I’m sure I’m the one being blonde… But I still don’t understand. Following that route would put power on the cathode side of the diode.
Guess we are being Dumb and Dumber, Blonde and Blonder
To me that path takes you from Pin5 to the anode of D1…
But, shouldn’t the current stop at the cathode of D2? I believe you’re right but for some reason my thick skull just can’t understand why. I removed all the LEDs but one and now that one works perfectly (after I bridged 3V3 & 3V3_B) and it’s brighter than before which tells me it’s being lit by the 3V3 now and a pin before. What would you recommend to solve this problem?
Attached is the path as you describe.
Justin’s probably sleeping. He already gets a free module. First person to suggest a working fix get’s one also!
Can you pull out R6 and check current flow with your multimeter through there, and prove (or not) that it’s your “source”? (in the example you highlight)
I can only think you’re going to need a real reverse protection diode. Placement: TBA (or wait for Justin )
I shall mull it over…but remember it’s Friday show and tell…
I woke up this morning with the same thought. I totally forgot that LEDs don’t provide reverse protection. I was expecting it to do that and of course it doesn’t. I don’t want to add 7 more diodes to this module. So, I’ll probably just nix the idea of disabling the LEDs unless someone has a simple solution to this problem. If someone doesn’t want the LEDs then they could just remove them.
@ ianlee74 -
I didn’t realize exactly when LEDs are illuminated. You see LEDS on as soon as you attach the gadgeteer connector ? Is it the same with high/low condition on the Y pins ?
To diagnose I think is better to connect the board to an Extendeder and try small piece of circuit at time, 3.3V first. The leds are passing through 0.6V on reverse side and I don’t think you can light the other led in this way.
Have you the same problem with 1 led attached ?
Only the correct LED associated with the pin that is grounded lights. However, this shouldn’t happen unless LED_EN is bridged with solder. The problem is that it lights even if LED_EN isn’t bridged. If I remove all the LEDs except one then this works perfectly. So, I’m fairly confident that the power is definitely coming through the other LEDs. If this doesn’t make sense, I’ll make a quick video tonight.
@ ianlee74 - You can easily trace them back to the socket thru the other LEDS hence by it works as expected once the others are removed…
To diagnose I think is better to connect the board to an Extendeder and try small piece of circuit at time, 3.3V first. The leds are passing through 0.6V on reverse side and I don’t think you can light the other led in this way. [/quote]
You nailed it. I put a Rewire module between the module and the mainboard and didn’t connect any of the pins to the board. Instead of the power coming from the pins, it’s coming from the 3V3 source in nearly the same route that Justin described. So, you both win. Please send your mailing address to ian at house of lees dot net and I’ll send you a module once the final batch arrives in a couple weeks.
I don’t see a way to keep the feature I wanted of being able to disable the LEDs in a single junction w/o inserting a diode in front of each LED. I don’t think the feature is important enough to add the cost of adding 7 more diodes and re-routing the board again. So, I’m going to just permanently connect 3V3 & 3V3_B Of course, if anyone has any brilliant ideas for an alternate way to achieve the original goal w/o adding more than 1 or 2 parts then I’m listening. I’ll probably send off for the revision tomorrow night.
Justin, of course, you get a module too. I’ll contact you. I have some other things to discuss.
Maybe something like this might help?
It’s basically a switching chip (a mux) that can handle up to 8 independent circuits. Let’s say you have a wire which we will call “AB”. This mux basically chooses whether “AB” is connected to another wire called “A”, a different wire called “B”, both “A” and “B”, or neither “A” nor “B”. Two control pins determine how the wires are routed. And it can do this for up to 8 sets, so basically up to 8 LEDs.
If I were to put this in your TripWire module, here’s what I’d likely do:
1.) Leave 3V3 permanently connected to 3V3_B so that positive voltage is always being fed to the LED anodes.
2.) Connect the cathode of LED D1 to mux A1 (D2 to mux A2, D3 to mux A3, and so on). The mux will basically determine whether the cathode is floating (connected to nothing) or is connected to the trace that it originally is supposed to meet up with in your schematic (U$1, U$2, U$3, etc).
