Hello,
I am hoping to get some advice on connecting my Filtrabox Compact fume extractor to my new OMTech 60W laser via the laser interface port on the fume extractor.
I am planning on using the STATUS output on the Ruida controller to activate the extractor but want to check to see how to properly wire things up to make sure I don’t fry something.
The extractor has a DB9 connector and states pins 4 and 5 need to get connected by the controller to activate/deactivate the extractor. It also states not to exceed 200mA on the input to the extractor.
That suggests they want a relay / switch between those two pins, rather than an external current source / sink, so the Ruida’s output transistor isn’t appropriate.
Driving a small relay with the Ruida output and connecting the normally open contacts (typically Common and NO) to the extractor pins should do the trick.
The outputs are all active-low transistors, so you connect the relay coil between the +24 V supply and the controller output pin. When the output is active, current flows from the supply through the relay to the controller pin and thus to common.
You should add a flyback diode across the relay, unless the one you pick already has an internal diode. Some controllers include a similar diode at their output transistor, so check the doc for whatever controller you have.
All those relays are grossly overqualified for the job, but they’re also big enough to work with and that’s more important.
As a quick sanity check, make sure the fan turns on when you short those two pins, perhaps by the crude expedient of just jamming a wire into the holes.
Ok, I think I am good to go but just to make sure I understand correctly, I am planning on running 24V + to terminal 8, STATUS to terminal 7, pin 4 of the DB9 connector to terminal 5, and pin 5 of the DB9 connector to terminal 3. Sounds good?
I guess, I’m lost… I don’t know what those pins represent. Which is negative? I always cringe when I hear about mechanical relays…
Generally with an input, you are driving something like an optical isolator or a coil for mechanical responses… The device should only draw what it needs at it’s operational voltage… hopefully 24V, but maybe it’s 5V?
Your mechanical relay can’t control the current any more than any other device… so I don’t know where the 200mA comes from…
I’d stick a meter across it and measure the current draw…
The internal diode is an LED that will turn on when the relay is active, but it does not act as a catch / suppressor / flyback diode.
You should add a power diode, like the 1N4007 mentioned elsewhere, connected the other way across the input terminals: cathode = bar end to pin 8, other end to pin 7.
At the moment when the relay opens, the diode carries the full coil current, so it must be rated for at least that much.
These little relays tend to be on the order of a few hundred milliamps, so a 1N914 “signal” diode will work just fine. The 1N400x “power” diode series have fatter leads that IMO are easier to work with and less likely to break off, but it’s mostly a matter of using whatever you happen to have on hand.
I thought so also, then I spoke to one of the head EE people where I worked and he explained that the current is much less than you’d expect, it only lasts 10’s of milliseconds.
He advised I use a small signal diode… He computed the current from the highest voltage spike… can’t create or destroy energy…
He showed me on his scope, so I believe him… What you find in there is the lowest cost diode available… That’s what convinced me with your argument.
Been meaning to advise you that when I changed out the tube, I tried the 1kHz pwm at 50% and the transition pulse wasn’t present… I guess it also has to do with the gas?
Probably because everybody thinks the current jumps to a crazy-high value along with the voltage when the transistor / switch / relay turns off. For an inductor, however, the current is continuous, so it’s basically constant during the voltage spike.
This is a 24 V blower fed by a simple MOSFET switch:
The top trace shows the drain voltage at 50 V/div with a 120 V spike when the MOSFET shuts off and the drain voltage rises. The lower trace is the motor current at 200 mA/div, which simply drops from 250 mA to zero without any fuss.
The catch diode just shunts the current from the opened switch into the coil, so it can circulate as the coil resistance dissipates the energy as the magnetic field collapses. Might take a few tens of milliseconds, but it’s not very exciting.
Even small signal diodes have a maximum current rating around an amp, so they’ll work fine for the relatively low currents in small relay coils.
This is a good watch… the reason you have such a high voltage is the current has no where to go, the voltage then goes up…
My understanding of basic laws of physics, I think argue with that. The wattage is the same, one goes up, the other goes down… if it didn’t work that way, there would be no electronics… or a negative resistance device, which a coil is not. Or you could make a perpetual device… like they show on youtube
The key value is the energy stored in the inductor’s magnetic field, which is created / set by the current flow; they’re inextricably linked. That energy will be gradually dissipated in the circuit as the current continues to flow through the diode, with the power set by the product of the current and the voltage at the inductor terminals.
As the switch turns off / relay opens / transistor shuts off, the effective resistance rises. Because the stored energy remains constant and is defined by the current in the inductor, the voltage across the switch rises until something breaks down to continue carrying the current.
The diode is normally held off, because it’s normally reverse biased, but it begins conducting as the voltage on the anode end rises above the supply voltage when the switch opens. The voltage across the coil gets clamped at the diode’s forward drop, so it’s maybe half a volt over the supply, and nothing exciting happens.
Without the diode, at about 16 minutes into the video, you’re seeing the switch / transistor voltage hit 700 V while the current is dropping, but it continues to flow through the transistor’s body as the voltage decreases.
Devoting some time to killing transistors while watching scope traces is definitely educational, because the interaction between the voltage and the current depends on a bunch of factors we normally ignore while thinking switches just go click.
:
