Hi there,
this is a newbie question.
I'm looking for a method to calculate quiescence current on a vacuum tube circuit.
Let me try to explain better:
if I have a circuit: say a marshall JCM800 preamp and I know anode voltages..how can I determine quiescent current on the tubes?
Usually the procedure is...choose a quiescence current and the anode voltage and then draw the load line on the tubes curves..
Now I want to do the opposite...I have the Voltage, I have the cathode resistor and the plate resistor..and I want to determine the quiescence voltage...
I suppose I'm missing the Load resistor..but how can I calculate on a complex circuit...
My example : JCM800 first triode...suppose Anode Voltage:350V, Rk suppose 820R, Load = ??? (how can I calculate it if I have many stages) quiescent current ?? (on the first triode)
I would like to know a practical mode (whit an excample if possible on that schematic ) to do this calculation..if any.
Thanks a lot for your help
Krusty
this is a newbie question.
I'm looking for a method to calculate quiescence current on a vacuum tube circuit.
Let me try to explain better:
if I have a circuit: say a marshall JCM800 preamp and I know anode voltages..how can I determine quiescent current on the tubes?
Usually the procedure is...choose a quiescence current and the anode voltage and then draw the load line on the tubes curves..
Now I want to do the opposite...I have the Voltage, I have the cathode resistor and the plate resistor..and I want to determine the quiescence voltage...
I suppose I'm missing the Load resistor..but how can I calculate on a complex circuit...
My example : JCM800 first triode...suppose Anode Voltage:350V, Rk suppose 820R, Load = ??? (how can I calculate it if I have many stages) quiescent current ?? (on the first triode)
I would like to know a practical mode (whit an excample if possible on that schematic ) to do this calculation..if any.
Thanks a lot for your help
Krusty
krusty75 said:
Do you mean you know the voltage AT THE ANODE, or the SUPPLY VOLTAGE (HT), which is not the same thing, of course?
If you know the voltage AT THE ANODE, and you know the anode resistor (Ra), then the quiescent current current is simply:
Ia = HT - Va / Ra
Similarly, if you know the cathode voltage (Vk), and the cathode resistor (Rk), then the quiescent current must be:
Ia = Vk / Rk
However, it sounds more like you know the HT, and you have a particular cathode resistor in mind, and you want to choose a load resistor? There is no particular load than corresponds to any particular cathode resistor, unfortunately. A rule of thumb used by lazy designers in the old days was:
Rk = Ra / mu,
or,
Ra = mu x Rk
which would yield 82k in your case, but this tells you nothing about how the stage might operate. 100k is the usual value for an ECC83 in a guitar amp. Perhaps you could explain your problem in more detail?
Yes you are right...I know Supply voltage not the voltage at the anode.so this is why I want to know a way to calculate quiescent current.
What I want to do is this:
I want to analyze an "already made" guitar preamp.
To have a practical example to work with I choose Marshall JCM 800 (first triode) so we can have the same schematic under our eyes.
So...I just want to analyze the preamp (because using the power supply I can "extract" current draw from the tube").
So I have this preamp..with plate resistor fixed (say 100k) and cathode resistor fixed (say 2.7K (but any other value it's ok..it's just an exercise for me)).
What I would like to know (learn how to calculate is)
1) How much is the quiescent current drown by the tube given those parameters. 2) How can I calculate the load seen by this triode? (this is just the first stage of the amp)
Hope my explanation is clearer now.
thanks in advance.
Krusty
Here is the basic quick and dirty method to determine your tube scenario....
I will post a photo of the graphical 12AX7 curves and load-line...
First get the 12AX7 Plate curves..... I vs V ........
Plot athe 100K Load-Line starting at 350V ...
This is 350V/100K = 3.5mA intersect at the Y-AXIS , current axis..
Now the rest is a trial and error, when backwards figuring the circuit.... Since this is a process of convergence.... To do this in one shot, you would need to derive a bunch of non-linear equations and set them equal to each other....but thats for another post....
Working only along this load-line....you would trial-and-error the fit for the cathode resistor... ( Bias voltage/2.7K) and you will get a current, check this to see if agrees by intersection of the load-line, if not then try again.... If the current is too low, then lower the bias voltage... If current is too high thn increase the bias voltage....you should converge in a few iterations.....
For example....2V/2.7K = .74mA, looking on the load-line you would have roughly 1.1mA at -2V ..... so you first guess is too low....
