1
\$\begingroup\$

how do I design a fan fail circuit (open circuit or fan not moving etc) for a 115VAC two wire fan and illuminate a 115V Panel indicator?

\$\endgroup\$

3 Answers 3

2
\$\begingroup\$

Depends, there are many ways, each good in is own way.

I'd probably prefer a differential pressure detection or airflow detection scheme rather than looking at the fan blades or fan motor current - both of which are valid schemes in their own way.

My preference is to look as far down the chain towards "why do I care if the fan is running" as possible and measure THAT parameter, rather than one further up the chain. So I want to see air moving, and I don't care why it moves, as long as it moves enough, and in the right direction.

As it happens, simple (KISS is my mantra) paddle type airflow switches are dirt common.

Paddle flow switch (image from web, not associated with nor endorsing source of image which was www.dwyer-inst.com)

You can either find one that will light your light directly, or one that will allow a relay to light your light.

If KISS is not your mantra there are a multitude of methods to overcomplicate it and get a micro-controller involved, such as building a hot wire anemometer or a spinning anemometer. In a few situations those will actually be better, not just more complex. In most situations, they will just be more complex.

\$\endgroup\$
2
  • \$\begingroup\$ +1 Even if the fan blades fall off this will work. \$\endgroup\$ Commented Feb 15, 2016 at 4:48
  • \$\begingroup\$ I like this idea, KISS is my mantra, just not sure I can find one small enough for the application. Fans are 60mm x 60mm in a 4U chassis. Think this is a great idea but probably more for an HVAC solution. I had a quick look but cannot find anything small enough, if you know of any "small / tiny" switches like the above then that would be a great solution for me. \$\endgroup\$
    – Douglas
    Commented Feb 15, 2016 at 22:12
0
\$\begingroup\$

Two ideas:

  1. Use a current transformer to get current through the fan. Rectify/average so you have average current. Pass that to a window comparator; if the average drops below a threshold or rises above a threshold, you could assume something not good has happened.

  2. Optical sensor monitoring the fan blades. Convert the pulse train to a voltage with an astable multivibrator, and into the window comparator again. Of course, if you want to make things really easy, use an MCU and it can handle the pulse train directly (and even inform you of RPM).

\$\endgroup\$
2
  • \$\begingroup\$ do you have any more info on the window comparators? ie where to get etc? easy sounds good to me, I would like to keep it simple and reliable, those are the keys for me. \$\endgroup\$
    – Douglas
    Commented Feb 15, 2016 at 22:15
  • \$\begingroup\$ You can buy window comparator ICs from various manufacturers, or you can use two bog-standard comparators and an AND gate (or something of that sort). \$\endgroup\$
    – uint128_t
    Commented Feb 15, 2016 at 22:22
0
\$\begingroup\$

You can get two devices that behave as thermisters that are the same .These could be diodes or KTY88 or whatever is convenient .Connect them in series and put some current through them such that they self heat like being warm to touch .Because these devices are small and therefore have high thermal impedences you wont be needing high currents .The expected voltages across the two series connected devices will be the same because thier heating current is the same.Place a small semicircular shield around one of the thermisters and the voltages will barely change .Now put your thermisters in the flow of your fan and you will see a very significant voltage imbalance occurring because the unshielded device is not being fan cooled and the shielded device experiences the cold blast of the fan .If you get this right you will find it easy to run a simple compariter like a LM393 or use a spare microprocessor input to implement a reliable GO/NOGO lamp .

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.