1
\$\begingroup\$

I have a Dewalt Metal cutting abrasive saw that tends to open a breaker on starting up. It draws 75A on startup tripping the magnetic part of the breaker and opening it.

This circuit is only to stop the breaker from opening during the inrush period when the saw blade is totally still. Once it is rotating inrush should be manageable and not as close to the breaker-open current.

The mechanism would essentially be trigger saw -> wait a second for rotation -> close light switch and get full power.

Are there any drawbacks to this approach? I don't think it could damage the saw, but I could be wrong.

I use the saw occasionally and wouldn't be doing a huge amount of repetitive cuts, so I don't see ohmic heating being much of an issue.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Are you sure that it's got an induction motor? Some power tools use series-wound brushed motors. Brush mounts (screw-like things at the back of the motor) are a dead givaway, as are brushes in the replacement parts list or exploded diagram (if they still do that sort of thing). \$\endgroup\$ – TimWescott Mar 6 at 1:21
  • \$\begingroup\$ @TimWescott Just checked the manual, it does have brushes. \$\endgroup\$ – Sarah Szabo Mar 6 at 1:28
  • \$\begingroup\$ I suppose that if you don't close the switch the resistor might get hot enough to start a fire unless it's enclosed somehow. \$\endgroup\$ – Spehro Pefhany Mar 7 at 20:06
  • \$\begingroup\$ @SpehroPefhany The resistor that I ordered from DigiKey has an aluminum housing and is rated for the current that I will be drawing. It's a chassis mount style which I will mount on a heat sink. So no worries here :) \$\endgroup\$ – Sarah Szabo Mar 7 at 21:45
3
\$\begingroup\$

Given that it is a brushed motor, your idea should work. I would be tempted to put a light bulb in parallel with the resistor, to give you a visual indication that the saw has spun up.

(Note that the fact that it's a universal motor means that it's going to pull a lot of current on startup, or when it's stalled -- a too-fast circuit breaker would definitely pop).

\$\endgroup\$
  • \$\begingroup\$ I agree, but 3 ohms would be closer to what you need than 30. \$\endgroup\$ – Charles Cowie Mar 6 at 3:43
  • \$\begingroup\$ I agree with @CharlesCowie. Considering the current flow, 3 ohms at 300 watts would choke back the inrush, but should allow enough current for the motor to have torque, or it will stall. \$\endgroup\$ – Sparky256 Mar 6 at 5:03
  • \$\begingroup\$ With 75A surge at 120V or 9000W , on a 15A breaker you need at least a 300W to 1kW Halogen light in series.for startup and then a run bypass timer relay. I suggest a 2kJ industrial ICL \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 6 at 7:27
  • \$\begingroup\$ @CharlesCowie I would agree, but a resistor that could take 120V/3 ohms = 40A -> 4800W (Instant) -> 1800W (Saw blade rotating) would be rather expensive. The specs I could get from DigiKey suggest that their resistors be +- 5% of the max value. I don't know if they can stand that current in a short < 3 second burst to get the saw started. \$\endgroup\$ – Sarah Szabo Mar 7 at 21:49
  • \$\begingroup\$ The motor does limit the current somewhat, so the peak current may be more like 30A, but the peak power will still be about 2700 W. A 1000 W resistor may be required. It is difficult to determine how much power a resistor can withstand for a short time. Different types of resistors have widely differing characteristics for short pulses of energy. \$\endgroup\$ – Charles Cowie Mar 7 at 22:39
1
\$\begingroup\$

If it trips the breaker 1 out of 5 times then this indicates that a zero-crossing Triac could prevent this. But that is not a cheap solution. A clever solution might be a ZCS triggered DC relay that switches exactly at 1 cycle later. But that may not reliable.

Proper method for choosing an Inrush Current Limiter (ICL)

  1. \$R_{25'C}[Ω]=V_{pk}/I_{surge}\$ for Vpk=1.41*Vrms and \$I_{surge}\$ = max. allowable surge
  2. \$E[Joules]=V_{rms}*\int{I_{rms}(t)}dt\$ , Energy to startup
  3. \$Iss\$ [Amps] = steady-state, Rated current on tool label.

Then go to www.Ametherm.com and consider MM35-DIN Industrial series

\$\endgroup\$
0
\$\begingroup\$

Just to throw this out there, the symptoms are also those of having too small of an extension cord feeding the saw. The cord is creating a voltage drop, which causes the universal motor on the saw to pull MORE current on start-up and it trips the breaker. Before messing with this by redesigning the saw circuit, go buy a heavier gauge extension cord and see if it fixes the problem.

\$\endgroup\$
  • \$\begingroup\$ I had it directly plugged into the wall as a test of voltage drop. This didn't change the outcome, although since some shops have very long runs of wire, this might be a good suggestion for other people. \$\endgroup\$ – Sarah Szabo Mar 7 at 21:43
  • \$\begingroup\$ Well that eliminated that possibility then... \$\endgroup\$ – JRaef Mar 7 at 21:50
  • \$\begingroup\$ 75A, if looked at as "normal" inrush of 600% of FLA, puts this at 12.5A, which SHOULD be good for a standard 15A (or 16A if you are not in North America) circuit breaker. The instantaneous trips on a 15A breaker are probably close to 150A, 75A all by itself should be fine if that's the only thing on the circuit. Is there anything else on that same circuit? If so, can that be turned off? If this is a dedicated circuit, your breaker may be bad. Repeated tripping makes a breaker need less and less to make it trip again. \$\endgroup\$ – JRaef Mar 7 at 21:54
  • \$\begingroup\$ That was what I saw flash across the multimeter once when I measured it, it could have been higher, I didn't have the "hold peak" button pushed in. It very well could have been higher. There is nothing else on the circuit that I know of. \$\endgroup\$ – Sarah Szabo Mar 8 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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