Well I’ve just lashed up the LED indicator using a 3.3pF capacitor, two junk box signal diodes and a random LED.
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The 3.3pF takes a feed from the hot side of my dummy load.
With the diode in free air the LED does not light, but if I place a finger next to it then a dim glow is seen. The LED I’m using is not particularly efficient; on diode test on my meter the glow is similar and I know that high efficiency LEDs will light up nicely on the diode test (1mA?).
So with a high efficiency LED mounted into a grounded metal case I’m quite certain that it will work.
An interesting diversion.
On 17 Mar 2021, at 19:57, John Pagett via groups.io <john.pagett@...> wrote:
You are right, that there can’t be a single ended indicator. But what the circuit diagram shows and what is reality can differ. Let’s assume there’s some capacitance to ground at the LED. To charge this capacitance, current would flow through one diode (with a smaller amount through the junction capacitance of the other diode). To discharge the stray capacitance, the current will then pass through the other diode, passing through the LED on the way.
I’m not saying that this is what happens, just that it’s a possible explanation.
Tomorrow afternoon happens to be a “do some electronics / radio” session over Zoom with a couple of friends. I’ll lash something up and see what results I get.
Pagett’s axiom - When reality and theory differ, reality takes precedence.
On 17 Mar 2021, at 17:19, Alan G4ZFQ <alan4alan@...> wrote:
There can be no "single ended" indicator. If it indicates it is consuming power that has to be returned, just like an antenna.
So chucking a "single ended" indicator to it is quite safe, it when you make it a "double ender" that it becomes a problem.
I would think that any harmonics would be low but they could well be there.
73 Alan G4ZFQ