# Design Of Pfc Device

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### #1 shaha

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Posted 04 September 2006 - 09:09 AM

Hello. I am new to this forum and first I want to excuse me for my probably bad english.
I am working on a device for automatic PFC (using just capacitors, not filters). My job in this project concerns generally weak current parts (microcontroller, indications, control, software). But now I have to read a lot of information about PFC. Unfortunatelly I found in this forum almost only one thing - Everybody repeats and explains that adding capacitors will not change the electricity bill. Well .
Could somebody give me links to some more specialized and detailed (designer's) information. For example at this moment I am looking for information about capacitors switching /Tips and tricks, possible problems, ways of realization and algorithms/.

### #2 jraef

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Posted 04 September 2006 - 06:27 PM

Correction: They will not save on you electicity USE as measured in kWh, but may in fact save money on your BILL. Because a poor power factor causes additional losses at the supply, the utilities often recover those losses in the form of penalties or charges based on power factor. If that is the case, improving the power factor will reduce those penalties or charges on the BILL.
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### #3 raman

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Posted 05 October 2006 - 04:40 AM

Hi,

First why need of pfc,

1.To reduce the harmonics frequency and delivered the fundamental frequency to load input.harmonics frequencies generaly doesn't effect the load, its effect the mounting socket(like heating),circuit breakers,
For examble one CB rating is 15A,and its normal working current is 12A(80% of the full load) without pfc powersupply,s power factor is 0.65 normaly and its efficiency is .85.so the total power deliver to the load is 230*12*.65= 1794w.when using the pfc the power factor has improved to upto .99.so the total power cosumable is 230*12*.99=2732.4w. total difference is 938w.

two types of pfc is there.1.passive pfc .this type of pfc is lesser coast and acive 0.8-0.9.its required some inductace(L)

2.active pfc>this type of pfc required higher coast,switching ic's,switches, x capEtc.this type we can acive upto 0.99.

QUOTE(shaha @ Sep 4 2006, 02:39 PM)

Hello. I am new to this forum and first I want to excuse me for my probably bad english.
I am working on a device for automatic PFC (using just capacitors, not filters). My job in this project concerns generally weak current parts (microcontroller, indications, control, software). But now I have to read a lot of information about PFC. Unfortunatelly I found in this forum almost only one thing - Everybody repeats and explains that adding capacitors will not change the electricity bill. Well .
Could somebody give me links to some more specialized and detailed (designer's) information. For example at this moment I am looking for information about capacitors switching /Tips and tricks, possible problems, ways of realization and algorithms/.

### #4 kens

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Posted 05 October 2006 - 09:52 PM

QUOTE(raman @ Oct 5 2006, 05:40 PM)

For examble one CB rating is 15A,and its normal working current is 12A(80% of the full load) without pfc powersupply,s power factor is 0.65 normaly and its efficiency is .85.so the total power deliver to the load is 230*12*.65= 1794w.when using the pfc the power factor has improved to upto .99.so the total power cosumable is 230*12*.99=2732.4w. total difference is 938w.

Raman, we need to be careful here . You are working at the problem from the wrong end. In your first calculation you have a 230V load drawing 12A from the supply. This equals 2760 VA ( 230V * 12A). Now we put in the pf .65 so 2760 x .65 = 1794W. If we improve the power factor we will have the following calculation 1794W / .99 = 1812 VA or 1812 / 230 = 7.8A. Improving power factor will reduce the VA but not the power (W). The only power saving will be due to lower losses from the reduced current.
Ken
An expert is one who knows more and more about less and less until he knows absolutely everything about nothing

### #5 raman

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Posted 06 October 2006 - 05:13 AM

do you know about boost inductor design,how to find out the ripple of inductor,
raman.

QUOTE(kens @ Oct 6 2006, 03:22 AM)

Raman, we need to be careful here . You are working at the problem from the wrong end. In your first calculation you have a 230V load drawing 12A from the supply. This equals 2760 VA ( 230V * 12A). Now we put in the pf .65 so 2760 x .65 = 1794W. If we improve the power factor we will have the following calculation 1794W / .99 = 1812 VA or 1812 / 230 = 7.8A. Improving power factor will reduce the VA but not the power (W). The only power saving will be due to lower losses from the reduced current.
Ken

### #6 kens

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Posted 09 October 2006 - 08:20 PM

Hi Raman sorry for the delay in my reply. I dont know about the design of these components.
Kens
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### #7 marke

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Posted 09 October 2006 - 11:16 PM

Hello shaha

Reading through your post, I think that you are asking about a different type of power fqctor correction than is gnereally being covered by this forum.
This forum is about topics oriented to motor control and power factor correction is commonly applied to reduce the displacement power factor due to the inductive nature of the motor.
I believe that you are looking for power factor correction as applied to rectifiers in swichmode power supplies etc.
If you are looking for information on power factor correction as applied to motor control, then refer to http://www.LMPhotonics.com/pwrfact.htm

Best regards,

### #8 marke

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Posted 09 October 2006 - 11:33 PM

Hello raman

Reading through your posts, I think that you are asking about a different type of power factor correction than is generally being covered by this forum.
This forum is about topics oriented to motor control and power factor correction is commonly applied to reduce the displacement power factor due to the inductive nature of the motor.
I believe that you are looking for power factor correction as applied to rectifiers in swichmode power supplies etc.
If you are looking for information on power factor correction as applied to motor control, then refer to http://www.LMPhotonics.com/pwrfact.htm
If you are wishing to find out more about the power factor correction of rectifiers, then your question should be posted elsewhere. Perhpas I need to set up a section for switchmode power supplies??

