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Help with a battery powered tube preamp

Discussion in 'Power Electronics' started by Cernnunos5, Sep 4, 2013.

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  1. Cernnunos5

    Cernnunos5

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    Sep 4, 2013
    Total newbie!

    I've taken on an interesting project as I've always had an interest in electronics, and felt this would be a great way to learn. This is my first day delving.

    I'm designing and building a small battery-powered, tube-amplified musical instrument preamp. The signal from which will eventually be used to power a small external 12vdc automobile power amplifier and out to a pair of 4" speakers. The whole thing will be built into a suitcase.

    The reason I chose to use a tube as an amplification medium as opposed to a FET, JFET, MOSFET or whatever else relates to both sound and selling power. Tubes sell in this industry, and they do tend to sound better, due to their clipping mechanics and the way it produces harmonics.

    I have picked up a book entitled "Designing Tube Preamps for Guitar and Bass" by Merlin Blencowe. Excellent book, but I am attempting to adapt what is given to work for my application. I am sure I will have more questions but right now these are my primary obstacles;

    I have selected for this project a 12AJ6 tube, and a 12vdc / 6.8a LiPo battery pack. The 12AJ6 is designed to run at 12vdc. Minimum current to function correctly is about 4a. Now, according to the book, a load line is a requirement on a circuit like this, which runs in series with the tube from the HT or battery. I'm still not entirely sure how a load line works, but working with the equation given, I believe what I am trying to do is keep the tube from seeing more current than it can handle? (That was question one)

    If that is the point of the load line and I am on the right track, according to the equation, if I wanted to have 4a of maximum current at 12v, I would want a .003kohm resistor. 12/.003=4,000ma or 4a. (Note: I think I actually want a 5a or higher max current, but I'm just running numbers) V/R=I

    Right now I am seeing an odd relationship with the numbers that I don't fully understand, and am patently convinced I am doing something wrong. According to this, if I where to run a .001k resistor (assuming there is such a thing), I would see 12a in current? Would that mean I would be forcing/allowing the batteries to discharge faster than is safe or would it mean that the .001k resistor would essentially do nothing, as it wouldn't "clip" the current until 12a, and my HT/B+ is only capable of 6.8a? (That was question 2)

    My final question is; if all of this is correct and I really do need a .003k resistor on my load line to reach a 4a current, in my shopping for resistors I have also realized that I am apparently running way more wattage from the battery than any .003k resistor can handle? As 12v x 4a = 48 watts? The highest handling .003k resistor I can find is about a watt. (That was question 3)

    As a final note, I found a schematic of something somewhat similar to what I am designing. Powering a similar tube with a 12vdc battery/supply, although in this case powering an entire amp. 2 Preamp tubes / 2 Power tubes. According to the schematic, it calls for a 1M resistor on the load line between HT/B+ and the tube. Why? Whats the math here? Why am I so far off? (Question 4)

    Thanks in advance for any and all responses.
     
    Last edited: Sep 4, 2013
  2. john monks

    john monks

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    Mar 9, 2012
    Can you post the schematic or tell me where to find it?
     
  3. Cernnunos5

    Cernnunos5

    4
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    Sep 4, 2013
  4. Cernnunos5

    Cernnunos5

    4
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    Sep 4, 2013
    Ahhh I found the datasheet for this tube (couldn't find it before, was using the wrong search term) and I see where I went drastically wrong. The current to run the tube is only about 150ma, not 4a. I'm reworking my math right now.
     
  5. Cernnunos5

    Cernnunos5

    4
    0
    Sep 4, 2013
    Okay, so at 150ma current to run the tube, I would need a .08k resistor on the load line. I would also only be running 1.8 watts across the resistor, correct?

    I still don't understand why the example I posted uses a 1mohm resistor from the HT to the tube, but the other math makes much more sense.
     
  6. Solidus

    Solidus

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    4
    Jun 19, 2011
    The 1M resistor is called a plate loader resistor.

