How do the diodes protect the OT?

[trobbins]

Getting there! Just stepping through the simplest scenario of no speaker loading with Alan. Adding a speaker load certainly makes for dramatic V-I excursions, and complicates any discussion (and simulation).

Even that simple scenario perhaps provides some insight in to why the plate locus in your scope clips gets to some regions - eg. the region on the zero current axis with negative plate voltage - which I would suggest is when that plate is starting to come out of cutoff (ie. high resistance plate to cathode) and the other plate is being forced to turn off quickly but has a high current in its primary half-winding (and the inductive dI/dt pushes that plate voltage high).

With respect to pursuing any testing of the 'no speaker load' situation, imho such a test would only be done with some form of known over-voltage protection in place - then the performance can be related back to the protection. Such a test would also be most appropriate for comparing the performance of different protection devices and placements, as having no speaker simplifies the observed waveforms to allow easier assessment.

Ciao, Tim

[Alan0354]

Alan and trobbins, neither of you seem to show the plate voltage swinging negative like my scope clips show.

Absent from this conversation is someone running a simulation. There are always problems with the models used and they might be incomplete, not properly accounting for negative plate voltage, leakage inductance, winding capacitance et al. But I wish someone would make the effort.

I've purchased some vintage amps that had arched the tube sockets. But I don't know if the arch was caused by big overdrive or an open load. Could it have been a bad tube? Who's to say? I really don't want to send a signal to an amp with an open load to find out which will fail first, the socket or the OT.

I did show in C in red that it dip way below 0V. That's the key, that is induced by the other tube flying high as show in the arrow pointing from graph of D to the graph of C. As the plate of the first tube goes low, the plate of the other tube fly high, that is the reason the plate of the first tube being forced to go below 0V way into the negative region.

[jazbo8]

Here are some simulation results to mull over, I could not duplicate loudthud's scope'd results, the parasitic parameters used for the simulations were probably much better than the actual transformers used by loudthud, but I think they still show what the clamping diodes are doing - it's not far from Tim's drawing. In this case, the excess voltage is what kills the tube/OPT - not the excess dissipation per se. I believe the components could very well be killed by "over-dissipation" but only after the shorts caused by the arcing resulted in excessive current flow. Do correct me if I am talking nonsense...

Simulated results for un-loaded secondary, EL34 PP, B+=Eg2=400V, Ra-a=4k, Eg1=-36V:

[Alan0354]

Alan, I did state no secondary load conditions, and I did indicate that flyback diode clamping caused D to E characteristic (without going in to affect of leakage inductance causing some additional overshoot). And yes the locus is for one of the PP valves only (the positive grid driven valve), as that allows someone looking at a valves V-I characteristic curves to pencil in what happens.

Yes the example scenario was driving grid to 0V, for convenience of explanation, but the same characteristic occurs for any grid drive above idle bias voltage - i.e. the locus effectively transitions towards the y axis (plate current) with little increase in current above idle, and then moves along the grid drive voltage characteristic curve (which is a little more complex to appreciate if the grid voltage is increasing as part of the signal cycle).

I suggest you have wrongly interpreted what is happening once the driven tube reaches its saturation position near the graph origin. That drive valve is acting like a switch (but with the plate characteristic curves of effectively a constant resistance region from the origin, up to the knee of the grid voltage characteristic curve, and then close to a constant current source) - and as such becomes the 'loading' on the OT. Eg. the incremental resistance can be down at 100 ohm loading for an EL34 if grid voltage had reached 0V. Hence, the driven valve clamps the windings during that initial stage if the input cycle.

Also note that if no diode clamping was present, then yes the voltage during D to E portion would fly higher (but not to infinity) as stray capacitance soaks up more and more energy, and possibly leakage breakdown starts to contribute (eg. across valve base terminals or in the OT).

I think we all understand the effect of diode clamping and we all agree here that it works. This is I call indirect clamping that prevent the negative side from going more negative than -0.7V, which indirectly clamp the other side from going over 2B+. Our biggest difference is using MOV and TVS to clamp the plate from going beyond 2B+ directly.

