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I know the "D" mean diode, so it has a built in diode. I would not think that the addition of a built in diode would lower the current rating by that much. Also, if I were to use a l293, where would the diodes go and in which orientation (I am driving a bipolar stepper)?

Also, would I be able to increase the current rating of the l293D somehow? I really don't think putting them in parallel works, even though many people say it does. The datasheet doesn't say if it is a FET or transistor based system, but I suspect it is transistor based.

electricviolin
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I agree that it is surprising that it is lower for the D version - the data sheet doesn't explain why although possibly it is because they use the same die and use some of the output transistor area for the diodes in the diode version - thus reducing the current capability.

The data sheet (TI L293 data sheet)shows that it is a bipolar part.

It is not usually desirable to parallel bipolar parts as they don;t share current well having a negative temperature coefficient.

The data sheet on page 8 shows where the diodes are - basically one diode fro each output to VCC2 and ground such that they only conduct when there is inductive kick from the load.

Kevin White
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  • I just found something interesting... see this other post: http://electronics.stackexchange.com/questions/23702/can-i-wire-the-two-sides-of-a-l293d-dual-h-bridge-together-if-i-only-need-one-h - it suggests that they can be put in parallel with a few wires moved around... would this work for a stepper motor? – electricviolin Jul 28 '15 at 14:38
  • Good find - that looks like it should be OK. It is important to parallel outputs in the same package as they will match better than between two different packages both from component variation and temperature point of view. Probably two outputs of a L293D would be good for about 2amps max. – Kevin White Jul 28 '15 at 15:26