• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

dual power supply

The Poor Man's Guide to OTLs.
 

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Ah, it's not that novel though. Driving a 800ohm speaker is a heck of a lot easier than an 8ohm speaker. The current requirements are 10x lower and the voltage requirements 10x higher which makes your problem 100x easier. Who owns 800ohm speakers anyway?

When I started to look into OTL amps it dawned on me that driving an 8ohm speaker at any decent power (>10W) was difficult unless you have a lot of tubes or you push the tubes incredibly hard in terms of peak current and power dissipation, which apparently results in arc-overs and unreliability. It's not just having a low output impedance but actually being able to drive the current into the load. With an ample amount of feedback even a single pair of EL84s can have near 0ohm output impedance but that doesn't mean it can drive an 8ohm load without running out of current in quick order.
Thankfully the power rises as the square of the output current, so if you double the number of output tubes you quadruple the output power. At some point the idle power becomes unacceptable due to the amount of tubes. Somewhere around 6-8 6AS7's per channel is a good place to be imo, getting you about 20Wrms output into 8ohm at negligible distortion if you allow them to peak at a few hundred mA each. With 3 or less 6AS7's the output power is too wimpy to be worth the effort. More than 10 and the idle power is becoming ridiculous (>200W per channel including filaments).
Once you come to accept that, you realise that you need to design the thing so it heavily enters class-B to get the peak output current you need, while not having a ridiculous bias/idle current due to the amount of tubes you have that makes the amp more useful as a space heater than an amplifier. Designing any tube amp that can enter Class-B heavily and not have grossly unacceptable distortion, or blow up your speakers or itself when you look at it the wrong way is a challenge in itself.

You could design an OTL that puts out decent power into say 64ohm fairly easily but then you'd be stuck with an amplifier and set of speakers that are largely incompatible with other amps/speakers and that seems like a bit of a copout to me from an engineering standpoint. I guess a practical solution may be an array of speakers that you can quickly configure from a low impedance to a high impedance by swapping the connections of individual drivers so you aren't stuck with a pair of speakers that only work from your special low-impedance unfriendly OTL tube amp.

Also if you design an OTL that puts say 20W into 8ohm, not only will it work well with most 8ohm speakers but if you then design a speaker with 16 or 32 or 64ohm impedance the OTL will put out a heck of a lot more power into that without any modifications.
 
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Ah, it's not that novel though. Driving a 800ohm speaker is a heck of a lot easier than an 8ohm speaker. The current requirements are 10x lower and the voltage requirements 10x higher which makes your problem 100x easier. Who owns 800ohm speakers anyway?
I didn't write it's novel. And if TS is behind the company "Lotsmorebetterhifi", than TS sells loudspeaker enclosures with 800 Ohm loudspeakers.

Although local (on a world scale): There are still many 800 Ohm loudspeakers circulating in and around The Netherlands. Here is some info on them: Philips 800 Ohm speakers
 
That pretty much proves my point that the only speakers that exist for such an OTL amplifier are the ones that were specifically designed to accompany it, which is no different to picking some arbitrary impedance like how 64ohms would be most ideal for an OTL amp with several 6AS7 tubes. 8ohm is a design constraint as otherwise you're not designing just an amplifier, you're designing a whole speaker system.

Theortically, 8ohms would be optimal if you built an amp with about 20x 6AS7 tubes. It would put out 2-300Wrms into 8ohm or something like that, but draw about 4-600Watts just idling, per channel!
 
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OK, so it's not novel (but I didn't write it's novel) and according to you your point is proven (but I already wrote that OTL amplifiers for 800 Ohm speakers constitute a small niche within the world of OTL amplifiers).

All I did with my post #40 is to question your statement that: "OTL is worse at just about everything compared to a design with an output transformer ...". I thought that your statement was to broad since there are exceptions to it.
 
