When having a resistor(R) in series with a diode(D), i keep seeing that in order for the diode to be conducting electricity the voltage on the branch R+D should be higher than the diodes potential. But what happens to the voltage drop on the resistor, why doesn't it get calculated?
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\$\begingroup\$ I assume with diodes potential you mean forward voltage, don't you? \$\endgroup\$– CurdCommented Aug 9, 2017 at 22:41
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\$\begingroup\$ Yes, i think that answers my question. \$\endgroup\$– user159536Commented Aug 9, 2017 at 22:42
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You only have to make sure the voltage across R + D is higher than forward voltage of the diode, because the resistor has to take what is left; unlike the diode a resistor conducts even at voltages close to 0V.
