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I need to switch an AC fan by a DAQ board's analog output.

Here is the setup in my mind for the moment:

enter image description here

The DAC of the data-acquisiton board can output 0 to 10V DC and can deliver max 5mA current. When the relay is ON the mains will power an AC fan. Here is the fan: http://uk.rs-online.com/web/p/axial-fans/2781543/ According to the data sheet I assume the fan will drive less than 1A. So I'm planning to use the following solid state relay: http://uk.rs-online.com/web/p/solid-state-relays/0346895/ since the data-sheet says it can handle upto 10A.

I have couple of questions here:

1-) In the relay's data-sheet, there is a section on the input current range which is not fixed. So I want to be sure if DAC can supply enough current for this relay. Do I need a voltage follower between the DAC and the DC input of the relay?

2-) If I can manage to turn on and off the relay, would the switching action cause a sudden peak current and damage the fan motor? Is there a difference between turning on the power of the fan via the relay and via a manual switch? I think motor works quite ok with the manual switch. I'm just wondering if the relay's switching action might cause some harmful inrush currents for this fan motor.

3-) If I can achieve to turn on and off the AC fan succesfully by using the relay and DAC, is there a possibilty to sudden stop this motor as well? I mean I have seen a VFD AC motor speed controller which is controlling a 3-phase AC motor. There is a switch which powers on of and breaks that motor. This manual switch when switched for the "break" function motor very quickly stops rotating. Maybe it is the function of the VFD controller or the 3-phase AC motor itself and may not be applicable in my case. The reason I'm asking about this, when the fan is powered off it keeps rotating for long time until it stops.

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  • \$\begingroup\$ Why are you using an analog output to drive the relay? The relay will be either off or on, never part-way. Relays are normally controlled by a digital output. \$\endgroup\$ – Peter Bennett Apr 28 '16 at 18:57
  • \$\begingroup\$ There is no digital output so I will apply lets say constant 5V or 0 V by the analog output of the DAQ. \$\endgroup\$ – user16307 Apr 28 '16 at 18:58
  • \$\begingroup\$ "... the "break" function motor very quickly stops rotating." I think you mean brake function. \$\endgroup\$ – Transistor Apr 28 '16 at 19:40
  • \$\begingroup\$ yes exactly. How does it work? \$\endgroup\$ – user16307 Apr 28 '16 at 19:45
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  1. The max input current for this SSR is 12 mA, so you have to use a current amplifier. It can be for example any general purpose NPN BJT connected with a common collector. I suggest using 10 V power supply for this BJT and using 0 V / 7 V DCA output for switching. Do not go to +10 V! This may damage your DAC.

Alternatively, you can use a general purpose N channel MOSFET with "logic level" switching. Then you have to drive it by 0 V / 10 V from ADC.

  1. No, SSR switch operates "smoother" for its load then mechanical switch. The latest one causes several on/off transients (ringing) while SSR switches on only once. So SSR can not damage a fan.

  2. No, you can not force stop the fan by this circuit. You need a special motor and a special switching circuit with "break" resistor (or more complicated means to return energy) to stop rotation quickly.

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  • \$\begingroup\$ Ok it says: Maximum Input Current [12 mA]. But I don't get it. Is this input current related to the input impedance and input voltage? Or does that mean if 32V is applied 12mA. Is there a way to related this current to the applied input voltage and input impedance? \$\endgroup\$ – user16307 Apr 28 '16 at 19:55
  • \$\begingroup\$ OK, you can try on your own risk. Datasheets specify maximum currents (may be, under the worst conditions), and engineers design circuits to operate under these worst conditions. There is absolutely no way to guess this current. You can measure it. However, even if it never happens to exceed 5 mA in your measurements, it does not mean that it would not exceed 5 or 10 mA under some other conditions. \$\endgroup\$ – Master Apr 28 '16 at 20:24
  • \$\begingroup\$ Alright I though the input current = Input voltage / Input impedance. But u say it is random. Thanks \$\endgroup\$ – user16307 Apr 28 '16 at 20:44
  • \$\begingroup\$ The relay internally limits the input current to the range 7-12 ma. If you externally limit the current to less than 7 ma it may not turn on. Your driver circuit should be designed to be capable of 12 ma. The internal impedance is variable and undefined. crydom.com/en/products/catalog/s_1.pdf \$\endgroup\$ – Charles Cowie Apr 28 '16 at 22:00

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