I have just finished build my table saw and now planning for integrated vacuum system for dust collection.

I've designed the circuit to detect if saw motor is switched on and automatically switch on the vacuum pump. Current sensing resistor Rs is on neutral line of the saw. One side of this resistor is floated on DC common as shown in the schematic below.

The voltage drop at the other end of Rs is used to turn on NPN transistor which drives relay. The contacts of the relay will act as the switch to turn on the vacuum pump.

Care have been taken to clamp Vbe to safe values using 1N4004 diodes.

enter image description here

Before I build this circuit and test it, I need some expert advice for Rs. My saw motor is 1500 watts universal motor. It has no soft start feature, meaning it may consume very high starting current which Rs will have to bear.

As such, what should be the right value for Rs in terms of resistance and power rating?

PS; Based on comments on this question, i would like to clarify the reason for this current sensing auto switch is to accommodate few more power tools so that when any of those dust making tool is switched on will start the vacuum unit.

  • \$\begingroup\$ Why not just trigger a relay from the switched supply to the table saw, so when you switch the saw on the relays goes on and so does the vac... \$\endgroup\$
    – Solar Mike
    Oct 30 '18 at 14:57
  • \$\begingroup\$ As @SolarMike suggested, just use a relay. Be sure to get a relay specifically meant for AC. \$\endgroup\$
    – vini_i
    Oct 30 '18 at 15:05
  • 1
    \$\begingroup\$ rockler.com/i-socket-110m-tool-and-vacuum-switch \$\endgroup\$
    – mkeith
    Oct 30 '18 at 15:41
  • 2
    \$\begingroup\$ The problem seems to be that the vacuum should operate only when the saw is running. As the saw (probably) has a local switch to control it, the answer is to sense the current drawn by the saw. I would use a current transformer for this, as it gives total isolation from the saw power. When I did this for an electric shower and extractor fan, I used a zero-crossing triac module to control the fan. \$\endgroup\$
    – henros
    Oct 30 '18 at 16:36
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    \$\begingroup\$ There are three ways I can think of to sense current. First is with a current sense resistor (which is not a good idea in this case in my opinion). Second is with a Hall effect current sensor such as ACS711 (or similar). Third is with a current transformer. The isolation provided with the latter two seems very desirable for a line-powered device. Also, they have negligible series resistance in this application. \$\endgroup\$
    – mkeith
    Oct 30 '18 at 18:46

Some small improvements to your circuit. It can be made to work but do take suitable care with construction and casing design to keep you safe. Plastic case is safer.

  • Your live and neutral connections are swapped on the table saw socket.
  • Add a capacitor across your two diodes to filter the 50Hz.
  • Add a resistor from the capacitor node to the transistor base so it discharges slower than 50ms to hold the relay.
  • Flip the single diode from ground to your 10k resistor so it is in series.
  • Add 4 diodes across the sense resistor to clamp the voltage, the diodes in a properly rated bridge rectifier can be used here by shorting the + and - and connecting the AC terminals across your sense resistor.
  • Add a fuse or circuit breaker to your control box, perhaps 2

I would consider a current shunt as the sensing device perhaps connected via a bridge and capacitor to a slid state relay.

  • \$\begingroup\$ Totally acceptable recommendations. After considering every aspects, i finally ditched the 'current sense ' idea and opted to use LDR mounted flush on saw table very close to feed out area of the blade. When wood pass on the LDR, it blocks ambient light and the signal is used to control the vacuum pump. \$\endgroup\$ Nov 2 '18 at 0:14

I think I understand the intent now.

Sizing the resistor will depend on how much voltage out of it you will need.

For example, if your mains are 120V then a 1500W motor will draw 12.5V RMS. Suppose you want 6V output at full load. Then simply use ohms laws V=I*R (6/12.5) the answer would be 0.48 ohms. The second calculation that would be needed is the wattage. The calculation would be W=I^2*R (12.5^2*0.48). This gives you 75 watts. This is like a large light bulb and will be a hefty resistor.

The drawback is that this is with the motor at its absolute max consumption. If the motor is running at 50% output the voltage will also only be 50%.

  • \$\begingroup\$ The sense resistor must be as small as possible so that it's presence is almost transparent to the load. As such, the sense circuit which reads from the Rs logically will have to be sensitive to small voltage drop across it. In the circuit above this voltage is not more than 1 volt. \$\endgroup\$ Nov 2 '18 at 0:04

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