Biasing Preamp Triode Gain Stage Help

[ericnuke]

So I've gone back and have been studying the AX84 tube theory page and trying to really apply all that knowledge to one of my builds. Here's where I'm at so far:

First gain stage is standard triode.
-I've got 252v before plate resistor and 101v at the plate on the tube.

-Looking at the Plate Amps/Plate Volts chart for an AX7 I see that I need a cathode voltage of .8v to be half way between 1.7v(where plate current is 0 amps) and 0v (where the tube is in saturation)

-So doing all the math I find that I should be using a 533 ohm resistor on the cathode to ground to get this cathode bias of .8v

Math
252 pre-plate resistor voltage minus 101 plate voltage =151v drop

151v divided by 100k plate resistor ohms = .0015 amps

.0015 amps multiplied by 533 ohms = .8v

First off, is this all right or am I missing something?

Secondly, is a 470 ohm resistor be "close enough" since I don't know where I can find a 533 ohm resistor without paralelling a couple 1.5k's

Lastly, is only changing the cathode resistor the right way to do this or should I be changing the plate resistor instead or change them both to make the bias point right? I have an 820 ohm on the cathode now and while it does work fine I want to really understand this stuff and not just go off schematics I've found any more. Thanks!

[martin manning]

I think you are mixing some things up. I would do it like this:

1) 100k plate load from 252V B+ means 252/100=2.52mA at zero plate volts (Ohm's law).
2) Draw that load line on the 12AX7 plate characteristics plot, from 2.52mA at zero volts to 0mA at 252 volts.
3) The load line crosses the 0V grid line at about 1.75mA plate current, so a good bias point would be at half of that, or 0.88mA.
4) At 0.88 mA, the load line would be crossing the -1.3V grid line if there were one drawn at that voltage (I'm interpolating), so you need a cathode resistor to hit 1.3V at 0.88mA, or 1.3/0.00088=1477 Ohms, or 1k5 is close enough.

You chose the combination of plate load and cathode resistor to set the bias point, and they will be in roughly similar proportion to center bias for any B+ and plate load, as in 1k5/100k=0.015. Setting the bias a liitle high or low is according to taste. I think most people will choose the plate load first and then compute the cathode resistor value for the desired bias point.

I think Merlin's stuff is far superior to anything I've seen for a clear explanation of this stuff.

[sunnydaze]

You might find this spreadsheet helpful.

Written by the Wizard (Merlin Blencowe) himself. You can modify the R & C values for the common preamp type tubes and it will plot on a plate characteristics chart. His book on preamps does a great job of explaining why and how a lot things are done the way they are in guitar amps.

Thanks
mike

[martin manning]

Or this one: https://tubeamparchive.com/viewtopic.php?p=76299#76299
Written with a little help from the Wiz.

[ericnuke]

I think you are mixing some things up. I would do it like this:

1) 100k plate load from 252V B+ means 252/100=2.52mA at zero plate volts (Ohm's law).
2) Draw that load line on the 12AX7 plate characteristics plot, from 2.52mA at zero volts to 0mA at 252 volts.
3) The load line crosses the 0V grid line at about 1.75mA plate current, so a good bias point would be at half of that, or 0.88mA.
4) At 0.88 mA, the load line would be crossing the -1.3V grid line if there were one drawn at that voltage (I'm interpolating), so you need a cathode resistor to hit 1.3V at 0.88mA, or 1.3/0.00088=1477 Ohms, or 1k5 is close enough.

You chose the combination of plate load and cathode resistor to set the bias point, and they will be in roughly similar proportion to center bias for any B+ and plate load, as in 1k5/100k=0.015. Setting the bias a liitle high or low is according to taste. I think most people will choose the plate load first and then compute the cathode resistor value for the desired bias point.

I think Merlin's stuff is far superior to anything I've seen for a clear explanation of this stuff.

Awesome! Thanks so much.

Now I'm following up to point #4. Taking .88mA and then tracing over the plate voltage line to 252 plate volts, wouldn't that end up crossing the -2.3 line? Then taking 2.3/.0008=2613 (2.7k resistor then) I guess I'm confused yet how you determined the right grid voltage line to cross?

And i'm downloading these other documents right away and ready to dig in!

[martin manning]

Find 0.88mA on the vertical axis and then go horizontally over until you intersect the load line. Compare that intersection point to the grid lines that are on the plot, and you'll see that it would be at about -1.3V.

[ericnuke]

Really cool. I get it now. I didn't get using the load line as the second indicator. I was using the .88mA on the vertical axis and thinking the other point would be the 252v on the horizontal and find where those lines crossed.


This is so legit! I know it's still just scratching the surface, being able to bias a simple triode right, but I feel like I'm 10 feet tall right now. Thank you very very much Mr. Manning!

[martin manning]

Looks like you're on your way!

[jjman]

Don't be surprised if the results differ somewhat from the graph's prediction. Different tubes of the same type will idle (and amplify) differently. A good tube should be close enough.