Easy. Look it up on the datasheet.
Otherwise, the mu is the slope of plate voltage change plotted against grid voltage change at constant plate current. If you have the curves in hand (but for some odd reason can't turn back a page to get the mu), lay a horizontal line across the curves at the planned current. Find the curve that's closest to the desired no-signal plate voltage and note the reguired grid voltage. Step over one curve and read off the plate voltage and note the grid voltage. Take the difference between the two plate voltages and divide it by the difference between the two grid voltages. Voila! You have mu.
Otherwise, the mu is the slope of plate voltage change plotted against grid voltage change at constant plate current. If you have the curves in hand (but for some odd reason can't turn back a page to get the mu), lay a horizontal line across the curves at the planned current. Find the curve that's closest to the desired no-signal plate voltage and note the reguired grid voltage. Step over one curve and read off the plate voltage and note the grid voltage. Take the difference between the two plate voltages and divide it by the difference between the two grid voltages. Voila! You have mu.
Hi Nick,
When you have plate curves, you have all !
In attached one (It's a Russian 6N6pi)
Pick a first point on a grid line, f.e. the red one where Vg is -4V, Vp is +155V and Ip is 30mA
Move down vertically (constant plate voltage) till you cross another grid line (the blue dot).
Here Vg is -6V and Ip is 12 mA.
That means that plate current change is (30 -12) 18mA for Vg change of 2 Volts.
Thus the slope is (18 / 2) 9mA per Volt, the same that a transcondustance of 9000 if expressed in mmoh.
Then, move left horizontally (constant plate current) till you meet the original grid line (green dot).
The plate voltage is now 120V for a 2V grid voltage difference, thus the Mu, µ or amplification coefficient is (155-120) / 2 = 17.5
The Rp is figured by the slope of a straight line joining the red and the grid dot.
This is the same as dividing the Mu by the slope (in Ampere per Volt), here 17.5/.009 = 1K94.
<edit>
Oopps, misread the plate voltage !
</edit>
Yves.
When you have plate curves, you have all !
In attached one (It's a Russian 6N6pi)
Pick a first point on a grid line, f.e. the red one where Vg is -4V, Vp is +155V and Ip is 30mA
Move down vertically (constant plate voltage) till you cross another grid line (the blue dot).
Here Vg is -6V and Ip is 12 mA.
That means that plate current change is (30 -12) 18mA for Vg change of 2 Volts.
Thus the slope is (18 / 2) 9mA per Volt, the same that a transcondustance of 9000 if expressed in mmoh.
Then, move left horizontally (constant plate current) till you meet the original grid line (green dot).
The plate voltage is now 120V for a 2V grid voltage difference, thus the Mu, µ or amplification coefficient is (155-120) / 2 = 17.5
The Rp is figured by the slope of a straight line joining the red and the grid dot.
This is the same as dividing the Mu by the slope (in Ampere per Volt), here 17.5/.009 = 1K94.
<edit>
Oopps, misread the plate voltage !
</edit>
Yves.
Attachments
Sy you would like the tube I got yesterday then,it's a 5762 which is a triode made by rca for vhf tv transmission it ha a 4kw anode diss. But the nice this about it is it's super low anode resistance which at 4 kv b+ is just over 1k.
It's not speced for class a but at 4kw anode dissipation and with low b+ I bet it could handle it just fine.
It's a pricy tube new though just shy of 2 grand.
Nick
It's not speced for class a but at 4kw anode dissipation and with low b+ I bet it could handle it just fine.
It's a pricy tube new though just shy of 2 grand.
Nick
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