Hi there
Hopefully a simple question which googled has struggled to answer. I'm building an opto controller board for a pinball machine. One of the common failure modes of the circuit is the transmitter led (940nm 5mm ir led) failing and stopping the detector working,though there are other failures (I'm the detection car unit side) to ease troubleshooting, I'd like to add a visible led, to show if the ir led is working. The ir led is powered by 12v, through a 270 ohm 2w resistor. I'd considered simply putting a visible led in series, which would turn off if the ir led were to fail open circuit. However if it were to fail short this would kill the visible led too. Is there a simple circuit I could use to detect whether the ir led is working?
Thanks very much
Rich

 

What's the maximum current rating of the two LEDs?
It sounds like you may be overdriving the IR LED, which could be causing the failures.
Normally such an LED should not fail during the lifetime of the product.

 

Hi there
Hopefully a simple question which googled has struggled to answer. I'm building an opto controller board for a pinball machine. One of the common failure modes of the circuit is the transmitter led (940nm 5mm ir led) failing and stopping the detector working,though there are other failures (I'm the detection car unit side) to ease troubleshooting, I'd like to add a visible led, to show if the ir led is working. The ir led is powered by 12v, through a 270 ohm 2w resistor. I'd considered simply putting a visible led in series, which would turn off if the ir led were to fail open circuit. However if it were to fail short this would kill the visible led too. Is there a simple circuit I could use to detect whether the ir led is working?
Thanks very much
Rich

Putting the two LEDs in series isn't a good idea. The two LEDs must run on the same current.

Instead a current sensor might work better. I can't assign component values not knowing currents.

 

There is a good chance that you just simply replace the IR LED with a visible LED. I did that one one product and it made alignment of the sensor optics much easier.

 

Hi there
Thanks for the swift replies to this. To answer a couple of questions raised. The existing opto detectors are quite narrow in the range of frequencies they accept, visible leds don't seem to work and I'm loathe to change all the detectors to suit the transmitters. The specs of the ir led supply are, 12v supply, 1.5v forward voltage and 40ma current.
Thanks for the suggestion hp1729. What's the method of operation for the current sensing circuit? I see the transistor q1's Base is connected past a resistor. How would that sense current flow rather than just see the voltage and allow current flow through collected and emitter? If the ir led were to fail short wouldn't that still light d4?
Sorry for the simple questions, I'm still quite new to all this and just learning.
Thanks again for the help
Rich

 

The specs of the ir led supply are, 12v supply, 1.5v forward voltage and 40ma current.

Thus with a 12V supply and a 270 resistor you have (12-1.5)/270 = 38.mA through the LED.
If 40mA is the maximum LED current rating, then you are running it too close to the maximum for good reliability.
Can you reduce the LED current by increasing the value of the resistor and still get it to work properly?

If you want to detect both short and open failures of the LED then you will need a more complex circuit.

 

I'll look into that. Thanks for the tip. I'm effectively redesigning an existing pinball board but trying to add these extra features to improve fault finding. The ir leds are a common fail point so a diagnostic led would be handy to quickly pinpoint the error to the emitter rather than detector or another fault in the chain somewhere.
I'm definitely interested in detecting either short or open failure modes so the diagnostic led doesn't show a false positive. Any thoughts on how I could achieve that?
Thanks again
Rich

 

Do you have any idea what the circuit looks like? Is the 12 V regulated?

There's a couple of issues here, both I've experienced in a design, I used an opto-coupler to monitor a 50 V power supply. The circuit was an opto, resistor, diode and zener diode. Like you, the optos would die until I put appropriately sized ZNR's across the unregulated power power supply capacitors. These are essentially transient voltage suppressors.

This http://www.ti.com/product/LM334 and a programming resistor may work for you better than a single resistor. It is available in a TO-92 (transistor) like package. If you suspect a reverse voltage issue an extra diode can be used in series with the IR LED. A series resistor could be used too to take the stress of the current regulator.

If long wires are used, you may want the ZNR/TVS closer to the LED.

With your "new" board, you could have a test pint where you could deferentially measure the LED voltage, but really I'd try to fix the design.

I'd make sure that relay coils have the reverse biased diode across them.