    Tubes amplify current, not voltage. A potential (voltage) at the grid changes flow in the electron current from cathode to plate depending on the charge present. A resistor "loading" the plate is used to transform the current change to a voltage modulation. Higher values increase the voltage swing for a given current, which allows for feeding higher-drive-requiring stages. Downside is, you get a higher output impedance which needs to be compensated for at the next stage's grid.

    The reason the schematic says it requires more like a 4A transformer is that the heaters are a significant sap source of power. Your heater requires 150mA. The actual amplification circuit will only consume less than 1mA.

    The tube will not draw more power than is needed, and its main limiting factor is the emission ceiling of its plate. That is, the plate (anode) has finite surface area and will not be able to draw more electrons past a certain point. That is the range at which you enter 'overdrive' distortion. This is most often done intentionally to achieve the harmonic effects which tubes display when they cutoff on the waveform.
     
  7. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    The circuit refered to does not specify B+ voltage.

    The tube (USA) needs 12V on the heater, some will run on a B+ as low as 12V.

    1M resistor with 100V across it will pass 0.1mA. This is the maximum that could be passed. If the B+ is lower and there is voltage across the tube, then the current will be less.
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,270
    Nov 28, 2011
    Hi Cernnunos5 and welcome to the Electronics Point forums :)

    I'm surprised you chose the 12AJ6. It contains two diodes, which would not be useful, and one triode, and is not a common component. Valve (tube) audio preamps usually use the ECC83/12AX7, which has two triodes, which can be cascaded or used for left and right channels in a stereo application.

    As Solidus pointed out, the 150 mA specification applies to the heater. You need to apply 12.6V (nominal) (AC or DC) between pins 3 and 4 to make the filament glow; the valve will draw nominally 150 mA through these pins.

    Small-signal valve-based circuitry generally runs with a B+ voltage around 200V. Valves can be used at lower voltages, but running B+ at 12V is pretty unusual - at least, for standard valves like these, and I would expect the amplifier's characteristics to change quite a lot at that voltage.

    A lot of the famous "valve sound" is due to distortion in the output stage, which occurs when an output transformer is used; the transformer is often driven by two or more valves in push-pull configuration. I think you would be hard pressed to find anyone who could tell the difference between valve-based and semiconductor-based preamplifiers in a double-blind listening test, unless they were hearing limitations caused by the low B+ voltage!

    If you really want valves in your signal path, you can generate a 200V B+ supply voltage from 12V using a boost or flyback converter. The anode current will typically only be in the milliamp range so the total power will only be a few watts.

    Switching power supply circuits do generate some interference but careful mounting and use of series inductors and decoupling capacitors should prevent any problems with noise getting into the signal path.

    If your concern is aesthetic, you could consider powering the filaments of the valves but not connecting them in the signal path. I imagine a lot of budget audio gear uses this trick, because "toobs are cool, don't you know", though I've never taken any apart to verify this.

    If your customers are slightly savvy, you might need to add circuitry to simulate the gradual warm-up of the valves and/or to detect when a valve is removed and kill the signal. In that case, adding a voltage booster and actually using the valves might work out simpler.

    The "load line" is not an actual part of the circuit; it is a graph that plots the anode voltage against the anode current. You need to study common cathode amplifiers to understand how the valve is used to amplify a signal, then you should be able to understand the purpose and meaning of the load line.

    Valves operate at relatively high impedances - at least, when compared to typical semiconductor circuitry. The circuitry the output of your valve stage feeds must have a relatively high input impedance (preferably higher, or significantly higher, than the anode load resistor value).

    Please do some more reading and research and get back to us with your ideas and questions.

    Edit:

    Oops I just looked more closely at the data sheet for the 12AJ6 and realised why you chose it - it's designed to operate with a 12V B+ rail. I guess if you can still obtain enough of them, they might be a reasonable option. I doubt anyone still manufactures them nowadays... What's the availability situation?
     
    Last edited: Sep 5, 2013
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