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Scratch Build, -from the beginning...

Discussion in 'Shock Brother's DIY Amps' started by Jewellworks, Sep 24, 2020.

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  1. Sep 24, 2020 #1
    Jewellworks

    Jewellworks Tele-Meister

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    i started a private conversation with andrewRneumann regarding some help with the https://www.vtadiy.com/loadline-calculators/loadline-calculator/
    he was a tremendous help in walking me through the load line calculations for a "Vintage" build im planning on, using a pentode as the V1.
    he suggested we take this conversation to the public forum so everyone can chime in, and/or gain some benefit from our discussions.
    so here it is, copied, slightly edited to keep the jibber jabber to a minimum.

    at the time of this post, we havent gotten too deep. but its a lot to post and read all in one shot.

    ==========================

    Yesterday at 12:10 PM
    [​IMG]
    Jewellworks

    andrew, youve been an enormous help with the Pentode calculations, and im hoping for some more guidance, since your familiar with this calculator.
    im looking at V2 now (6J5 =triode), but going back and forth from RobRobs page on how to draw load lines, and this calculator... which LOOKS simpler, im still lost.
    RobRob says you get your operating point by combining the value of both the Plate load resistor, and the Cathode resistor. that seems simple enough, but when i use the numbers i had on my design, (copied from something Printer2 posted for a 6J5 in the V2 spot) my load line is VERY flat. which -according to RobRob, will get more gain, and less headroom.
    the 6J5 calculator defaults to a Load Resistance of 12000. which may or may not mean anything, but it shows a much more vertical line, which indicated less gain, but more clean headroom. but if the 12000 DOES mean "something", then how do i know how much of that is the Cathode resistor, and how much is the Plate Load resistor? the "load" on the calculator is both, combined.
    and then theres the actual operating point. im not entirely sure what im looking at, or where to set it.
    in robrobs example on his website, the 0v grid line is the saturation point, and sets his operating point halfway between 0V grid curve, and the 250v at the bottom of the chart. in my trials (and errors) using 300v as my B+, i set my operating point at 200v, which is about halfway between 0v grid curve and cutoff at 300v, using 12000 as my Load Resistance. it gives me a reading of -5.92 Grid Voltage.
    what does that even mean? we were trying to keep it under -1 with the Pentode, assumingly because were using grid leak biasing. but what does it mean for a triode with cathode biasing? the grid voltage in my tests above is at -5.92. thats where the operating point is set. i guess what i dont understand is... is that good? or bad? too high? to low? or just a "thing".
    the whole point of this is to know what resistors to use, where, and ... well.. im lost. im sure there are calculations i need to make w Ohms Law, but i dont know what they are. i also dont know if my load lines are good, bad or otherwise.
    im not wanting you to do it for me. i just dont understand how all this works and need someone to hold my hand and explain it. you were awesome with the Pentode. i hate to impose on you again for this, -but i guess i am...

    Yesterday at 2:10 PM
    [​IMG]
    andrewRneumann
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    Hi @Jewellworks,

    I'm going to answer your question with another question! Sorry...

    Have you designed the power amp yet? There is definitely an "order" of design that makes more sense than doing it from input to speaker. One thing depends on another and if you design parts in the wrong order, you will end up redesigning previous stages. So what I suggest is actually to start with you speaker and work backwards to the input.

    What speaker do you plan to use?
    What output transformer?
    What voltage will the power tube run at?
    What will the bias point of the power tube be?

    Once you answer these questions, you can then determine a very important piece of information: What signal can the power amp accept before it reaches overdrive?

    This is important because now you have a target. You can say, "I want my preamp to take my guitar signal, amplify it, and end up with something that will drive the power amp to full power." What I'm saying is, how can you design the preamp if the power amp isn't designed? You don't know what your target is. So I suggest taking a step back from the preamp, and learning everything you can about single-ended power amp design. Design your 6L6 and figure out what kind of signal it will take to take it to full drive. Then we will come back and look at you 6J5 and even redesign the 6SJ7 if we have to.

