In a basic power supply like this one below, for a positive half-cycle the capacitor is charged up along with the rectified load current.
For negative half-cycle diode is "off", so no conduction from the input, but now the charged capacitor will supply current into load.
Now comes my question(confusion). When the capacitor is supplying current, doesn't it complete a path like this and turn diode "on" during negative half cycle?
When the capacitor is supplying current, doesn't it complete a path like this....
....and turn diode "on" during negative half cycle?
No it doesn't. Current cannot flow back thru the transformer secondary winding and then thru the diode because the diode will block this. The voltage on the cathode of the diode is greater than the voltage on the anode therefore, no current flow. A diode is not "turned on" by current, it conducts current only when the anode is more positive than the cathode.
Note - simplifications on diode characteristics have been used in this answer. Leakage currents when diode is reverse biased means a tiny current flows back thru the transformer winding on negative half cycles of the secondary voltage.
Current flow is from positive to negative terminal. Here it flows from positive terminal of the capacitor pass through load resistance and finally reaches negative terminal of the capacitor.
Assume current is now at the bottom (negative) end of the capacitor. Like you said it has two option (1). Go to negative potential terminal of capacitor which is resistance free path and capacitor is pushing too. (2). travel through Inductor, diode and reach positive terminal (current cannot flow from positive to positive) of capacitor. Think which path it will choose ..
Diode can be ideally treat as a switch. For the sake of understanding, please consider the following two circuits
simulate this circuit – Schematic created using CircuitLab
In left one, positive terminal of battery is connected to anode of diode and negative terminal of battery is connected to cathode of diode. Hence Anode relatively in higher potential than cathode. So when the applied voltage greater than barrier voltage of diode, diode is said to be forward biased.(closed condition) and hence a current flows through the circuit in the arrow direction.
In right one, negative terminal of battery is connected to anode and positive terminal is connected to cathode. So anode is relatively in lower potential than cathode. This makes diode in reverse bias mode, leaving the diode always in open circuit mode. So no current flow through the circuit.
Now consider our case. Here capacitor acts like a battery in negative half cycle.
So positive side of capacitor connected to cathode and negative side of capacitor is connected to anode. It is similar to the open circuit diagram (right side circuit in the first section). Since the anode is relatively lower potential than cathode, it is reverse biased mode. Hence no current flow through diode. Current will flow through load resistor only, as indicated by arrows.
Hopes this will make a good understanding. Thanks.
