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I have a circuit that I believe will work, but I was hoping someone could double check?

Removed image to make room for edit. - Original schematic

In my circuit, I use a MOSFET, a photo-interrupter, a CVCC, several resistors, a battery, and an LED.

The load has a resistance about 2.525Ω

Assuming that the CVCC is required, would my circuit work?

Edit: I edited my circuit design, could it work better now?

Image removed for space.

Edit: Edited schematic to reflect changes.

Also, the purpose of this circuit is switching. When the photo-interrupter beam is intact, the load gets no current. When the beam is broken, the load gets current. Or, that's the idea here.

Removed image for space: 3rd update schematic

Edit: Changed FET to suggested. Suggested FET

Removed for space: 4th update schematic

@CopperMaze, Thanks for the suggestion!

Edit: Edited schematic to reflect suggestions, and resolve mix-up.

Removed image for space

Edit: Added freewheel diode, re-added R3 to soften the inductor impact on the MOSFET.

Removed for space

Edit: Changed freewheeling diode to Shottky diode as suggested.

enter image description here

share|improve this question
Explain how it is supposed to work. Step us through what the user would expect to see (and do) when the power source is turned on. You have an opto-isolator that will always be on when power is applied. There's no actual isolation in the circuit, so why the isolator? –  Dan Laks Aug 25 at 4:16
@DanLaks, notice the optocoupler is labelled as "opto-interruptor" –  The Photon Aug 25 at 4:41
I assume isolator is sort of a switch when there is any interference of any object blocking the infrared light the circuit will be open/isolated. Is that true? –  dr3patel Aug 25 at 4:46
You've made some changes since we talked about this in chat. In chat you talked about a solenoid as the load, but here you're showing a resistive load. In chat you talked about a 12 V supply, here you're showing 2 V. Could you make sure the schematic actually shows what you're trying to do? –  The Photon Aug 25 at 4:51
@ThePhoton, ah, thanks for pointing that out. I made an assumption when looking at the circuit without reading the description carefully. –  Dan Laks Aug 25 at 4:59

4 Answers 4

It won't work, your MOSFET has a VGS(Th) of 3 V. Your supply is only 2 V.

What is this circuit for; if we know we can help you better?

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I'm sorry to say that your circuit will not work:

  • The MOSFET RDS(on) (2.8 Ω) is too big compared to your load resistance (2.5 Ω). If your load needs 2 V to work properly, be aware that it will only have around 1 V (assuming the MOSFET was fully on, which as mentioned above, it won't be).
  • The LED (indicating if the load is powered) has no current limiting resistor. You should add a small resistor (330 Ω will be ok)
  • Your circuit switches the power ON when your opto-interrupter is occluded. If you needed it to works the other way around, you may want to use a P-MOSFET.
share|improve this answer
The load requires a target current of 1A. As far as I know, the voltage shouldn't matter. The LED I can make switch directions easily enough(And thanks for telling what it does). And the circuit is supposed to me ON when the interrupter beam is broken. Also, with the Vgs(th) doesn't R3 bump the voltage back to about 6.3V? –  CoilKid Aug 25 at 14:28
"(And thanks for telling what it does)" sorry I didn't meant to be rude. I was just assuming it was the purpose of the led, so you could have corrected my answer in case it was not. Your schematic indicate a +2V power supply at the top. Were you meaning +12V ? If it is indeed +12V your circuit will work flawlessly. If it is only 2V, then it's impossible to have 6.3V anywhere in the circuit. Your CVCC will adjust the voltage to match the desired current, so maybe the supply voltage will be enough to drive the mosfet... Test it ! :P –  CopperMaze Aug 25 at 14:58
No reason to think you were rude. I honestly had no idea what it did. The schematic I am basing this off of had a diode, and a lamp. Turning on when the load is active would make sense. Thanks! –  CoilKid Aug 25 at 15:03
If your load is inductive, you will need a freewheel diode (I think it was the purpose of the Led in first place). A solenoid will push back energy into the circuit when shut off abruptly. –  CopperMaze Aug 25 at 15:14
Thanks! Just saw the fairchildsemiconductor just now. I guess I will have to get better at soldering. –  CoilKid Aug 25 at 18:31

In addition to the too low supply voltage there is another issue. If the LED is intended to light when the load is activated you will need to turn the LED around the other way and place a resistor in series with the LED to limit it's forward current. In the end the LED may not even light if the supply voltage is kept lower than the forward voltage drop of the LED.

