Help With Power Supply

[automan]

Let's preface this with: I don't know much about electricity. I can rewire most any guitar, follow most any wiring diagram, etc... but I don't fully understand electricity.

That said, I am in the process of building my pickup winder. I got a bathroom fan motor (since I can plug it directly into the wall), Cub 5 with the add on power supply (so I can also plug this directly into the wall) and the IR optical switch.

My problem is that the optical switch is 1.8V and I don't know how to run power to it. I dont want a power cord for the motor and counter AND a wall wart to power the switch. So I need to either build a 1.8V PSU and wire it into the A/C line that will support the motor and counter or hope that maybe the counter can power the switch.

Any ideas would be greatly appreciated.

[Joe Gwinn]

Originally Posted by automan

My problem is that the optical switch is 1.8V and I don't know how to run power to it. I dont want a power cord for the motor and counter AND a wall wart to power the switch. So I need to either build a 1.8V PSU and wire it into the A/C line that will support the motor and counter or hope that maybe the counter can power the switch.

Will the optical switch work on 1.5 volts DC? If so, a single D cell in a battery holder will do it.

[automan]

I'll try that to get up and running, but I really want to have it all powered with one cord to the wall.

On a side note, what part of Boston are you in? All the Guitar Center guitar techs in MA and NH work for me. I live close to the MA/NH border in North Andover. You mind if I pick your brains sometime since you are local?

[belwar]

Many of the counters will power thier own sensors. I bought a PAXLCR meter from Red Lion and a Thru-Bore encoder. The PAXLCR powered the encoder through the wires.

Red Lion is EXTREMELY helpful. I was a total nub. I called and asked about wiring questions and they guided me through the whole thing. Phone them and tell them that you are thinking about buying a CUB5 (Even though you already have) and that you have a 1.8v optical sensor.. Ask them what you will need to make it work with the CUB5. My guess is nothing and that the unit will power it.. Though a quick glance through the manual shows that the output seems to be 4 - 28 vdc. Also ask them what wires go where.

Super cool people and I would buy from them again just based on thier customer support.

b.

[chevalij]

Originally Posted by automan

Let's preface this with: I don't know much about electricity. I can rewire most any guitar, follow most any wiring diagram, etc... but I don't fully understand electricity.

That said, I am in the process of building my pickup winder. I got a bathroom fan motor (since I can plug it directly into the wall), Cub 5 with the add on power supply (so I can also plug this directly into the wall) and the IR optical switch.

My problem is that the optical switch is 1.8V and I don't know how to run power to it. I dont want a power cord for the motor and counter AND a wall wart to power the switch. So I need to either build a 1.8V PSU and wire it into the A/C line that will support the motor and counter or hope that maybe the counter can power the switch.

Any ideas would be greatly appreciated.

Which optional power supply did you get? The 12VDC or 24? It doesn't really matter as you can build a voltage divider for either with a resistor network. A voltage divider does literally what it says. It divides a single voltage into other voltages. For instance, if your power supply puts out 12vdc (volts direct current) and you put it into two identical resistors in series, then between the resistors the voltage will be 6 volts. If there were three in series, then the first junction would be 8 volts, then the second junction would be 4 volts.

BUT, if you put the 12vdc into a 5K resistor then a 1K resistor, the voltage between them would be 2vdc. It basically uses 5/6ths of the voltage across the 5k resistor. 2vdc is probably within spec of the optical switch.

[RedHouse]

While this isn't your sensor, this diagram illustrates how to interface a sensor with a CUB counter:

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

[automan]

I have read every bit of your website. It is what made me decide to build a pickup winder. I found it a while before I found these forums. Lots of the pages are broken, but some of them you can correct the url to get working.

[RedHouse]

Originally Posted by automan

I have read every bit of your website. It is what made me decide to build a pickup winder. I found it a while before I found these forums. Lots of the pages are broken, but some of them you can correct the url to get working.

Can you PM me and tell which pages you see as broken? I moved the stuff a while back off ISP.com over onto a hosted site with GoDaddy.

[-Elepro-]

Originally Posted by chevalij

For instance, if your power supply puts out 12vdc (volts direct current) and you put it into two identical resistors in series, then between the resistors the voltage will be 6 volts.

yes but without load..... but if you put a load (the IR optical switch) beetwen they (and gnd), the current in first resistor is not same that in second resistor and then tension is not V/2.....

but what is the model of the ir switch?

[automan]

Originally Posted by -Elepro-

but what is the model of the ir switch?

Optek OPB830W55Z

[-Elepro-]

Originally Posted by automan

Optek OPB830W55Z

....and how many voltage is PSU of cub5?

[-Elepro-]

connect:

black & green to PWR COMMON
white to INPUT A
red to +9-28VDC cub5 inpout voltage trough a resitor (needs voltage value PSU for resistor value)

[chevalij]

Originally Posted by -Elepro-

yes but without load..... but if you put a load (the IR optical switch) beetwen they (and gnd), the current in first resistor is not same that in second resistor and then tension is not V/2.....

but what is the model of the ir switch?

But any solid state device should only draw micro amps as current and shouldn't effect the voltage drop across the resistors. This is the IR supply, not the switched voltage.

[-Elepro-]

Originally Posted by chevalij

....micro amps

the IR emitter consumption is 20mA = 20000 uA.... there are no led (IR emitter) that turn on with micro amps

Originally Posted by chevalij

BUT, if you put the 12vdc into a 5K resistor then a 1K resistor, the voltage between them would be 2vdc. It basically uses 5/6ths of the voltage across the 5k resistor. 2vdc is probably within spec of the optical switch.

this solution is totally wrong.... if you put a 5k resistor in that circuit can not be more than 2,3mA..... and the ir emitter don't work...

if you search you never will see a voltage divider for supply a led (the ir emitter) but always a single resistor....

