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19th June 2007, 01:25 PM  #1 
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tube basics: computing mu and transconductance
these 2 words has been mentioned millions of times. But when I have a look at the datasheet, they are nowhere to be found.
Are these 2 values computed? maybe it is measured using a tester? if the values can be computed, can somone explain using an example? thank you very much ps. maybe you can point me to site? I've looked but nothing interested came up. 
19th June 2007, 01:45 PM  #2 
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Hi,
It is very unusual to not find u and Gm in datasheets, all datasheets that I have seen have u and Gm for some typical operating point. If you have anode curves it is possible to derive it from there, in Morgan Jones book he mention how to do it, it is very simple but difficult to explain without a set of curves so I suggest that you look in his book or some other basic tube electronic book like RDH. Which tube do you need u and Gm for? Regards Hans 
19th June 2007, 02:34 PM  #3 
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Actually, computing mu and gm from curves is fairly simple for triodes, and is derived directly from the definitions of the terms.
mu is the change of Vpk per unit of change of Vgk, at constant plate current. Therefore, pick an operating poit Vp, Ip on the curves. Then draw a horizontal line through this point. Take a convenient change in Vg from the one at the operating point, and move along your horizontal line to this different (delta) Vg, then check on the X axis (Vp) how much the Vp changes with your chosen Vg change. Divide the two (delta Vp / delta Vg) and you have mu. Typically, you chose an operating point on one of the Vg curves. This is because it makes it simple to see what the change of Vp is for the adjecent Vg curves on the left and right, along a horizontal line through the operating point. gm is the shange in plate current Ip with change in grid voltage Vg, at constant plate voltage Vp. Therefore, chose an operating point on the curves. This time draw a vertical line through the operating point, epresenting a constant Vp. Again, move up and down on this line for a convenient change in Vg, checking on the Y axis of the curves (Ip) how much Ip changes. Divide the change in Ip with Vg and you have gm. Again, chosing an operating point on one of the Vg lines in the curve diagram siplifies things considerably. 
19th June 2007, 05:32 PM  #4 
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also, if you have 2 in the sheet, you can work the other one out.
its oddly a little similar to ohms law u = Gm rP rp = u / gM ignore cases 
19th June 2007, 11:58 PM  #5  
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Quote:
In simple terms mu is also referred to as "amplification factor." mu is a pure number not expressed in any "unit"(micromhos, ohms, etc.). mu is a ratio of plate voltage changetocontrol grid voltage change with a given constant current. So say that I want (x) change in anode current; I can either, 1.Change B+ to the anode, or, 2. Change the control grid bias voltage. In a tube with a high mu value you would be able to get the desired current change with only a few volts in bias change as opposed to, say, 100 volts in B+ change. For this reason we say that mu is an indicator of grid efficiency. I.E. A tube with a high mu value will produce a relatively large change in current with a smaller bias change. Gm=Transconductance=mutual conductance is related to mu, but not the same. Gm is a ratio, but unlike mu, it is expressed in a unit; micromhos. Gm is a ratio of grid voltage changetoanode current change when B+ is constant. So, unlike mu, if you know what the Gm value is you know exactly how much the anode current will increase or decrease respectively with a given bias voltage. So, say a tube has a Gm value of 1000 uMhos(this could be simplified, obviously, to read "one millimho," but tube Gm values are always given in uMhos) this means that a onevolt change in grid bias will correspond to a onemilliamp change in anode current. u=/\eb//\ec Where eb=anode(plate) voltage, ec=grid voltage Gm=/\ib//\ec Where ib=anode current, ec=grid voltage *"/\" or "Delta" means 'a change in'
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20th June 2007, 12:00 AM  #6  
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Quote:
can you point me to the exact graph for getting the mu? there are several in the datasheet. same question for transconductance. Thank you very much for the help 

20th June 2007, 12:29 AM  #7  
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Quote:
Here is an example. Let's say we chose an operating point at Iplate = 1.2mA and Vplate = 200V. You will find that this point sits exactly on the 1.5V grid voltage line, which is why I chose it  simple convenience. Now, draw a horizontal line through this point, these are all points in the diagram where plate current is 1.2mA. Moving to the next grid voltage line left and right of the 1.5V along this line, and reading what the plate voltage is for a +0.5V grid voltage change will give you the plate voltage change for that constant plate current, 1.2mA, now divide the plate voltage change with grid voltage change and you get mu. In this example, you get Vplate = 155V @ Vg = 1V and Vplate = 252V @ Vg = 2V, both following the constant current line of 1.2mA. (252  155)/(1 (2))=97, and this is the mu of the 12AX7 at the chosen operating point. For gm, we do a similar thing, but drawing a vertical line through the chosen operating point, to represent all points in the graph at which Vplate = 200V. Again, we move to the next grid voltage line this time up and down following the 200V line, and read the currents we get: Iplate = 2.13mA @ Vg=1V, and Iplate = 0.53mA @ Vg=2V. (2.13  0.53)/(1(2)) = 1.6 mA/V, and this is our gm. In reality, the figures are approximate  if you look carefully, the error is readily apparent for the gm calculation  the distance between the 1 and 1.5V grid voltage line and 1.5V and 2V grid voltage line along the Vplate = 200V line is not the same, and the calculation assumes that it is. This can often be the case as it is impractical to draw many grid voltage lines on the plate characteristics diagram because the places where they are close would become too cluttered to make any usefull reading. Still, unless you are looking at gross asymetry as you move to the next grid voltage line over, you will get very usable results, the error will likely be smaller than actual tube tolerances, anyway. If you also look carefully, you will notice that for the given operating current of 1.2mA, the grid voltage line intersectios with the 1.2mA line are very evenly spaced. This very graphically shows the linearity of a tube used as a pure voltage (plate current is constant as in the mu calculation) vs pure current (plate voltage is constant as in the gm calculation) amplifier. The 12AX7 was made for linear voltage amplification, and this clearly shows. The procedure for deriving mu also shows you why a triode with a current source in the plate circuit has an amplification factor equal to mu  because such a connection drives the triode exactly the way mu is defined  with a constant plate current. 

20th June 2007, 12:34 AM  #8  
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20th June 2007, 07:18 AM  #9 
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Jartel, look at page 4 of that datasheet. You will see that mu remains fairly constant with varying plate current but both Gm and Rp vary a lot. The lower the plate current, the lower the Gm and the higher the Rp. Page 4 shows you the relationships. With a triode, you cannot simply compute Gm from mu and Rp, you have to take into account the operating current.
When Gm and Rp are quoted in datasheets, they are quoted for specific operating conditions. You will see, at the foot of Page 1 under CHARACTERISTICS AND TYPICAL OPERATION, that different values are given for both Gm (1,250 and 1,600 umho) and Rp (80,000 and 62,500 k) at two different plate currents (0.5 and 1.2 mA.) Mu, however, is 100 at both currents. You might also find this Patrick Turner link interesting. It explains how to figure out the Gm, mu, etc. 
28th January 2012, 08:12 PM  #10  
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Quote:
Ok, I know this is an old thread...but for us hard heads...can someone use the indicated formula and work out the answers. Ip is 21ma and plate voltage is 250V. 

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