Anyone working with Op Amps or DC amplifiers has used dual power supplies.
Easiest way to get there is a full bridge rectifier hung on the secondary of a CT transformer.
I've used it several audio amps. Usually some common +ve HV, say 250 to 450 volts
& negative 150V. No problems at all.
I am still having difficulties keeping the cathode resistor cool on V2-- it overheats making it too hot to touch though it is rated at 50W. I don't understand how negative voltage going into one end of a resistor can add to the total voltage to drive the tube on the one hand, and NOT heat up a resistor on the other. In other words, why/how does adding a negative voltage to a cathode resistor "work" to drive a tube without heating up the resistor at the point where it enters the circuit? Can anyone explain this to me please?
 
Still you didn't post measured voltages (see your posts #8, #13, and #16) and still you didn't post the schematic of your entire set-up (see your post #23). At the time (January 11th) you wrote that you were behind in your normal working routine. But a few days later you did have the time to start and participate in another thread.

Looking at the little information you did post, the current through the cathode resistor is 105 mA. The voltage drop over the cathode resistor is V = I x R = 0.105 x 550 = 57.75 V. So the power dissipation of the cathode resistor is P = I x V = 0.105 x 57.75 = 6.1 Watt. Passing 105 mA at Va = 674 V (= 492 V + 182 V) and Vg = - 57.75 V is in line with the data and curves in the datasheet for the GM70. So if the cathode resistor, having a power rating of 50 Watt, is overheating, than there must be something wrong in your set-up. But this is basically a repetition of what other forum members already wrote last January.

As long as you refuse to post the as built schematic and the measured voltages of your entire set-up, we can only keep on guessing.
 
Truer words were never spoken. I have had many distractions since my last post here and your generous answers. What remains is that your calculations are correct and as of this evening, the bias voltage at the cathode of v2 is way off. So yes, there is something else wrong with the circuit. I will endeavor to rectify it soon after I have completed my other projects which is not limited to this very, very frustrating amplifier.
Given that the p/s is doing its job (now correctly), my attention has been diverted to what it is that is creating this incorrect bias voltage. My next thing is to isolate the dc voltage on the filaments of v2 by using a/c to see if it is the dc voltage is interfering with either the voltage on the tube and/or the rise in the resistor temperature.
I will keep you posted.
Thanks for your patience and your willingness to help.
 
The most difficult issue is that the output valve's input signal is referenced to a power supply (the negative voltage supply) rather than the comparatively clean signal ground. Otherwise, it would always be done this way.

One line of attack is to bypass the output valve's cathode to the driver's positive voltage supply. Jack Elliano calls this Ultrapath, and it makes a lot of good sense in your circuit. The goal is to make the output valve's grid-to-"cathode" voltage a function of the voltage across driver anode resistor.

edit: rather than voltage across anode resistor plus difference between positive and negative supplies.

All good fortune,
Chris
Could you explain this more, please? What are the advantages/ disadvantages to this? And would any of them be "audible"?
 
Still you didn't post measured voltages (see your posts #8, #13, and #16) and still you didn't post the schematic of your entire set-up (see your post #23). At the time (January 11th) you wrote that you were behind in your normal working routine. But a few days later you did have the time to start and participate in another thread.

Looking at the little information you did post, the current through the cathode resistor is 105 mA. The voltage drop over the cathode resistor is V = I x R = 0.105 x 550 = 57.75 V. So the power dissipation of the cathode resistor is P = I x V = 0.105 x 57.75 = 6.1 Watt. Passing 105 mA at Va = 674 V (= 492 V + 182 V) and Vg = - 57.75 V is in line with the data and curves in the datasheet for the GM70. So if the cathode resistor, having a power rating of 50 Watt, is overheating, than there must be something wrong in your set-up. But this is basically a repetition of what other forum members already wrote last January.

As long as you refuse to post the as built schematic and the measured voltages of your entire set-up, we can only keep on guessing.
As it stands, the voltages in the amp are now as I have them in the schematic. It took some time for me to figure out the mistakes I had made in the wiring of the amp, but now I am closer to my schematic as I have ever been, in design and "by the numbers". It has been a slog.
However, one interesting thing remains: the 7905 does not draw much in the way of current, but when I remove it form the working circuit the a/c current from the wall (I have an a/c current meter on my variac) drops by half.
What gives?
And where should I look to determine the problem?
 
