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  • #16
    Originally posted by Dave H View Post
    The voltage divider equation only works if there's no current taken from the junction of the two resistors and there's always current taken from each node of the power supply chain so you have to use Ohms law. For example the junction of the 10k and 22k has to supply the 6V6 screen current. Can you calculate the screen current?
    This would essentially equate to a parallel Load across R2 in a voltage divider, right?
    If I have a 50% chance of guessing the right answer, I guess wrong 80% of the time.

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    • #17
      "
      The 10k? Same deal. The same 2ma from the 12AX7 goes through it too, plus the power tube screen grid current - about 2ma. That makes 4ma. What resistor will drop 40v at 4ma current? Ohm's Law says 10k"


      This is where I get dropped. How do we know the current of the screen grid? Is it another rule of thumb?
      It's weird, because it WAS working fine.....

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      • #18
        Look at the schematic voltages.

        The 22k resistor drops 320 to 280vDC. 40v dropped. Ohm's LAw tells us then 40v/22,000 ohms = 2ma. The only thing feeding off that 280v is the two triodes of the 12AX7. So 1ma each for 2ma.

        Now look at the screen node, the 320v. To get there we dropped also 40vDC (360v - 320v = 40v), and that was through the 10k resistor. Ohm's says 40v/10,000 ohms = 4ma.

        Now recall the 12AX7 had 2ma, and those milliamps flow through the 10k along with the screen current. SO of the 4ma through the 10k, 2ma of them were for the preamp tube. And that leaves the remaining 2ma for the screen. Screen is the only other thing drawing current through the 10k.

        SO not a rule of thumb, but calculated from the voltages provided on the schematic.


        But having said that, 2ma is a good average number to use in the absence of better numbers. So in this case, the power tube cathode current includes that 2ma along with whatever the plate is handling. In larger amps like a Twin Reverb, I usually just assume about 5ma per screen just to have a number. In that amp I could easily measure voltage drop across the screen resistor, but I don't care enough.

        MY way of looking at it is this: IF I calculate some dissipation for my power tubes using the cathode current, I know it includes some screen current. But if I ignore that, the fact that the true plate current is really a couple milliamps lower, it just gives me a little pad, my amp runs a hair cooler. This is like how my wife sets her car clock ten minutes fast to give herself a pad against being late.
        Education is what you're left with after you have forgotten what you have learned.

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        • #19
          I understand that everyone is trying to help, and I'm really not trying to be rude here, but this has become very frustrating because it's almost as if nobody has read the question all the way through.

          I KNOW it's ohms law. I KNOW it's volts divided by amps. What I can't seem to get anyone to see is that in order to do the ohms law calculation, aside from the voltage value, I need a current value. We have the voltage values all day long on this schematic, which was only an example so everyone could see what resistors I was talking about. And I know that on this schematic, we can work backwards to get the current value. But I needed to know the other way around. I'm designing an amp. I know I need a dropping resistor. What value resistor do I use?

          That aside, I have actually been able to piece together the answer to my question which was if I'm starting with a blank sheet of paper, how would I calculate the value of those resistors. I know that Fender, Marshall, and all the others didn't just throw darts and voltage and current values, they started from somewhere. Well, turns out the answer is called a load line calculation. I'm still getting my head around it, but you have to go to the data sheet, look at the graph that shows supply voltage vs plate current based on the grid input, decide what voltage and current values you want to use (that's where I'm still a bit hazy) and draw a line between the two. Then from that you can decide what plate voltage you want to use, which then gives you the difference in the supply voltage vs the plate voltage, and since you already know the current, you can figure the load resistor value. And since we now know the supply voltage we want, we know how much we want to drop it from the last voltage on the B+ rail, and using one thing that has been stated, add up all the current from that and all other downline stages.

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          • #20
            tp, listen, I am glad you are where you are on this, but right above my last post was someone who still didn;t quite KNOW those things. So I want him to be as up to speed as the rest of us. My post was aiming to help Randall get it too.

            Fender took their circuits right out of the RCA book. RCA provided those basic circuits along with the typical resistors to use to make designing their tubes into things easier for engineers. If RCA used load lines, so be it, but you can grab that 100k plate and 1.5k cathode resistor right from the chart and have an excellent starting point. I would not be so sure Fender ever ran a load line for a 12AX7.

            To calculate anything you need two of the variables. You have a target current you want to design around? What power supply do you intend to use? Then the resistor value is simple. And that is why the tubes have families of curves instead of just one curve. I have to consider that the input stage sees up to a volt, but the next stage might see 10-20v or more of signal. That affects the bias point I want for my tube.
            Education is what you're left with after you have forgotten what you have learned.

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            • #21
              Originally posted by tpaairman View Post
              I'd say 4 mA based on the 40 volt drop across the 10K resistor.

              But here's where my hangup is. In this example, we're going with the cathode current to work back from, which makes sense. And as Enzo said, we can use 1mA as a rule of thumb for the 12AX7. However, if I was designing my own amp from scratch, where would I start with a value to work with? Or put another way, when the folks at Fender designed this amp, how would they have come up with those dropping resistor values?

              I apologize if I'm coming off as a pain in the rear on this one. It's just that I really want to learn so I can pick up the data sheet and know how to do it.
              If you were designing the amp from scratch you would know the currents and voltages required by each section of the amp.