3.) Connect U$1 to mux AB1, U$2 to mux AB2, U$3 to mux AB3, and so on.
4.) Leave all the “B” pins on the mux disconnected (B1, B2, B3, etc). We dont want them to play any role. We’re only concerned with the “A” pins.
5.) Permanently tie mux /BEN to 3V3 (High). This further ensures that the “B” pins play no role whatsoever.
6.) Create a two-state jumper that feeds into mux /AEN. The two possible jumper states should be 3V3 or GND.
Now this is where the magic happens:
If the jumper is connected to 3V3 (mux /AEN is High), that means all the “A” pins are disconnected from the “AB” pins. This means your LEDs are disabled. If the jumper is connected to GND (mux /AEN is Low), this means each “A” pin is connected to its respective “AB” pin (there is continuity between the LED cathode and its original trace). This means the LEDs will be enabled.
We’re basically using the mux as a solid-state version of up to 8 SPST switches.
You’ll have to read the truth table in the datasheet to get a clear picture.
P.S. The mux’s Vcc pin should be powered with 5V.
@ Iggmoe That’s brilliant. I’m going to sleep on it, though. My module’s already crowded enough. I’d most certainly have to re-route the whole thing again and probably drop down a wire size or two on everything to fit in the extra wires required for that chip. I’m not sure it’s worth it at this point considering the risk that it would add and the possibility of yet another revision being possible.
Does anyone have any strong feelings about this feature existing or not?
if it uses not much power, it’s hardly worth the ability to switch it off. Or take a soldering iron to them, or ask for a do-not-place set up front where Ian doesn’t solder on the LEDs
Ian Thank you so much, but the discover is coming from justin!
Interesting solution the mux chip, but I will consider transistor npn like 2n2222 or similar. They have collector-emitter very high impedance and probably you will solve. I’ll make a test circuit on breadboard to check.
Don’t worry, Justin will be taken care of. I want to give you one also. Please send me your address.
I’m really curious to see what you have in mind here. Are you thinking it could be done with a single transistor?
@ ianlee74 -
Ian I’m thinking about transistor use for each led on the board, so 7 led you win 7 transistor. This is a bit an overdose for manual soldering …
I can’t write down any schematic with this damned ipad in this moment., but as I’m in lab I’ll show it.
But making a step backward and looking your target, why not use an optocoupler double-buffered input ? I mean in this way to have absolute protection from both side of the application making it very safety. The complication is that you need also separated VCC on the two sides or other way to drive the fotoled on the input side. I’m thinking to HPCL2630 chip for example (need a 3.3V version). Optoinsulators are a bit slow, but I don’t think you are input driving very high frequency signals, if you talking about switches and buttons.
Yea, I can see how that would work although it really seems overkill. Couldn’t I achieve the same thing by adding a Zener on the cathode side of every LED? I thought maybe you had come up with an interesting way of switching a single transistor.
That’s a great idea. However, adding two 4-channel versions of that chip to the module to cover all the pins would more than triple the cost of the module. I think I’ll keep that in mind for a future revision if there’s interest. I really wanted to be able to sell this one for <= $10.
@ dobova - I think I may have messed up my DigiKey query I did earlier. The cheapest part I found then were $5.50 each. I will need two for each board… This part looks like it should do the job. Whatcha think? I really don’t want to be responsible for burning up someone’s mainboard and I think you’re right that an optoisolator is the best way to prevent it. My little module keeps getting bigger…
@ ianlee74 -
Sorry for the delay but I was away. Optocoupler is for sure more expansive and you need to add $1 for port, so minimum expense increase by $7. HCPL2630, for example, is made by a lot of brand (Fairchild, Vishay and so on) and it is 2 port for about $1.50/each, and for 8 port it gets $6 more or less.
I made an example schematic with eagle in a while.
After looking a bit at the solution with transistor I’m not really convinced, I will try deeping into.
Just to give an idea I atteched a simple schematic.
As you can see on the input side (X2, X3) there is no pull-up or puldown. I missed also leds on the port.