You should notice that the opto-isolator LEDs are connected to three wires labeled `FWD`, `REV`, and `ENA*. ` These wires deserve a bit of explanation. The FWD, REV, and ENA* lines are the interface between the bridge and the microprocessor. You will notice their is no `ground` signal. When you connect these pins to a BASIC stamp or a 68HC11 or what-ever, combinations
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of 1`s and 0`s on the line turn on different pairs of transistors. The following table lists all possible combinations of input. This works because the processor pins become a connection to ground when they are outputting a logic 0. Thus when FWD is 1 and ENA is 0, the lower right sink is getting current from FWD which it is returning through the pin connected to ENA. At any given time the pin must be able to supply enough current to turn on two LEDs and when set to zero sink the current of two LEDs. With the 560 ohm resistors and a 5V processor like the PIC this is not a problem. From the data sheet on the opto-isolator, each LED has a forward voltage drop of 1. 2V, so (5 - 1. 2) / 560 is 6. 8mA per LED or 14mA of load total. The PIC is specified to be able to drive 20mA. You can replace the 560 ohm resistors with 680 ohm resistors to reduce the current through the LEDs still further. However the LED voltage drop can be as high as 1. 4V and the PIC`s operating voltage can be as low as 4. 5V, in that condition with 680 ohm resistors you would only put 4. 5mA into each LED which is below the 5mA specified in the datasheet. The 560 ohm resistor is the better choice as it maintains sufficient margin. So why go to all this trouble The interface as designed gives you access to all of the interesting combinations of sinks and sources enable, while not allowing for any "illegal" states. An illegal state in a full quadrant...
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