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Huh.. That's something to think about... –  CoilKid Aug 25 at 14:29

There is no way to tell whether you circuit will "work" since you haven't said what it should do. "Working" means the actual behaviour matches the specification. Obviously that requires a specification, which is missing in your case.

What this circuit will do is to blow out the LED when the light is interrupted. Whether that is the desired behaviour and it is therefore "working", we can't tell.

Some observations:

  1. R1 is very low. LEDs in opto-interruptors usually have around 1.2 V forward drop at rated current. 100 Ω for R1 sets the LED current at 38 mA in that case. Are you sure that is within spec? Even if it is, does the light really need to be that strong?

  2. R3 is pointless. Replace is with a short.

  3. R2 seems rather low, but I haven't checked the opto-interruptor datasheet (that's your job). For the opto output transistor to turn on fully, it has to sink about 19 mA. Can it do that? Does it really need to? If R1 were 1 kΩ instead, the transistor would only need to sink about 5 mA. The only drawback would be a little longer rise time in turning on the FET when the light is obstructed.

  4. There current thru the output LED is not limited. If the FET turns on fully, which should happen when the light is interrupted, most of the 5 V supply voltage will be applied to the LED, which will damage it. If this is a typical green LED, then figure it has about 2.1 V accross it when lit. A 150 Ω resistor in series with the LED will allow about 20 mA thru it, which most LEDs can handle. Check the datasheet of your particular LED though.


You have now changed your circuit again. Since this seems to be a moving target, I'm including your latest version here so that it's clear what I'm answering:

The LED now shouldn't blow up, and the useless FET gate resistor has been removed, but there are other issues:

  1. Assuming the opto-interruptor has a typical IR LED which drops 1.2 V at useful current, the LED is now being driven 15 mA. That's fine.

  2. With the new value of R2, the output transistor in the opto must sink about 48 mA to bring its output about as low as it can go. I don't see a need for such large current. Low impedance gate drive will make faster edges, but I doubt the opto is so fast that going to this low a impedance makes any difference.

    Another issue is that now you are requiring the opto to have a current tranfer ratio of (48 mA)/(15 mA) = 3.2. I suppose that's possible, but it would be rather unusual for a opto-interruptor. These often have current transfer ratios less than 1.

    As I said before, I'd make R2 1kΩ. That would require the output transistor to only sink 4.8 mA, which requires the otpo to have a current transfer ratio of only 1/3, which a lot more opto-interruptors can do.

  3. I meant to say this earlier, but somehow it slipped thru the cracks. You need a kickback catch diode accross the load since it is inductive. For such low voltages, I'd use a Shottky. The diode is connected in reverse accross the inductor so that it doesn't conduct when the inductor is on. When a inductor is suddenly switched off, it will maintain the instantaneous current somehow. The diode provides a nice safe path for that current. Without that, the inductor will create whatever voltage is necessary to maintain the same current in the short term, possibly blowing out the LED and/or the FET in the process.

Update 2:

Added point 3 to the list above.

share|improve this answer
As I believe I have already said somewhere here, its supposed to run current through the load if the interrupter is blocked, and If it helps I can put in in the OP. My photo -interrupter runs at a max of 20mA for the diode, and 50mA for the transistor. At least, that's what I got from the specs. Also, I need R3 to adjust the gate capacitance on the FET(or so I've been told). I will probably take your advice about the status LED, as its good advice. –  CoilKid Aug 25 at 18:01
Sorry, had those specs backward. The interrupter diode can take 50mA and the transistor can take 20mA. –  CoilKid Aug 25 at 18:11
@Coil: No, you don't need R3 to "adjust the gate capacitance on the FET". That makes no sense anyway. There are sometimes situations to put a resistor in series with a FET gate, but this not one of them. You probably heard someone say that it is necessary because he always does it for religious reasons without thinking about what purpose it is actually serving. –  Olin Lathrop Aug 25 at 18:16
Huh. Okay, I guess I can get rid of it then. –  CoilKid Aug 25 at 18:17
Also, I changed the resistors around because I mixed them up. –  CoilKid Aug 25 at 18:24

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