    Guitar -> 6SJ7 -> Tone Stack -> Volume -> 6J5 -> 6L6
    -17dBV +? dB -? dB -? dB +? dB Target dBV

    Here is what I think you have for a signal chain. We are going to work in dBV instead of volts. A 200mV humbucker strumming all out is -17dBV. That's on the left. Target dBV to amplified by the power tube is on the right. How do we get from -17dBV to the target if we don't know what the target is?

    Once we know the target, we can make estimates for how many dB your tone stack and volume control (it will depend on knob rotation) will cost you. Then we can go in and figure out what we need gain wise from the 6SJ7 and 6J5--these tubes add dB.

    Read everything on Merlin's valvewizard website. Also ampbooks.com is very good. Aiken amps has some amazing information. I've probably used those three sites more than anything.
    This is a learning process, it should be fun, but it takes work and a lot of reading and repetition before it starts making sense.

    Andy Neumann
     
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  2. Sep 24, 2020 #2
    Jewellworks

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    Yesterday at 3:02 PM

    [​IMG]
    Jewellworks

    you have the basic building blocks right for this design. my plan was to basically copy the Power section of a 5F1 Champ, but use a 6L6G instead, and bias as needed to keep the plate dissipation right at the limit, or just a tad over. i wont be touring stadiums with this lil thing. just bedroom plunking, and probably some recording. i have an 8 ohm OT that came from another SE amp with a 6L6GC. i thought id keep the power around 350v, but my PT can do close to 450v! so that could change, as needed... i dont have any idea what the dBv is.
    ive read some of Merlins pages, but its confuses the hell out of me. it might as well be from Mars.

    and i still dont know how to extrapolate the information i need from the vtadiy calculator

    Yesterday at 3:50 PM

    [​IMG]
    andrewRneumann
    Tele-Holic

    Ok, start with your power transformer. Figure out the actual filtered voltage you are going to have available to the 6L6. You say 450V is available, but was that peak AC voltage? If so, then the DC voltage after power supply filtering will be lower. If 450V was RMS voltage, then your DC may be higher than 450V. Either way, we need to know this. My experience with single-ended designs is a choke for the power supply is advisable, or else you will end up with 120Hz hum. Do some reading on Merlin's website about power supply design, smoothing and filtering!

    This is to determine one thing: B+ available to the 6L6 plate and screen. Then if you have the OT specifications, you can draw a load line for the 6L6. Once you have that, we can talk about what the target dBV (output signal level) is for your preamp.

    Steps Recap:
    1. Actual filtered DC voltage available with your PT and rectifier.
    2. Load line for 6L6 with your OT.
    3. Figure out what signal level you need from your preamp.

    This needs to be done before you worry about the nitty gritty details of the preamp. Show me the 6L6 load line!
    You've got the 19W 6L6 or the 30W 6L6?

    Yesterday at 4:27 PM

    [​IMG]
    Jewellworks
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    https://frank.pocnet.net/sheets/127/6/6L6G.pdf
    10.8 watt
    mine is Sylvania branded, but this is easier to read than French...

    power transformer
    VOLTAGES: using a variac, i set the input to the primary at 120VAC

    These PT tests were on the bench. Not in any circuit, no filter caps, nuthin

    2ndary: HT: 862 VAC (!!)
    HT-a to C/T: 432
    HT-b to C/T: 429

    6.3v Filament: 7.53
    F-a to C/T: 3.77
    F-b to C/T: 3.73

    5v Heater: 5.77

    in my 5F2a variant, i put a 22uf/450v filter cap at the 1st node of the rectifier, before the transformer to filter it. At first, i put it directly through the transformer and it hummed like mad. the 22uf cap fixed that. then i dropped the voltage w a 22k/2w and another 22uf filter cap to the screen. i just figured id do the same for this Vintage style amp.
    otherwise... what can be done?
    i dont have any OT specs. i just know that it came from an amp with an 8 ohm tap. i labeled everything before i took it out.



