Counter Idea

[RedHouse]

I wondered about that as well. I got that number from a similar config:

http://classicamplification.net/winder/OPTEK-CUB4.pdf

....

These resistor values are derrived from the Optek application note 213 which describes in detail the requirements of the LED and collector current for the transistor. They work, I use the setup daily.
(with a CUB counter)

I couldn't figure out the requirement for the counter...

I Looked back over these pages in this thread but couldn't see exactly which counter you bought, there were two posts asking if a certain one was any good but I didn't see which brand/type you actually bought?.

Anyway that's why the CUB4 and CUB5 are really well suited for our application, they have Schmidt trigger inputs and can handle up to 4-28V signals, and have a "high speed" counting input for up to 5kHz (not the CUB4L8 though) which is fast enough to count the turns on a Dremel tool.

Winding at a rate of 2000 RPM is actually 33.3333 revolutions per second, which re-stated is a mere 33.3Hz signal and a walk in the park for the CUB4/5. The ongoing debate over duty-cycle within these counter threads is not critical with the CUB counters and optical switch, CUBs trigger on a rising (or is it falling?) edge of the input signal, not the apeture of duty-cycle time.

On my 2nd winder (the one folks refer to where I showed the Optek/CUB thing) I used the circuit shown which gives the specified collector current for the sensor to function correctly (see app note) and I used what equates to a 20% window or duty cycle, the counter handles this quite easily at that machines max speed of 1500 RPM. On my latest (3rd) winder I used a 25% window just because it was easier to mill out.
(my most recent winder has the opto-vane built right into the aluminum bobbin flange)

[MistaT]

Well here is the deal for calcualting power needed for a resistance.

P = I * E

Where:
P = Power in watts
I = Current
E = Voltage

So lets take ur example with a 2k resistor. Since you calculated 2ma, I will use that.

P = 0.002 x 5 = 0.01 watt <== This is how much will be dissipatted by the resistor.

With resistors, its a good idea to take your calculated dissipation and double it (safety factor) then go to the next highest wattage.

OK all that to say that 1/4 resistors are just fine. But at least now you know the thought process in the future.

BTW, I'm glad this is a learning experience for you, and I am glad to help out. Like I have said before, I get a lot out of this forum, and I am thrilled to be able to give back a little.

[tonedeciple]

Honestly I am surprised at their recommendation of 10k, but if that is what they recommend, then I guess it should work.

Of course they don't know what my counter requires. Elepro stated 2mA should be enough which I calculate at 5V to be about a 2k resistor. Are 1/4-watt enough? Or should I opt for 1/2-watt.

I have to say this whole experience has been a lot of fun It has reall blown the dust of the left side of my brain!

[MistaT]

Well basically there is a lot of room here for a very wide resistance range. Honestly I am surprised at their recommendation of 10k, but if that is what they recommend, then I guess it should work.

Like I posted earlier for getting my Cub5 counter going, I used a 560 ohm with a 12V power supply.

So like you said, start with a 10k if you are comfortable with that and see if it works. You certainly won't blow anything up for sure. If your circuit doesn't work, then cut the resistance in half. Keep doing that until it does work. But do your ohms law calculation so you know how much current is flowing through your circuit and keep the current below all of your devices max. allowed current, and you should be fine.

[tonedeciple]

That pretty much will work, but I am a bit concerned about the 10k resistor.

I wondered about that as well. I got that number from a similar config:

http://classicamplification.net/winder/OPTEK-CUB4.pdf

...and this recommendation:

QRB1114 IR Sensor (SKU: 35140) - HVW Technologies

I guess I can't start low and work up until I have enough of a pulse to trigger the counter accurately.

[MistaT]

That pretty much will work, but I am a bit concerned about the 10k resistor. At 5 volts, you will only draw 0.5 ma. I am not sure that is enough current flow through the sensor for it to "turn on" hard enough for the counter to "see" a voltage low.

If it were me, I would make that something like 270 ohms. This would give you enough of a current flow through the opto to it would turn on hard enough for the counter to see the pulse.

[tonedeciple]

with 4k7 at 12volt you have more than 2mA.... it is enough for counter...

