• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

Tubelab SPP first timer build

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#561
Just to write some boring information for the readers.
The famous R1, which was replaced in my case using a choke.
As we know, i had a high B+, so a resisitor/s was fitted from the centre tap of the secondary to ground.

I was 'playing' with values to get just over 320v.
I am happy to say that my resistors are now 150 ohm + 330 ohm = 103 ohm, giving me a B+ of 322v.

As for the voltage across R1 (103 ohms) we have 22.16 v.a.c.
Yes Francois, you were so close with your recommended 100 ohms.
So i think it's 22.16v X 22.16v ÷ 103R = 4.78 watts
Out of curiosity, how do i work out the wattage per resistor?


Based on the B+ 322v i measured these voltages: -
V102 Cathode = 11.07v - therefore = 0.034 A (11.07v÷ 322v = A)
V101 Cathode = 10.86v - " " = 0.034 A
V202 Cathode = 10.83v - " " = 0.034 A
V201 Cathode = 11.17v - " " = 0.035 A

Also measured (for power dissipation): -
R117 / V102 pin 9 = 319.5v (319.5v - 11.07v = 308.43) X 0.034 = 10.48 W
R113 / V101 pin 9 = 319.7v (319.7v - 10.86v = 308.84) X 0.034 = 10.50 W
R217 / V202 pin 9 = 319.5v (319.5v - 10.83v = 308.67) X 0.034 = 10.50 W
R213 / V201 pin 9 = 319.3v (319.3V - 11.17v = 308.13) X 0.035 = 10.79 W
 
#562
calpe said:
Yes Francois, you were so close with your recommended 100 ohms.
So i think it's 22.16v X 22.16v ÷ 103R = 4.78 watts
Out of curiosity, how do i work out the wattage per resistor?
Click to expand...

😎 Lucky guess.🙂

Since both resistors have the same 22.16 V across them you can simply use the same formula with the individual resistor values as divisor. I.e. 22.16 x 22.16 ÷ 150R = 3.274 Watts.

When you are ready please l us know how it sounds, and perhaps a picture or two of what your final product looks like.
 
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#563
Hello calpe,
calpe said:
Based on the B+ 322v i measured these voltages: -
V102 Cathode = 11.07v - therefore = 0.034 A (11.07v÷ 322v = A)
V101 Cathode = 10.86v - " " = 0.034 A
V202 Cathode = 10.83v - " " = 0.034 A
V201 Cathode = 11.17v - " " = 0.035 A
Click to expand...
Unfortunately, an incorrect value was used in the calculation. For the total dissipation, the value of the resistance of R112/R116 (300 ohms) must be used.

... for the current flow (Ohm's law)
V102 => 11.07 V / 300 Ω = 0.0369 A
V101 => 10.86 V / 300 Ω = 0.0362 A
V202 => 10.83 V / 300 Ω = 0.0361 A
V201 => 11.17 V / 300 Ω = 0.0372 A

... and for power dissipation
V102 => (322 - 11.07) * 0.0369 = 11.47 W
V101 => (322 - 10.86) * 0.0362 = 11.26 W
V202 => (322 - 10.83) * 0.0361 = 11.23 W
V201 => (322 - 11.17) * 0.0372 = 11.56 W

calpe said:
Also measured (for power dissipation): -
R117 / V102 pin 9 = 319.5v (319.5v - 11.07v = 308.43) X 0.034 = 10.48 W
R113 / V101 pin 9 = 319.7v (319.7v - 10.86v = 308.84) X 0.034 = 10.50 W
R217 / V202 pin 9 = 319.5v (319.5v - 10.83v = 308.67) X 0.034 = 10.50 W
R213 / V201 pin 9 = 319.3v (319.3V - 11.17v = 308.13) X 0.035 = 10.79 W
Click to expand...
For the calculation of the dissipation of g2 (max. 2.2W) the value at pin 9 is necessary, but also the current flow at this point.
The current flow can be determined by measuring the voltage across the relevant resistor R113/R117 (100 ohms).
In each case the measuring probe at the ends of the resistor.

Best regards 😉
 
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#564
Francois G said:
Since both resistors have the same 22.16 V across them you can simply use the same formula with the individual resistor values as divisor. I.e. 22.16 x 22.16 ÷ 150R = 3.274 Watts.
Click to expand...
Francois G said:
When you are ready please l us know how it sounds, and perhaps a picture or two of what your final product looks like.
Click to expand...
150 & 330 ohm 10W Resistors, I had to order via my nearest component shop in Algeciras (Spain), 20 min drive. Should be here by Saturday.
Once fitted etc., I'll happily post finished pictures
 
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#565
stonegreen said:
Hello calpe,
For the calculation of the dissipation of g2 (max. 2.2W) the value at pin 9 is necessary, but also the current flow at this point.
The current flow can be determined by measuring the voltage across the relevant resistor R113/R117 (100 ohms).
In each case the measuring probe at the ends of the resistor.

Best regards 😉
Click to expand...
Trust me to balls up the calculations, thanks for the corrections.
I will check the other voltages across R113, 117, 213 and R217 as soon as i can.
 
#567
From post #566 would i be correct in saying: -
P = V x I (voltage dropped across screen X Screen current)

V102
0.49v (across R117) ÷ 100 (R117) = 0.0049A (4.9mA)
319 (+V on pin 9 V102) X 0.0049 = 1.56 watts

V101
0.48v ÷ 100R = 0.0048A
319v X 0.0048A = 1.53 watts

V202
0.49v ÷ 100R = 0.0049A
319v X 0.0049A = 1.56 watts

V201
0.48v ÷ 100R = 0.0048A
319v X 0.0048A = 1.53 watts

How does this appear?
 
#569
jcalvarez said:
Close, but you need to subtract the cathode voltage from the screen grid voltage before multiplying by the current, for example:
V102
0.49v (across R117) ÷ 100 (R117) = 0.0049A (4.9mA)
319-11.07(+V on pin 9 V102 - V102 Cathode voltage) X 0.0049 ~ 1.51 watts
Click to expand...
Thank you kindly for the corrections. Trust me!

What does the 1.51 watts indicate?
I've looked at the technical sheet of the Sovtek EL84, didn't see anything.
 
Last edited:
#570
calpe said:
I've looked at the technical sheet of the Sovtek EL84, didn't see anything.
Click to expand...

Datasheet here (P = power, a = anode, g2 = grid2)

1687291414386.png
 
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