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I have a fan motor which is a permanent split phase single phase capacitor motor: http://uk.rs-online.com/web/p/axial-fans/2781543/

I currently control this fan's speed with the following controller: http://uk.rs-online.com/web/p/fan-speed-controllers/6685345/?origin=PSF_438361|acc

As you can see the speed-controller is manually controlled by a knob(by hand).

I need to control this fan's speed by a computer software which uses a DAQ board. The DAQ board outputs voltage in 0-10V DC range. So I need to vary the fan speed by a DC input.

Here is the full photo of the controller circuit(They don't share schematics): enter image description here

As you can see this small controller uses W06 silicon bridge rectifier. The rest of the circuit consists of a diac, resistors, capacitors, a 220K poti, a toroid inductor, a fuse and an adjustable component(next to +MIN SPEED and I couldn't figure out what it is). UZ and U goes to the fan. N, L and PE are for the AC mains input.

There is a triac-like component in the copper side. Here is the other side of the circuit:

enter image description here

First thing confused me was that there is only bridge rectifier without a triac and I'm wondering how this works without any PWM signal? And what happens when the potentiometer is turned? Here is a short video when I turn the poti(I can only show the upper part of the voltage since the scope cannot plot all): https://sendvid.com/5xr3yn4l The scope shows the voltage between motor's terminals(UZ and U). As you can see the freq. remains constant but the waveform changes and the RMS value of the voltages also changes(I checked with a voltmeter).

I was planning to interact with this circuit for my aim(to control it with a DC voltage input), but it seems not an easy task.

Either I need to build a new circuit or buy another controller. I couldn't find any speed-controller in the market for my case where one can control this AC fan with a DC input in 0-10V range. I think I need something like a dimmer which is controlled by a DC input and can supply this fan motor.

I would be glad to hear some circuit suggestions or any such controller in the market. If I need to build one, do I really need a uC for this purpose?

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  • \$\begingroup\$ even if the fan moves , which I doubt, I predict the fan will burn out. \$\endgroup\$ – Tim Spriggs Apr 21 '16 at 17:54
  • \$\begingroup\$ @TimSpriggs What do you mean? In which case it will burn? \$\endgroup\$ – user16307 Apr 21 '16 at 17:55
  • \$\begingroup\$ because it wasnt designed for that kind of voltage (low and direct current). It will be a slow burn, but it will burn. \$\endgroup\$ – Tim Spriggs Apr 21 '16 at 17:57
  • \$\begingroup\$ First thing: is the pot working as a 3-terminal potentiometer or as a 2-terminal variable resistor. If it's the latter you may be able to replace it with a light dependent resistor (LDR) and use an LED to generate the (variable brightness) light. Major advantage with this is opto-isolation between your control circuit and your mains circuit. \$\endgroup\$ – Transistor Apr 21 '16 at 17:57
  • \$\begingroup\$ @TimSpriggs I think you didnt understand my question. I'm talking about "controlling" the AC by a 0-10V DC, not supplying the motor with DC. \$\endgroup\$ – user16307 Apr 21 '16 at 18:00
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From our discussion in the OP comments we have established that the potentiometer is wired as a two-terminal variable resistor rather than a three-terminal potentiometer. This gives the possibility of replacing it with an LDR (light dependent resistor).

The first LDR I found on a web search is the NORP12 / NSL19-M51 available from RS.

Table 1. Basic specification of NORP12 / NSL19-M51 LDR.

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Replace the potentiometer with the circuit on the left.

Try the circuit shown in figure 1.

enter image description here

Figure 2. Spectral sensitivity.

enter image description here

Figure 3. 550 nm on visible light spectrum.

It looks like a yellow or green LED would be most suitable for the LDR.

Safety

The LED / LDR will be the opto-isolation between your micro and the fan controller. The LDR leads should be treated as live. Remove the pot, solder in some leads to the LDR and mount it securely slightly off the board. Mount the LED in close proximity and shield the combination from stray light. An opaque tube such as a pen or marker might suffice. Make sure that the control wiring will never come in contact with the LDR or PCB.

Test with a 9 V battery and a variety of resistors to figure out what LED current gives you the minimum and maximum speed you require.

Control

Your DAC can output 0 - 10 V. I presume that you have full control over the output so that if, for example, you can get the full range of speed control with a particular LED - LDR optical coupling (positioning) in the range of 2 to 7.3 V you won't have a problem implementing that scaling in your software. In that case minimum speed (0%) might be 2 V out and maximum speed (100%) might be 7.3 V.

On second thoughts you can minimise risk of damage to the controller by turning the pot to maximum resistance and adding your test resistors or LDR in parallel with the pot. When the LED-LDR goes completely dark it will have a 1 MΩ resistance which will make hardly any difference to the pot. You could also use the pot as an override should the DAC system fail.

schematic

simulate this circuit

Figure 4. 5 mA max current directly from the DAC. Figure 5. Emitter follower gives 20 mA (or more if you decrease R2). The emitter will be 0.7 V below the DAC output due to base-emitter voltage drop. Multiple LEDs can be added in series to increase light output, if required.

See Figures 4 and 5 for ideas on how to drive the LED. Note that neither will turn on until about 1.5 V across the LED.

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  • \$\begingroup\$ I am impressed with the amount of detail in your answer. \$\endgroup\$ – Mark Apr 21 '16 at 22:22
  • \$\begingroup\$ @user16307 - I think this will work. Keep in mind that the LED may need at least 20 ma, and the output of the DAC may be unable to supply that much current. If you need and amp, wire it for a current output to make the LED's output more stable. \$\endgroup\$ – Mark Apr 21 '16 at 22:26
  • \$\begingroup\$ @user16307: "But the resistor varies from 0 to 160K . Do u (sic) think photo resistor still work?" (from your comment to my suggestion of this scheme on the OP). The LDR goes down to 400 Ω at high light levels. Power off the circuit, adjust the pot to 1 kΩ, power up and see how if the speed is what you require. You could put two LDRs in parallel to get the resistance down to 200 Ω. They're cheap. \$\endgroup\$ – Transistor Apr 21 '16 at 23:31
  • \$\begingroup\$ @transistor Many thanks for your great helpful suggestions, I will try it in some days and give feedback as soon as possible. But I'm kind of confused at the last point. I need resistor can go up to around 160K. But in dark I think these resistors go to 1MegaOhm. How can I adjust the upper limit (when LED is dark) to 160K and lower limit to 200ohm. I would be glad if you can draw what you mean as you did in your answer. Thanks.. \$\endgroup\$ – user16307 Apr 22 '16 at 0:20
  • \$\begingroup\$ @Mark DAQ does not have a current output. It has 0-10V analog voltage output. This is the DAQ board I will use: mccdaq.com/pci-data-acquisition/PCI-DAS6036.aspx Do you think it would be enough current for this application. If not what kind of interface would you recommend for current? \$\endgroup\$ – user16307 Apr 22 '16 at 0:59

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