170628-2011 EDT
lbmcse:
I am glad you are hanging in.
I suggest that you not worry about the internal working of a diode, whether a semiconductor or tube type. Ideally it is simply a switch that is closed when the applied voltage polarity across the diode is in one orientation. and open with the opposite polarity.
Don't try to analyze a circuit's operation by what particles or charges you think are moving in the circuit. Just accept the convention that in an external circuit (not the source of energy) that positive current flow (whatever it may be) flows from a more positive point to a more negative point.
In any circuit the sum of the voltages around any closed loop is zero.
In any circuit the sum of all the currents at a point (a node) is zero.
In the simple circuits you are looking at you have only one series loop. Thus, at any point in the loop at an instant of time the current is the same, and in the same direction at any point.
At an instant of time the voltage drop across the load is equal in magnitude to the voltage drop across the voltage source, but one has a polarity that has opposite the other such that when you add voltage drops going around the loop in the same direction the sum is zero.
I second GoldDigger's comment.
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