Bitten by the bug

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers' started by devbro, Sep 20, 2006.

  1. devbro

    devbro Member

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    I want to learn how to work on my amps so bad but I cant seem to grasp it. I have read so much my head feels like its going to explode. It's like dying of thirst but can't find a drink.

    I bought Dave Funk's Tube Amp Workbook and it pretty much flew right over my head. I've read everything I can find on the internet. It's strange because I can't explain the cycle of fifth's or music theory but I can play guitar. Must I know electronic theory to work on my amp? I can halfway read a schematic and a layout but I can't seem to connect the dots when I try to correlate them with my amp. Damn its frustrating:mad:.

    Is there a simple book someone can recommend?
     
  2. 52ftbuddha

    52ftbuddha Member

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    Try and find one of the DOD electronic training courses like the AF CDC's I suspect that you just need a basic grounding in electronics. The Air Force has gone to a computer based electronics traning for its basic electronics. See If you can find someone who has a copy.
    Rob
     
  3. brad347

    brad347 Member

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    i say just keep at it.

    On the fourth or fifth time reading it it will start to make sense. Try Gerald Weber's books, too. They are NOT perfect, but things are often broken down in a way that is very easy to understand for the layman.

    That coupled with getting your hands dirty (once you know the basics of discharging caps, etc) will get you up to speed in short order. I can do fairly complex repairs now and i've only been at it about a year, off and on.
     
  4. Sparky6string

    Sparky6string Member

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    Perhaps building a basic amp kit would help? Something like a tweed champ or ax84 type.
     
  5. TheAmpNerd

    TheAmpNerd Member

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    Yeah, that is just what we want. Confusion is a great
    learning process, really! It shows your brain is churning.
    You are learning, thinking, trying to put it all together.

    Buy a kit and have at it. Start with something simple that you
    will have a success in building, like one of these:

    http://www.guytronix.com/home.html

    Check out his site, people are building these amps
    from all over the world too. Heck, it looks fun, I might
    even buy one just to have something simple around the house.

    Plus, he did raffles etc and gave to the Salvation Army for the Katrina folks. It doesn't get much better than that. Heck Rich even posted
    pics showing he actually wrote the checks. My only real beef with it
    is Rich should learn how to use a computer and software. I mean
    come on, we are living in the digital age. Learn to digitally airbrush the names and account numbers out, don't just cover them up with scrap
    pieces of paper! :AOK

    Back to amps
    Then in the process of building it, lights will start to come on
    in your brain, the stuff you are reading about will start to become
    understandable.

    You only think you got it bad now, wait until you get a few
    under your belt, then you'll really be screwed.


    Enjoy, and have fun too, it is what it is all about.
     
  6. donnyjaguar

    donnyjaguar Member

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    Start with Ohm's law. If you don't understand the rudiments of electric current there's no way you can understand more advanced subjects. I think you're probably trying to put the cart before the horse. The group is an excellent resource with some very clever people. Its also largely self-correcting too, which is excellent.

    Here's your first test:

    Q1: A tube has a filament voltage of 6.3V and it draws 300mA. How much power is the heater dissipating?

    Q2: You measure 450Vdc on one terminal of your multi-capacitor. From this point to the next section of the capacitor there is a 10k resistor and you are measuring 400Vdc. What current is passing through the resistor?

    Q3: Your 2 output tubes have their cathodes tied together and a 330 Ohm from this point to ground. You are measuring 33Vdc at the cathodes. What is the current through each tube?


    :)
     
  7. BBQLS1

    BBQLS1 Member

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  8. devbro

    devbro Member

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    Alright, I'll bite.

    Q1: A tube has a filament voltage of 6.3V and it draws 300mA. How much power is the heater dissipating?
    A: Assuming power is watts, then watts equals volts times amps (6.3 x 3.33 = 20.9) correct?


    Q2: You measure 450Vdc on one terminal of your multi-capacitor. From this point to the next section of the capacitor there is a 10k resistor and you are measuring 400Vdc. What current is passing through the resistor?
    A: Amps equals volts divided by ohms (450 / 10000 = .045) Correct?

