For an enclosure, I will control the power at the terminals of a panel mount AC socket by using an SSR. The AC mains socket will have this indicator lamp in parallel to indicate whether the socket is live/powered. The indicator lamp according to the datasheet draws only 3mA current at 230V AC.
As shown above the system worked but the lamp was extremely dimmed when the relay was ON. Then I connected a voltmeter across it(across A and B) and nothing changed. Except I saw 230V AC even relay is OFF due to capacitive coupling.
But when I set the voltmeter to low impedance setting, the indicator started to glow as it supposed to. And the fake ghost voltage at OFF times also disappeared.
So I went back to the SSR’s datasheet and at its output characteristics it says “Min. Load current to maintain on: 50mA”
My conclusion was the problem is: if the load draws less than 50mA I will have problem; either with the indicator lamp or any such weak load.
To solve this issue can I connect a resistor across the node A and B in my diagram which draws 50mA current. To make sure the weak loads like the lamp would work fine when the SSR is ON.
If I use a resistor across A and B for 50mA the minimum resistor becomes 1000*(230/50) around 47k. And power for such resistor is 1.25 Watt.
I never used a resistor for such purpose for AC power application. Can any 3W 47k resistor be used? Do I need any heatsink? The relay might be ON more than six months. Can this be a solution or what else can be done?