[RedHouse]

Originally Posted by automan

Optek OPB830W55Z

Really? then you sure can use my pdf as a guide to hook your's up.

Look at the picture in the middle of the page, you will see how.

Right above the part that says "78Lxx" you will see a small chart, you get to pick which 78Lxx device you want to use.

Here's how, if you have (or want to use) a 78L09 which is a 9V regulator, you can see on the chart that your resistor values will be:

R1 will be 385 Ohms
R2 will be 15K

These values will give you the correct LED and Ic currents for the Optec. The final piece you need is just to power the circuit with a wall-wart which puts out a DC voltage of at least 3 volts more than the 78Lxx device you decided to use.
(ie; get a 12VDC wall wart)

[automan]

Originally Posted by -Elepro-

....and how many voltage is PSU of cub5?

12VDC @ 400mA

It's the MLPS1000:
https://store.redlion.net/store/come...idProduct=1681

[automan]

Originally Posted by RedHouse

Really? then you sure can use my pdf as a guide to hook your's up.

Look at the picture in the middle of the page, you will see how.

Right above the part that says "78Lxx" you will see a small chart, you get to pick which 78Lxx device you want to use.

Here's how, if you have (or want to use) a 78L09 which is a 9V regulator, you can see on the chart that your resistor values will be:

R1 will be 385 Ohms
R2 will be 15K

These values will give you the correct LED and Ic currents for the Optec. The final piece you need is just to power the circuit with a wall-wart which puts out a DC voltage of at least 3 volts more than the 78Lxx device you decided to use.
(ie; get a 12VDC wall wart)

Awesome! I won't even need the wall wart since my motor uses 110VAC which i can use to power the Cub 5 PSU which will put out 12VDC to send through your diagram to power the switch. Thanks so much for your help, everyone. I'll just have one power cord coming out of the winder, which is what I was shooting for.

[-Elepro-]

Originally Posted by -Elepro-

connect:

black & green to PWR COMMON
white to INPUT A
red to +9-28VDC cub5 inpout voltage trough a resistor (needs voltage value PSU for resistor value)

with 12 vdc PSU the resistor value would be 550 ohm (like R1 in redhouse pdf)
the dip switch 1 on cub5 must be in logic position
the dip switch 2 in sink positi

(note that when dip-switch 2 is in SINK position inputA of cub5 already has a pull-up resistor then the R2 in Redhouse pdf doesn't need)

[RedHouse]

Originally Posted by automan

Awesome! I won't even need the wall wart since my motor uses 110VAC which i can use to power the Cub 5 PSU which will put out 12VDC to send through your diagram to power the switch. Thanks so much for your help, everyone. I'll just have one power cord coming out of the winder, which is what I was shooting for.

Baddda-bing-badda-boom!

[chevalij]

Originally Posted by -Elepro-

the IR emitter consumption is 20mA = 20000 uA.... there are no led (IR emitter) that turn on with micro amps




this solution is totally wrong.... if you put a 5k resistor in that circuit can not be more than 2,3mA..... and the ir emitter don't work...

if you search you never will see a voltage divider for supply a led (the ir emitter) but always a single resistor....

Okay, may have grabbed the wrong value resistors for current flow, but my winders been working fine with a voltage divider for over two years? Then again, are you talking about a discrete optical switch, or a two component LED and sensor setup?

[-Elepro-]

Originally Posted by chevalij

Okay, may have grabbed the wrong value resistors for current flow, but my winders been working fine with a voltage divider for over two years?

your sensor work well because you are a lucky man.... what is the switch model? what are the resistors values?.....

Then again, are you talking about a discrete optical switch, or a two component LED and sensor setup?

there is not difference..... discrete optical switch is just a box with a led and a sensor inside.....
here an example datasheet (note the max current in emitter is 50mA=50000uA... and it is a SMD component) and application note...... they don't recommend a voltage divider for supply.... but a single resistor

bye

[chevalij]

Originally Posted by -Elepro-

bye

Mine works, Fairchild H21LOB. I don't have any pics of the resistor board handy, but it's a mess I know that.

Next iteration will use either same sensor, or Honeywell SS49E hall sensor. Being fed into Labjack U3HV, software written in Labview 8.5 developer.

90VDC Leeson motor driven by Minarik Motor controller, reversed through Grayhill H-Bridge relay board ( all in turn driven by Labjack). Of course, I have to use an op-amp for the speed control because the labjack only puts out 20ma (20000ua because we apparently have to convert now). Linear stepper driven by labjack as well through stepper controller. All software controlled with infinitely variable traverse rates and wind speeds up to about 1750 rpm. No limit switches, will use an initial rig point and then wind count multiplied/divided by traverse speed. Should be a cool project if I'm smart enough to plug it into the wall that is....

[-Elepro-]

Originally Posted by chevalij

Oh god, your one of those people that corrects everybody to show how smart he isn't ..... I work with one of those, nobody can stand him.

.........................

...... i'm speechless.....ciao

[rockin roy]

I read all this pretty quick so I may have missed some important details but if you have 12vdc and the unit needs 1.8v and pulls 20ma then a series resistor should be enough.
10.2/.02=501, stock values are 470 and 510. I squared R is like .2 so a quarter watt should do. It is just like a diode right?

[RedHouse]

Originally Posted by rockin roy

I read all this pretty quick so I may have missed some important details but if you have 12vdc and the unit needs 1.8v and pulls 20ma then a series resistor should be enough.
10.2/.02=501, stock values are 470 and 510. I squared R is like .2 so a quarter watt should do. It is just like a diode right?

Pretty much.