    Q3: Your 2 output tubes have their cathodes tied together and a 330 Ohm from this point to ground. You are measuring 33Vdc at the cathodes. What is the current through each tube?
    A: Volts divided by ohms (33 / 330) equals 0.1 amps. 0.1 amps divided by .5 (2 power tubes) equals .02. Correct?

    Sadly, even if I'm correct (which I doubt), I don't know what it means or how to apply it. I still won't know how to find my screen resitors or why I can find no plate resistors on V1 or why there is a resistor soldered across the pins only on V1, etc, etc, etc. :( :(



     
  9. donnyjaguar

    donnyjaguar Member

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    That's the spirit!

    Okay, watch those decimal places, 300mA = 0.3A so...
    6.3 x 0.3 = 1.89W

    Valiant effort, but you need to take the difference between the 2 voltage readings, which is 50V. Therefore the current through the resistor must be:

    50 / 10,000 = 0.005A (aka 5mA)

    Okay, you have the current correct at 0.1A (aka 100mA). You have to divide this by 2 (for the 2 output tubes) which will equal 0.05A (aka 50mA) per tube.

    We'll get there. Baby steps my friend. Let's apply the above learnings to a simple triode voltage amplifier. If you need a picture just look at the first section of most every musical instrument amplifier made after about 1960. In this example let's put a 1k resistor from the cathode to ground and a 24k resistor from the plate to the supply voltage. Let's put a 1M (megaohm) resistor from the control grid to ground. If the supply voltage is 300Volts and you are measuring 1V on the cathode, tell me the following:

    1) current through the tube
    2) voltage on the plate
    3) power dissipation of the tube
    4) total power dissipation factoring in 6.3V filament @ 300mA

    (BONUS) Now tell me if a 12AX7 could handle this kind of configuration. :)
     
  10. 52ftbuddha

    52ftbuddha Member

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    Now this is a forum.
    Rob
     
  11. devbro

    devbro Member

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    Here's where I go south. I think I'm missing some vital understanding of what's really going on inside the tube. Here goes.... the anode heats up from the heater circuit like a light bulb and causes current to "jump" accross to the cathode (plate) after it passes through a screen. The amount of current sent to the screen controls how much current passes through the screen that the cathode accepts. Somehow, somewhere in all this, a guitar signal is burried and is louder when it leaves the cathode. It then goes to the next stage and does the same thing.:crazy

    1) Assuming current equals voltage divided by resistance, then the current through the tube would be 300V / 24K = 0.125amps. This would be the current at the anode creating the heat yes?

    2) I think the voltage on the plate is somehow affected by the grid which I think is the bias. So, if the anode current is 0.125 and voltage equals current x resistance then the plate voltage is 0.125 x 1,000,000 = 1250 becasue the the current passed througha 1 megaohm resistor.

    3) I dont understand dissipation or it's purpose
    4) ditto
    Bonus) I suspect this is too high a voltage for a 12AX7 but I'm guessing.

    This really is a board. :dude
     
  12. donnyjaguar

    donnyjaguar Member

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    Okay, I'm busy at work today resolving multiple issues so I'll answer as much as I can as I sip my coffee on a break. As a bit of background I'm a terrestrial microwave engineer, which is like the extreme sports of analogue electronics. :)

    Okay, you've got the formula down, but you need to apply it in the right manner. In the end it'll all make sense (I promise). Current doesn't change in a circuit. If you take the voltage reading on the cathode you are measuring the voltage across the 1k resistor. If this is 1 V, then the current is 1/1000 = 0.001A (aka 1mA).

    Now we know the current through this resistor, through the tube and also through the plate resistor. Let's calculate the plate voltage, but before we do that we need to calculate the voltage drop across this resistor. We know the plate resistor is 24k and the current is 0.001A. Therefore 24,000 x 0.001 = 24V. So the plate must be the supply minus the drop: 300 - 24 = 276Vdc.

