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Determining necessary bias voltage from tube data sheets

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  • Determining necessary bias voltage from tube data sheets

    I have no idea how to interpret tube data sheet graphs, or draw lines to determine current and bias points and all that jazz.

    I'm building a submini express using 5899 output tubes. I want to know where I should bias them at initial startup, so I can determine the circuitry for my bias tap. I'm using back to back 12V transformers and I've got a HV+ of 140V with no load. My existing bias tap circuit gives me a max negative voltage of about -12V. I don't think this is low enough so I'll put together a voltage multiplier to get me in the right range.

    But I need to know that range first. How do I figure out my bias voltage from the datasheet info?

  • #2
    ...it's a miniature semi-remote cutoff pentode, but the first question should be "...how are you planning on using it?" and, then these:

    1) which: triode or pentode operation?

    2) which: small signal or 'power' amplification?

    3) which: single-ended (SE) or push-pull (PP)?

    4) which: class-A or class-AB1?

    5) which: cathode bias or fixed bias?

    6) what plate- (Vp) and screen-voltages (Vs) to be used?

    ...from the data sheet, it looks like you'll need a "fixed" bias of about -1.0 to -1.5Vdc with Vp = Vs = 100Vdc (EbIb chart on page 6) for pentode operation; or, about -2 to -2.5Vdc if triode operation at Vp = 150Vdc (EbIb chart on page 9).
    Last edited by Old Tele man; 09-27-2009, 05:29 AM. Reason: spelling correction
    ...and the Devil said: "...yes, but it's a DRY heat!"

    Comment


    • #3
      Hi Teleman , this is interesting.

      Maybe you could say a bit more about these datasheets ?

      e.g. Transconductance.

      Does 4200 umhs mean 4.2 mA/V ? Is it the same ?

      And the platecurrent number in mA means how "strong" the tube is when it's new . I mean how much current it draws then ?

      For people like me who haven't had a technical education this is quite difficult sometimes.


      Thanks , Alf

      Comment


      • #4
        Howdy, Alf!

        Yes, the transconductance value of 4500 µmhos is the same as 0.0045 A/V...however, in europe, it's defined as Siemens (S), so the value could also be called 4500 µS or 0.0045 S. Notice that the 'old' USA units of mhos is simply ohms backwards, named so because transconductance (A/V) is simply the inverse unit of resistance (V/A).

        The plate current value shown as Ip = 7.2 mA is just the value at the stated operating condition: Vp = Vs = 100Vdc, Rk = 120 ohms, and Is = 2.0 mA, it is NOT a value indicating tube strength...that's the transconductance value = figure of merit or goodness.

        With that tube, the two design-controlling (limiting) conditions to be mindful of are: (1) the LOW plate and screen dissipation wattages of 1.1W and 0.55W, respectively; and, (2) the "absolute" maximim Cathode Current value of 16.5 mAdc. The best way to do these are to: (1) draw-in a 1.1W dissipation curve over the EbIb curves (triode and/or pentode) and (2) also draw a straight line from the 16.5 mA point on Y-axis (Ib) horizontally across the EbIb curves...together, they show you the tubes operating boundaries because we need to keep both idle and maximum power points at/below BOTH lines for Class-A operation.

        With push-pull Class-AB operation, however, you can approximately double the 16.5 mA value because with PP operation each tube conducts on alternating halfs of the input signal and rests during the other half, so the "average" current remains at 16.5 mA, like this:

        Ik(avg) = (2*16.5 mA)/2 = 16.5 mA

        ...where the first "2" is because you've doubled the maximum value, while the second "2" value is because its occurs on alternate cycles...so, they effectively cancel.
        Last edited by Old Tele man; 09-27-2009, 04:00 PM.
        ...and the Devil said: "...yes, but it's a DRY heat!"

        Comment


        • #5
          Thanks man for the reply , hope I can "bother you" some more whenever I don't understand these things.

          Regards, Alf

          Comment


          • #6
            ...to add a 'final' answer to the orginal question--Yes, there *IS* an way to estimate idle bias voltage, but it's only a rough approximation (±20%)!

