Could anyone post a picture of a finished Interface board - i need to check orientation of the PSU and Q1 connectors.
This is the beta version, but those parts should be the same:
For the MOSFET and the voltage regulator (they're both in the same TO-220 package), the little outline on the board is pretty helpful. Like everything else, it's basically an outline of the top-down view of the component package when installed. For the TO-220's, the thick line on the outline represents the metal fin. So just line it up with the metal fin on the same side as the thick line drawn on the board.
There are no sockets on the mouser list for the ICs and the Optocouplers - is it wise to solder them directly to the board? I would prefers getting sockets for them, i think.
I don't use sockets myself. It won't do the chips any harm to solder them, but sockets do make it easier to get the chips out if you installed them backwards or if you ever needed to replace one. You'd want to check that there's room, though; the board is packed pretty tight and the sockets usually take up a little extra room.
Also, it occurs to me that it might get too tight vertically for the chips installed under where the KL25Z goes.
what I'd like to confirm are whether I need to run a separate +5V line with a separate resistor inline to each RGB contact or I just need to connect one for each resistor value and I can daisy chain them?
Sorry to say that daisy-chaining of the resistors won't work. You need a separate resistor for every LED element individually. So for the 5 flashers, you need 15 resistors.
You can put the resistor on either side of the LED, so there are two basic ways to wire this:
1. Daisy chain the +5V line to one end of each resistor. The other end of a given resistor goes to the "+" terminal of its LED. The "-" terminal of the LED goes to the output port on the expansion board.
2. Daisy chain the +5V directly to the "+" terminal of each LED. The "-" terminal of the LED goes to one end of the LED's individual resistor, and the other end of that resistor goes to the output port.
Hope that makes sense! I can sketch out a schematic if that would be more helpful.
In case you're interested, I'll try to explain why you can't daisy-chain the resistors. In an LED circuit like this, we use the resistor to "set" the amount of current (amperage) that's running through the circuit. An LED is almost like a piece of wire in that it'll allow almost unlimited amounts of current to flow through it. (It's a wild oversimplification to say it's "like a piece of wire", of course, but it's pretty accurate as far as this feature of allowing unlimited current.) That's why we need these resistors - they serve to limit the amount of current that can flow so that we don't overload the LED or melt wires. For resistors, there's a simple linear relationship between voltage, current, and resistance known as Ohm's law. The linear relationship means that if you know any two of the quantities, you can calculate the third. In an LED circuit, we know the voltage and the target current, so we can calculate the resistance. That's what you did when you plugged the various values into the LED calculator to figure the size of resistor you needed. And this is where the daisy-chaining problem comes in. When you calculated the resistor size you needed, you used the current flowing through ONE LED. If you hooked up two LEDs in parallel to a single resistor, you'd have two separate LED paths for the current to flow through, and each path would need that same amount of current. The paths converge at the resistor, so TWICE the current would have to be flowing through the resistor. But the voltage hasn't changed, so going back to Ohm's law and that linear relationship between voltage, current, and resistance, we'd have to HALVE the resistance to accommodate that double current. If you hook up three resistors, you'd need 3x the current and thus 1/3 the resistance. And so on. But wait! It gets worse! For our situation, it's not just a matter of dividing by N if you want to hook up N LEDs in parallel. With the flashers, each one turns on and off individually. So when one flasher is on and the other four are off, we need the original resistance. When two flashers turn on at the same time, we need 1/2 the resistance. When 3 turn on at the same time... Anyway, this is completely unworkable because we can't have the resistance change on the fly according to how many flashers are operating at any given time. So the only way around this is to dedicate a resistor separately to each current path.
Edited by mjr, 01 July 2016 - 09:59 PM.
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