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How can I calculate voltage over cathode resistor?

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My head is getting stuck....

Please give me a quick rough guide to calculate the DC voltage where the cathode resistor and the valve meets...
 

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Sweet SY, that was very handy...

Now if I know the B+ is 250 and there is no plate resistor and a 56k on the cathode to ground, and 144V on the grid would I need load lines?....

I had a tutorial once which got me to understand it, unfortunatley I am 10 second Tom...memory of a nougat bar.

I am sorry to be such a bother, I am not trying to design something, just to calculate appropriate component dissipations to not start fires.... when building an existing design.
 
That's an easy one, too. You can do a graphical construction to get a precise answer. Or you can get really, really close by making some approximations.

If you've got 144V on the grid, the cathode will be a couple of volts north of that. Call it 146V, just for fun. That means the tube's current is 146V/56k = 2.61mA. If it was really 147V or 145V on the cathode, you can see that the current will still be pretty close to 2.6mA, so you don't need to sweat it.

Ohm's Law is a pretty powerful tool.;)
 
the one right next to the GND point...
The schematic is not kind enough to state any current or cathode voltage so to calculate it you must do it graphically as mentioned in earlier posts. Draw a 220 K load line, and on each grid line put a point where current equals grid voltage divided by 1.5K ohms. Join those points to form a line, and where that line intersects the load line is the operating point.

Or, you can just notice that the only difference between this triode and the other ones is 1.5K instead of 2.2K cathode resistor, so I would estimate about .6 mA instead of .5 mA, 110 volts on the plate and .9 on the cathode. See how they fit with the plate curves and adjust accordingly.

As for calculating the stage gains to get more accurate values for dissipation, I personally would not bother for this low level type of circuit. I would just take the calculated DC disipation, double it, and then pick the next higher power rating for resistors.

But I could be wrong, and if you want to calculate the gains , notice that with the exception of the last stage the triodes are all cathode bypassed amplifiers, with gain given by mu * Rl/(Rl +rp). For 12AX7 I would use mu = 100 and rp = 75000, and for Rl remember it is the parallel combination of the plate resistance and the following input resistance. I wouldnt bother with the frequency dependant stuff, the tone control caps that is, just go with the volume control resistance, say in mid position.

Ignore the 68K and 480K grid stopper resistors on the first two triodes, do not ignore the 480k resistor on the third triode, it is part of a voltage divider.

The last stage is a cathode follower and has gain of just under 1.

I will leave it to you to do the actual calculating.
 
Originally #5 posted by SY
That's an easy one, too. You can do a graphical construction to get a precise answer. Or you can get really, really close by making some approximations.

If you've got 144V on the grid, the cathode will be a couple of volts north of that. Call it 146V, just for fun. That means the tube's current is 146V/56k = 2.61mA. If it was really 147V or 145V on the cathode, you can see that the current will still be pretty close to 2.6mA, so you don't need to sweat it.

Ohm's Law is a pretty powerful tool.;)

Hi SY,
not that easy, @ 100Va-->k 0V g-->k you get 2mA.

Hi Nordic,
is distortion caused by this effect desired?

Kind regards,
Darius
 
oldeurope said:


Hi SY,
not that easy, @ 100Va-->k 0V g-->k you get 2mA.

Hi Nordic,
is distortion caused by this effect desired?

Kind regards,
Darius

I think the point being made here is, by the time you get 144V on the grid, you will probably already be into positive grid bias voltage and grid current with this tube? Because, for about 144V on the cathode (plus or minus 5 volts, which is all you will ever have for bias voltage, in any reasonable situation with most small signal tubes in CF), the resistor WILL be drawing close to 2.6ma, for a close to 144 volt drop. Ohm's law will not be circumvented...

If that means that the grid must be positive to the cathode, so be it... the tube WILL try to stabilize with the cathode within a few volts of the grid...

However, this would potentially be a distortion problem in the making, unless the source driving the stage is near-zero DC impedance...

Regards,
Gordon.
 
Erm, yes follow the input it goes through a triode buffer after which there is a switch selecting either a distorted channel or a "clean" channel with only volume control.

The distortion is for use as a guitar amp pre... I will likely try to increse the distortion levels...

The idea is to amplify the valve type clipping in the next half of the amp which is solidstate... to emulate the sound of output tubes being driven into distortion...

It is nice to have a nice oragnic transition between a "clean" sound and some distortion by simply playing a bit harder.... something the all transistors can not do without some serious computer intervention.
 
Nordic said:
Erm, yes follow the input it goes through a triode buffer after which there is a switch selecting either a distorted channel or a "clean" channel with only volume control.

The distortion is for use as a guitar amp pre... I will likely try to increse the distortion levels...

The idea is to amplify the valve type clipping in the next half of the amp which is solidstate... to emulate the sound of output tubes being driven into distortion...

It is nice to have a nice oragnic transition between a "clean" sound and some distortion by simply playing a bit harder.... something the all transistors can not do without some serious computer intervention.

Ah, thanks for explaining.:)
now I understand what is going on in the circuit.

Kind regards,
Darius
 
SY said:


No, the point being made is irrelevant to this discussion. The bias is not positive nor zero. oe didn't read (or if he did, didn't comprehend) the post he cited.

I went back and ran this alignment through TubeCad (cathode follower, 250V B+, 56K cathode resistor, 2.6ma current).

SY is absolutely correct. This tube will have about 145v on the grid, in that alignment... and the grid bias is still -0.4v. So, the cathode is at -144.6v. No grid current, at least in theory... though, with some 12AX7s, this might be getting close to "trickles" on the grid...

BTW: in case it is important, the gain of this stage is .95, or -.294dB.

HOWEVER- there is one problem. With 2.6ma on the plate, you've exceeded the plate current limits of the 12AX7 (which are 2.5ma). You may want to look at a different cathode resistor (higher resistance) if you want to use that tube, or a different tube if you want that current... something like a 6SL7/5691 would be able to handle up to 5ma...

Regards,
Gordon.
 
Gordon, there's no question- at those sort of low grid bias voltages, 12AX7s show significant grid current. The higher the plate voltage and current, the worse the issue. It's a matter of geometry- the grid is of necessity near the cathode in a high mu tube and has a high cross-section.

I didn't read this part of the question as involving a 12AX7, and you're right, at 2.6mA, that's not the tube to use.
 
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