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    ptheisen
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Smart Module2 dimming output

Hello . . .

we have a strange problem that maybe someone has seen before

we are using a SM2 in a FOH position for 4 S4 575s ... when two of the channels are on (@ FULL) they seem fine .. when we add the other two channels we notice that the output of the first two channels dims a bit . . .  not much .. but enough to notice

mathematically we are within the wattage ratings for the SM2 and nothing is tripping . .

they are getting power from a CC20 card and not a dimmer parked or set to switched or non dim . .

any ideas?

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  • Hello, 

    I think what you are experiencing here is voltage drop due to an increase in current. U=IR

    For example, if we have a U of 110v at the Sensor module, and a 100ft (30m) run of AWG14 (2mm_2) to the SM2.

    AWG14 has a resistance of 2.525Ω/1000ft. (2.525/1000)*100=0.2525Ω for a 100ft run.

    Combine this with an assumed internal resistance of 0.2Ω with in the SM2, we get R=0.4525Ω

    We already know the wattage of the fixtures, and as they are purely resistive loads P=IU or I=P/U.

    With two S4 we have I=(575x2)/110 = 10.455A

    With four S4 we have I=(575x4)/110 = 20.91A

    So combining this all together, the voltage drop with two S4 is:

    U=IR=10.455x0.4525=4.731v or roughly 94.8% of full

    with four S4:

    U=IR=20.91x0.4525=9.462v or roughly 89.6% of full

    In reality there are many more factors to take in to account, but this does demonstrate the effect of current on a single circuit to the perceived light output.

    (N.B. The above values are for example only, and I normally work in metric!!)

    Regards,

    Marcus

Reply

  • Hello, 

    I think what you are experiencing here is voltage drop due to an increase in current. U=IR

    For example, if we have a U of 110v at the Sensor module, and a 100ft (30m) run of AWG14 (2mm_2) to the SM2.

    AWG14 has a resistance of 2.525Ω/1000ft. (2.525/1000)*100=0.2525Ω for a 100ft run.

    Combine this with an assumed internal resistance of 0.2Ω with in the SM2, we get R=0.4525Ω

    We already know the wattage of the fixtures, and as they are purely resistive loads P=IU or I=P/U.

    With two S4 we have I=(575x2)/110 = 10.455A

    With four S4 we have I=(575x4)/110 = 20.91A

    So combining this all together, the voltage drop with two S4 is:

    U=IR=10.455x0.4525=4.731v or roughly 94.8% of full

    with four S4:

    U=IR=20.91x0.4525=9.462v or roughly 89.6% of full

    In reality there are many more factors to take in to account, but this does demonstrate the effect of current on a single circuit to the perceived light output.

    (N.B. The above values are for example only, and I normally work in metric!!)

    Regards,

    Marcus

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