Re: [CAD_CAM_EDM_DRO] Re: SSR rating ?

So will a 230v device work with 110v?

Charles from Oz

Kevin Martin wrote:

>You are confounding the power the unit can control with the power the
>unit can dissipate. Ohm's law has nothing to do with it.
>
>When the switch is "on" and the load is drawing 10 amps, the switch will
>have a small but nonzero voltage across it, maybe 1 volt (just guessing
>here), and the remainder of your supply voltage will be across the load.
>In this case the unit is dissipating 10A x 1V = 10 watts of power, which
>is being turned into heat and must be dissipated into the air (directly
>or through a heat sink).
>
>At the same time, your load is getting 229V (with 230V supply) or 119V
>(with 120V supply) and consuming 2290W or 1190W. Note that some of this
>power is also turning into heat due to losses in the load, but most of
>it will be turned into mechanical energy (assuming the load is a motor,
>of course).
>
>The 10 amp rating for the SSR would be based on 2 things: One is the
>power dissipation of the unit, as mentioned above. The designers of the
>part know what voltage drop there will be (I guessed 1 volt), what
>temperature the unit can take, and choose the current limit for a
>particular heat sink capacity and ambient air temperature.
>The other basis for the 10 amp rating is physical limitations within the
>unit, either due to thermal sinking issues within the case, or nonlinear
>effects in the semiconductors.
>
>Generally this means that no matter what voltage you are switching or
>how much heat sink you use you should not exceed the 10 amp limit. If
>the unit and heat sink are in air hotter than the design air temperature
>you either must reduce this current limit ("derating") or use a better
>heat sink that can dissipate the same Watts of heat despite the warmer
>air.
>
>The voltage limit comes into play when the switch is "off" and the
>entire supply voltage is across it, and is based on the limits of the
>electrical insulation between the terminals and the breakdown voltage of
>the semiconductor. No amount of extra cooling will increase this.
>
>There is likely also a limit on switching rate if it is used for PWM
>control. Each time it switches on or off, there is a very short but
>nonzero time where it must dissipate a substantial fraction, perhaps 1/4
>of the load (around 500 watts in the above example). This peak happens
>at the point when half the supply voltage is across the switch and half
>across the load. Although the switch does not spend very long in this
>transient state, the power it must dissipate during these times can add
>up to a substantial fraction of the total power dissipation if the
>switching rate is too high or if the control signal has rise and fall
>times that are too slow.
>-Kevin Martin
>
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