I'm trying to achieve the minimum load current for this AC SSR which is 50mA. My previous question revealed that zero crossing SSRs need minimum load current operate. I didn't want to ask many different questions to the same answerer so I decided to open a new question. Here at the bottom there is something called a bleeder resistor.:
Solid-state Relays typically have less than 10 mA or leakage current at 200 V. To prevent the leakage current from causing load reset errors, the reset current is estimated to be at least 10% of the rated current, and the minimum load current is specified at 0.1 A.
In my case, that minimum load current is specified at 50mA.
My circuit basically look like this:
Above 68k represents my load which draws only 3mA rms current. 23k is the impedance of the open SSR's snubber at 50Hz to represents the 10mA leakage current.
So the problem is my load which draws 3mA which means less than 50mA will not work with this SSR. I cannot change the SSRs so I need to find a workaround to draw around 50mA more through the SSR. I might also use low power applications so I really need to draw more current through this relay.
The only options I came up with is adding another load in parallel with the original load to enable the SSR.
But for 50mA if I add only a resistor this will be like 4.7k minimum. And this will dissipate around 12W.
But if I add RC series in parallel with my load as shown:
Then I get the following plots:
Above the power dissipated by the resistor is less than 1W.
So in terms of heating and active power RC solution is better. But in my previous question I was recommended to chose 100 Ohm and 5uF.But in that case the power dissipated and the drawn current is too much.
My questions are:
1-) Is my RC values fine for this application? Is it fine to use a 480V AC rated capacitor with a 100 Ohm 10W resistor in my case?
2-) In this forum some are warning about using capacitor for zero crossing SSRs. Does anybody have experience with that?