    We'll get to that. But before we do I have to point out that the grid does not draw any current. Its the voltage on the grid that determines what comes out of the collective circuit. More later when my phone stops ringing...

    Update...

    Okay, let's determine the power dissipation of the tube. Basically any analogue device is less than perfectly efficient but the reason we care about this is because each tube has a limit to how much power it can dissipate and we must not exceed this under normal operating conditions. Okay, we know now that:

    - the grid draws no current, so therefore it doesn't add to the dissipation.
    - the plate voltage is 276Vdc
    - the cathode voltage is 1Vdc

    Therefore the voltage across the tube is 275Vdc. We know the current is 0.001A (because its always the same anywhere in the circuit). So the power dissipation will be Voltage x Current = 275 x 0.001 = 0.275W

    Most small signal tubes (12AX7, 12AT7 etc) are good for about 1Watt so we're in good shape here.

    Let me know if this all makes sense before I move on and possibly confuse you. :)
     
  13. brad347

    brad347 Member

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    a simple layman's description, suspending the math:

    A tube (valve) is a component that allows a SMALL voltage to control a LARGE voltage.

    For sake of simplicity, we'll use the example of a triode and describe it thusly: Electrons that boil off the cathode are attracted to the plate like mad. Really strong attraction. The grid in the middle influences that attraction, slows it down a bit, making it fluctuate in proportion to the signal on the grid. The grid is where the audio signal is applied. In this way, a small voltage on the grid can control the large voltage on the plate. I think this is the basic premise of amplification.

    But my description may be lacking, i'm just a hobbyist. :)

    For the way my brain works (very right-brained individual, not mathy) I find it most helpful to understand the general concepts of how something works before delving into the math. I like to be able to visualize a square's shape before knowing that its angles add up to 360 degrees :D
     
  14. TheAmpNerd

    TheAmpNerd Member

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    Blad, you lealned well weed hoppa;
    all thlee sides of squales angles add up to 360 deglee.
     
  15. boobtoob

    boobtoob Member

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    I just read the above entries, and I don't have a clue what ya'll are talking about...yet I built several amp kits "by the numbers"...still never been shocked...still never burned myself on the soldering iron (too badly).

    So, it is possible no matter how techie you are.
     
  16. devbro

    devbro Member

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    I'm pretty much lost although I beleive I'm on the verge of an epiphany. I think I too need to know how it works before why it works. I can do the math but I'm missing some major piece somewhere.:crazy Here's my take....The cathode gets hot and causes current to jump to the anode. This is DC voltage. The guitar signal comes in on the grid as AC current. How does the DC current flying through the grid and the screen "pick-up" the AC signal and take it to the anode and make it louder? I assume that the signal leaves the tube through the anode and then the coupling cap somehow filters out the DC current so just the amplified signal goes to the next stage. Somehow the screen controls how much how much current gets through and this is the same as bias.

    Am I a complete idiot or what.........:confused:
     
  17. brad347

    brad347 Member

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    it doesn't "pick it up and make it louder."

    It is CONTROLLED BY it.

    The small voltage controls (modulates, or causes changes in) the large voltage.
     
  18. devbro

    devbro Member

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    So if the small signal voltage (AC) is modulating on the grid - like a guitar signal would - and it somehow increases the large voltage (DC) traveling between the plate and the anode, how does the AC signal find its way to the anode? And why would and increase in DC voltage at the anode make the AC signal louder?
     
  19. callaway_1

    callaway_1 Member

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    What a great thread! what a bunch of great guys too! I have been looking for this kind of info for a while!

    Clayton
     
  20. brad347

    brad347 Member

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    As voltage applied to the grid (a small voltage relatively speaking) varies from negative to positive, the amount of electrons flowing to the plate vary accordingly.

    Imagine the influence of magnetic polarity. You can certainly influence the pull of a magnet by the presence of another magnet, and if that magnet flipping polarity then it will alternately make stronger and weaker the pull of the 'main' magnet.
     

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