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Old 14th September 2006, 02:27 AM   #1
xasm is offline xasm  Canada
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Default Plate Voltage on 12ax7a

Okay, bear with me as this is my first Tube circuit as of yet. Hopefully there will be many more in the future.

The most basic of all tube circuits is already confusing me. It is a simple self-bias grounded cathode amp.

Two things that I can't seem to figure out:

1) Plate voltage
2) Gain of the circuit

Let's say I'm running a 230V B+ voltage through a 100k Ohm resistor on the plate and I need to determine the DC voltage at the plate in order to find the value of the Cathode-to-ground resistor for Biasing purposes.

Also, for arguments sake, I have a 1Meg pot connected at the output (plate) to a cathode-follower (buffer). I think pot value will affect the Gain of the circuit, but not neccesarily the Plate voltage. Perhaps this is incorrect.

At first glance, I would have considered the plate to have a very high input impedance such that the plate voltage would be the same as B+, but I do not think this is correct anymore as I'm prett sure there is a voltage drop across the plate resistor at DC.

I know that for a given Plate and Grid voltage, there is a determined plate current form the tube charts. However, is this the current that flows through the plate resistor as well?

As far as the gain goes, I've seen formulas, but no real explanation. Yes, you could just change the values in the formula to acheive the desired gain, but would be nice to know where the gain formula came from...

Any help is greatly appreciated.

Cheers,
Tim
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Old 14th September 2006, 02:33 AM   #2
poobah is offline poobah  United States
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Do you mean "grid-leak" bias?



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Old 14th September 2006, 03:20 AM   #3
SY is offline SY  United States
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What you need to do is learn about load lines. VERY valuable and easy way to get all the info you need. In brief, you know that if the tube is turned off completely (zero current), the plate voltage will equal the B+. And you know that if the tube were turned on completely (can't happen in real life, but this is a model), the plate voltage will be zero, and the current will be B+ divided by the load resistor.

Now, on the chart of the tube's plate voltage vs plate current curves, you've got two points already figured out. Draw a line between them; that's the load line. Pick a spot more or less in the middle of the line (we can refine things later). That will tell you what the grid-to-cathode voltage will need to be. It also will tell you what the plate voltage and plate current are under these conditions. Cool! Now you can use Ohm's Law to calculate the cathode resistor.

I'll work an example next.
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Old 14th September 2006, 03:33 AM   #4
SY is offline SY  United States
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Let's use the GE data sheet for the 12AX7.

http://frank.pocnet.net/sheets/093/1/12AX7A.pdf

On page 3, you'll find the Ip/Vp curves. To make my math easy, I'll assume a 300V rail and a 100k plate load resistor. So, one point is at i = 0, V = 300. The other is at V = 0, i = 300V/100k = 3mA. Draw a line between these points.

Now, we notice that if we want to run 1mA through the tube, the plate voltage will be 200V, and that it will take about -1.6V grid bias to get that. Thus we can calculate the cathode resistor, Rk = 1.6V/1mA = 1.6k.

Et voila!

Now, this is the DC load line. The AC load line will include the effects of any load capacitively coupled to the plate, e.g., the grid leak of the next stage or the volume control. You'll need the AC load to calculate gain. If the cathode resistor is bypassed, the gain will be mu times the total load resistance divided by the sume of the load resistance and the plate resistance. If the cathode is unbypassed, the denominator must have (mu +1) times the cathode resistance added in.

At 1mA, the curves on p4 show the plate resistance to be 60k, mu is 100. So the gain will be (100)(100k)/(100k + 60k + 1.6(101)k), which is about 32.
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Old 15th September 2006, 06:51 AM   #5
xasm is offline xasm  Canada
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Thanks so much for the info. I've been running back and forth on this for a few days now.

Okay, I understand the DC circuit now after running through the curves many times, so it's just the AC section. And yes, I will bypass the cathode resistor in order to acheive a higher gain, however, if I do this, does this remove the inherint negative feedback?

The gain that you have calculated must be based on a certain freq, or perhaps it was DC gain that you worked out, maybe if we use frequency domain we can get a dependent gain for the circuit. Of course, we'd probably need some values for the de-coupling (or is it coupling...?) capcitors between our gain stages, except we only have one gain stage

I was thinking of having a 2 stage amp, both of this same type, with a cathode follower inbetween used a buffer circuit. What are your thoughts on this?

I would like to have an input gain and an output level (or volume) or the circuit, not too sure where to put them. For the input gain, I was thinking of just padding it right on the input. For the output gain, maybe after the second gain stage??

Thanks again,
Tim
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Old 15th September 2006, 12:37 PM   #6
SY is offline SY  United States
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If you bypass the cathode resistor, you'll indeed reduce the feedback. Distortion will be higher, but output impedance will be lower.

The gain you get from this IS the AC gain also, assuming that you plug in as a plate load the parallel combintation of the actual plate resistor and the load resistance.

Plug/Broken Record time: All of this stuff is explained in great detail and in a very readble style in Morgan Jones's book "Valve Amplifiers." If you want to start playing with tube stuff and understand what you're doing, this book is essential and incredibly useful.
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Old 15th September 2006, 05:02 PM   #7
xasm is offline xasm  Canada
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Sweet, I had a quick look at some reviews of "Valve Amplifiers" and it looks good, so I tihnk I'll pick that up as well.

Thanks for the info!

Cheers,
Tim
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