Estimating G2 current

[rzenc]

Hi!
I'm reading Richard Kuehnel 'Guitar Amplifier Power Amps' and got curious about the use of a formula on Page 16 (4th printing):

Ip2=Ip1*(((Vg+(Vs2/Us))/(Vg+(Vs1/Us)))^(1.5))

Where:
Ip1: published plate current at published screen voltage;
Ip2: 'new' plate current;
S2: desired screen voltage;
S1: data sheet voltage;
Vg: grid voltage;
Vgc: cut off grid voltage at published G2 voltage;

Us=|(Vs1/Vgc)|


The author mentions this method as accurate when desired screen voltage is CLOSE to published data sheet voltages. However, I did not find any 6550 with screens at 400V.
Do you think this method would be reasonable with screens around 450V?? Data sheet show screens running at 310V. The difference I understand is great (450-310=140). But would it worth the effort to plot new curves in order to determine DC operating points and load?
Are you aware of other methods to do it?
Would the rate between plate current and screen current be applied in such case?
There is a constant factor inside the main formula, which is Us. It relates published screen voltage with cut off voltage at the same screen voltage.

The example given on the book makes use of a 6L6GC with published screen at 400V and the author 'up's' it for 430V. The difference is much smaller (30V against 140V on 6550 case).

Hav you tried this?
Thanks in advance,
Best Regards.
Rzenc

[Pleximan]

I guess it’s the difference between “accurate” and an “estimate”

Here’s your Home work

1)Down load the 6L6GC data sheet and take a look at the
curves for 250 volt screens and 400 volt screens

2) Apply Kuehnel formula to the 250 volt screens and
check you answer on the 400 volt screens and judge for yourself.

Plotting new curves using K’s formula can become quite a tedious task

Fortunately, you only need know peak current to calculate the load.
For a symmetrical load you can shoot for the knee of 0 curve
In which case you can simplify K’s formula when re-scaling the
0 curve to Ip2=Ip1 *(Vs2/Vs1)^1.5

Don’t forget that your screen resistor will drop a bit of voltage depending on the value
creating a sliding screen voltage so you need to estimate your new screen voltage of 450 less the voltage drop across the screen resistor at peak current.

[rzenc]

Many thanks Pleximan!!!

I will try my best mit homework! Thanks for pointing me on this direction!!

---If I'm not mistaken, some time ago there was a topic about centering cathode current instead of centering bias swing but I could not find it ---

Fortunately, you only need know peak current to calculate the load.

G.E. data sheet quotes maximum DC cathode current at 190mA, M.O. quotes KT88 at 230mA. So I believe it would be the maximum plate + sceen current That I will be allowed to drag?

TIA
Best Regards

Rzenc

[Pleximan]

G.E. data sheet quotes maximum DC cathode current at 190mA, M.O. quotes KT88 at 230mA. So I believe it would be the maximum plate + sceen current That I will be allowed to drag?

No, That's the max constant DC current rating

I’m talking about the peak current when the grid is driven to 0 volts as taken from the data sheet curves.

I think we have to back up a bit here.
Before you can re-scale anything you need to know how to calc the load
from the datasheets curves first.

Let assume you want to build a push pull amp with a pair 6550 tubes
with the plate voltage 450 and the screen voltage a rock solid 300 volts

I’m looking at the 6550 data sheet which shows the curves for 300 volt g2

http://www.triodeel.com/6550ap4.gif

As beam pentodes have pretty sharp knees you can just shoot for the knee of the 0 curve
for symmetrical output

Draw a load line from the 450 volts (at 0 current) to the knee of the 0 curve.
Now you can just read the peak Ip from the curves.
From the curves the peak current is 410mA and the plate voltage swings from 55 volts to 450 volts

So the plate load is just the voltage swing divided by the Ip (450-55)/0.410 = 963
The plate to plate load is 4 * this which works out to a 3.8K load a-a

Power = voltage swing * Ip/2 = 81 watts rms

Your title – “estimating G2 current”
You can see from the curves at peak Ip, the peak screen current is 80mA
The average screen current is roughly ¼ of this making the screen dissipation
0.02 * 300 = 6 watts which is also the maximum screen dissipation for 6550

The point I was trying to make in my first post that is you can just re-scale the peak current to work out the plate load with out re-scaling a complete set of curves

[rzenc]

No, That's the max constant DC current rating. I’m talking about the peak current when the grid is driven to 0 volts as taken from the data sheet curves.

Sorry... my bad.. I misunderstood you here...
I will re-read the posts and try to make myself clear about it...

Be back soon!!


plotting formulas Sometimes its just as easy to use a spreadsheet to plot numbers.

I think it might be easier to plot number on spreadsheet and drawing curves by hand in such case??

Thanks!!

Best Regards
Rzenc