Electronics 1b - Diodes as rectifiers Flashcards
What is Half-wave rectification?
When a diode is connected in series with an AC power supply, only allowing the positive voltage (converting it to DC), this creates a fluctuating voltage output but one that is only positive.
This voltage output is slighly lower by 0.7V as the diode requires 0.7V across to allow current flow.
What is full-wave rectification?
When four diodes are connected into a specific configuration to allow current to flow in both the positive and negative output of the AC power supply, however the output voltage is also only positive (converted to DC).
There is a slight drop in the output voltage as two diodes in each direction are required so 0.7V + 0.7V = 1.4V drop from origional AC voltage supply.
The output voltage still fluctuates from 0 to Vout, but now the frequency has doubled compared to input frequency
How do we smooth and reduce ripples in the output voltage for a full or half-wave rectifier?
Add a smoothing capacitor that is parallel to the load.
This means that there is a minor reduction but Vout doesn’t entirely go back to 0 volts.
This difference in the voltage at its peak and when it at its lowest once discharging throught the load is the ripple voltage amount
How do we calculate the ripple voltage (ΔV) and the average voltage in the full-wave rectifier with a smoothing capacitor?
ΔV = i(DC) / 2fC
Vavg = Vp = i(DC) / 4fC
How do we calculate the value of the smoothing capacitor required for a half-wave rectifier?
Since the output frequency is the same as the AC frequency, then we need to increase the capacitor value by a factor of 2.
