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If the diode is in the right direction, then the current will flow through it, if not, it won't, correct?
I'm thinking...burnt orange...
And normally, a diode works this way, but in the drawing above, the current would technically not flow in the diode.. |dw:1343337901676:dw|
Keep in mind that "conventional current" is reverse to what the electrons actually do. :-)
Diodes are like one-way valves in a water pipeline system. They put up extreme resistance if you go the wrong way.
"conventional current" = the direction of positive charge flowing around a circuit
|dw:1343339611472:dw| So the 'real current' flows like this?
Was there a question in all of this ? I'm confused :P
But how do you explain this? D:
You have an LED with a current limiting resistor and it's forward-biased, so it should light up when the NPN transistor is on :)
As long as the anode is about 0.6V-0.7V more positive than the cathode, current will flow through it from anode to cathode. Is that what you're asking?
Well the two diodes are in different direction and current still flows in both, that's what I'm not getting :(
Current won't flow through the one on the left...since the cathode is more positive than the anode.
Here's the circuit: http://www.oomlout.com/oom.php/products/ardx/circ-03
If the current doesn't flow through it, why did they put it there?
The diode on the left is a protection/blow-out/flyback diode to protect the motor...it's not meant to have current running through it
It's pretty common to see that diode configuration in solenoids and small motors....they prevent voltage spikes (from the collapsing field of the inductor) from damaging the motor.
I probably didn't word that very well but in essence it's a protection device for the motor.
I see, I thought that diode was use to block current or something, thanks!
I haven't studied diodes in many years so I'm a little weak on the full theory behind it...but it's there to prevent voltage spikes from damaging the load when the field of the inductor collapses. In normal operation, there should be 0 current going through it.
Alright, I think I got the concept, thanks again!