paralleling the plate resistor with the input resistance of the next stage for dynamic (AC) signals, because these can pass through coupling caps, making the plate of the triode and the plate resistor, coupled with the input resistance of the next stage. This resistance will usually be the grid leak resistor of the next stage, or the input resistance of some other device that is being driven by your tube, say a power amp. There are standards and specs for this, so you can look it up, and calculate the actual resistance the tube plate sees  (Rplate x Rload )/(Rplate + Rload). Then you draw a load line for this resistance through the static operating point. The easyest way to do this is construc the angle of the dynamic load line, by using, say, 100V on the X axis as a reference. Say your totalload is calculated as 10k, the current through this at 100V would be 100V/10000ohms = 10mA. Find 100V on the x axis and 10mA on the Y axis, and put your ruler so that both these points lie on it. Then, move the ruler keeping it at the same angle, until the static operating point is on it. Now draw a line along the ruler, through the static operating point  and this is your dynamic load line.
The thing you need to check now is similar to when you were chosing the static load line  that intersection with various grid voltage lines remain nice and even, and that for a maximum peak grid voltage swing left and right of the static operating point along the dynamic load line, the actual operating point remains within permissible plate current and voltage and plate dissipation limits. IF this is so, you are done! It is permissible for the dynamic operating point to swing slightly over the maximum dissipation limit or the maximum current limit. This is because the maximum parameters are given for static, or very long term average conditions. Because all AC voltages in audio are swing symetricaly to + amd  from zero, for a reasonably long term average (but much shorter than that used to plot the tube curves and derive maximum permissible voltages and currents!), as the signal swings, as much as the dissipation or current goes over the max on a positive swing, it will go under the max for a negative swing. If your static operating point is well within the limits, the long term average will also remain well within the limits and you are fine. Even so, you will find that violating max current or dissipation usually means the operating point is close to the maximum limits, which invriably means shorter tube life. Very often, and quite ironically, the most linear operating points are found for conditions that do not favour longest tube life, so what you actually chose is always a compromise. If however, checking out the dynamic load line turns out to produce gross violations of limits or results in nonlinearities, you are back to finding a suitable operating point that does not. Keep in mind that it is entirely possible you can't find a good operating point for a given tube and desired usage  which means your conditionas are too strict, or most likely, you are attempting to use the wrong tube for that particular job. Again, a good eye is needed when plotting a load line  along with checking that it intersects with grid voltage lines at even intervals, chack that the grid voltage line spacing is nice and even 'around' (leftright, abovebelow) where you intend to put your operating point. Soon you get the hang of it and will be able to gauge by eye the various distances depending on graph svale and input and output voltages you want. So, there you have it.... PS  using a constant current source (CCS) as a plate load, considerably simplifies finding a static load line, as it is always parallel with the x axis and passes through the y axis at a value equal to the CCS current. The actual position of the oprating point is moreorless freely chosen by shoosing an appropriate grid voltage (this is acyually the 'grid bias voltage'). You will find that intersections of grid voltage lines along this sort of load line tend to be very evenly spaced. However, this still needs to drive a real life load, which is a resistor. Often a nice CCS load line tends to show poor performance once a dynamic load line is plotted too (the dynamic load line under these conditions is plotted for the load resistance only, no need to calculate anything in parallel as CCS 'resistance' is essentially infinite, anything in parallel with an infinite resistance remains the same anything). 
19th August 2008, 04:59 AM  #7 
diyAudio Member
Join Date: Jul 2008

WOW! Thank you very much. I will try and follow your advice.

17th September 2008, 02:28 AM  #8 
diyAudio Member
Join Date: Jul 2008

noisemaker
Well, I got the amp working and encountered a few surprises. In the offchance that the more experienced will stumble across this string, I'll try to describe what happened.
I tried to design a load line for the 6ak5 driver stage that would have the quiescent point at about 150 volts. It mimicked the 6al11 output tube connections that Peter Millett described in his Audioxpress article and website. The schematic is at http://www.pmillett.com/midget.htm. I used .075 uf orange drop caps both at the input jacks and between the driver and output tubes. (The Millett schematic does not use a capacitor at the input jack, and uses a .22 uf cap between the driver and output tubes.) I dropped a couple of components because I am lazy. They included the resistor that followed the capacitor between the driver and output stages and the electrolytic capacitor after the connection of the power supply to the output transformer (both on the Millett schematic). I used a standalone lab power supply and transcendar 5K SET transformers. I also used a stand alone DC power supply for the filaments (which draw more than two amps, primarily because of the 6al11s.) To make a long story short, the "design" voltage was supposed to be at 240 volts. It worked, but had a horrible hum. When I turned down the voltage to about 140, the hum became almost inaudible and it sounds great. Emmylou Harris recordingswhich has exposed flaws in some of my other motley collection of ampssounds like a real, human, female voice. It seems to draw a lot of currentabout 100 maeven at 140 volts. The power supply keeps going to constant current instead of constant voltage. Well, I didn't electrocute myself and got the amp to make noise, so I guess it's a success. If I can figure out how, I will draw the schematic and post it. If you know what I did right or wrong, and are willing to let me know, I would be in your debt. 
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