Full Wave Rectifier Efficiency: Why is it More Efficient?

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A full wave rectifier is generally considered more effective than a half wave rectifier due to its ability to utilize both polarities of the AC signal, resulting in a more consistent output. While a half wave rectifier may have a slight edge in power efficiency when considering power out versus power in, it discards half of the AC signal, leading to higher current demands on the diode and increased heat. The full wave rectifier, particularly in bridge configurations, conducts less current per diode, reducing heat and allowing for smaller filter capacitors. This results in better performance and lower ripple in the output voltage. Overall, the full wave rectifier is favored for its effectiveness in delivering a stable and efficient power supply.
dnyberg2
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Is it true that a full wave rectifier is more efficient than a half wave and if so why?
 
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dnyberg2 said:
Is it true that a full wave rectifier is more efficient than a half wave and if so why?

In simple terms, in the half-wave rectifier you throw away half of the AC signal by only usiing conduction in one polarity. The advantage is using fewer components (lower cost). The disadvantage is that for a given output voltage the diode must be able to conduct a larger load current (heat).
 
dnyberg2 said:
Is it true that a full wave rectifier is more efficient than a half wave and if so why?

This is a trick question as stated, IMO. You need to give more context to the question, and give some definition of "efficiency".
 
If you took efficiency as just power out divided by power in, the half wave rectifier would be slightly more efficient than a bridge rectifier (if we used a bridge rectifier) because the bridge rectifier has two diode voltage drops while the single diode just has one.
Otherwise, there is not a lot of difference because the input power to a half wave rectifier is supplied in half wave bursts and most of this power is delivered to the output.

However if they really meant "how effective are the two methods?", then the full wave bridge circuit would win every time.
The diodes of a bridge rectifier are only conducting half as much current as a half wave rectifier.
The peak current used for charging the filter capacitor is lower because pulses are delivered twice as often.
The filter capacitor can be smaller for the same hum-out level.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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