Best regards,

### #9 joshua

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Posted 20 November 2006 - 06:29 AM

QUOTE(kens @ Oct 6 2006, 05:52 AM)

Raman, we need to be careful here . You are working at the problem from the wrong end. In your first calculation you have a 230V load drawing 12A from the supply. This equals 2760 VA ( 230V * 12A). Now we put in the pf .65 so 2760 x .65 = 1794W. If we improve the power factor we will have the following calculation 1794W / .99 = 1812 VA or 1812 / 230 = 7.8A. Improving power factor will reduce the VA but not the power (W). The only power saving will be due to lower losses from the reduced current.
Ken

I'm new to this field and i would like to ask some questions for further clarification. As you mentioned above so after correcting the power factor, the current reading should be 7.8A. Just wanted to check so now the calculation for real power would be 230V*7.8*0.99 = 1776W. It looks like there is a saving of 18W. Can this be true or did I make a mistake in the calculation ?

Thanks.

### #10 jOmega

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Posted 20 November 2006 - 01:01 PM

QUOTE(joshua @ Nov 20 2006, 12:29 AM)

I'm new to this field and i would like to ask some questions for further clarification. As you mentioned above so after correcting the power factor, the current reading should be 7.8A. Just wanted to check so now the calculation for real power would be 230V*7.8*0.99 = 1776W. It looks like there is a saving of 18W. Can this be true or did I make a mistake in the calculation ?

Thanks.

The problem is in rounding off the decimals in your calculations.

Consider:

1794 W / 0.99 = 1812.1212121212121212121212121212 VA

then ...

1812.1212121212121212121212121212 VA / 230 v = 7.8787878787878787878787878787879 amperes

so that ...

230v x 7.8787878787878787878787878787879 amperes x 0.99 = 1793.9999999999999999999999999998 watts

which is so close to 1794 that it can be considered to be virtually equivalent thereto.

Ok ?

### #11 joshua

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Posted 21 November 2006 - 02:02 AM

QUOTE(jOmega @ Nov 20 2006, 09:01 PM)

The problem is in rounding off the decimals in your calculations.

Consider:

1794 W / 0.99 = 1812.1212121212121212121212121212 VA

then ...

1812.1212121212121212121212121212 VA / 230 v = 7.8787878787878787878787878787879 amperes

so that ...

230v x 7.8787878787878787878787878787879 amperes x 0.99 = 1793.9999999999999999999999999998 watts

which is so close to 1794 that it can be considered to be virtually equivalent thereto.

Ok ?

oh ok... now I get a clearer picture. So this so called "energy saver" or power factor correctors do not have any benefits from the consumers point view eh ? It only saves when there is a surcharge or penalties in poor power factor rite.

thanks alot for the great explanation.

### #12 kens

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Posted 21 November 2006 - 02:14 AM

Hi Joshua, PFC can save a SMALL amount of ENERGY from losses in cabling etc. It can at times save a substantial amount of costs depending in how the electricity delivery charges are structured.
Ken
An expert is one who knows more and more about less and less until he knows absolutely everything about nothing

### #13 joshua

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Posted 21 November 2006 - 02:51 AM

QUOTE(kens @ Nov 21 2006, 10:14 AM)

Hi Joshua, PFC can save a SMALL amount of ENERGY from losses in cabling etc. It can at times save a substantial amount of costs depending in how the electricity delivery charges are structured.
Ken

For the household we pay for KWH only. How about for smaller scale industries ? I have seen some bills which they charge for KVARH. So will there be more savings in this cases ?

Thanks

### #14 kens

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Posted 21 November 2006 - 03:45 AM

Hi Joshua, that is correct, inproving the power factor will reduce the kVAr penalty charges. Also if a portion of the bill is charged on kVA demand then this will also be reduced.
Ken
An expert is one who knows more and more about less and less until he knows absolutely everything about nothing

### #15 joshua

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Posted 21 November 2006 - 04:02 AM

just wanted to check, KVAr charge and PF penalties are 2 different things rite ? or is it the same thing ?

### #16 jraef

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Posted 21 November 2006 - 04:41 AM

The penalties are based on excessive kVAR demand, but are the result of poor power factor, so that is why PFC reduces the penalties.
"He's not dead, he's just pinin' for the fjords!"

### #17 AB2005

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Posted 22 November 2006 - 02:55 AM

Dear Joshua,

I would like to guide you to the literature on Power factor. You should read this literature and improve your sense in power factor and its correction. I too, had no more knowledge/experience in power factor correction. But I have improved myself by reading threads in this very useful forum. I also studied the literature in the links bellow.
1 - http://www.ibiblio.o...l#xtocid2508324
2 - http://www.lmphotonics.com/pwrfact.htm
3 - http://www.tvss.net/pq/pf.htm
4 - http://www.nokiancap..._Correction.pdf

"Don't assume any thing, always check/ask and clear yourself".

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