I had the same idea too. You and I might share part of our brains or something. I don't suppose you left part of your neocortex on the north side of the US border?
I used to live in Los Angeles until a few years ago, and I've been in Thousand Oaks several times. Maybe I stole some of your brain cells one day on a drive through Thousand Oaks?
Adding extra (small) resistors in series with the LEDs will help to stabilize current through the LED a bit. But also increase the voltage drop across the (LED+resistor) combination a bit. (That might not be a problem, as long as you still have enough voltage left for the relays to operate reliably.)
I also tried to think of a way to add a couple of small-signal transistors to act as current-sinks in series with the LEDs, stabilizing LED current. Or to add a transistor in parallel with the LED/resistor pair, in a way that clamps the voltage across the LED to a constant value. So far, I haven't come up with a way to do either of these things. I'm zero for two.
-Gnobuddy
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That's a great idea!
A possible downside: the LED won't go out immediately when you hit the footswitch, because of the stored charge in the cap. This may not be a problem if the cap is sized just right: big enough to latch the relay, small enough to let the LED go out so that the guitarist can be confident the LEDs actually show the current state of the relays.
-Gnobuddy
Sorry but I'm a bit confused now...
thanks everyone for all the help
- In the datasheet of my relay here (page 2, standard 12V), "Must operate voltage" is the hold voltage or the pickup voltage?
- we are talking about the two leds inside the footswitch (D3) right?
- so you are suggesting to add a resistor in series with the led AND a cap in parallel? (should I also keep the reverse parallel diode in this eventuality?)
- if so, what range of values do you suggest for resistor and cap?
thanks everyone for all the help
Akkar, let's keep it simple: it takes more voltage to close an already open relay (because the moving metal part is further away from the electromagnet when the switch is open.)
So if your relay is currently open, it's guaranteed to close if 75% of the nominal voltage is applied. 9 volts, in the case of the 12V relays you're considering.
Once the relay has closed, the movable magnetic part is in contact with the electromagnet. The magnetic field is much stronger at this very short distance, so you can decrease the current through the relay coil a lot before the magnetic field finally becomes so weak that the relay opens. According to your datasheet, the relay is only guaranteed to open once the voltage across it has dropped to 5% of the rated voltage (0.6 volts, for a 12 V relay.)
Yes.
I suggest that you first build the circuit exactly the way you drew it. You may need to tweak R3 a little bit (by trial and error) to make it work well. Keep it simple, see if it works!
IF - and only if - you find that you just cannot get the LED brightness right, then you might consider the additional complications we've been discussing here. I think some experimenting will be called for, as I can't think of a way to calculate exactly what additional resistors or capacitors you might need.
In any case, I suggest you always have those protection diodes wired directly in reverse-parallel with the LEDs. Most LEDs will break down and be damaged with just a few volts of reverse voltage, and a relay coil can generate short spikes of hundreds of volts at turn-off without clamping diodes. So I think it's cheap insurance to use those protection diodes - all four of them, two across the actual relay coils, two across the two LEDs, exactly as you have them in your original schematic.
-Gnobuddy
OK. Without messing with numbers I was thinking 100uFd (depending).
Tomato tomato. But not a 1uFd pea or a 5,000uFd watermelon.
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