    Yesterday at 9:27 PM

    [​IMG]
    andrewRneumann
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    Please answer the following:

    I just want to make sure you used a voltmeter that measures RMS voltage. If you measured the primary voltage at 120v then, it is most likely RMS. Can you confirm this?

    Also, what did this PT come from? Any idea of how much current it can handle? Does it have any markings or anything on it to help identify?

    What rectifier do you plan to use? Valve or solid state?

    If you can afford to buy a choke, I would recommend planning one in your design, but we can probably make do without. We will get into sizing the resistors and caps later. Sizing the resistors is one of the last things to do.



    Use your variac to determine the winding ratio of your OT. Apply 1v (measure it will your voltmeter--don't trust the variac label) to the secondary winding and then measure the volts across the primary winding. We will need this to draw a load line for the power tube.

    Andy Neumann
     
  3. Sep 24, 2020 #3
    Jewellworks

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    @andrewRneumann -i have a Fluke 77 digital multi meter. i confirmed the voltage going in and coming out of the PT with the meter, and didnt even look at the reading on the Variac, as i wouldnt trust it. i DO trust my meter.
    as for the OT, again, this was the original OT in a lil 6L6GC/SE amp that had a 70v, 25v and an 8 ohm tap on the 2ndary. im 100% certain it is what it says it is. i wont be able to do any measuring until later tonight, but the resistance across the primary is 270ohms. im positive it is what it says it is. i wont be using the 70v and 25v taps, and ill cap them so they dont short anything.
    the PT came from an old Ampro projector using 3, 6V6GTs, the 6SJ7 and 6J5 in my Vintage build, and im using the 5Z4 rectifier that was in the Ampro. looks like from the 40's, 50's...
    i really dont want to buy a choke for this amp. ive had good results using a 22uf/450v filter cap in my other 5/10w SE build. i want to use all leftover parts i have already on hand. especially the big pieces
     
  4. Sep 24, 2020 #4
    andrewRneumann

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    Great. I wish I had a Fluke! So we can rest assured you measured Vrms which is good. From here on out when talking about AC voltages we will just assume RMS unless specifically noted otherwise.

    I’m unfamiliar with OT’s that have windings with voltage ratings like that. No worries, we will ignore those. What we need to know is the winding ratio of the secondary that you have designated as 8-Ohm. Since OT’s step down voltage, I suggest using the variac set at 1v (measured with the fluke), connected to the secondary winding—the winding that would normal go to the speaker. Then measure vac across the primary winding and report back. It should be something higher than 1, like 30. Once you report back we will use that to determine something called an impedance ratio.

    So if it ran those 6 tubes, then it should run 4 even if one is a 6L6. Is that the thinking? I’m just thinking about the total current draw. Sounds like we are good.

    Ok we will make do without. If your PT is putting out way more voltage than we need, we have more room to filter it with resistors instead of a choke. Good plan.
     
  5. Sep 24, 2020 #5
    Jewellworks

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    i measured 10volts into the 2ndary of my OT, and i got 203v on the primary. did the math and thats a winding ratio of 20.36. squared, thats 414.52 impedance ratio. looking at the 6L6G data sheet, class A load resistance is 5000 (@ 270v). 5000/414.52 is 12.06. -hhhmmmm...
    then i looked at the 6L6GC, which is what was originally tied to that particular OT. its load resistance (class A) is 4200. do the math, its 10.13.... hhhmmm again...
    then i looked at the 6V6GT load resistance, Class A, =8500. math = 20.5! what?!?
    then for giggles, i looked at 6L6G in AB mode. 3800. that comes out to 9.16.

    none of these is 8 ohms, which is what the OT was originally used, and labeled as.

    im going to have to find a Tube with a Load Resistance of around 3300. -or a different OT, with an impedance ratio of 625 to use with this 6L6G i want to use. -which sucks. especially since transformers dont come with that sort of information.
    :mad:
    the only thing that might fudge the numbers is the volts listed for that load resistance is almost 100V lower than what i was planning on putting on the plate. so "maybe" i can still get away with using this OT? o_O

    or, maybe its "off" because of the other 70v and 25v taps? i cant see how, but... maybe?
     
