Ampbooks SLO 12AX7 Bias Calculation

[Gregwor]

After a lot of reading, thinking, and tinkering I simply cannot figure out how on earth this calculator determined that "the DC grid bias is -1.5V".

Here is a screenshot from the site Circuit Analysis of the Soldano Super Lead Overdrive Channel



Can anyone please show me how to determine the bias of a triode by looking at the schematic.

Thanks in advance!

Greg

[sluckey]

Bias is defined as the difference in voltage between the grid and cathode. The voltage on the grid is zero. The voltage on the cathode is +1.5V. That makes the grid -1.5V in respect to the cathode.

[Gregwor]

Bias is defined as the difference in voltage between the grid and cathode. The voltage on the grid is zero. The voltage on the cathode is +1.5V. That makes the grid -1.5V in respect to the cathode.

Thank you for your reply! I fully understand the concept of cathode biasing. If you try the ampbooks 12AX7 calculator, you'll see that you have to input the current and grid bias, so I cannot figure out how they figured out that it is in fact -1.5V. I didn't sleep last night trying to sort this out so I appreciate any help! I hope I'm just missing something super simple! I love how Fender schematics label the cathode voltage.

Thanks!

Greg

[2deaf]

If you try the ampbooks 12AX7 calculator, you'll see that you have to input the current and grid bias, so I cannot figure out how they figured out that it is in fact -1.5V.

I had to go to a different online calculator to get the numbers and then come back to Ampbooks and enter those numbers. 

[Gregwor]

I had to go to a different online calculator to get the numbers and then come back to Ampbooks and enter those numbers.

I did that too but different online calculators gave me different results. Which site did you use?

Greg

[Leevi]

I think the cathode resistor is an essential parameter in the calculation when counting the bias and it could be given as an input.
Now the calculator uses the characteristic of the tube and defines the cathode resistor and bias based on the plate voltage and plate resistor.


Furthermore the tubes are not identical with each other.


/Leevi

[2deaf]

I did that too but different online calculators gave me different results. Which site did you use?

I went to trioda first and got numbers that worked, but they weren't quite the same as Ampbooks.  Then I went to VTADIY and got the same numbers that Ampbooks used.   

[Gregwor]

I think the cathode resistor is an essential parameter in the calculation when counting the bias and it could be given as an input.
Now the calculator uses the characteristic of the tube and defines the cathode resistor and bias based on the plate voltage and plate resistor.


Furthermore the tubes are not identical with each other.


/Leevi

For sure.

I went to trioda first and got numbers that worked, but they weren't quite the same as Ampbooks.  Then I went to VTADIY and got the same numbers that Ampbooks used.   

Care to share what values you entered in order to get VTADIY to match up? I tried this yesterday and got these "different" results. Here is a screenshot of what I did.



Thanks for the help so far guys!

Greg

[tubeswell]

After a lot of reading, thinking, and tinkering I simply cannot figure out how on earth this calculator determined that "the DC grid bias is -1.5V".

Here is a screenshot from the site Circuit Analysis of the Soldano Super Lead Overdrive Channel



Can anyone please show me how to determine the bias of a triode by looking at the schematic.

Thanks in advance!

Greg

First, find out the tube current with 360v B+ and a 220k plate load resistor: 360V/220k = 0.0016A (1.6mA). This is what the tube would conduct if all the HT voltage was being dropped across the tube and nothing across the plate resistor. This is where the (red) loadline hits the vertical axis.   So your DC loadline runs between 1.6mA on the vertical axis, and 360V on the horizontal axis.


Now, if you know you want 1.5v bias with 1.6mA tube current, the cathode resistor required would be:

1.5V/0.0016A = 937R


Alternatively, if your schematic shows (say) a 1k cathode resistor and you know you have 1.6mA tube current, then the bias voltage would be:

1,000 x 0.0016 = 1.6V


In the plate characteristics chart above, we can see that at an idle voltage of 175V, the load line shows 0.83mA tube current (where the 1.5V grid curves crosses the loadline). Therefore the cathode resistor required to a achieve bias voltage of 1.5V is:

1.5v/0.00083 = 1,807R


Or if we wanted an idle voltage of 200v, this corresponds with the point along the load line (between the 1.5v and 2.0v grid curves) that shows about 1.8v @ 0.75mA, so the cathode resistor would be:

1.8v / 0.00075A = 2,400R

[Williamblake]

The bias is not drawn in the picture, only the grid curves are. If you enter your bias point it calculates your cathode resistor. You would have to chose the bias point yourself and would probably aim for the most voltage swing at the output. Aka the middle of the loadline.

