A few questions regarding de-clicking a relay...

[trixdropd]

Alright, I've been somewhat successful at using the rc network described on this page: http://www.geofex.com/article_folders/rmtswtch/rmtsw.htm

The problem is that instead of the 12 volts shown there, I'm using 5 volts.

What I'm doing is taking 9 volts into a 5.1v zener diode and powering the setup. I'm also using an omron g5v-2-h1 relay. It has 166.7 ohms on the coil, with a current rating of 30 m/a. datasheet here: http://www.omron.com/ecb/products/pdf/en-g5v2.pdf

The issue I encounter is that my led stays dimly lit upon bypass. I am sure the resistor values of the r/c network need changing, but I'm not a math wiz. What I've done is throw a large value resistor across the led legs and this "drains" the voltage I imagine. I wanna get the circuit proper to begin with.   

Can anyone help me out?

[R.G.]

What I'm doing is taking 9 volts into a 5.1v zener diode and powering the setup. I'm also using an omron g5v-2-h1 relay. It has 166.7 ohms on the coil, with a current rating of 30 m/a...
The issue I encounter is that my led stays dimly lit upon bypass. I am sure the resistor values of the r/c network need changing, but I'm not a math wiz. What I've done is throw a large value resistor across the led legs and this "drains" the voltage I imagine. I wanna get the circuit proper to begin with.   

LEDs only glow when there is current running through them. Ergo, there is still current running through the LED. So the question is - *exactly* how do you *really* have the LED, dropping resistor, zener and switch transistor wired?

Just as some observations,
- you do need a resistor between 9V and the 5.1V zener. I'm guessing you have this and didn't mention it.
- that resistor value is important, as is the power rating of the 5.1V zener; the resistor has to let through 30ma or the relay won't make
- if you have attached the LED to the 5.1V zener, then the series resistor also has to let through enough current to run the LED and the LED resistor has to be appropriately smaller.
- the zener performs only the function of "eating" any excess current that the coil is not using; if the coil is switched off, the current the zener is handling goes up by 30ma; if you have an assortment of relays across it, the zener eats 30ma for each relay that's off; if it's a 1/2W zener, then it can only eat I = 0.5W/5.1V = 98ma, or three relays' worth of current. Four relay currents will make it overheat when they're all off.
- You can run a relay from a higher voltage if you put a resistor in series the coil to keep the current down to spec; in this case, you can run your 5V relays from 9V by limiting the total current to 30 ma with R = 9V/0.03 = 300 ohms, of which 167 ohms is in the relay, so you can put a 140 ohm resistor in series with the relay coil and run it from 9V directly.

[trixdropd]

I don't have my exact layout handy at the moment. I do not have the resistor on the zener. Your post was informative.
I should point out that the led glows dim only after you engage the relay and dis-engage. when the cap is drained, the led doesn't glow. I was thinking the tranny is letting some voltage through.

[R.G.]

I don't have my exact layout handy at the moment. I do not have the resistor on the zener.

The zener's life is likely to be short. I think it and the 9V are fighting it out for what the voltage should be. If you did that with a raw supply that has lots of current available, like maybe a car battery or something, you'd create a light emitting zener. Not to mention smoke emitting, and maybe glass-pieces-emitting.   

I should point out that the led glows dim only after you engage the relay and dis-engage. when the cap is drained, the led doesn't glow. I was thinking the tranny is letting some voltage through.

Yep. You gotta get that power supply thing fixed though before you can tell much, though.

[trixdropd]

Thanks for the insight. I'll breadboard it up and see what I get.

[trixdropd]

I don't have a 140 ohm resistor handy today so I gotta postpone the experiment. What value resistor would I be wanting to use in series with the zener and why?

The relays will work fine with no zener as their max voltage is 75% of rated voltage which is 9v. This is pushing it but i haven't had an issue there.

My thinking was that by using less voltage at the relay itself, I would make the potential for popping that much less, thus making it easier to eliminate. Is this not the case? I could just as easily get the 9v version of the relay and halve the current use per relay. 

[R.G.]

What value resistor would I be wanting to use in series with the zener and why?

