Originally posted by Alan0354
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How do the diodes protect the OT?
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I set it up, The diode is 2 1N5408 in series connected exactly like the diode in the Prosonic in the first post with a 10 ohm resistor in series to the ground. I use scope probe to measure the voltage drop across the 10 ohm resistor.........I drove the high gain channel with 4 gain stages full tilt, I got mostly square wave into the PI. So I am driving with clipped wave to simulate the worst case. I use the guitar to play a note at a time.
I got nothing, I cranked up to half power, got nothing across the 10 ohm resistor!!!
I checked the scope probe, it'a alive. I plugged in the dummy load, the OT is alive with signal driving the dummy load!!!
So, the plate never hit negative voltage. This implies the plate never hit above 2B+!!! Luckily I have the sense to try it before setting the video camera!!! I think this is as far as I would go at this point. Maybe I should do a voltage divider and measure the plate.
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Thanks Dave. This shows imho the generic situation with gross overdrive where the valves effectively act as switches, and during the transition, the loading of each switch on their respective winding is not sufficient to maintain the continuous flow of inductive current by providing sufficient combined plate current through those windings. As such, the excess current (over what is being conducted by both valves) is being soaked up by the effective capacitance across each winding (ie. C.dv/dt voltage rise). The plates aren't both infinite resistance during the transitions, and so do pass some of the inductive current that is wanting to remain constant.
Would you be able to show one of the plate currents on a plot?
I think frequency of the waveform will likely change level of plate current (compared to the 400Hz sim). Also the transition time (ie. level of overdrive) will likely have some impact on peak voltage.
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Originally posted by Dave H View PostIf you can believe it this is what LTSpice says my 2 x EL84 amp does with no load (top plot) and no load + plate clamp diodes (second plot).
[ATTACH=CONFIG]30314[/ATTACH]
[ATTACH=CONFIG]30315[/ATTACH]
I don't simulate (old dog yadda yadda) but my scope says the same.
A transformer has two or more windings on a core. An inductor has one. How many windings on a PP OT, even when secondary winding is excluded?
Secondary winding, if open, is irrelevant.Juan Manuel Fahey
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Originally posted by J M Fahey View PostThat's the way it works.
I don't simulate (old dog yadda yadda) but my scope says the same.
A typical 50W Fender transformer: 1900 turns plate to plate.
Secondary winding, if open, is irrelevant.
You are the expert in transformer, you have a good explanation? I don't mean the see saw thing you explained, that's understood already. My question is why the MOV or TVS don't burn when use as the high side clamping. I predicted the spike will shoot up and down with a lot more vigor, but apparently it does not. I don't know what to think.
If you observe the wave form is like what David H's simulation, OT burn out or even tube burn out should not even be a problem even if you forget to plug the speaker in. I just don't think 1.2KV is going to burn anything. Even 1.5KV is not too bad. It's when you get pass 3KV that things get dicey.Last edited by Alan0354; 09-03-2014, 05:31 AM.
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Originally posted by Alan0354 View PostI got nothing, I cranked up to half power, got nothing across the 10 ohm resistor!!!
I checked the scope probe, it'a alive. I plugged in the dummy load, the OT is alive with signal driving the dummy load!!!
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Alan, there are many variables that could cause an OT to fail due to playing an amp with unconnected speaker. Only a good understanding of how and under what circumstances the transients occur, and what factors influence the level of transient, can a good start be made to assess the performance of particular protective devices. And even then, different amps will impose different operating conditions, and protective devices may be designed more to protect an OT from other failure mechanisms, and if the device fails itself but protects the OT then that can still be a good outcome for certain failure mechanisms.
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Alan, yes you should check the plate voltage (with care as suggested by Jaz). If the voltage waveform shows clipping, then perhaps if you change to a 100 ohm or 1k sense resistor, and check through your cro specs to see if it can trigger or capture the expected transient current wave shape, as the duration may be quite small. Also you are making assumptions about what is 'worst-case' - perhaps if you put an investigatory hat on.
Have you disconnected the output, or is it as shown in your first post schematic?
Is the schematic exactly what your test amp is?
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Originally posted by Alan0354 View Post.........I drove the high gain channel with 4 gain stages full tilt, I got mostly square wave into the PI. So I am driving with clipped wave to simulate the worst case. I use the guitar to play a note at a time.
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I'll have a go at a description...
As the plotted valve gets to the end of cut-off duration, the plate voltage is being pushed negative by the other plate flying high. As such, the winding is conducting current that is charging the capacitance across that winding (aka I=c.dv/dt), but the current is going in to the winding capacitance not the plate, and is a negative polarity (compared to current polarity that would charge up a capacitor from plate to 0V). Then the plate effectively snaps in to conduction and the current in the winding wants to go positive polarity - which causes a large di/dt transient in that winding with voltage snapping back to 0V and the current ringing away (in the closed loop of plate, cathode, supply capacitor, primary half-winding).
Plate current rises during the 'on' period of the valve from the voltage across the winding inductance, and then gets forced to zero by the valve turning off, with the stored inductive current then being diverted in to the winding capacitance (and the other winding capacitance), with the result that the plate voltages fly off again.
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Originally posted by trobbins View PostPlate current rises during the 'on' period of the valve from the voltage across the winding inductance, and then gets forced to zero by the valve turning off, with the stored inductive current then being diverted in to the winding capacitance (and the other winding capacitance), with the result that the plate voltages fly off again.
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