[Gregwor]

So what I'm understanding here is that there is no definitive way to figure out the bias voltage the designer chose?

Like tubeswell implied, it's easy to choose a bias voltage during the design stage. Reverse engineering it seems to be a guessing game. I assume that's why digital amp designers like Cliff Chase state that in order to emulate an amp they need one physically in front of them.

I have not dug into LTspice much yet but I'm assuming it could help in this matter? Spice sure has changed quite a bit since I used it 17 years ago during my EET schooling!

Greg

[2deaf]

Care to share what values you entered in order to get VTADIY to match up? I tried this yesterday and got these "different" results. Here is a screenshot of what I did.

-1.39V divided by 0.87mA equals 1.598K.  1.8K is the target cathode resistor, so 0.87mA isn't what we want.

I adjusted the plate current until the grid bias voltage divided by the plate current equaled 1.8K.  It's not visible, but the plate current is 0.8349mA giving a cathode resistor of 1.797K at a bias voltage of -1.5V.

[2deaf]

I must be missing something with the Amp Books calculator because it seems to give a bunch of scenarios that aren't possible.  Even the example given has a plate current that doesn't match the cathode current.  The DC plate voltage should be 177.4 instead of 175.9 in order for the currents to be equal. 

Here is an example that gives a cathode resistor of 1.8K, but the other numbers are not possible.

[Gregwor]

I must be missing something with the Amp Books calculator because it seems to give a bunch of scenarios that aren't possible.  Even the example given has a plate current that doesn't match the cathode current.  The DC plate voltage should be 177.4 instead of 175.9 in order for the currents to be equal. 

Here is an example that gives a cathode resistor of 1.8K, but the other numbers are not possible.

Which numbers don't check out?

HT = 360V
Plate Voltage = 113.8V

Means there is 360V - 113.8V = 246.2V drop across the plate resistor.
Current through the plate resistor should be 246.2V / 220kΩ = 1.119mA. Ampbooks says is should be 1.11mA. The difference seems to be a rounding error.

For the cathode, their calculation of 1.802kΩ with their 1.11mA current claim should result in a voltage of:
V = 1.11mA x 1.802kΩ
Cathode voltage = 2.00022V

I can choose any Vgk along the DC loadline and give you a current and cathode resistor to work with it. It seems all of ampbooks math works out, but again, we have not yet solved how to determine the actual bias voltage of the circuit.

Again, thanks for trying to sort this out guys. It's been eating at me and making me feel defeated in my tube amp design endeavor. Hopefully we can get to the bottom of this. Maybe one of you knows the dude behind ampbooks and could ask them where they're getting their figures from?

Greg

[PRR]

...find out the tube current with 360v B+ and a 220k plate load resistor: 360V/220k = 0.0016A (1.6mA). This is what the tube would conduct if all the HT voltage was being dropped across the tube and nothing across the plate resistor.

I think you typed that wrong. 360V/220K is "all voltage across resistor, zero voltage across tube". Which is not a workable operating point, but is the limit if the tube could be driven all the way to a short.

> Amp Books calculator because it seems to give a bunch of scenarios that aren't possible.

Yes. I too can force "impossible" answers, including negative plate voltage.

However I am not sure why we need calculators? Conditions can always be plotted on tube curves, and approximated on a matchbook. Quite large deviations from expected op-point make little difference to max output and almost none to gain.

[2deaf]

Which numbers don't check out?

HT = 360V
Plate Voltage = 113.8V

Means there is 360V - 113.8V = 246.2V drop across the plate resistor.
Current through the plate resistor should be 246.2V / 220kΩ = 1.119mA. Ampbooks says is should be 1.11mA. The difference seems to be a rounding error.

