Hi everyone!
I wonder why the classic shorts/leak test using neon bulb doesn't kill some tubes due too high voltage between electrodes (particulary grid-cathode and cathode-heater).
As one picture is worth a thousand words, here's a simple model schematic and waveforms.
This is the situation when grid is on the hot side and all the other elctrodes are in the cold side, so to speak.
If using 115 Vac power source, about 280 V peak voltage are present between grid and cathode. Isn't this too much?
Vgk max for some signal triodes is 50 V according to datasheet.
What am I missing here?
Thank you!
I wonder why the classic shorts/leak test using neon bulb doesn't kill some tubes due too high voltage between electrodes (particulary grid-cathode and cathode-heater).
As one picture is worth a thousand words, here's a simple model schematic and waveforms.
This is the situation when grid is on the hot side and all the other elctrodes are in the cold side, so to speak.
If using 115 Vac power source, about 280 V peak voltage are present between grid and cathode. Isn't this too much?
Vgk max for some signal triodes is 50 V according to datasheet.
What am I missing here?
Thank you!
#1) The grid-cathode diode shown in your sim does not exist in vacuum tubes with cold heaters; there's just nothing between electrodes, not even electrons.
#2) In a vacuum with cold heaters Field Electron Emission is the only mechanism which could cause "vacuum breakdown";
from: https://en.wikipedia.org/wiki/Field_electron_emission: "Field emission in pure metals occurs in high electric fields: the gradients are typically higher than 1 gigavolt per metre"; That is 1 million volts per millimeter ...
#3) Vgk max in datasheets is for normal operation, i.e. hot heaters.
#4) If however a tube has become "gassy" (loss of vacuum or too many gas molecules escaped from metals) much lower voltages - like 100V or so - can cause ionization of those gas molecules and trigger a gas discharge current - which is indicated when that neon bulb lights up.
#2) In a vacuum with cold heaters Field Electron Emission is the only mechanism which could cause "vacuum breakdown";
from: https://en.wikipedia.org/wiki/Field_electron_emission: "Field emission in pure metals occurs in high electric fields: the gradients are typically higher than 1 gigavolt per metre"; That is 1 million volts per millimeter ...
#3) Vgk max in datasheets is for normal operation, i.e. hot heaters.
#4) If however a tube has become "gassy" (loss of vacuum or too many gas molecules escaped from metals) much lower voltages - like 100V or so - can cause ionization of those gas molecules and trigger a gas discharge current - which is indicated when that neon bulb lights up.
Totally agree with #2, #3 and #4.
But most tester do the neon leakage test after powering the heater/filament. To make sure, I just checked the operation manual of four of five of them, and they all describe similar process. EICO 625, for instance:
So, there's space charge while doing the neon test.
But most tester do the neon leakage test after powering the heater/filament. To make sure, I just checked the operation manual of four of five of them, and they all describe similar process. EICO 625, for instance:
- Set the switches.
- Insert tube.
- Power on.
- Line adjust.
- Short test.
- Then proceed with emission testing
So, there's space charge while doing the neon test.
If you use AC on the control grid with heaters on the whole test becomes meaningless because the neon will always be on.
So either negative DC and heater on, or AC and heater off for G1 shorts test.
quote: The EICO 625 is fairly unique in having its own 6H6 diode tube to rectify the 30V filament voltage. It provides DC for the neon short-indicator bulb. :unquote.
So either negative DC and heater on, or AC and heater off for G1 shorts test.
quote: The EICO 625 is fairly unique in having its own 6H6 diode tube to rectify the 30V filament voltage. It provides DC for the neon short-indicator bulb. :unquote.
No, it won't be always on. The series capacitor prevents the neon from lighting.
The red trace is neon current. It remains off (except while capacitor is charging up).
As soon as we simulate a leakage path, the neon turns on.
Anyway, there's almost 300 V peak on the grid-cathode when shorts/leakage testing on a good tube, as I showed in the first post.
It is sure that this kind of testing works perfectly fine, as it's been used for ages. But I don't know how this is possible without destroying some tubes. Any idea?
AFAIK the 6H6 is used for the line voltage adjustment only.
The quotation is from here: https://www.amateurradio.com/eico-model-625-tube-tester/
it also says: quote: It provides DC for the neon short-indicator bulb : unquote
but maybe you just answered your question yourself:
if the neon doesn't light up then there is obviously no current;
no current, no energy, no damage ...
it also says: quote: It provides DC for the neon short-indicator bulb : unquote
but maybe you just answered your question yourself:
if the neon doesn't light up then there is obviously no current;
no current, no energy, no damage ...
The current is limited anyway by the resistor , cap , neon bulb . Not the voltage kills the tube but the arc energy or instant current , in this case not enough
Thank you. I was having a "semiconductor approach", where voltage can kill a device right away even if no current flows.
Most junctions can touch breakdown if no current flows. Even very small current may be OK.
But usually when we break-down a junction, 'infinite' current tries to flow. This is bad.
As Depanatoru said, the shorts Neon invariably has a resistor or other impedance. Indeed if it didn't, the neon's violent break-down would blow it.