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  6. Sep 24, 2020 #6
    Jewellworks

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  7. Sep 24, 2020 #7
    andrewRneumann

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    Sweet, that sounds about right. You may already understand this, but what this is saying is that whatever speaker you decide to use, that impedance will carried through the OT (some call it “reflected” impedance) and magnified by that number—415. So... say you pick an 8-Ohm speaker. That will look like an 8 x 415 = 3320-Ohm load to the power tube. Congratulations you have found yourself a 3.3K:8 OT. Notice, that if I change the speaker impedance, I change the reflected impedance, but the ratio remains the same. So this same winding is also a 1.6K:4 or a 6.6K:16 depending on the speaker. The different speaker impedances show up differently to the power tube.

    Ok quick basics on what the datasheet is telling you. First, there is usually a section on some basic general information, then there is a section of limits/maximums. The limits and maximums are not targets per se, and many amps exceed one or more published maximum, but you have to know what is safe to exceed and what needs to be adhered to strictly.

    Then comes the part that I think is throwing you off. The tube manufacturer will throw in some sample designs. These might be used by a designer who just wanted to drop the tube into a system and be reasonably assured it will work. The sample includes performance of then tube under those conditions so the designer doesn’t have to get into the weeds of the graphs unless absolutely necessary. How does this apply to you? Well, we aren’t going to follow those sample designs. We are going to have a different plate voltage, a different screen voltage, and a different bias point. They are just examples, not requirements to follow. The only requirements are listed in the general and maximum ratings sections.

    If anyone else would like to comment on the other taps, go for it. I have no idea what 70v and 25v are doing in an OT.

    We are one step closer to designing this power amp. What we need now is to estimate the voltage available from your power supply. For that I refer you to Merlin. Read the whole page before doing any math. The section at the end about 2 phase rectification applies to you, so you need to do it that way.
     
  8. Sep 24, 2020 #8
    andrewRneumann

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  9. Sep 24, 2020 #9
    andrewRneumann

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    I should say that the typical operation or sample designs the tube datasheets give can be very helpful in estimating values where precision isn’t required. We did that earlier with the transconductance of your 6SJ7. You might find yourself doing that again when you find the need to estimate the total current the amp will draw. You can’t know this precisely until you design the amp, and you can’t design the amp until you know this... so you are caught in the middle. The best thing to do is a make a reasonable guess (say based on the published typical conditions in the datasheet) and proceed from there. Keep track of your educated guesses and when the design comes together if any guess was way off, you may need to go through another design iteration to get things close. My limited experience tells me this is rarely required for something as imprecise as a guitar amp.
     
  10. Sep 24, 2020 #10
    Jewellworks

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    i thought once you divided the load resistance by the impedance ratio, that gave you the ohm reading for that 2ndary. 12.06 is an odd number with a 6L6G, and the amp it came from, with the 6L6GC comes to 10.13... -still not 8! or 7 something... so, did i miss a step?

    ill read the merlin article you linked and get back to you
     
  11. Sep 24, 2020 #11
    andrewRneumann

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    I think what you’re describing is a common misconception.

    YOU pick the load impedance (resistance) by your selection of OT and speaker. It is not hard wired into the tube. Does that make sense? A tube will have a certain range of loads that it is optimized for, but it’s a pretty big range and there’s plenty of room in there for variation. So, do you want a 1.6K load, a 3.3K load, or a 6.6K load. You get to pick. Once we get into load lines you will see the consequences of your choice and the best choice will probably present itself at that time.
     
