RE: [CAD_CAM_EDM_DRO] Re: SSR rating ?

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

-----Original Message-----
From: CAD_CAM_EDM_DRO@yahoogroups.com
[mailto:CAD_CAM_EDM_DRO@yahoogroups.com] On Behalf Of turbulatordude

--- In CAD_CAM_EDM_DRO@yahoogroups.com, Charles Anderson
<charlesanderson@...> wrote:

>
> If the device is rated at 230 v, will it work with a lesser voltage

(yes

> it is a noob question).

being a noob now and again, it seems that power is not always so simple.

if it is rated at 10 amps at 230 volts, then the unit could be rated
at 2,300 watts. At 120 volts, that could mean 20 amps if Ohms law
were the single factor.