Question
Question: In half wave rectifiers a p-n diode with internal resistance \[20\Omega \]is used. If the load resis...
In half wave rectifiers a p-n diode with internal resistance 20Ωis used. If the load resistance of 2kΩ is used in the circuit, then find the efficiency of this half wave rectifier.
Solution
Hint It is given that a P-N junction diode has an internal resistance of 20Ω and there is a load resistance given. Directly substitute the values in the efficiency of the half wave rectifier formula and calculate the percentage value.
Complete Step By Step Solution
A Half wave rectifier is a type of rectifier which is tuned to allow only one half cycle of alternating transformation wave to pass and block the other half , which converts one half AC voltage into DC voltage. This rectifier requires only one single diode to achieve this operation.
Rectifier efficiency is defined as the ratio of the amount of DC power obtained by the AC input voltage given. It is mathematically given as ,
η=Pdc/Pac=rf+RL0.406RL
Where RLis load resistance value and rfis the internal resistance value of the diode. In our given question, rfis said to have a resistance of 20Ω and Load resistance of 2kΩ. Substituting the values in the above formula we get,
⇒η=(20+2×103)0.406(2×103)
Further simplifying we get,
⇒η=2020812
⇒η=0.401
Taking the percentage value, we get
⇒η=40.1%
Therefore the efficiency of the given half wave rectifier is 40.1%.
Note
A rectifier is a device that is predominantly used to convert alternating current that reverses its direction in periods , to DC current, that is a straight flow of current. A rectifier circuit requires a set of p-n junction diodes based upon the rectification scale. There are two types of rectifier namely, half wave rectifier and full wave rectifier. A full wave rectifier consists of two power diodes that are connected to a single load resistance, with each diode functioning in each periodic cycle of the alternating current.