calculating gain
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 31st December 2008, 12:05 PM #1 overdrajv   diyAudio Member   Join Date: Dec 2007 calculating gain Dear diyers! I've been trying and trying to understand the corelation between math and tube data charts but obviously, I still have some basic lack of knowledge... OK, first here is the chart : http://www.drtube.com/datasheets/ecc83s-jj2003.pdf I was calculating from this: http://www.ampwares.com/schematics/princeton_5f2.pdf So if we have 150V at the plate and 1.5V bias, we can get the internal plate resistance by drawing a tangente trough the operating point and calculate the differences between the Ep and Ip. We get around 93k. Now if we make the same on the transconductance chart we get Gm is around 0,95 mA/V The mu should be 100, but it's actually 0,95 X 93 = 88 Did I make everything OK? I calculated the gain of the stage with the tone control on half is 18.8 So if we have 0,1Vrms signal from a typical guitar that is 0,14V peak, we get 18,8 X 0,14V = 2,6V peak at the grid of the second gain stage, while the bias is 1,5V. That would be real heavy distortion for such a gentle bluesy amp. To make things worse the gain of the second stage is also approximately 18 so the peak voltage at the grid of the power tube is 47V while the bias is 19V... Where did I go wrong? Sorry for the long and cumbersome question, but I couldn't ask it more simple than that... Thanks! Miha
 31st December 2008, 01:00 PM #2 Robert McLean   diyAudio Member   Join Date: May 2005 Location: Stittsville, Ontario, Canada Your calculation of stage gain = 18.8 is basically correct. Dont worry about small discrepancies taking slopes off charts, they are not that accurate to begin with. But you are forgetting that the cathode resistor provides local feedback, and the 22k resistor provides feedback to the second stage. For the moment assume the second stage is open loop, no 22k resistor connected. It is wrong to thinking that 2.6 volts at the grid results in distortion just because the grid bias is only 1.5 volts. Dont forget that you have a 1.5K cathode resistor providing negative feedback. the 2.6 volts is from grid to ground. The voltage swing from grid to cathode is much less than 2.6 volts. The grid-cathode voltage is going to be -1.5 +/- 1 for example ( I am guesstimating the 1 volt swing ), in other words varying from -2 to -.5 volts. The grid will not be driven positive. In addition, the feedback from the 22K resistor means that the gain of the second stage is not 18.8, so you are not going to have 46 volts at the output grids. __________________ Robert McLean
 31st December 2008, 03:36 PM #3 Robert McLean   diyAudio Member   Join Date: May 2005 Location: Stittsville, Ontario, Canada It bothered me a bit that I was "guesstimating" the grid-cathode voltage swing in the previous post, so I did the math, and it turns out the grid to cathode voltage is Egk = Ein * [1 - ( u * Rk ) / ( Rl + Rp + (u+1)*Rk) ] = Ein * [ 1 - |gain| * Rk / Rl ] so in your example with Ein = 2.6, u = 88, Rk = 1.5K, Rp = 93K, and Rl the total ac load = 61.5K, Egk = 1.4 volts This little more than my guess of 1 volt, but still less than the bias of 1.5 volts. Although I would probably like a little more than a .1 volt margin with a 12AX7. A real life 12AX7 is probably starting to draw grid current at this point. Note the 68K grid stopper in the circuit. __________________ Robert McLean
 3rd January 2009, 09:04 AM #4 overdrajv   diyAudio Member   Join Date: Dec 2007 Thanks for your effort! I was playing abit with your equotions, but I dont really understand where did you get the Rl (total ac resistance) from.. I thought maybe for the voltage generated acros the cathode resistor you are actualy calculating the gain of the cathode follower, thus Rl is Rk. What do you think? Cheers, Miha
 3rd January 2009, 01:40 PM #5 Robert McLean   diyAudio Member   Join Date: May 2005 Location: Stittsville, Ontario, Canada Rl is the total AC load resistance, in other words the 100 K plate resistance in parallel with the input resistance of the first stage. This is the resistance the AC "sees" since at audio frequencies the impedance of the coupling capacitor is virtually 0, and the impedance of the power supply is virtually 0. At least negligible compared to the other resistances in the circuit. In the case of the second 12AX7 this is 100K in parallel with 220K which is 68K. In the case of the first 12AX7 it is 100K in parallel with the resistance presented by the tone control. This is frequencey dependant as well as dependant on the settings of the tone control pots. I derived 61.5k by working backwards from the fact you said the gain was 18.8, and using your stated mu and rp. I just wanted the example to match your example. One thing to point out about the equations is that they are based on small signal analysis, but the problem being analysed ( distortion due to clipping) is a large signal one. So to be a little more accurate, you should use the values of mu and rp that apply at the point in question, in other words Vg = 0. Looking at the spec sheet in the link provided it looks like at Vg=0 that rp= 50K and mu = 90, so plugging those values into the equation for Vgk I get 1.18 volts Vgk for a 2.6 volt input. __________________ Robert McLean
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Quote:
 A real life 12AX7 is probably starting to draw grid current at this point.
A real life 12AX7 starts drawing grid current at about 0.7V grid to cathode. That's the source of excess distortion in many 12AX7 circuits.
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Miles Prower
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Join Date: May 2005
Location: USA
Re: calculating gain

