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  • More voltage or more current/better or worse

    I have been building some 5E3s lately using salvaged non guitar amps. The voltage from the PT is a little high. My question is: without lowering the overall voltage, what is better, to raise the cathode resistor for less current, or to lower it for less voltage?

    I was interested to see the interaction between these to elements and to compare the sound of each. So, I used a a rotary switch with several values that could be switched in parallel with a main larger resistor so i could see what each value did without having no resistor for the fraction of a second while switching. Because of the parallel thing, the values aren't ideal but you get the gist. Watts are calculated via the weber bias calculator.

    Here's what I got:

    Res.
    Ohms__mA___Volts__Watts

    266----40.2----383----15
    354----33.8----390----13
    498----27------400----11
    572----24.6----403----10

    Before the test, I have been biasing around 13.5 watts dissipation. To get down to 12w puts almost 400 volts on the plates. 13.5 brings the mA about to about 35. which is worse? And what suffers? What makes for shortened tube life? What makes for shortened transformer life? Any expert opinions available? More volts less mils, or the reverse?

  • #2
    How are you determining the cathode current? Are you measuring the voltage drop across the cathode resistor and using it's value to calculate I? If I used my amp's numbers and Weber's calculator, it says I'm running 36 watts, which should redplate my tubes. But all is well, or so it seems. I've got my numbers posted in my thread above, about my 5E3 getting first voltage...We do have to remember to divide the total by 2 which would be 18 watts for me, still 50% over the max output of 12 Watts, so I know that I'm making an error somewhere...sorry I can't be of more help. I'll get back after I get some solid numbers in my head. I've had the flu for 3 or 4 days and I'm not thinking squarely...
    Thermionic vacuum devices rule.

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    • #3
      previously I've done the calculation for dissipation using the method where one measures the voltage drop across the resistor, divides that number by the cathode resistor ohm value. Then takes the plate voltage, subtracts the voltage drop and multiplies by the number determined in the first calculation - then divide by two for the two tubes.

      Recently I made a "bias probe" type device using old tube sockets as described at the hoffman site. It has allowed me to measure the current directly.

      Then I plugged the numbers into the weber calculator.

      Comment


      • #4
        Originally posted by atmars View Post
        I have been building some 5E3s lately using salvaged non guitar amps. The voltage from the PT is a little high. My question is: without lowering the overall voltage, what is better, to raise the cathode resistor for less current, or to lower it for less voltage?

        I was interested to see the interaction between these to elements and to compare the sound of each. So, I used a a rotary switch with several values that could be switched in parallel with a main larger resistor so i could see what each value did without having no resistor for the fraction of a second while switching. Because of the parallel thing, the values aren't ideal but you get the gist. Watts are calculated via the weber bias calculator.

        Here's what I got:

        Res.
        Ohms__mA___Volts__Watts

        266----40.2----383----15
        354----33.8----390----13
        498----27------400----11
        572----24.6----403----10

        Before the test, I have been biasing around 13.5 watts dissipation. To get down to 12w puts almost 400 volts on the plates. 13.5 brings the mA about to about 35. which is worse? And what suffers? What makes for shortened tube life? What makes for shortened transformer life? Any expert opinions available? More volts less mils, or the reverse?
        You don't need a bias calculator, with a cathode biased amp you can just divide the cathode voltage by the resistance to get the cathode current. Then subtract the cathode voltage from the plate voltage and multiply the result by the cathode current for total dissipation.

        With a cathode biased amp you can safely run it higher than the typical 70% of max plate dissipation, because of the bias shift that occurs as current increases. The tubes will run fine at 14 watts quiescent dissipation.

        Comment


        • #5
          In another thread a friend pointed out that the original spec for the 5E3 was 325-0-325 on the HV secondaries. I made a big mistake in trusting Weber Speakers and bought a $60 Heyboer that their website claims has correct voltages for Deluxes, Deluxe Reverbs, and something else. I didn't do my homework, bought it, installed it, then measured it after my amp ran crazy hot. Lo and behold! Weber specified a 370-0-370 secondary. No kidding. So, a dig through my diodes drawer turned up a 50 volt zener. I put it in the HV center tap and that dropped the B+ to around 375 or so, and a 100 ohm 10 watt wirewound in series brought me to 349 volts at the OT plates. Yay!!! I just did the calculations (Thanks hasserl, now I remember the formula) and my amp is at 12.05 watts and sounds so sweet!
          A note of interest may be that before I put in the zener and resistor, I put the amp on my variac and played it as I ran the voltage up and down. There is definitely a sweet spot at Vp of 350. More, above 360 or so, sounds strident and pushed, and as you go down from there you lose volume and punch.
          Thermionic vacuum devices rule.

          Comment


          • #6
            You don't need a bias calculator, with a cathode biased amp you can just divide the cathode voltage by the resistance to get the cathode current. Then subtract the cathode voltage from the plate voltage and multiply the result by the cathode current for total dissipation.
            Thanks - this is what I used before I made the bias probe to measure current directly. I wanted to see what kind of numbers I would get with direct measurement.

            14 watts has been what Iv'e been shooting for, in general.

            Before the test, I have been biasing around 13.5 watts dissipation. To get down to 12w puts almost 400 volts on the plates. 13.5 brings the mA about to about 35. which is worse? And what suffers? What makes for shortened tube life? What makes for shortened transformer life? Any expert opinions available? More volts less mils, or the reverse?
            I'm guessing getting the right dissipation is paramount so it doesn't matter that much if voltage or current is higher or lower as long as it's not off the charts? Is there any danger in running big currents or big voltage other than reduced tube life?

            Comment


            • #7
              the original spec for the 5E3 was 325-0-325 on the HV secondaries.
              It's 325V at full load. Most manufacturers quote unloaded voltages, with most PT's spec'ed at 370 to 380V, their windings show quite a high Z resulting in 325 to 330V at full load.

              Comment


              • #8
                ANd in response to the original question, it isn;t about voltage OR current, it is about both. You want the tube dissipation to be right.

                Whenever you lower the current, the B+ voltage will rise, that is normal. It is the combination you should follow, not one or the other.
                Education is what you're left with after you have forgotten what you have learned.

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