03-01-2026, 04:52 PM
OK, there is something in your wording that I do not fully understand.
So,
>>Volts on strong station tuned in is 4 volts when move off station I see the voltage slowly climb to 9 volts.
Current is .002 on station to .004 off station. Takes same amount of time for current to increase as volts increase.
Are you measuring V/I across / through the shadograph or something else? The shadograph acts as regular resistor, so the Ohms law applies. If the V increases, so does I, and vice versa.
>>>Going from off station to on station the results of both readings is immediate.
Does that mean that slow change is only when you de-tune from the station?
ALso, "off station" means "not tuned to the station" and "on station" means "tuned"?
So,
>>Volts on strong station tuned in is 4 volts when move off station I see the voltage slowly climb to 9 volts.
Current is .002 on station to .004 off station. Takes same amount of time for current to increase as volts increase.
Are you measuring V/I across / through the shadograph or something else? The shadograph acts as regular resistor, so the Ohms law applies. If the V increases, so does I, and vice versa.
>>>Going from off station to on station the results of both readings is immediate.
Does that mean that slow change is only when you de-tune from the station?
ALso, "off station" means "not tuned to the station" and "on station" means "tuned"?
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