Greetings,

For my research I am supposed to redesign an already functional board to have a visual LED array to serve as an on/off indicator for the solenoid array.

Basically there are 20 solenoids currently wired to a DB25 power input connector and in parallel with 20 fly back resistors. I plan to use 2 10 LED bar graph arrays to have one led per solenoid. The solenoid specs are .35W (15 mA for our 24 V input) and the led specs are here http://www.digikey.com/product-detail/en/MV57164/1080-1184-ND/2675675

Attached are pictures of the original board, as well as my progress on the multisim schematic for the new one.

My trouble is figuring out how to integrate the leds into the board. I was thinking of putting the leds in parallel with their corresponding solenoid, then have the grounds of all the leds go through a voltage lowering resistor (step down for 24 to 2v). Does this work? or do i need a resistor per led, or should i do each led in series with its solenoid?

I've never done anything like this (or used multisim) and i want to make sure my schematic is perfect before i move on to the physical board design.

Thanks in advanced for your help.

Here are the pics:
original board:

multisim schematic:

 

Unless there is some other form of current limiting, you will need one resistor per LED. There are resistor arrays in SIP format designed to use with bargraph LEDs. You can put one resistor in series with one LED, and place that in parallel with the solenoid coil. And I assume you meant flyback diodes and not flyback resistors.

 

There seems to be something missing here? The rest of the picture!
The cathodes of D1 ~D20 are all connected to what I take is the P.S. common? if so they would all conduct in that configuration.
Where is the rest of the circuit?
Max.

 

There seems to be something missing here? The rest of the picture!
The cathodes of D1 ~D20 are all connected to what I take is the P.S. common? if so they would all conduct in that configuration.
Where is the rest of the circuit?
Max.

Probably drawn backwards by the OP...easy mistake to make. Although on close exam of the photo, it looks like the cathodes may be toward the common bus.

 

Haha sorry flyback diode is what i meant. I was thinking i needed a current limiting restistor per led lince each goes to an independent solenoid, but someone at my school told me otherwise...Ill look into the resistor array. How would i go about sizing the resistors? Using ohms law I calculated about 1k per LED considering 24 volts, and the led 30mA max.
As far as the circuit, the picture is all I have. There was a DB25 connector input that is somehow connected to a 24V 2A power supply. My schematic is likely wrong if you think you are misunderstanding something. It was my attempt to copy what was in the picture.

On a side note, does anyone know how to connect several pins to ground? The top side of that 40 pin connector all goes to ground.

Ill rework out the schematic including a resistor array and check back.

Thanks All!!

 

With a 24V supply, a 1k resistor will allow about 20mA through the LED, which will be plenty bright. I would probably use a 2.2k and reduce the current to about 10mA. I think the LEDs would still be very visible at 10mA.

In the attachment, the red arrow points to where the multiple connector pins are connected. And, it looks to me like the red wires are connected to the common trace. Are the red wires or the black wires connected to the common trace?

The blue arrow points to where the multiple diodes are connected. Are the anodes or the cathodes of the diodes connected to the common trace?

 

As far as the circuit, the picture is all I have. There was a DB25 connector input that is somehow connected to a 24V 2A power supply. My schematic is likely wrong if you think you are misunderstanding something. It was my attempt to copy what was in the picture.

Looking at drawing the way you have it the diodes are not only backwards but not across the loads.
Restudying the photo's It appears you have the diodes the right way but the cathodes should show connected to +ve Not common?
This +ve bus also shows going to the red conductors, the black ones would then be the outputs to the various loads and the other side of these loads would go to the P.S. common.
This would then seem to make sense.
Max.

 

With a 24V supply, a 1k resistor will allow about 20mA through the LED, which will be plenty bright. I would probably use a 2.2k and reduce the current to about 10mA. I think the LEDs would still be very visible at 10mA.
In the attachment, the red arrow points to where the multiple connector pins are connected. And, it looks to me like the red wires are connected to the common trace. Are the red wires or the black wires connected to the common trace?
The blue arrow points to where the multiple diodes are connected. Are the anodes or the cathodes of the diodes connected to the common trace?

ok ill use a 2.2k and up it to 1K if they seem to dim.

after a closer look it looks like the red are in fact on the common trace, as are the cathodes of the diode. I can confirm this tomorrow in the lab.

Looking at drawing the way you have it the diodes are not only backwards but not across the loads.
Restudying the photo's It appears you have the diodes the right way but the cathodes should show connected to +ve Not common?
This +ve bus also shows going to the red conductors, the black ones would then be the outputs to the various loads and the other side of these loads would go to the P.S. common.
This would then seem to make sense.
Max.

ok i think i got it. I'll redo the schematic to reflect that.
Thanks

 

Here is the updated schematic. I did the resistors on the V+ bus which should be fine right?

 

Looks more like it.
Max.

 

I would probably use a 2.2k and reduce the current to about 10mA. I think the LEDs would still be very visible at 10mA.

Ditto. No reason to burn out your LEDs. Also, using less current in each will reduce the load on the power supply in the inevitable (Murphy's law) event they are all turned on at once.