 

Hi there,
thanks for that. The current circuit is far from ideal in a few ways, but i'm struggling to come up with a relatively simple alternative. The 12v supply is unregulated and actually is around 14v. The circuit I have powers 7 of these IR LED's. Each is powered separately, with a 270ohm 2w current limiting resistor (which does get rather toasty). I'd looked into using a voltage regulator, but the heat dissipation is a bit of a worry. Looking at that regulator, looks ideal apart from the current max of 10ma and the leds I'm using are 40 (though I could reduce that a little as suggested above). Would it not also get very hot dissipating that kind of power? Please correct me if I'm wrong on this as my understanding isn't that detailed

 

This is the whole circuit I'm trying to replicate/improve, but the specific bit I'm on about here is in one corner. Leds and resistors shown on their own and separate from the lm339 setup for the detectors, which I was just going to use pretty much as is, except adding visible leds to show when a detection is triggered.
Thanks
Rich

 

My point is, if you can reduce the LED current to make it more reliable, you likely won't need a circuit to detect a failure, since such a failure should be extremely rare.

 

hi there,
thanks for that, i see your point now, I've looked up the specs of the ir led used (QED123, 880nm rather than the 940nm i thought. The absolute max specs are 1.7v and 100ma so hopefully running a 270 ohm resistor puts me well within the safety zone (even if the voltage is up around 14v, which being unregulated it often is). Even given this opto transmitter failure is relatively common on these parts, which is why i was after a visible solution. I guess being packed into a pinball machine and all the vibration and electrical noise causes early failure? Would a voltage regulator on the opto board to regulate the unregulated 12v (14v) down to 12v regulated perhaps help with the longevity of this?

 

Sorry about the 10 mA limit. My eyes must have been deceiving me. This http://www.linear.com/product/LT3092 part is good to 200 mA.

The fact that it's unregulated is basically your problem. Some decent switchmode power supplies are here: http://www.trcelectronics.com/Meanwell/power-supply.shtml It may solve a LOT of issues.

For some of the linear regulator to work, you need at least a 3V differential.

 

There are a few issues with your circuit.
1) There's no hysteresis, so comparitors could oscillate
2) the output diodes should be unnecessary because the outputs are open collector anyway.

@crutschow:
Any good ideas how to add about 10 mV hysteresis? LM339 datasheet is here: http://www.ti.com/lit/ds/symlink/lm339-n.pdf

 

@crutschow:
Any good ideas how to add about 10 mV hysteresis? LM339 datasheet is here: http://www.ti.com/lit/ds/symlink/lm339-n.pdf

What circuit are you referring to that uses the LM339?

 

Here's a three transistor circuit that monitors both open and short conditions of D1 (IR emitter).
D2 illuminates unless D1 is open or shorted.
Q2 monitors D1's current and shuts off if D1 opens.
Q3 monitors D1's voltage and shuts off if D1 shorts which also shuts off Q1
Q1 is needed to add gain (forming a Sziklai pair) so that R2 can be sufficiently large so its current (and thus the current through R1) is low enough when D1 opens to keep Q2 turned off.

 

@crutschow

The screen shot in this (post #10) http://forum.allaboutcircuits.com/threads/hopefully-simple-led-question.128749/#post-1054193 post.

i.e.: http://forum.allaboutcircuits.com/attachments/screenshot_20161027-173639-png.114346/

 

To add hysteresis you need feedback from the output to the positive input, so you would need to add a resistor in series with each of the plus inputs along with a resistor from each of the outputs to the plus input.

 

That's what the datasheet suggests (10 mV ), but not sure how to "calculate" it. Comparitors without hysteresis are problematic.

the OP/TS should design a more robust circuit rather than a band-aid. I think he clearly has an overvoltage condition. A number of solutions present itself with some of these options.
1) TVS/ZNR
2) A real 12 V power supply for this or most of the machine.
3) Remove unnecessary diodes
4) Replace the resistors with a Linear technology current source and a programming resistor.
5) Add Hysteresis.

The OPTO should last a long time. Generally LED's degrade over time. Ideally, that should be the failure mode.

 

Hi guys, thanks very much for this. I think I've got a bit of reading and research to do to improve this design to the point where it's a bit more reliable. I'll get looking into all the suggestions you've made and report back with where I'm up to
Thanks for taking the time to help, I do appreciate it.
Rich