Help With Power Supply

[automan]

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

12VDC @ 400mA

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

[automan]

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-]

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]

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]

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-]

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]

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-]

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]

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.