I'm new with capacitors. I see many benefits in using them.

What I don't know is alot, but striving to understand.

I know they work as a short term battery. I see they build up current that my low powered solar array can not supply to run my TSO. I now know how to wire them to run in multiple configuration.

Testing them for failure is something else...

What I'm hoping for is a thread that will not only teach me, but will help many more of us to understand and work with capacitors.

Old in age, not in mind, so 'Teach me something new'!

Capacitors , a practical guide ( in several postings # 3 )

Using the discharge from a capacitor to charge a battery

There are quite a few circuits that use solar panels to charge a capacitor which , once a certain voltage is reached is "dumped" into a battery.

A few assumptions have been made in this example , in the real world other factors will have an effect on the actual charge transfer ( the capacity amp/hrs , internal resistance of the battery etc ) so please treat as a guide only

A typical "dump" voltage seems to be approx 21 volts with a charge battery of 12 volts and the cap bank is in the order of 140000uF

Going back to part 1 , E(joules) = 1/2 xC x V^2

E(joules) = 1/2 x 140000 exp 10-6 x 21^2

E(joules) = 30.87 joules ( thats quite a "kick" ) per cycle

HOWEVER WE HAVE A PROBLEM !

We have calculated the amount of energy stored being fully discharged into the battery , this is not the case because the battery is not at zero volts

lets assume the battery is at 11 volts

so voltage difference is 21 volts down to 11 volts

So how much energy is "dumped" into the battery per cycle

= E(joules , fully charged voltage ) - E(joules , at battery voltage )

E(joules) = 1/2 x C x V^2

E(joules) = 1/2 x ( 140000 x exp 10-6) x 11^2

E(joules) = 1/2 x .14 x 121

E(joules) = 8.47

So total energy dumped into battery is

30.87 joules - 8.47 joules = 22.4 joules

Now we can see why the battery takes longer to charge , also as the charge on the battery increases the terminal voltage also increases and the charge likewise drops off.

PLEASE NOTE I MADE AN ERROR IN THE CHARGE DIFFERENCE RATIO THIS HAS NOW BEEN CORRECTED , SORRY

This is great! I'm kinda scared to ask anything right now, cause I'm working to keep up..

Just the way I want it!

But... If I may.... size matters?

I can look at a capacitor the size of my finger nail vs the size of my fist and be the same voltage and uF. Okay, maybe a bit exaggerated, but you get the thought. Larger size means longer duration of release? Smaller means quicker, but more often? Virtually meaning the same thing?

Old in age, not in mind, so 'Teach me something new'!

I would venture to say that, for example, if you are looking at 2 capacitors, both 75V, 10,000 uF, and they vary largely in size... you are looking at different construction technique, and not a whole lot more.

Also, keep in mind there are low ESR electrolytics, polarized and non (AC or DC), ultracapacitors, and supercapacitors. And others. In these cases they do indeed have different operational characteristics.