No.  1.119mA rounds to 1.12mA for the same number of significant digits as Amp Books' DC plate current.  1.12mA does not round to 1.11mA within that same number of significant digits.

The plate voltage needs to be 115.8V in order for the cathode current to equal the plate current.  113.8V certainly doesn't round-off to 115.8V.

You can keep going after the decimal place, but he only shows 1.11mA.  Since it is possible to keep going, there is at least some justification to take 1.11mA as 1.1100000mA.  Even if you run it on out, it still gives you the wrong plate voltage.

 

[PRR]

> Conditions can always be plotted on tube curves

This is a basic skill.

I took TubeBook's curves. We plot 360V and 220k. Unsurprisingly this lays right on top of the calculator's line.

Now we plot the cathode. Because the tube is not linear this line is not straight. The curve is mild and might normally be neglected, but youse guys are posting decimal numbers so let's be fussy. "1.8k" was specified. Figure some grid voltages across 1.8k as cathode currents. Connect the dots.

The intersection of the two lines is the operating point, very nearly(*). 186V 0.83mA, 1.5V.

BTW the "AC load" is not part of the DC bias condition. Here it may serve as a place marker.

(*) Strictly we should deduct the 1.5V from the 360V and iterate with 358.5V supply. On high Mu tubes the error is very small. 0.5% here, and we know all tube-work is +/-20% precision.

[Gregwor]

 

Thank you guys! I wish ampbooks would just include the cathode resistor in it's calculator and plot the cathode load line. I pictured the cathode load line in my head over and over thinking about this and simply didn't draw it out because I'm a dork!

Thank you again!

Greg

[2deaf]

For the cathode, their calculation of 1.802kΩ with their 1.11mA current claim should result in a voltage of:
V = 1.11mA x 1.802kΩ
Cathode voltage = 2.00022V

Yeah, but look at the -2.0V grid curve.  The operating point isn't anywhere near it.  The 1.8K cathode load line is also nowhere near the operating point.   

[Gregwor]

Lastly, do you guys know of any site or excel program where I can enter values over grid curves? The less I have to deal with paint type programs or printing grid curves out the better.

Greg

EDIT: Paint is actually wicked easy to plot on 

[2deaf]

I can choose any Vgk along the DC loadline and give you a current and cathode resistor to work with it. It seems all of ampbooks math works out, but again, we have not yet solved how to determine the actual bias voltage of the circuit.

You cannot determine the bias voltage by simply plugging in numbers on the Amp Books calculator.  You have to already know the answer in order to plug in the proper numbers.  If I were going to determine the SLO quiescent conditions, I would pull out one of my paper charts and put a ruler to it. 

[Gregwor]

Yep! Manually plotting on the grid curves matches the bias voltage of the ampbooks second SLO gain stage as well so this is obviously the sound method 

Greg

[PRR]

> Paint is actually wicked easy to plot on

+1.

You can also Xerox the tube chart and draw on it with a pencil. No sin in doing it the Old Way.

[HotBluePlates]

... Manually plotting on the grid curves matches the bias voltage of the ampbooks second SLO gain stage as well so this is obviously the sound method ...

Calculators & Sims are great when you want to arrive at numbers that are tedious to figure by-hand some other way.  Like "Gain in Decibels" from one of Kuehnel's other calculators.  Or anything to do with plotting frequency response of a tone stack for arbitrary knob-settings.

But the manual method using a set of curves, plot the line for the plate load resistor, plot the "line" due to the cathode resistor, see where they cross...  well, that's just the easiest & fastest way to garner the answers wanted.

* Calculators & spreadsheets give "false precision."  Those curves will be somewhat different for the individual tube you plug into the socket.  Which will be somewhat different from the next-tube you plug into the socket.  This doesn't matter too much because other than how quickly the tube distorts, the differences are "fraction of a dB" and perhaps measurable but not obviously audible.

But it takes some time tinkering & calculating to get confident about where the line is between "sloppy" & "good-enough" in circuit calculations.