How to set bias on X100b

Discussion in 'Fisher' started by brad44, Jul 1, 2017.

  1. dcgillespie

    dcgillespie Fisher SA-100 Clone Subscriber

    Ball Ground, GA
    Brad -- Some pointers to help you out:

    1. Adjustments should be made with no signal applied, and the amplifier properly loaded.

    2. If you installed individual 10Ω cathode resistors, then correctly installed, one end of each resistor is connected to pin 3, and is the only thing connected to pin 3 at each tube. The other end of these four resistors all connected together, and then connect to the four items originally connected to the daisy chained pin 3 terminals: i.e., R76, C4, R71, and pin 5 of V4. With the individual 10Ω resistors installed, the point where these four resistors and the four component leads just mentioned all connect together is called the CATHODE BUSS.

    3. You can:

    A. Bias the amplifier as you were, by setting the voltage between the Cathode Buss and ground to 40 vdc. When setting the bias this way, you can then measure the voltage between the Cathode Buss and pin 3 of each output tube (effectively placing your meter across each 10Ω resistor, one at a time). THIS reading will let you know how much current each individual tube is passing, based on which 10Ω resistor you are measuring. Whatever the reading, divide it by 10 to convert the reading into AMPS. Therefore, if the reading for a particular tube is .50 volt, then that tube is passing .05 AMP. Since a mA is a thousandth of an Amp, you can multiply .05 by 1000 to show that the tube is passing a cathode current of 50 mA. While this exercise shows you how a reading of .50 vdc indicates a current flow of 50 mA, it might be easier for you to just take any reading across the 10Ω resistors and multiply it by 100 as a one step process of converting the voltage reading into a mA current reading.

    By comparing the readings across each 10Ω resistor, it will let you know how closely your tubes are matched under quiescent (no signal) conditions. A decent match would have the current flow of all four tubes within 10% of each other. And excellent match will have them within 5% of each other. In this amplifier, it is very important that the tubes are closely matched under quiescent conditions.

    B. If you want to target a specific current flow for the output tubes -- say 35 mA each -- then working backwards by dividing 35 mA by 100, that means that for each tube to draw 35 mA, it will produce a voltage reading of .35 vdc across each 10Ω resistor (that is, as measured between the Cathode Buss, and pin 3 of each output tube). The four readings will not be identical, so adjust the bias so that the average reading of all four tubes is .35 vdc. Then, whatever the voltage is between the Cathode Buss and Ground is whatever the voltage will be. It is important only because if it becomes too low (< 40 vdc), the 12AX7 preamp tubes will not receive enough power to properly light their heaters, and if it is too high (> 42 vdc), then the output tubes will be overly stressed under quiescent conditions. BTW, using this scenario in a stock X-100B, you will find that a current draw of 35 mA per tube will not produce nearly enough current flow to properly light the 12AX7 heaters in your amplifier.

    Using method A lets you target a proper 12AX7 heater voltage, and then see what each tube is drawing to produce that heater voltage. Using method B lets you target a particular average output tube current draw, and then see what 12AX7 heater voltage that current draw produces. You can graph both approaches to learn how one affects the other. In any event, with the individual 10Ω resistors installed, you will be able to see how well the individual current draw of one tube matches the other, regardless of which way you bias the amplifier.

    4. As for the output tube dissipation (stated in watts), for any one tube, measure the voltage between pin 3 and 9, and multiply it by the cathode current draw of that tube in AMPS. Therefore, using the schematic values, the voltage between pins 3 and 9 of each tube is 396 vdc. A voltage of 42 vdc between the cathode buss, translates into a total cathode current draw of 210 mA, or .21A, meaning that if each tube is perfectly matched to one another, they are individually passing .052A. From the above exercise, you know that this will produce a reading across each 10Ω resistor of .52 vdc. Now, multiply the voltage across the plate and cathode element of each tube (396 vdc) by the current flow through each tube, so that 396 vdc X .052 A = 20.59 Watts as designed. When the screen grid current is accounted for, it means that the tubes in this design at operating with a plate dissipation of 19 watts, which is the absolute maximum allowed for with these tubes. What this shows is that the design operates the output tubes very hard, meaning that they must be tightly matched to prevent any of them from becoming a current hog which would spell certain doom for the it, and that a fan would be very highly recommended for this model.

    No doubt you will need to re-read this a number of times to gain a clearer understanding, but understand that the two most important take away points are: (1) Do not let the Cathode Buss to ground voltage go below 40 vdc or above 42 vdc. A 40 volt target would be optimum. (2) Use the readings from the Cathode Buss to pin 3 of each tube to ensure that all four horses are pulling an even amount of current. If the match is not close, it will be very hard on the tubes carrying the lions share of the load.

    I hope this helps!

    Last edited: Sep 3, 2017
  2. brad44

    brad44 Active Member

    thank you very much for this reply, its spectacular. I am going to print out a copy of it and keep it with the manual.

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