            ...using the tubes' published transconductance (gm), triode & pentode amplification factor (µ1 and µ2) values, and the intended idle plate (Vpq) & screen (Vsq) voltages, it's possible to backsolve the Child-Langmuir 3/2's Law equation to determine an approximate control grid idle bias voltage (Vg.dc):

            Vg.dc ~ 0.9*[(3/2)*(Ipq/gm)] - [Vsq/µ1 + Vpq/µ2]

            ...so, assuming these values and Class-A pentode operation:

            Ipq = 7.3 mA (so, Ppq = 1.1W)
            Isq = 1.9 mA (so, Psq = 0.3W)
            Vp = 150 Vdc
            Vs = 150 Vdc
            gm = 0.0045 A/V
            µ1 = 30 (guessed value, not given value)
            µ2 = 1170 (260K*0.0045A/V)

            Vg.dc ~ 0.9*[(3/2)*(0.0073A / 0.0045A/V)] - [150Vsq/30 + 150Vpq/1170]
            Vg.dc ~ 0.9*[0.0219A / 0.0090A/V] - [5V + 0.13V]
            Vg.dc ~ (2.19V - 5.13V) = -2.94 Vg.dc

            ...or, you could use the charts on page 10 and page 11, which also show Vg ~ 2.9-3.0Vdc.
            Last edited by Old Tele man; 09-27-2009, 07:14 PM.
            ...and the Devil said: "...yes, but it's a DRY heat!"

            Comment


            • #7
              Old man tele, thanks so much for the explanations! I never would have come across those equations on my own I think.

              To further elaborate, this amp will be class AB push-pull. The 5899 power section will be fixed bias, pentode operation, with plate value of around 130? I'm guessing because I don't know how far my unloaded 140V supply will pull down when it gets loaded. Screen voltage should be about 10V lower I think.

              To bring up another question, how do I determine plate to plate load resistance? Other tube datasheets I've looked at list this value, but this datasheet does not. Is there a way to figure this out?

              Comment


              • #8
                ...the "rule-of-thumb" for push-pull pentodes is plate-to-plate impedance (Zpp) is approximately 1/6th to 1/8th of the tubes dynamic plate resistance (rp), so that'd be something between:

                Zpp ~ 260K/6 = 43.3K ohms
                Zpp ~ 260K/8 = 32.5K ohms

                ...that's way out of the typical OT region, so you might have to look for a center-tapped differential line-balancing transformer or something similar?!?

                ...also, with PP operation (the Vg.dc example above was for SE Class-A), the idle plate current will be lower, but the idle plate voltage MUST be kept at/below 165Vdc for continuous operation...hence, the "recommended" use of 150Vdc as a maximum intended voltage in all the specs and curves.

                ...so, if we arbitrarily assume Ip.max = 16.5 mA (for now) and Vp.q = 150, that'd be an effective plate load resistance (Ra or RL) of 9090 ohms so the Zpp value (which is 4*RL) would be about 36K ohms, which is roughly in-between the above 32.5K-43.3K estimated values. Since Vs should be well filtered (but Vp isn't so critical due to OT cancellation of 60/120Hz ripple), it's common that Vs WILL be lower than Vp due to "smoothing" (either choke or RC filtering). Which do you plan to use?

                ...now, to estimate idle plate current, lets assume 70% plate dissipation: Ppq = (0.7*1.1W) = 0.77W or about 0.8W, so at Vp.q = 150 Vdc, that'd be:

                Ip.q = (Pp.q / Vp.q) = (0.77W/150V) = 0.0051A, or about 5 mA ...and Is.q would be roughly 1.3 mA (~¼th of Ip.q)

                ...with "fixed" bias, Vg.dc = -3.8 Vdc, or with "cathode" bias Rk ~ 302 ohms, shared by both tubes.

                ...screen grid voltage CAN be the same as plate voltage (for maximum tube efficiency and power), but yes, Vs certainly can be lower than Vp, but doesn't HAVE to be. The "danger" is when Vs is HIGHER than Vp, which can lead to current runaway (and screen 'melt down') during periods of high power output.
                Last edited by Old Tele man; 09-27-2009, 07:55 PM.
                ...and the Devil said: "...yes, but it's a DRY heat!"

