>
>Rich wrote:
>>
>>>
>>>Rich wrote:
>>>>
>>>>>
>>>>>Rich wrote
>>>>> The glitch R needs to be in the positive HV lead.
>>>>>
>>>>>Piece of education needed for me here. The glitch resistor is there to
>>>>>limit fault current in event of a flash-over. The only source of these
big
>>>>>currents is from the filter capacitor and the only route for the current
>>>>>back to that cap is via the 25R resistor. So why does the resistor need
to
>>>>>be in the +ve line.
>>>>>
>>>>The typical positive filter cap ckt is insulated for several kV.
>>>>The negative filter cap/cathode circuitry/fil. transformer CT is not.
>>>
>>>The negative end of the filter cap can easily be insulated to those
>>>standards - you just have to remember to do it.
>>>
>>One does not have to remember anything if one puts the glitch R in the
>>positive HV lead.
>>
>Except to insulate *both* ends to full HV.
Ian plays the Wild Card.
>A possible advantage of
>placing the glitch resistor in the negative return is that one end is
>always close to chassis potential.
>
? for sure - the glitch diodes limit the drop to c. 1.5v.
>>>The cathode circuitry and filament transformer (assuming we're talking
>>>about GG triodes) would not see a spike caused by a direct short or arc
>>>from B+ to grid or chassis - they are protected by the glitch diodes.
>>
>>Limiting the V-drop across the glitch R to 1.5v will hardly allow it to
>>limit current.
>>
>Stop and think again. The cathode circuitry, meters and transformer are
>protected by the diodes, but the glitch resistor must NOT be shunted by
>the diodes - it sees the full HV.
>
In order to avoid the glitch diodes across the grid current meter shunt,
the glitch R needs to be in the positive lead.
>>> If an arc inside the tube penetrates through the grid to the cathode,
>>
>>I have autopsied dozens of tubes. I have never found a penetrated grid.
>>
>Maybe it depends on the types of tubes... or maybe you've led a
>sheltered life :)
>
3-500Z, 8874, 3cx800A7, 8877.
>I have seen it on an autopsied GS35b tube, which has a flat oxide
>cathode and a domed mesh grid. The grid had about six arc marks on it.
>Some marks were on the grid alone, with no mark on the cathode
>underneath. One (probably the one that ended the tube's career) had
>burned a hole right through the grid and then burned the cathode
>underneath.
>
>A couple were especially interesting - there were marks on the cathode
>directly below the marks on the grid, but they had done that *without*
>burning through the grid. My speculation about these would be that when
>the arc hit the grid, it pulled that local area of the grid very
>positive for a few nanoseconds (which could happen because the grid mesh
>has a finite self-inductance) and that stripped the oxide coating
>directly underneath.
>
>(Note to GS35b users: that tube had been operated at close-on 5kV.
>Normal users don't need to worry.)
>
>>
>>>those
>>>components are not protected any differently by a glitch resistor in the
>>>B+ or the B-minus. As Andy said, it's the same surge current flowing
>>>around the loop, ...
>>
>> It can not be the same current if the Glitch R drop is limited to 1.5v
>>by the glitch diode.
>
>You're imagining the diode in the wrong place, shunting the glitch
>resistor.
>
see diagram
>>---------
>>In "The Almost Perfect Amplifier" (*QST*/January, 1994), I recommended
>>using a rheostat to adjust filament V. During the grate parasitics
>>debate, at least five members of the Rauchian camp argued long and hard
>>that a rheostat can not be used to control filament V.
>>
>Forgive me, but I'm not seeing the relevance of that statement to the
>present discussion.
>
No surprise there. Do you feel that a rheostat can be used to control
filament V?
end
- Rich..., 805.386.3734, www.vcnet.com/measures.
end
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