The difference between me and you in the no load situation is you claim the MOV or TVS will work, and I claim you will burn the MOV and TVS if you do positive clamping because too much voltage across the MOV and TVS when trying to discharge the energy ( current) from the transformer with no load. I thought you are going to tell me why the MOV or TVS will not blow in a PP amp with OT.

[trobbins]

Would you be able to post the sim circuit and describe the simulation period being plotted - to help everyone appreciate what the plot is of.

[trobbins]

Imho, jumping to conclusions too quickly is prone to missing an appreciation of what is actually going on. A vetted simulation is a start, followed by some bench waveforms. Then going back to a simulation will allow the energy per cycle to be calculated for some example drive signal amplitudes and frequencies. If simulation doesn't look valid, then bench testing could be done and temperature rise noted, as that could be reverse designed back to approximate power dissipation for some test condition. Only then would a comparison of techniques be reasonably substantiated.

[Alan0354]

Imho, jumping to conclusions too quickly is prone to missing an appreciation of what is actually going on. A vetted simulation is a start, followed by some bench waveforms. Then going back to a simulation will allow the energy per cycle to be calculated for some example drive signal amplitudes and frequencies. If simulation doesn't look valid, then bench testing could be done and temperature rise noted, as that could be reverse designed back to approximate power dissipation for some test condition. Only then would a comparison of techniques be reasonably substantiated.

I only have RF and LTSpice simulation program, don't think these will help. I am not going to try on real amp with no load and I can burn the OT and the tube.

I thought you said there are reasons in the PP circuit the OT will not dump a lot of energy into the MOV or TVS and it will not burn the MOV. It is my understanding that OT without load behave more like an inductor that store energy once you have current going through it. Energy stored in form of W=1/2 LI^2 will be dumped (in form of current) through the MOV or TVS if you use it to clamp it. At 1000V across the MOV or TVS, you don't need much current and much time to heat and burn the device. eg. 25mA dropping 1000V across is 25W!!!! Only takes a split second passing 25mA through the MOV to burn it up.

The time it takes to drain the energy can be calculate as time constant T=L/R where R is the on resistance of the MOV or TVS. Say the L=10H, R=10 ohm. T=L/R=1 second to drain 60% of the energy!!!! 25W for 1 second can burn the small MOV. If the MOV has higher impedance, T is going to be shorter, BUT voltage across will be higher. The result is the same.

So the bottom line is if you can prove the OT is not behaving like an inductor with no load, then it will not dump energy through the MOV. But if the OT behaving like say a 10H inductor, it will burn the MOV or TVS. This is because the event is not one shot, if you strum the guitar for a few seconds before realizing it, you will smoke the MOV or TVS.

[jazbo8]

Would you be able to post the sim circuit and describe the simulation period being plotted - to help everyone appreciate what the plot is of.

It's a totally generic circuit, for the XY plot - X-axis is the voltage at the top primary winding and y-axis is the current to the top EL34's anode, V4 is 72Vp-p, 1kHz sine wave, you can ignore V6 as it's only there to take the current reading:


Here is the OPT model in case you want to play with it, e.g., make the parasitic parameters worse to emulate a real guitar OPT...

*
*   TAMURA OPT F685
*
.SUBCKT F685 P1 B1 B2 P2 S8 S0
* Primary Inductances (p-p 8kohm 200H)
L11 P1 12 50H
L12 13 P2 50H
* Primary Parameters
R11 12 B1 202
R12 B2 13 202
C11 P1 B1 1150p
C12 P2 B2 1150p
* Secondary Inductance (8ohm)
L13 S8 16 0.2206951H
* Secondary Parameters
R13 S0 16 0.382
* Coupling Parameters (can be used to change the parasitic inductances)
K1 L11 L13 0.99998
K2 L12 L13 0.99998
K3 L11 L12 0.99998
.ENDS

[Alan0354]

What simulation program is this? Can you draw out the circuit model of the OT and how does it look if the secondary is open?

[jazbo8]

What simulation program is this? Can you draw out the circuit model of the OT and how does it look if the secondary is open?

