Help with achieving minimum load current for SSR for weak loads

[doncarlosalbatros]

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. Besides the indicator lamp my other loads could be low power as well. So I need to achieve an extra 50mA through relay output pins.

Below is the diagram of the system:



I'm trying to achieve the minimum load current for this AC SSR which is 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. I guess at this site here at the bottom there is something called a bleeder resistor for such purpose:

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 I was recommended to chose 100 Ohm and 5uF. But in that case the power dissipated and the drawn current is too much.

So 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? What type of capacitor and resistor recommended?

2-) In this forum some are warning about using capacitor for zero crossing SSRs. Does anybody have experience with that?

(I have many of these relays with weak loads so I really would appreciate a suggestion without changing the SSR relays.)

 

[BradtheRad]

The capacitive drop method has become popular because it is a cheap way to get small current from mains AC.
1 uF is a reasonable value.

There's a chance the SS relay is activated just by brief inrush current (50 mA). Then you may find a weaker load is sufficient to keep it energized. Try putting a diode in series with the RC. The cap charges after a few cycles of inrush current. (Quickly current is blocked.) If you experiment with R & C values you might get a few cycles of 50-100 mA inrush current.

You may need two such diode-RC networks (pointing in opposite direction. You may need to install high-ohm resistors to reset (discharge) the capacitors during power-off.

 

[KlausST]

Hi,

The first that comes into my mind is: replace the SSR with a more suitable one, but I see you don't want this.

The next is to use an indicator on the input side of the SSR....but I guess you don't want this.

Then there is the idea with resistive load. Too much dissipated power.

Then the RC load. This is at least worth to try. But here I expect a problem with the phase shifted current. I assume it is a triac operated SSR, thus the current shrough the triac will fall to less than 50mA soon after the mains voltage peak (90°, 270°)...and cause the triac to release.

I don't know how Brad's solution with the diode will work. I expect the current to decrease very fast because the capacitor will be charged to a DC voltage.

Thus I don't have a true solution with the existing SSR.
Maybe you don't want to see when the SSR is excited to put a LED on the input side of the SSR...maybe you really want to see the existance of mains voltage....
Thus my workaround is: to use a second SSR (or any other low current AC switching device, without the minimum load current problem) in parallel to the first SSR that just controls the indicator. Then the indicator is ON only when the SSR input is activated AND the mains power is available..

Klaus

 

[BradtheRad]

An ordinary 12 volt relay might pull in at 10V, and release at 7V. If you're lucky then the solid state relay behaves in a similar manner. If so then you can activate it by drawing 50 mA briefly, then reduce to a lesser amount for normal running. Lower watt ratings can be used.

Simulation of what I had in mind.

The capacitor-resistor-diode networks provide inrush current in a controlled manner.



Values are chosen so inrush current is sufficient to activate a mechanical relay or SSR.

The capacitors are charged from DC (via diodes in opposite directions). Over the next few cycles Ampere level drops to a lesser amount. This running level depends on the resistor divider maintaining a certain charge on the capacitor. It needs to be high enough to keep the relay activated.

 

[KlausST]

Hi,

Values are chosen so inrush current is sufficient to activate a mechanical relay or SSR.

In opposite to a mechanical relay an SSR needs to be activated twice a fullwave and will drop twice a fullwave.
Thus I expect the declining current maybe makes the SSR to operate correctly for the first halfwaves but then the current will be too low to keep the SSR activated.

And for a mechanical relay you may use a higher current to activate the relay and then you may lower the current....but mind: the coil current (here the DC side), not the load current...

Klaus

 

[BradtheRad]

Reading the previous post by Klaus, could it be sufficient to send 50 mA pulses of current for just a milli-second per each cycle? That would be less heating effect, and it might be detectable by the SSR.
Could the pulses be of one polarity (pulsed DC), and not AC?

I admit a better comparison would have been a thyristor (scr or triac), as needing minimum current draw to stay activated like the SSR.

480V AC rated capacitor with a 100 Ohm 10W resistor in my case?

I made a simulation with your schematic. 1 uF and 100 ohm resistor. It resembles your plots. It draws 103 mA peaks. The resistor gets peaks of 1.07 W. It seems feasible.
The resistor value can be anywhere within a wide range. Its chief purpose is to limit inrush current.



Your text mentions 5uF which draws 1/2 A peak. This causes 26 W peaks in the resistor. It does seem wasteful.

- - - Updated - - -

some are warning about using capacitor for zero crossing SSRs. Does anybody have experience with that?

It probably has to do with power factor error. Notice the AC supply voltage waveform does not coincide with the current waveform. The capacitor creates current lead to almost the 90 degree theoretical maximum. This may or may not matter to the SSR. You may need to increase the resistor ohm value in order to improve power factor. Then you'll also need to raise its watt rating.

 

[crutschow]

The RC load should be fine.
Even if it drops out during part of the cycle due to the phase-shift, that still should be enough to illuminate the indicator lamp, which I assume is the reason you want the relay to be on for no load.

 

[doncarlosalbatros]

The RC load should be fine.
Even if it drops out during part of the cycle due to the phase-shift, that still should be enough to illuminate the indicator lamp, which I assume is the reason you want the relay to be on for no load.

Thanks for the answer. My load will not be inductive the loads will always be an SMPS power supply. I guess an SMPS is more a capacitive load.

For the RC;

I used this 1uF motor cap:
https://uk.rs-online.com/web/p/polypropylene-film-capacitors/1728182/

And 100 Ohm 15W resistor:

https://uk.rs-online.com/web/p/panel-mount-fixed-resistors/0159792/

I made a test and it worked(I didn't use a diode) and the lamp works now(goes ON or OFF properly). I didnt try with any load other than a resistive one. Are the concerns for capcitor only for inductive loads?

 

[crutschow]

I didnt try with any load other than a resistive one. Are the concerns for capcitor only for inductive loads?

It could conceivably resonate with the inductance.