Quote:
 Originally posted by overdrajv Dear diyers! I've been trying and trying to understand the corelation between math and tube data charts but obviously, I still have some basic lack of knowledge...
Lemme guess: started out as a sand circuit designer amirite?

Unlike SS, where everything is connected together, and therefore have definite realtionships that allow for "design by calculator", VTs have all their parts just hanging out there free and unconnected, so each operates independently and does its own thing. That makes "math" much less useful, and didn't bother to memorize various equations since I hardly ever use them. I design with VTs by using loadlines and plate characteristics. So far, I've had errors between design nominal values and measured values coming in with less than 5% errors.

Quote:
 So if we have 150V at the plate and 1.5V bias, we can get the internal plate resistance by drawing a tangente trough the operating point and calculate the differences between the Ep and Ip. We get around 93k.
That's a real PITA, I prefer to calculate the g(m) and u from the characteristics, then solve for r(p) than trying to eyeball tangentiality.
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overdrajv
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Join Date: Dec 2007
Re: Re: calculating gain

Quote:
 Originally posted by Miles Prower I prefer to calculate the g(m) and u from the characteristics

How do you calculate them?

Thanks,

Miha

 5th January 2009, 05:47 AM #9 ray_moth   diyAudio Moderator Emeritus     Join Date: Jan 2004 Location: Jakarta Unfortunately, you can't just read off Gm and Ra from a triode data sheet, because they vary according to the operating point. Some data sheets include curves for Gm, Ra and Mu against plate current and plate voltage. These are very useful and show, among other things, how Ra falls with increasing current ,while Gm rises and Mu remains reasonably constant (as you'd expect, because Mu is a product of Gm x Ra).
Miles Prower
diyAudio Member

Join Date: May 2005
Location: USA
Re: Re: Re: calculating gain

Quote:
 Originally posted by overdrajv How do you calculate them?
Look at the green lines Loadline. The horizontal line is a constant current line. Since each grid line differs by 0.5V, it spans a total of 1.0V. Drop down to read off the corresponding plate voltages and subtract to get the delta Vpk. Since: u= deltaVpk / deltaVgk, that tells you what the u-factor is.

The vertical green line is a constant Vpk line that also spans a deltaVgk of 1.0V. Read off the corresponding plate currents, and solve: g(m)= deltaIp / deltaVgk. Since:

u= g(m) X r(p) you can solve that for r(p).

When doing such calculations, it's best to use a small delta for best accuracy.
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