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Okay. I'm completely at a loss here. I was trying to design a switching circuit for a 3A load (Solenoid) using a photo-interrupter. I got to about here: enter image description here

But I have no idea what to look for in my parts. I plan to use these as the current limiters, and I have to use this as the photo-interrupter. I was thinking I could set one of the current limiters to the specifications of the photo-interrupter, but I don't know if it would work. (I really hope it would.)

As for the transistor, I was thinking of finding an pnp switching transistor that can carry 3A without overheating, but I've never been transistor shopping before, so I don't know what to look for in the specs. Also, I don't know what to set the current for the... Transistor collector? I was thinking of looking for something that switches at the same input current for the phototransistor, so when current stops flowing through the phototransistor, it goes through the collector.

I plan to have a coil of wire as the load, with a total resistance of about 1.213Ω.

Can anyone tell me if my plan will work? Also, if so, where to look for parts? I tried Ebay and Amazon, but neither have any interactive spec chart for Transistors.

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  • \$\begingroup\$ Is it a coilgun? \$\endgroup\$
    – venny
    Aug 22, 2014 at 21:48
  • \$\begingroup\$ Gasp Yes, and no. Yes, I am building an accelerator(well, this one circles back, so) and no. I am not building a coilgun/Gauss rifle etc. What I am building is a mock-up version of the accelerator at CERN. My design accelerates ball bearings and smashes them together. (A major difference from CERN being, I am not smashing protons, and my accelerator ring is 4' across) \$\endgroup\$
    – CoilKid
    Aug 22, 2014 at 21:52
  • \$\begingroup\$ I think the name "coilkid" was a clue \$\endgroup\$
    – Andy aka
    Aug 22, 2014 at 21:53
  • \$\begingroup\$ Lol, yup. This is for a group of people I help with physics, and also I get the added benefit of being able to say I have an accelerator in my basement. \$\endgroup\$
    – CoilKid
    Aug 22, 2014 at 21:54
  • \$\begingroup\$ Digikey has my favorite search engine for parts. (no comment about the rest.) \$\endgroup\$ Aug 22, 2014 at 23:50

3 Answers 3

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It's not a great circuit. The PNP transistor would not switch on fully so I'd use an NPN with the emitter grounded - this way, current into the base turns the transistor on to about 0.1 volt across collector and emitter and wastes less power.

Having said all that I'd use an N channel MOSFET that maybe switches on to a few millivolts.

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  • \$\begingroup\$ If I were to use an N channel MOSFET, would it work well? \$\endgroup\$
    – CoilKid
    Aug 22, 2014 at 21:55
  • \$\begingroup\$ It'd work a whole lot better than the PNP but your question is very subjective because it might work so well it could fry your 1.213 ohm solenoid from an 11.4 volt supply if the current limiter didn't work. \$\endgroup\$
    – Andy aka
    Aug 22, 2014 at 21:58
  • \$\begingroup\$ Also, where could I find this part? Like I said Amazon and Ebay weren't very helpful. \$\endgroup\$
    – CoilKid
    Aug 22, 2014 at 21:58
  • \$\begingroup\$ Well the limiter should work. I asked several questions here on EE.SE. The stuff I came up with would have wasted several W of power, but the current limiter I linked, uses switching at several mH to control current/voltage. \$\endgroup\$
    – CoilKid
    Aug 22, 2014 at 22:00
  • \$\begingroup\$ A switching current limiter will work pretty good. Also don't forget a reverse biased diode across the coil to kill off the energy spike when the transistor switches off. Requesting recommendations on actual components are a bit frowned upon. \$\endgroup\$
    – Andy aka
    Aug 22, 2014 at 22:03
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(This should perhaps just be a comment, as it only provides a partial answer, but I wanted the drawing...)

You don't need one of the LED power supply modules you lined to for the photo-interrupter.

Electrically, the photo-interrupter is two independent parts - the IR LED, and the photo-transistor.

You should treat the IR LED as you would any other LED - just use a series resistor to limit the current. Unfortunately, the data for the opto-interrupter doesn't give a forward voltage for the LED, but a 4N35 opt-isolator has a typical forward voltage of 1.3 volts, so I'll use that. The maximum LED current for the photo_interrupter is 50 mA, so I'll use 40 mA (I don't like using maximum values...). since you have an 11.4V supply, the series resistor will be (11.4 - 1.3)/0.04 = 252 ohms - next higher standard value is 270 ohms, which will dissipate .4 watts, so you need a 1/2 watt resistor, or two 130 ohm 1/2 watt in series.

What you do with the photo-transistor depends on what follows it to switch the actual coil current. I've shown two possible connections for the photo-transistor.

schematic

simulate this circuit – Schematic created using CircuitLab

Since the photo-trasistor will conduct when it is illuminated by the LED, OUT1 will be normally low, and will go high when the path from the LED is blocked.

OUT2 will normally be high, but will go low when the light path is blocked.

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  • \$\begingroup\$ Thank you very much. I really appreciate the assistance. I have a design in mind for my circuit, and I believe all that is left is for me to check if it works. \$\endgroup\$
    – CoilKid
    Aug 25, 2014 at 3:40
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Another option might be a phototransistor going into a comparator driving a MOSFET (sourcing the solenoid).

With this solution, you could adjust the reference on the comparator to change the sensitivity.

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