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              • #22
                Originally posted by the fatch View Post
                If you were designing the amp from scratch you would know the currents and voltages required by each section of the amp.
                (posted incomplete comment, I don't know how) As I was saying, the current in each dropper resistor is the sum of the current required by the stage it is supplying and the sum of the currents in the previous stages. Knowing the current flowing in each dropper and the voltage to be dropped across it, finding the required resistance is a simple application of Ohm's Law.

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                • #23
                  Tpaairman said " I need a current value. We have the voltage values all day long on this schematic, which was only an example so everyone could see what resistors I was talking about. And I know that on this schematic, we can work backwards to get the current value. But I needed to know the other way around. I'm designing an amp. I know I need a dropping resistor. What value resistor do I use?"

                  I may not get your question, but when I think about designing an amp or stage I think about what tube(s) to use, considering gain and impedance and linearity, and that choice tells me how much current I'll need. Choose plate resistors for the gain and distortion you want, with higher values less distortion and more gain. I then decide what voltage I want to run the tube at, considering voltage swing needed and the linearity of the plate curves, and that sets the B+. Adjusting the cathode resistor size gets the tubes into the bias I want. The filter effect of each RC stage lowers the power supply noise, too, and that's a consideration towards big caps, but big caps sag less, if that's important. In the push/pull part of the amp, the effect of PS ripple is somewhat lessened, but the preamp triodes want it quiet, so I try to find a balance of high enough voltage and little noise. Sorry if this is too trivial.

                  Dan

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                  • #24
                    Originally posted by tpaairman View Post
                    I understand that everyone is trying to help, and I'm really not trying to be rude here, but this has become very frustrating because it's almost as if nobody has read the question all the way through.
                    We HAVE read your question and answered it, including "you should study a little more":
                    Now you are frustrated because you donīt "get it" .... but you didnīt show due diligence studying.

                    This included actually downloading and reading 12AX7 datasheets, where the typical design examples are shown.

                    Hereīs the GE datasheet "example chart".

                    The typical 100k - 1k5 - 250V combo is in the RCA one but this is what I found and will use:
                    Click image for larger version

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                    I marked one choice, the one closest to typical Fender use, and which you might have picked for your design.

                    I chose 180V +V .
                    Why?
                    Because I liked it, all 3 values work but in general itīs safer to choose a "midrange" option and not go to extremes, unless you have a very definite reason to do otherwise.

                    This example uses 100k plate resistor, a 1100 ohms cathode one, and is happy driving a 240k load.

                    It uses an easy to achieve +180V rail and can put out some 20V RMS "clean" .
                    Actually 5% distortion but by Tube standards that passes as clean.

                    I KNOW it's ohms law. I KNOW it's volts divided by amps. What I can't seem to get anyone to see is that in order to do the ohms law calculation, aside from the voltage value, I need a current value.
                    *ITīS-SHOWN-IN-THE-DATASHEETS*

                    ...... where? .......... I donīt see it!!!

                    You need to apply Ohmīs Law:

                    1) rail voltage is +180V

                    2) usually plate voltage is about half that, so signal has the same "room" going upwards than going downwards, for symmetrical clipping and maximum Vout.
                    No need to spell that out, itīs a common sense choice.

                    So plate voltage, even if they donīt *specifically* show it, must be around +90V .

                    So you are dropping some 90V across the plate resistor.

                    Applying Ohmīs Law: 90V/100000 ohms=0.9mA ...... which is very close to 1mA estimated by Enzo, simply based on his ample experience.
                    But now you know where it comes from.

                    As of dropping resistors: suppose you have a, say, 300V supply, and want to tame that down to 180V .
                    R=V/I so (300-180)V/0.9mA=133k .

                    This is just an example, in a full design you choose every single gain stage from the chart as you wish, calculate each triode current needs, add them up as needed in different nodes (you donīt feed them all straight from main supply but in general make a string of RC filters to reduce hum and possible instability) and calculate each section dropping resistor based on available voltage - desired voltage and current needs.

                    In fact I suggest as "homework" you choose any Amplifier schematic you like (Iīd start with a Champ just to make it quicker, then it applies to any other) , download the RCA datasheet which is what Leo actually used, calculate all currents at different spots, repeat Fender Voltage choices and calculate the dropping resistors, which was your original doubt.

                    I would NOT be surprised if what you calculate roughly matches what Leo used


                    We have the voltage values all day long on this schematic, which was only an example so everyone could see what resistors I was talking about. And I know that on this schematic, we can work backwards to get the current value. But I needed to know the other way around. I'm designing an amp. I know I need a dropping resistor. What value resistor do I use?

                    That aside, I have actually been able to piece together the answer to my question which was if I'm starting with a blank sheet of paper, how would I calculate the value of those resistors. I know that Fender, Marshall, and all the others didn't just throw darts and voltage and current values, they started from somewhere. Well, turns out the answer is called a load line calculation. I'm still getting my head around it, but you have to go to the data sheet, look at the graph that shows supply voltage vs plate current based on the grid input, decide what voltage and current values you want to use (that's where I'm still a bit hazy) and draw a line between the two. Then from that you can decide what plate voltage you want to use, which then gives you the difference in the supply voltage vs the plate voltage, and since you already know the current, you can figure the load resistor value. And since we now know the supply voltage we want, we know how much we want to drop it from the last voltage on the B+ rail, and using one thing that has been stated, add up all the current from that and all other downline stages.
                    See above explanation

                    And yes, the next higher stage is to dismiss the example chart and cook your own, straight from the graphic curves ... but as-is the "simpler" method has served well to tons of Designers.
                    Last edited by J M Fahey; 05-20-2018, 11:00 AM.
                    Juan Manuel Fahey

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