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Control your motors with L293D and Arduino

Step 2Advanced implementation

Advanced implementation
This is the correct Implementation (with the capacitors), and note that pin 8 is feeded by unregulated voltage. This means that if your motors need more than 5V, you should power this pin with that amount of voltage, and the rest of the circuit with 5V.

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7 comments
Apr 28, 2009. 11:07 PMticapix says:
Thanks for the article :) Can you explain or give to link which explain why you put capacitors ? (and why this way ?) Diodes aren't good ?
Apr 28, 2009. 11:45 PMWyle_E says:
The capacitors from the motor leads to ground absorb noise generated by the motors. Motors are notorious for generating spikes that interfere with nearby logic and sensor circuits. The big capacitor across the battery smooths the supply voltage, keeping the internal resistance of the battery from dragging the bus voltage down when the current into the circuit suddenly increases (like when you start a motor). The parallel pair of capacitors from +5V to ground filter the logic-circuit supply. The big electrolytic capacitor soaks up low-frequency trash, but isn't so good at handling high-frequency noise, like that generated by fast-switching logic. The .1 microfarad capacitor is a ceramic or polyester type that handles the hign frequencies.
Jan 21, 2012. 3:18 AMbooze.net says:
This post has been a great help to me. I had numerous problems affecting my logic as I hadn't implement the ceramic capacitors detailed in your post. I was tearing my hair out and temporarily abandoned my project. Thank you very much.

Paul
Apr 29, 2009. 12:09 AMticapix says:
thanks Wyle_E :)
May 15, 2010. 4:33 PMmoose4987 says:
 so, say i were to use a motor to turn a solar panel... if i used the voltage from the solar panels which vary from 6-9 volts to power the arduino, could i also use that unregulated voltage to power the L293? 
Apr 28, 2009. 3:07 AMxchip says:
"This means that if your motors need more than 5V, you should power this pin with that amount of current, and the rest of the circuit with 5V." you seem to mix the concepts "voltage" and "current"

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