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I came across this formula for rectification efficiency of half wave rectifier:

η= (40.6 *resistance of load resistor )/ (resistance of load resistor+ resistance of secondary of transformer + resistance of semiconductor diode)

I couldn't understand how is it derived, although I am familiar with the derivation in which no resistances are considered and efficiency is obtained as 40.6 %.

Can someone help me with the logic behind this formula?

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  • \$\begingroup\$ An ideal diode is 100% power efficient so, what do you mean when you talk about efficiency? \$\endgroup\$ – Andy aka Mar 10 at 14:11
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    \$\begingroup\$ efficiency is obtained as 40.6 % In my opinion that is nonsense. If you understand how a half wave rectifier works, it is evident that the efficiency scales with voltage. Rectifying 2 V AC is much less efficient than rectifying 200 V AC due to the influence of the diode's forward voltage, which doesn't vary much. I've said it before and say it again: electronics isn't about formulas, it is about understanding how things work, that can then be described by using formulas. \$\endgroup\$ – Bimpelrekkie Mar 10 at 14:15
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    \$\begingroup\$ That is a poorly written article! They claim that the efficiency is 40.6% but fail to clarify how they calculated the \$P_{dc}\$ and \$P_{ac}\$. In a proper article they would have mentioned \$P_{dc}\$ and \$P_{ac}\$ first and show how their values are determined. There is no mention of \$P_{dc}\$ and \$P_{ac}\$ before mentioning that efficiency so in effect what they write lacks context and is meaningless. The effciency can be 40.6% but only under specific conditions. The fact that that is not mentioned says enough, whoever wrote this doesn't fully comprehend what is going on. \$\endgroup\$ – Bimpelrekkie Mar 10 at 14:25
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    \$\begingroup\$ Adding to bimpelrekkie: If you consider efficiency as output power divided by input power, you have to ask "where was the power lost?" It's typically lost as heat, where current times voltage is that power. In a half wave rectifier with an ideal diode, with no resistances, then anyaka has a good point - where is the loss? Just because it's not conducting doesn't mean it's lower efficiency. For example, a class c amplifier is kind of like a half wave rectifier in that in conducts only part time - yet is more efficient than class A which conducts all the time. \$\endgroup\$ – KD9PDP Mar 10 at 14:41
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    \$\begingroup\$ I’m voting to close this question because it requests the justification of a false premise. \$\endgroup\$ – Charles Cowie Mar 10 at 14:52
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Half wave is 50% max with no load

So 40% means 80% of peak or a drop of 20% from impedance ratio with ripple and DC drop. Optimum depends on ripple tolerance and transformer losses typically design for 10% for some acceptable ripple .

Since charging duty cycle is limited to only when AC-pk>DC out, the peak current/avg DC current is about the same ratio in order to maintain <=10% ripple, this brings down efficiency another bit with cap ESR and diode Rs.

But how they defined these assumptions determines the optimum ratio.

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