SSR & LED

[ptonsparky]

Ref SSR vs Mechanical thread. I did a small shop experiment. I removed power at the start of each lamp change. Electric unit heaters may have cycled at some time during test.

Voltage = 124.9 Control is SSR 12vdc

ssr 60 w Lamp

L|-------][--------------------INC-------------|N

Apply L-N=124.9

Open Ssr=124.9 Lamp=.78mv

Closed Ssr =1.18 Lamp=123.9 (On)

Open Ssr=124.4 Lamp=.94 & going down (Off)



Removed all power, changed Lamp to 60 watt LED non dimmable

Apply L-N=124.9

Open Ssr=86 Lamp=40

Closed Ssr=1.02 Lamp=123.7 (On)

Open Ssr=86 Lamp=40 (Off)



Remove all power, changed lamp to 5.5 watt LED dimmable

Apply L-N=124.7

Open Ssr=122.7 Lamp=7.06

Closed Ssr=1.4 Lamp=124 (On)

Open Ssr=31.9 Lamp=103.8 (On, dim)

My question is where did the increased 11 volts come from ? (No, not a neutral issue)

 

[synchro]

...
Remove all power, changed lamp to 5.5 watt LED dimmable

Apply L-N=124.7

Open Ssr=122.7 Lamp=7.06

Closed Ssr=1.4 Lamp=124 (On)

Open Ssr=31.9 Lamp=103.8 (On, dim)

My question is where did the increased 11 volts come from ? (No, not a neutral issue)

The "open" SSR and the dimmable LED lamp are in series, and so their voltages must add up vectorially to the applied L-N voltage. They are both conducting the leakage current that's passing through the open SSR. However, the voltages developed across the SSR and across the lamp depends on their impedance, which can have a reactive and/or resistive component. The voltage developed across a reactive component will be at 90 degrees from that developed across a resistive component with a same current passing through them. Unless the voltages developed across the open SSR and the LED lamp have the same phase angle, their magnitudes (what your meter measures) will not add up to the applied L-N voltage. And if they are not at the same phase angle, the sum of their magnitudes will have to be greater than the applied voltage.

 

[ptonsparky]

Ok.

The characteristics of my SSR must allow it to stay On even though control voltage has been removed. A different SSR would react differently?

SSR is Opto22 120D10.

 

[ptonsparky]

The SSR has 7ma of leakage current, so enough current to maintain lamp illumination at a lower level, but not enough to start it. FWIW, that's enough to keep two On.

 

[wwhitney]

Apply L-N=124.7

. . .

Open Ssr=31.9 Lamp=103.8 (On, dim)

My question is where did the increased 11 volts come from ? (No, not a neutral issue)

So if the meters are accurate and the discrepancy is entirely due to phase angle, then we can use the law of cosines to determine the phase angle theta between the Voltage across the open SSR and across the Lamp:

|A+B|² = |A|² + |B|² + 2 |A| |B| cos theta
124.7² = 31.9² + 103.8² + 2 * 31.9 * 103.8 * cos theta
cos theta = 0.567
theta = 55.4 degrees

That means that the two impedances have phase angles that differ by 55.4 degrees. I.e. if one were resistive, the other would be have a leading or lagging power factor of 0.567.

Cheers, Wayne

 

[ptonsparky]

So if the meters are accurate and the discrepancy is entirely due to phase angle, then we can use the law of cosines to determine the phase angle theta between the Voltage across the open SSR and across the Lamp:

|A+B|² = |A|² + |B|² + 2 |A| |B| cos theta
124.7² = 31.9² + 103.8² + 2 * 31.9 * 103.8 * cos theta
cos theta = 0.567
theta = 55.4 degrees

That means that the two impedances have phase angles that differ by 55.4 degrees. I.e. if one were resistive, the other would be have a leading or lagging power factor of 0.567.

Cheers, Wayne

 

[ptonsparky]

FWIW the leakage current of my SSR lights three strands of LED Xmas lights. Dimly but very obvious in a well lit shop.