Now try 2.5V/2.7K = .926mA.....looking at load-line you would have .78mA at -2.5V.... Now you are too high....So now you know the answer is between -2V and -2.5V , it just requires a few more iterations to converge..... The solution is roughly -2.35V at .87mA...But keep in mind that this is based on the data sheet...tubes will vary within reason and resistors are with 5% and 10% so never take this type of analysis to heart.....
Technically the cathode resistor is added to the plate load resisitor for the load-line, but that is non-sense in this case, since the cathode resistor falls in much smaller than the tolerance of the plate resistor and the tolerance of the 12AX7.....
Also, the true Quescent point moves when the amp is running at full power output....and those are the "real" voltages you should be designing to... For example lets say your Marshall has 450V on the plates at idle and 350V at the source side of the 100K load resistor for 12AX7....
At full power output your B+ will drop and could be down at 400V, thus your 12AX7 load-line analysis may need to be redone at 300V ..or ect... or whatever it droops to at full power....
Keep in mind that the 820 ohm resistor in some Marshall amps is shared by both halves of the same 12AX7.... So the analysis need to be done at 1.6K for one tube plate curves....
Best Regards
Chris
I will post a photo of the graphical 12AX7 curves and load-line...
First get the 12AX7 Plate curves..... I vs V ........
Plot athe 100K Load-Line starting at 350V ...
This is 350V/100K = 3.5mA intersect at the Y-AXIS , current axis..
Now the rest is a trial and error, when backwards figuring the circuit.... Since this is a process of convergence.... To do this in one shot, you would need to derive a bunch of non-linear equations and set them equal to each other....but thats for another post....
Working only along this load-line....you would trial-and-error the fit for the cathode resistor... ( Bias voltage/2.7K) and you will get a current, check this to see if agrees by intersection of the load-line, if not then try again.... If the current is too low, then lower the bias voltage... If current is too high thn increase the bias voltage....you should converge in a few iterations.....
For example....2V/2.7K = .74mA, looking on the load-line you would have roughly 1.1mA at -2V ..... so you first guess is too low....
Now try 2.5V/2.7K = .926mA.....looking at load-line you would have .78mA at -2.5V.... Now you are too high....So now you know the answer is between -2V and -2.5V , it just requires a few more iterations to converge..... The solution is roughly -2.35V at .87mA...But keep in mind that this is based on the data sheet...tubes will vary within reason and resistors are with 5% and 10% so never take this type of analysis to heart.....
Technically the cathode resistor is added to the plate load resisitor for the load-line, but that is non-sense in this case, since the cathode resistor falls in much smaller than the tolerance of the plate resistor and the tolerance of the 12AX7.....
Also, the true Quescent point moves when the amp is running at full power output....and those are the "real" voltages you should be designing to... For example lets say your Marshall has 450V on the plates at idle and 350V at the source side of the 100K load resistor for 12AX7....
At full power output your B+ will drop and could be down at 400V, thus your 12AX7 load-line analysis may need to be redone at 300V ..or ect... or whatever it droops to at full power....
Keep in mind that the 820 ohm resistor in some Marshall amps is shared by both halves of the same 12AX7.... So the analysis need to be done at 1.6K for one tube plate curves....
Best Regards
Chris
For basic analysis you can use thenormal curves, but add a new voltage scale, indicating thje cathode-to-ground voltage, see:krusty75 said:
http://www.freewebs.com/valvewizard1/dccf.htm
However, the curves themselves are altered by the feedback extant in a cathode follower, so you need to take care how you interpret the load line, see page 3: http://www.thermionic.info/cathode_ray/CathodeRay_CathodeFollowers.pdf
krusty75 said:
The follower is a different beast all together.... It's all feedback...
Here are simple first-order quick and dirty methods that provide close approximations...
The most important voltage to note is the GRID voltage, since the CATHODE will FOLLOW it..... Lets say you have 350V on the plate and the GRID has 190V on it.... The your CATHODE will have roughly 192V minus the bias voltage... which is around 2 volts, give or take, in the applications you have.....since yo have the direct coupled version..... The DC voltage drop, plate voltage, of the gain stage before provides the grid voltage for the follower...which is a 100K ....and the follower uses a 100K in the cathode....which is fine....
You have a 100K in the plate load....this programs the current in the follower....so 190V/100K = 1.9mA ........ You will always get roughly 190V on the CATHODE regardless what value cathode resisitor you use.....so if you use a 56K you would get 3.39mA ....
The proper way to do cathode follower design is to start out with the IV "PLATE" curves..... then there is a way to redraw over them and come up with the "CATHODE" curves.... This way you can see the cathode curves and how to load them and you can see the large voltage swing at the input grid....
If you use too small of a cathode resistor you start to loose gain as it drops below unity....
Chris
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