Thanks again for your reply. My question is about the voltage. Please review the attached to determine if it would/should work. If not, why not? Let me say at this point, that I have put it into the circuit and so far, my 50W cathode resistor heats up and starts to smoke and I don't know why. Thanks for any help
Looking at this schematic again I'm trying to reset & discover exactly what it is you are trying to do.
The circuit as you have here is a natural smoke producer as well as deadly to life. 🙂
 
As it stands, the voltages in the amp are now as I have them in the schematic. It took some time for me to figure out the mistakes I had made in the wiring of the amp, but now I am closer to my schematic as I have ever been, in design and "by the numbers". It has been a slog.
However, one interesting thing remains: the 7905 does not draw much in the way of current, but when I remove it form the working circuit the a/c current from the wall (I have an a/c current meter on my variac) drops by half.
What gives?
And where should I look to determine the problem?
Check your wiring, again and again, since it took you some time to find multiple wiring mistakes in the first place.

Out of curiousity: Are you the person that is doing the modifications at Lotsmorebetterhifi ?
 
I would worry about your main circuit topology before power supply filtering. Ask yourself this. Discounting OTL or circlotron designs, how many dual supply single ended designs have you come across? Here's a hint. If you need a 1000 volt supply you don't need to worry about 1000+ volt capacitors.
 
I have experimented with dual power supplies, and it works. But the easiest option to use lower voltage capacitors is to connect them in series.
Hi,
I have what I hope would be a simple question for someone who has worked with a dual power supply in a tube amp.
Normally, one determines the cathode bias resistor based on the voltage going through the tube at the plate and the current requirements from the tube characteristic curves, assuming that there is only one B+.
But with a dual supply, where the voltage at the plate is actually less, how do you calculate the bias resistor? Or does the fact that the voltage is really the same in a dual supply (even though half of it does not come through the plate anymore), make it the same as if it was a single B+?
Thanks in advance.
 
Hi,
I have what I hope would be a simple question for someone who has worked with a dual power supply in a tube amp.
Normally, one determines the cathode bias resistor based on the voltage going through the tube at the plate and the current requirements from the tube characteristic curves, assuming that there is only one B+.
But with a dual supply, where the voltage at the plate is actually less, how do you calculate the bias resistor? Or does the fact that the voltage is really the same in a dual supply (even though half of it does not come through the plate anymore), make it the same as if it was a single B+?
Thanks in advance.
Is not just the plate voltage, it depends on the voltage between anode, cathode and the grids, and de desired quiescent current, given the tube curves. It does not mater how many power supplies you use to reach the goal.
 
I'd advise you todo some modelling to understand what happens with tube failures etc. Last thing you want is 1KV at the speaker jack or the B+ being dragged negative (with polarised electrolytic on the B+).

Voltage across the tube. Not "plate voltage". (I apologise if I miss read that point)

I had tube CCS'd cascode frontend like the M60 designed. +320V and -320V supplies with ~155V across each tube. The Issue I had is that I then needed two heater power supplies one elevated to +50V and the other referenced to -150V. It doesn't matter where the tube sits, relative to itself it still sees 160V across it. You're probably better positioned with a directly heated tube.

Current play thing is a hybrid circlotron - so it has four power supplies (9 in total) and due to both sides being within 5V at any point of the waveform output, I can use a single heater power supply and a single tube. It doesn't change if it's a tube or SS - it's the voltage across not simply "at the plate", even more so with a circlotron's floating system.

Caps - keep them across the the B+ and return of the same supply. If you need more voltage then caps in series and then use balancing resistors. The same for series resistors to get you up to higher voltages.

Although I am concerned with the question and 1KV mention in the same sentence.. (maybe I've missed something)

EDIT: JC beat me to it.
 
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