I couldn't figure out the requirement for the counter. At any rate, here is what I've come up with by relying on everyones help here and elsewhere. It may be overkill, but I think it'll work. I have to order a few components, but I've got the power supply regulated at 5VDC now.

[-Elepro-]

This all makes sense, except for one thing; how did Elepro come up with the 4k7 R2? I need to know this since I want to drop the power source to 5VDC.

with 4k7 at 12volt you have more than 2mA.... it is enough for counter...

[tonedeciple]

There's nothing wrong with the 16.6V, just put a 1K or 5K pot in series with any old LED and put your meter in series as well in DCA mode. Turn the pot from max down until you get to 20ma. Measure the DCR across the pot and substitute that value of resistor.

That's a good idea. I'm going to keep the 16.6VDC and do either this or add a 7805 regulator and some capacitors. I know for sure I'll order at least five of those QRB1114s just to play it safe.

[David King]

There's nothing wrong with the 16.6V, just put a 1K or 5K pot in series with any old LED and put your meter in series as well in DCA mode. Turn the pot from max down until you get to 20ma. Measure the DCR across the pot and substitute that value of resistor.

[tonedeciple]

Hope this makes sense.

You'll have to forgive me in this area. I mostly carve wood and inlay, but I've decided to take more control of my product by winding my own pickups.

This all makes sense, except for one thing; how did Elepro come up with the 4k7 R2? I need to know this since I want to drop the power source to 5VDC.

[MistaT]

Well the magic formula in all cases is good old Ohms law:

E = I * R.

So in calculating the desired resistance you would change the above formula to this:

R = E / I

Where:

R = Resistance
E = Voltage
I - Current

Now the question is, what current to pick? Something less than the max. current of the sensor (IIRC 50 ma?), and something greater than what is required to trigger the counter. (IIRC 600 uA? which equals 0.06 mA)

So for me, I would go for something in the 20 ma range.

So if you were to use a 5 volt supply, then

R = 5 / 0.020 = 250 ohms. So a standard value of resistor would then be 270 ohm resistor.

Hope this makes sense.

[tonedeciple]

Hmmm.. something fishy here.. First of all, if you are reading 35 vac, then I wonder if all they did in the wall wart to create DC was to use just a diode.

I made a mistake when I tested the voltage earlier. It's actually 16.6VDC, which I think may have been way to high for the emitter even with a 560 Ohm resistor.

I'm thinking of dropping down to a 5VDC regulated wall wart. I know how to calculate the resistor for the emitter, but what is the formula for calculating the sensor side resistor? Elepro's schematic for 12VDC shows a 4K7 resistor, but what would it be for a 5VDC power supply? Of course, I'll test the voltage from the 5VDC wall wart first before I calculate the values.

[MistaT]

Hmmm.. something fishy here.. First of all, if you are reading 35 vac, then I wonder if all they did in the wall wart to create DC was to use just a diode. Which is sorta DC only because its rectified AC, and is missing the "negative" half of the wave. (yet another reason not to use wall warts..)

Off the top of my head, a way to test this out is to get an electrolytic capacitor of pretty much any value, say 100 uF or greater (with say a 150 vdc voltage rating), and hook it across the wall wart. Be careful and make sure you hook neg. of the cap. to the neg. of the wal wart.

Electrolytic caps. want to blow up if they are in the circuit backwards. Then measure the voltage again on the AC scale. If it drops a lot, like down to 0.5 volt or something, then I would say what you have is a DC wall wart w/o AC filtering.

Its my guess that the current rating of the sensor was exceeded and has blown. ESP. if you have un-rectified AC of 35 volts.

I would either make absolutely sure that you have good DC (and no AC) voltage coming from that wall wart, and that it is in the 12 VDC range.

One final thing. If you do in fact do the electrolytic capacitor check, after you power off the circuit, make ABSOLUTELY SURE that it is completely discharged. Do this by jumpering across the cap. Be ready, as it might (probably) give a BIG spark!

[tonedeciple]

why ACV?? you have to use DCV scale...

It's too early in the morning!! 16.5 volts. Still high, though.