A zener is a device which conducts no significant current at all as long as the voltage across it is below it's breakover voltage. When the voltage across it tries to go above its breakover voltage, it starts conducting heavily. In theory, it will conduct any amount of current at all. In practice, it can only conduct a current which makes the power it dissipates (that is, voltage across it times the current through it: P = V*I) increase until it burns up. This is why there is a power rating on a zener. A 1/2W zener can only get rid of one half watt without destroying itself. That is a function of the package, not the zener voltage, so if you have a half-watt 5.1V zener, it can only conduct I = 0.5W/5.1V = 98 milliamps the possibility that it will burn up. If you had a 12V zener in that same 1/2W package, the current it can stand is again I = 0.5W/12V = 41.6ma.

Remember: zener diodes do NOT regulate voltage. They conduct current suicidally if the voltage across them goes too high. Something else has to limit the current to save them.

Zeners by definition always die unless something outside the zener limits the current it can have. If you're supplying 9Vdc to a 5.1V zener, the power supply is trying to push 9V into it. The zener is conducting all the current it can, since the 9V supply is trying to force it to 9V, 3.9V higher than it can go. I'm not sure why your zener is still alive at all. It's probably that the 9V power supply has some kind of internal impedance or current limit and is backing off on the current the zener gets to let the zener stay at 5.1V.

The resistor you put in series with your zener can be no smaller than the excess voltage of the raw supply (that is 9V in this case) minus the zener voltage, which is 5.1V in this case, divided by the maximum current the zener can withstand. That means that
- (1) you have to know the power rating of your zener; half watt is common, as are bigger ratings, but the common glass body zener is usually 1/2W maximum:
- (2) you have to calculate the maximum current by dividing the power rating of the zener by the zener voltage to find the maximum current it can take:
- (3) you have to then divide this maximum current into the excess voltage (3.9V in this case)  to get the resistance to put in series with the zener.

This is the simple case, where the external load after the zener may go all the way down to zero amperes out. If there is a minimum load current you're sure of you can size the series resistor to let through more current, but that is how zener-beginners burn up zeners, getting this calculation wrong.

You accidentally have a 9V supply that will not kill your 5.1V zener without a resistor. You will not always be that lucky. 

The relays will work fine with no zener as their max voltage is 75% of rated voltage which is 9v. This is pushing it but i haven't had an issue there.

Actually, if they are 5V zeners, 9V is 9V/5V = 1.8, or 180% of their rated voltage.  They will work fine on 9V, until the coil heating of 180% of their designed coil  current overheats the coil wire insulation.

Actually, that "soft" 9V supply is what is saving the relay coils too. A 167 ohm coil with 9V across it lets through 9V/167 ohm = 53.8ma. They are rated for 5V/167 ohms = 30ma. The rated power in the coil is 0.030*0.030*167 = 150mW. The actual power if your power supply keeps 9V across them is 0.485W, or over three times their rated coil power. How long they will take this depend on how long it takes them to heat up and whether the final temperature happens to be below the wire's melting temperature or not. It may take a long time to overheat. Or it may limp along for a long time and just die slowly one day. In any case, it's not good for the relays.

My thinking was that by using less voltage at the relay itself, I would make the potential for popping that much less, thus making it easier to eliminate. Is this not the case?

To some extent, yes. It cuts the size of the flyback pulse down by 2:1. But slowing the speed of the flyback pulse down can be even more effective than making the voltage across the relays lower. It's unknowable which is better unless you know the parasitic capacitances of the relay you have. This is not necessarily a bad idea, but in practice, the extra complication of how to drive the relays and deal with the extra voltage may be more trouble than it's worth. My approach would always be to buy one relay, try it, and see what I had to do to make it quiet. But using lower coil voltages for smaller flyback voltages is a subtle point. Making the relays easy to drive and reliable is a big point.

I could just as easily get the 9v version of the relay and halve the current use per relay. 

Current per relay isn't really a big deal unless you're planning to run this from batteries. Very few pedals run from batteries these days, so it's much less of an issue than it was a decade or so ago.

[dklimbani1]

I want to replace faulty de-clicking relay in my nebulizer machine, but I am not able to find it anywhere online, can you guys guide me where can I buy it? Which model is best for a new nebulizer, I am going to buy one recommended from this site: best nebulizer machine

Any suggestions?

Thanking You

[PRR]

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