  12. Sep 25, 2020 #12
    Jewellworks

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    im a senior level AV tech, and i understand what the 70v and 25v taps are for. this OT came from a very small PA amp (only 10 watts) and those taps are for running whats called a "distributed" audio system. these kinds of systems are used in areas where there are multiple speakers, such as airports, shopping malls, theme park audio... you can string several speakers together, each one having its own transformer tapped at a certain wattage that it will use off the main feed. in this case, there were only 10 watts available, so my guess is they used 3 speakers at 3 watts each. or 8 or 9 speakers at 1 watt each, leaving a little headroom. its also how the telephone company phones work. 100v line with a 200 watt amplifier and your only running a telephone speaker tapped at 1/8 watt or less... you can string a LOT of phones together... the voltage on the line determines how long the run can be. in our industry, we use 70v all the time.

    still deciphering Merlin's algebra... my brain is boiling :confused:
     
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  13. Sep 25, 2020 #13
    Jewellworks

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    i "think" i got the first part of his Wizardry figured out...

    "The source resistance is equal to the resistance of the secondary coil, plus the resistance of the primary coil when 'reflected' to the secondary:
    Rs = Rsec + Rpri/(Vpri/Vsec)^2
    Where:
    Rsec = measured resistance of the secondary winding.
    Rpri = measured resistance of the primary winding.
    Vpri = RMS primary voltage (e.g., mains voltage).
    Vsec = RMS secondary voltage measured with no load.

    For a two-phase rectifier the voltages can be predicted in exactly the same way as above, except that we only need to consider half the transformer's secondary winding."

    R2ndary= 217 (c/t to one side of the 2ndary. so half the winding)
    Rprimary= 5.9 + 60? (=66)
    "    Also, if a valve rectifier is used, then its internal anode resistance also forms part of the total source resistance Rs, as shown. This is not constant, so just use a ball-park average. This can be obtained from the valve data sheet"
    the 5Z4 data sheet (https://frank.pocnet.net/sheets/127/5/5Z4.pdf) says the "Total Effective Plate Impedance Per Plate" shows 30 ohms. since there are 2 plates, thats 60. but this data sheet (https://frank.pocnet.net/sheets/049/5/5Z4.pdf) shows a Minimum of 50, and a Maximum of 125 ohms, PER PLATE. sooooooooo... o_O thats a huge difference.

    anyway... 217 + 66 = 283
    Vpri = 120 / V2ndary = 430 = .28
    283/.28 = 1010
    squared, thats 1021514, or 1.2M
    RS = 1.2M

    thats as far as i got, and i have no idea if this is even close.

    the 2nd part, is complete gibberish

    "The amplifier circuit (or whatever happens to be draining current from the reservoir capacitor) can be represented as a single load resistance, Rl. This can be estimated simply as:
    Rl = Vdc / Iload
    Where:
    Vdc = DC supply voltage after rectification (i.e., the voltage across the reservoir capacitor).
    Iload = average DC current demanded by the amplifier circuit.

    If the amplifier runs in class A then its current demand does not change much with drive level, so as far as the power supply is concerned it looks like a constant resistance. A class AB amp, on the other hand, will demand less current at idle than at full drive, so its apparent resistance is not constant. Therefore, when estimating the load resistance, use whichever level of current you are more interested in. If you're really keen you could do two calculations, one for idle, one for full drive, so with the following method you could accurately predict the amount of voltage sag you will get under sustained overdrive...
    Of course, to find the load resistance we need first to estimate what the DC supply voltage will be. If the circuit were ideal the capacitor would charge up to the peak AC voltage, or Vrms*sqrt2. This should be close enough as most amplifiers will adjust their idle current to suit the supply voltage anyway, so they are more-or-less resistive."

    -Valvewizard

    every single word of this is gibberish to my mind. i have no clue what im supposed to be doing.

    HELP!
     