                Comment


                • #9
                  If you look at the plate curves for Ec2 - 150V, a good load line that intersects the "knee" of the curves would put you around (130V/24mA) = 5.4K, so for push-pull class AB1 you'd want 4x that or around 22K p-p.

                  Comment


                  • #10
                    Originally posted by raiken View Post
                    If you look at the plate curves for Ec2 - 150V, a good load line that intersects the "knee" of the curves would put you around (130V/24mA) = 5.4K, so for push-pull class AB1 you'd want 4x that or around 22K p-p.
                    Randall brings up an excellent point, ie: running the effective RL load line through the "knee" is where optimum output occurs. The rule-of-thumb values aim more for voltage gain over power gain.

                    ...but (correct me if I'm wrong) I assumed Mr. Allen isn't familiar with "reading" the EbIb curves yet, so I just stuck with the basic equations.
                    Last edited by Old Tele man; 09-27-2009, 07:16 PM.
                    ...and the Devil said: "...yes, but it's a DRY heat!"

                    Comment


                    • #11
                      Originally posted by Michael Allen View Post
                      Old man tele, thanks so much for the explanations! I never would have come across those equations on my own I think.
                      Yup you can say that again. I find these posts most useful (if a little technical)
                      Building a better world (one tube amp at a time)

                      "I have never had to invoke a formula to fight oscillation in a guitar amp."- Enzo

                      Comment


                      • #12
                        Originally posted by Old Tele man View Post
                        ...but (correct me if I'm wrong) I assumed Mr. Allen isn't familiar with "reading" the EbIb curves yet, so I just stuck with the basic equations.
                        I wasn't intending to slight your equations! Just offering an alternative method of finding a good load impedance that maximizes output power - he's only going to get a watt or so...

                        Comment


                        • #13
                          Originally posted by raiken View Post
                          I wasn't intending to slight your equations! Just offering an alternative method of finding a good load impedance that maximizes output power - he's only going to get a watt or so...
                          ...I didn't feel you were slighting me at all, because, actually, you "jumped ahead" to the answer that I was (eventually) going to present--using the curve to find the numbers...using the "..thru the knee..." method.

                          ...I was the HARE and you were the RABBIT (ha,ha)!
                          ...and the Devil said: "...yes, but it's a DRY heat!"

                          Comment


                          • #14
                            Starting to get this I think. To draw the load line, you start with one point at 0 current, 150V. Where is the second point on the Y axis? At 16.5mA or 33mA? And then you just adjust that point on the horizontal axis until the line intersects the knee of where we want idle current to be?

                            So as Mr. Aiken suggests, adjust the horizontal voltage to 130V and draw a line to 33mA? Then at about 12mA would be the intersection of the -1.5V grid voltage?

                            Comment


                            • #15
                              Originally posted by Michael Allen View Post
                              Starting to get this I think. To draw the load line, you start with one point at 0 current, 150V. Where is the second point on the Y axis? At 16.5mA or 33mA? And then you just adjust that point on the horizontal axis until the line intersects the knee of where we want idle current to be?

                              So as Mr. Aiken suggests, adjust the horizontal voltage to 130V and draw a line to 33mA? Then at about 12mA would be the intersection of the -1.5V grid voltage?
                              I just did a quick visual approximation using your plate voltage and looked where it intersected the knee to get a rough estimate of the required plate resistance (which is multiplied by 4 for class AB1 push-pull operation). If you want a more detailed explanation check out the pentode section of these papers (although they all use different methods, so it may confuse the issue more - you may want to find a good textbook explanation

                              http://www.ax84.com/media/ax84_m225.pdf

                              http://www.freewebs.com/valvewizard1/pp.html

                              http://greygum.net/sbench/sbench102/pent.html

                              ...although they all use different methods, so it may confuse the issue more - you may want to find a good textbook explanation here:

                              http://www.pmillett.com/tecnical_books_online.htm


                              RA
                              Last edited by raiken; 09-27-2009, 09:06 PM.

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