# Cap input filter rectification efficiency

1. Feb 2, 2010

### bitrex

A question regarding power supply efficiency: The theoretical maximum efficiency of a full-wave rectifier circuit with a resistive load and no smoothing is around 82%. In the calculations to derive that the resistance of the source or the rectifiers does not enter into the equation, so the first question would be: If the loss does not take place in the resistance of the source or in the resistance of the rectifiers where does it take place? I had also assumed that a similar theoretical limit applied to a capacitor input filter, but have been informed that this is not the case and the efficiency can go to 100% with a rectifier and source with no resistance. Why is this the case? Thanks for any insight.

2. Feb 2, 2010

### The Electrician

Where did you read that the "The theoretical maximum efficiency of a full-wave rectifier circuit with a resistive load and no smoothing is around 82%."?

How do they calculate it?

Apparently, they are using the phrase "rectification efficiency" in a non-standard way.

A web search turned up this:

Down near the bottom of the page they say "Rectification efficiency measures how efficiently a rectifier converts AC to DC. It is defined as the ratio of the DC output power to AC input power, where DC output power is a product of the average current and voltage. ", but then they give a calculation for efficiency "A simpler way to calculate efficiency is with V_{DC}^2over V_{AC}^2." which is incorrect.

Then they say "Without smoothing, full-wave rectifiers have 8overpi^2 or 81% efficiency.", which is a number close to your 82%, but isn't the ration of output power over input power.

Here's another web page that makes the same calculation:
http://www.indiastudychannel.com/projects/8-Full-Wave-Bridge-Rectifier.aspx

And, here I see this same calculation:

http://engineering.wikia.com/wiki/Rectifier

I think they are using the word "efficiency" when they should use "effectiveness" instead. To me efficiency means "output power/input power", but they're not properly calculating that.

http://www.toroid.com/custom_transformers/technical_bulletin_1.htm [Broken]

part way down, you'll find the sentence "It is important to note that better efficiency of voltage conversion (as measured by h) can only be obtained at the cost of higher Form Factor, and conversely lower Form Factor can only be obtained at the cost of poorer DC load regulation." Here they use the word "efficiency" in a non-standard way.

So, I guess the answer to your query is that they are using the word "efficiency" in a different way that you expect.

Last edited by a moderator: May 4, 2017
3. Feb 2, 2010

### uart

Hi bitrex, as pointed out by "the electrictian" this is basically just a misuse of the term efficiency.

The terrm "efficiency" normally means the ratio of output power to input power. Here however they are referring to the ratio of DC output power to total output power. Let me sumarize the situation for an ideal FW rectifier.

Pout = Pin (100% efficiency)

But the output power can be split up into two components, one a smooth DC and the other one zero mean AC.

Pout = P_out_DC + P_out_AC (Output is the sum of a DC and an AC component).

The relative power in the DC component is 81% and the relative power in the AC component is 19%. So no power is lost, we simply have,

Pin (100%) = Pout (100%) = Pout_DC + Pout_AC (81% + 19%)

Hope that helps.

4. Feb 2, 2010

### Bob S

In general, the rectification power loss goes into the voltage drop across the diode rectifiers. For a typical silicon diode, the forward voltage drop is ~0.8 volts, so the power loss is 0.6 volts x diode current. If the rectifier is a full wave bridge, two diodes are always conducting the full current, so the power loss is ~1.6 volts x current. If the diodes are going into a capacitive filter, the power loss is even higher, because the current through the diodes is a smaller fraction of the voltage waveform and the peak current is higher, and the voltage drop across the diodes is higher for higher peak currents. Look at Figure 2 for the 1N4004 voltage drop vs current in

http://www.datasheetcatalog.org/datasheet/fairchild/1N4004.pdf

The voltage drop is 0.8 volts for 200 mA. and 1 volt for 2 amps.

Bob S

5. Feb 2, 2010

### The Electrician

Well, that explains it. However, the various sources I linked to have it wrong. For example, the wiki link:

http://engineering.wikia.com/wiki/Rectifier

"Rectification efficiency measures how efficiently a rectifier converts AC to DC. It is defined as the ratio of the DC output power to AC input power..."

They should have said something like:

"Rectification efficiency measures how efficiently a rectifier converts AC to DC. It is defined as the ratio of the DC output power to AC output power..."

But, even then, to get the numbers the OP refers to, and that you got in your example, they should say:

"Rectification efficiency measures how efficiently a rectifier converts AC to DC. It is defined as the ratio of the DC output power to total output power (DC + AC)..."

Further, I would choose to call it "rectification effectiveness". I think that using the word "efficiency" is likely to confuse the reader, as it did the OP.

6. Feb 3, 2010

### uart

Yes I agree, that's how they should have defined it.

7. Feb 3, 2010

### bitrex

Thank you for the clarification, everyone!