  14. Sep 25, 2020 #14
    Jewellworks

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    he doesnt say how to get the Current demand. only that its constant in a Class A circuit. he says the cap should charge up to the peak AC voltage, which should be 430, as thats half the voltage measured at the C/T. he also gives a gibberish algebra formula, but acts as it its the same thing as the half voltage. so why bother with the math? but he never says how to get the Iload or current draw. the rest of the page is gobbledygook
     
  15. Sep 25, 2020 #15
    andrewRneumann

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    In a previous post you said the primary resistance was 270 Ohms. This shouldn’t be a math problem. You should just measure the resistance across the primary with your ohmmeter.

    I assume 217 Ohms is what you measured directly with the ohmmeter on the secondary?
     
  16. Sep 25, 2020 #16
    Jewellworks

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    I was mistaken in that post. It's 217
     
  17. Sep 25, 2020 #17
    andrewRneumann

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    Ok I think I understand you that 217R is for one half the secondary winding. Now, what is the primary winding resistance you measured?
     
  18. Sep 25, 2020 #18
    printer2

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  19. Sep 25, 2020 #19
    Jewellworks

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  20. Sep 26, 2020 #20
    andrewRneumann

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    Nice. So I see an Rpri = 5.9 and Rsec = 217, but we will use 1/2 half of the secondary at a time, so Rhalfsec = 109 Ohms.

    What we are trying to figure out is how many volts we are going to lose due to the resistances of 3 things:

    1. the transformer secondary DC resistance, but we only use half since we are using full wave rectification and CT,
    2. the transformer primary DC resistance (reflected to the secondary, so must be divided by the square of the turns ratio—just like we did for the figuring out the OT reflected impedance!), and
    3. The internal anode resistance of the rectifier tube.

    We are going to add all 3 of these up and call them Rs, for “source resistance.”

    For 1., Rhalfsec we just determined is 109 ohms.
    For 2. use the formula Rpri / (Vpri/Vsec)^2 ... I will let you figure this out. You have all the numbers you need. Note Vpri/Vsec is the turns ratio.
    For 3. To get the internal anode resistance we turn to the datasheet and this is a little confusing for sure. Use this datasheet as it was one of the only ones I could find with a plate characteristic graph for the 5Z4 rectifier.

    4B866896-8066-4E3F-96DE-A41CAE3CA2FF.jpeg

    Merlin notes that the anode resistance of a tube rectifier isn’t constant, and he’s right because the line is curved. What this graph is showing is how many volts the rectifier will drop from anode to cathode for any given current. An ideal rectifier wouldn’t drop any voltage and the line would be vertical along the y-axis. But tube rectifiers definitely have an internal resistance, and it is used to contribute to the tube dynamic in Class AB push-pull designs aka “sag”.

    I chose a current of 80mA and it corresponds to a 15V drop between anode and cathode. Remember on these characteristics charts, the voltages are always measured relative to the cathode. What is the effective resistance this represents? Ohm’s Law states that the resistance is the voltage drop divided by the current. So let’s do the math 15V/0.080A = 188 Ohms.

    Ok, you should have enough to calculate the Rs. 3 resistances, add them up.

    Now I see you are struggling with Rl—the load resistance offered by the amp itself to the power supply. For this, we are going to make some educated guesses from the datasheets. You need to make an estimate of all the current your amp circuit will draw (not counting heaters/filament). This doesn’t have to be super accurate. You have 3 tubes, and 3 datasheets. Use the typical operating conditions to make an estimate of the total DC current each tube will draw under no signal conditions. Don’t forget that pentodes draw current from plates and screens, while triodes only draw current from their plates. Add them all up.

    Next we need to ballpark the voltage that will drive this current through the load resistance, and Merlin suggests just use the peak voltage. The keyword there is “peak”. That is 1.414 times the RMS voltage. Your RMS was 430v so I’ll let you figure out the peak voltage.

    Alright, if you have added up your total current, and you have a voltage, you should be able to use Ohm’s Law again to calculate the effective load resistance “Rl”.

    I’ve got to leave it there. Hope this helps you make some headway on Merlin’s power supply calculations.
     
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