Wait thats wrong, it should be
8 go to 12V
12 go to 5V
1-9 go to 5V
2-15 go to logic 1
7-10 go to logic 2 (opposite of logic 1)
6-11 go to coil side 1
3-14 go to coil side 2
4-5-12-13 go to GND
So is this correct?
Repost it with the label "Pin-" in front of the pin numbers and the word "Input" or "Output" where appropriate, so it reads
Input X => pin Y where "X" and "Y" represent whatever the number should be.
(VCC2) Pin 8 go to 12V
(VCC1) Pin 16 go to 5V
(1,2EN - 2,3EN) Pin 1 - Pin 9 go to 5V
(IN 1 - IN 4) Pin 2 - Pin 15 go to logic 1
(IN 2 - IN 3) Pin 7 - Pin 10 go to logic 2 (opposite of logic 1)
(OUT 1 - OUT 4) Pin 3 - Pin 14 go to coil side 2
(OUT 2 - OUT 3) Pin 6 - Pin 11 go to coil side 1
(GND) Pin 4 - Pin 5 - Pin 12 - Pin 13 go to GND
Correct.
If you have any small heat sinks and thermal paste you can put a heatsink on top of each chip.
Test it with a 100 ohm resistor as a dummy load in place of each winding first.
Measure the voltage across the dummy load resistors for ON and OFF and post that info
for feedback. If it appears to be working correctly shut off the power and replace the resistors with the motor windings.
You can also substitute a LED in series with a 470 ohm resistor in parallel with a duplicate set with the polarity reversed for each winding so the GRN led turns on for one direction and the RED led for the other direction. Do that for each winding (two leds , opposite polarity with resistors for each winding. Don't parallel the resistors, parallel the led -resistor series combinations.
I do have an old heat sink, but not thermal paste. I think it should be fine though because I am splitting the load among two ICs now (a single chip got very hot, but never blew).
I only have 1/4W resistors on hand - will that work for this test?
Yes
12V/0.020 A = 600 ohms (470 ok)
12/470= 0.025 A
P = I*V = 0.025 * (12V-2.2V) = 0.245 W (1/4 W)
I did the whole setup, and the motor turned. However, one side of one IC got very hot, while the others didn't. The motor was running, so both halves of all the ICs had to be doing something. I am not sure why only one half of one IC got hot.
Did you test the chips with resistor dummy loads and measure the voltage drop across them ?
I did, but it just fried the resistors... maybe they were 1/8 watt whoops...
Should I try to flip the signal wires or something (so that the bottom of the IC is getting the signal the top used to and vice versa). I am thinking I can then see if the bottom of the IC gets hot. I am assuming I would also have to flip the output wires then as well?
Can the L293 be used in parallel?
Well - as you've found, people have said they've done it with no ill effects...
That said, realize this: If the L293 could be used in such a manner, don't you think it would be mentioned in the datasheet? The fact that it isn't, is a big clue...
Here's the deal: The L293 is a bipolar transistor logic device. As such, you are not supposed to parallel such devices together normally - mainly because the devices won't generally share the load equally, and you'll get an unbalanced circuit, overload, then burnout of one or both devices.
The only way around this, is how the L298 is constructed - if you read the datasheet for the L298, you'll find that it is composed of two independent h-bridge drivers (for a single bipolar stepper, or two separate DC motors) - and that those drivers can be paralleled to allow you to drive a single DC motor. Even though the L298 too is a bipolar transistor based device, it has one important characteristic over the L293: Both drivers (well, actually the transistors which make up the drivers) share the same die. As such, they are highly matched in spec (having been essentially built at the same time), plus by sharing the same die, thermal differences are minimized, and are balanced. Now - you can't parallel two separate L298 ICs - so don't try that...
So - if you want to have a snowball's chance of this working (and there would still be no guarantee) - first make sure that both of your L293 ICs come from the same manufacturing batch. The best way to do this would be to purchase a factory sealed tube of the ICs - any two of the ICs sequentially should be very close to each other in spec. You might also be able to grade the devices with a testing rig, so as to determine which match closes to each other in operating characteristics.
Then - mount them both piggyback style, with some kind of copper strip between each pair, some heatsink grease or tape or putty - and make sure everything is tight together. Solder all of the legs together. Add a heatsink to the upper IC (more paste, etc) - then connect both sinks together (if using copper strips, use copper bolts - if aluminium - use aluminium bolts). You might want to also force cool them with a fan. The idea here is to basically heatsink them in such a way so that they will share the heat output load as much as possible. It isn't as ideal as being on the same die, but it is as close as you can get with these chips.
Don't be surprised though, if despite a ton of hard work, if none of that works or helps.
I think I am going to stick with the method mentioned previously as it seems to sort of be working.
The L298 modules are cheap enough that it really isn't worth the trouble to mess around with L293s when you need more current.
I am assuming I would have to use a heat sink and diodes with it?
I am eventually going to put the whole project on a printed PCB, so I am going to remake the circuit if needed rather than using a module.
If you use the L298 you need a heatsink or it will burn up as soon as you apply power.
You also need the diodes,they aren't present in the "chip".
You also need the diodes,they aren't present in the "chip".
The L298 modules have all that already.
Why don't you just get a module and mount it with standoffs and make an interface cable to connect to your arduino circuit ? Why would you try to make your own L298 pcb when you can get it premade so cheap ?