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View Poll Results: why the efficiency of FWR is double that of HWR.
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  #1  
Old 09-27-2009, 12:23 PM
ashuprasad1990 ashuprasad1990 is offline
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Post efficiency of half wave and full wave rectifiers

as we know that the efficiency of a half wave rectifier ~=40.6% and that of a full wave rectifier is 81.2%,nearly double that of the HWR...so why this is so...the reason i know is that in HWR only one diode is used and due to rev.biasing it does not conduct so it looses the negative part and in FWR there are 2 diodes(center tapped and also bridge 2 diodes conduct at a time) so its efficiency also doubles....but my professor isn't satisfied with this he wants to know something else.........help needed
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Old 09-27-2009, 01:37 PM
t_n_k t_n_k is offline
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Originally Posted by ashuprasad1990 View Post
as we know that the efficiency of a half wave rectifier ~=40.6% and that of a full wave rectifier is 81.2%
Where did you get these values and for what circuit conditions do they apply? Are you considering the rectifier in isolation from all other circuit components such as isolation transformers?
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Old 09-27-2009, 03:58 PM
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beenthere beenthere is offline
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Just as a thought problem - imagine a capacitor with a constant load drawing current. What difference might be between full wave and half wave charging off a transformer?
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Old 09-29-2009, 01:39 AM
ashuprasad1990 ashuprasad1990 is offline
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yes sir,in this case we are only concerned with their efficiency's in case of transformers
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Old 09-29-2009, 01:42 AM
ashuprasad1990 ashuprasad1990 is offline
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Exclamation eficiency in case of isolation transformers.

yes sir u r right,i want to know teh reasons for the case of transformers
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Old 09-29-2009, 02:28 AM
Skeebopstop Skeebopstop is offline
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The transformer efficiency depends on the output power and should be read from the transformer spec.

The rectifier efficiency can be easily calculated from the datasheet and input voltage.

For example, lets say Vf (forward voltage) of the diodes is 1V @ 4A RMS. If we consider an RMS in voltage of 100V, the efficiency becomes.

((100V*4A - 1V*4A) / 100V*4A)*100 => 99%.

Now consider a similar scenario, 4A RMS @ 10V RMS in.

((10V*4A - 1V*4A) / 10V*4A)*100 => 90%.

As you can see, the efficiency of a rectifier varies with loading. The same principle applies to Transformers but frequency must also be considered (easier to find a curve in the spec sheet).
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Old 09-29-2009, 02:32 AM
Skeebopstop Skeebopstop is offline
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With relevance to my previous post, you must define what you mean by efficiency. Efficiency is typically a term applied to power and not to 'rectification purity'.

The power efficiency between full wave/half wave won't be inherently obvious, however the 'ripple voltage' on the output side will be drastically different.

This question can become quite tricky unless you very accurately define it.
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