I use Micro-Cap, the output is basically un-loaded with 1 Giga Ohms on the secondary winding, since you can't have a true open circuit in SPICE.

[jazbo8]

eg. 25mA dropping 1000V across is 25W!!!! Only takes a split second passing 25mA through the MOV to burn it up.

So the bottom line is if you can prove the OT is not behaving like an inductor with no load, then it will not dump energy through the MOV. But if the OT behaving like say a 10H inductor, it will burn the MOV or TVS. This is because the event is not one shot, if you strum the guitar for a few seconds before realizing it, you will smoke the MOV or TVS.

The simulations above showed there is very little current when the Ep swings high during the flyback, i.e., no where near 25mA. I do not know if they match up to the real circuit though... And no one is saying that the OT does not behave like an inductor, of course it does, but how much stored energy can there be when there isn't a load? If someone has some actual data on that, please post the link.

[Alan0354]

The simulations above showed there is very little current when the Ep swings high during the flyback, i.e., no where near 25mA. I do not know if they match up to the real circuit though... And no one is saying that the OT does not behave like an inductor, of course it does, but how much stored energy can there be when there isn't a load? If someone has some actual data on that, please post the link.

I am not familiar with your simulation, so I cannot comment. The problem is not the simulation, the problem is how do you model the transformer particular if there is no load. Unless you have a good grapse in how to model the OT, it become speculation again and no better than what we talked about so far.

So the ultimate question is how big an inductor is the OT when the secondary is not loaded. then you just use the current driving through the MOV to calculate the power dissipation of the MOV using W=IV where V is the clamping voltage and I is the change of current.

25mA is idle current for typical 6V6. For EL34 and 6L6, it's a lot higher. How much current driving through the MOV is the difference in the plate current to the idle current. If you driver a very small signal and the tube only change by say 1mA, then only 1 mA is through the MOV. But if the input signal is larger, you can change the plate current more than 25mA, then the whole 25mA will be dumped through the MOV. If you clamp the voltage to 1000V, the OT ( as an inductor) will source 25mA until the stored energy ( in form of magnetic field) is exhausted.

I worked with electromagnet before that require 50A. We had banks of 200W carbon comp resistors for it to fly back. I don't remember the number, but it was more than 10 resistors. Power dissipation during flyback of the inductor is very real.

[trobbins]

Is this a plot of a period over a single cycle of 1kHz (1ms), after steady state operation was achieved? Would you be able to show a time plot of these parameters as well as screen voltage

Perhaps if screens were set at a typical datasheet level like 250V then that may change plot a bit.

Ta, Tim

[jazbo8]

Unless you have a good grapse in how to model the OT, it become speculation again and no better than what we talked about so far.

The OPT model is good, if fact too good as I mentioned, it needs to have the parasitic parameters changed so it is closer to the guitar OPT. But if your question is whether the OT model can be even trusted, then you can rest assure that it's accurate enough for our discussion purpose.

So the ultimate question is how big an inductor is the OT when the secondary is not loaded.

why would the inductance change whether it is loaded or not?

25mA is idle current for typical 6V6. For EL34 and 6L6, it's a lot higher.

you are talking about the DC current which does not flow in the primary, except for whatever small mis-match might exist in the real circuit, but that is not really relavent to the discussion here, think about how much AC current flows when the output is un-loaded...

[Alan0354]

The OPT model is good, if fact too good as I mentioned, it needs to have the parasitic parameters changed so it is closer to the guitar OPT. But if your question is whether the OT model can be even trusted, then you can rest assure that it's accurate enough for our discussion purpose.


why would the inductance change whether it is loaded or not?


you are talking about the DC current which does not flow in the primary, except for whatever small mis-match might exist in the real circuit, but that is not really relavent to the discussion here, think about how much AC current flows when the output is un-loaded...

Transformer ONLY behave like a transformer under normal working condition.....Within a range of load impedance, within certain bandwidth and within the limit of core saturation. When you have open load, the transformer no longer behaving as transformer. So the question is back to what is the model when there is no load.

The DC current do flow through the primary. BUT You bring up a good point that the DC current flow in opposite direction in both half. That might negate the energy.

I am going to bed.