Cheers

~~~~~~~~~~~~~ You wouldn't laugh at my igloo if you knew how cold my beer is!

Kyle is basically correct , construction technique / materials. Here's an example:-

High quality caps are made using metal foil strips seperated by a suitable insulator , the same uF value could also be made using a metalised film deposited on both sides of a suitable insulator ( quality not so good ). So whats the difference.? The biggest difference will be the ESR , the lower the better. ESR effects the rate at which the capacitor can be discharged and the operating frequency of the capacitor , also reduces the amount of heat generated in the cap when it's being "pushed" ( hence extensive use in switched mode power suppllies and high end audio amplifiers )

I also mentione "suitable insulator" this can range from Mylar film through to PTFE again the higher dielectric constant for the insulator the closer the "plates" and the lower the ESR

A capacitor has capacitance , resistance and inductance ( honest ! ) the design effects these 3 values and at what frequency they come into effect. Needless to say it's a big subject , but at the end of the day.

Firstly lets assume we have some perfect capacitors ( ones that don't exist but text books are full off ) Real capacitors will always have a leakage current.... no such thing as a perfect insulator/dielectric

Capacitors behave differently depending on if AC or DC application.

A capacitor will "block" the flow of DC BUT it will allow AC to pass/flow through it. The amount of current that the capacitor will allow to flow through is dependant upon frequency ( ie it's AC reactance / resistance varies with frequency )

If you charge up a capacitor with DC voltage it will hold / store the charge until either you "release" it or it falls away due to internal leakage current.

"Are capacitors AC/DC particular?" ... some types of capacitor can only be used in DC applications ( namely electrolytics ) other types such as polystryene , polypropylene , ceramic etc will work on both AC and DC

The role of the capacitor in the circuit is also a factor to be considered.

The formula tell's you the capacitive reactance for a given value of capacitance at a given frequency. the Ohms Law equation simply allow the calculation of current flow at a given voltage for that particular capacitance/ frequency combination.

But what about this combined capacitors working voltage ? the answer is you always pick the lowest working voltage value so in this example it's 25 volts

Series connected

C(total) = 1 . _________________________________

. 1 + 1 + 1 . _____ _____ _____

. C(1) C(2) C(3)

C(total) = 1 / ( 1/10 + 1/47 + 1/2.2 )

C(total) = 1 / ( 0.1 + 0.0212 + 0.4545 )

C(total) = 1 / 0.5757 = 1.737 uF

This time we add the voltages so our "new" capacitor is 1.737 uF at 175 volts ( well 175 volts at least in theory for safety assume about 40 volts or less ! )

PLEASE NOTE it is deemed bad practice to use caps of different working voltage in this arrangement ( especially when cap uF values are different ) because you cannot assume that each capacitor will "sort out" its own actual working voltage

It would be wise to work on about 1/4 of the above value ( ie approx 40 volts )

However if all three capacitors were identical in both uF and working voltage , then a value of working voltage of 2.5 times the working voltage could be assumed.

ALSO PLEASE NOTE it is normal practice to add a high value resistor across each capacitor to balance out the voltage across each capacitor (equipotential arrangement)

Kinda jumping ahead of myself, but... Lesson 4 has a point I would love to see gone into deeper.

"ALSO PLEASE NOTE it is normal practice to add a high value resistor across each capacitor to balance out the voltage across each capacitor (equipotential arrangement)"

I will happily await as not to mis-direct the course.

Old in age, not in mind, so 'Teach me something new'!

"Are capacitors AC/DC particular?" ... some types of capacitor can only be used in DC applications ( namely electrolytics ) other types such as polystyrene , polypropylene , ceramic etc will work on both AC and DC

This is a sigh of relief..

This though raises yet another question that I need to do some work with; "The formula tell's you the capacitive reactance for a given value of capacitance at a given frequency. the Ohms Law equation simply allow the calculation of current flow at a given voltage for that particular capacitance/ frequency combination."

I'm working with my Solar Panels. I never thought of testing frequency.. Now I see I must!

You asked..."hope that helps" I truly hope you know how much all of this does!

Old in age, not in mind, so 'Teach me something new'!

"ALSO PLEASE NOTE it is normal practice to add a high value resistor across each capacitor to balance out the voltage across each capacitor (equipotential arrangement)"

The best way of thinking about this is as follows :-

If you had a power supply of say 50 volts and you wanted to supply a small amount of current( say 10 milliamps = 0.01 Amps ) at 25 volts , you could use a voltage divider consisting of two resistors and since 25 volts just happens to be half of 50 volts then each resistor must have an identical value.

Likewise , 4 identical resistor ... each resistor would carry 1/4 of the supply voltage across it

When electrolytic caps are connected in series it is necessary to ensure that each cap has an equal voltage across it

Reason

1) very difficiult to get two identical electrolytic caps ( therefore voltage could be different across each cap )

2) electrolytic caps must never be allowed to "reverse potential" themselves ( its a chemical reaction that forms the dielectric between the plates of the cap , if this breaks down then so does the cap..... sometimes with spectular / explosive results !!! )

3) Dielectric breakdown will occur if the rated working voltage is exceeded ( see 2) above

So to prevent this situation a resistor is connected in parallel with each capacitor then each cap / resistor pair are connected in series to reach the cap / voltage value required. ( each cap has an identical charge voltage which is within its rated working voltage )

Also when the power is removed each cap discharges through its "own" resistor thus maintaning the correct polarity on the capacitor

Capacitive reactance ONLY APPLIES to AC , your solar panels only provide a DC output

"In guide #3 you go into figuring the cycle draw by substracting start from finish voltage in Joules.

Can the same be done for the amperage solution?"

Has with all things in life , nothings easy ! so here we go :-

In section 3 the amount of energy transferred from the capacitor to the battery was calculated and this energy was measured in joules. I also stated that 1 joule = 1 watt per second so yes we can approximate and give an average current flow per unit time. So why approx and average value ? The answer is that

1) a capacitor will discharge in an exponential fashon ( follows a decay curve )

2) the voltage on the cap is falling per unit time and likewise the voltage on the battery is increasing per unit time but most likely not like for like

3) as the voltage falls on the cap and rises on the battery the quantity of current flowing ALSO reduces with time

4) a capacitor , discharged through a very low resistance does so with a decaying oscillatory waveform ( ringing )

THIS METHOD IS ONLY APPROX AND SUITABLE FOR COMPARISION ONLY ( TOO MANY VARIABLES )

So given the above it is easier to simply measure the time the cap "dump" lasts

Lets assume a starting voltage of 21 volts down to 11 volts , difference 10 volts

And a 20 joule charge "dump" for a duration of 1 second

So 20 joules = 20 watt / seconds

since Watts = volts x amps ( and since it lasted 1 second )

20 watts = 10 volts x amps so amps = 2

So the average current flow was approx 2 amps ( REMEMBER , APPROX GUIDE ONLY ! )

Please remember, this is a 'work in progress' and needs to be tweeked by 'those in the know'.

This is built in Excel, but should also work in Open Office. If, for any reason you can not run it in Open Office, let me know and I will re-write in their program.

Old in age, not in mind, so 'Teach me something new'!

Ron_o brings out a very important point above, regarding the voltage divider across caps...

When you build a capacitive transformer, if for example it is a step down, the proper term for that is in fact "voltage divider"...

and, if you are doing this from mains power...

The explosive results he speaks of, can be pretty exciting.

Balancing is pretty crucial when playing with those circuits, which, can be, by the way, huge money savers, as discussed by Goerge Wiseman at Eagle Research.

Large Transformers are really pretty expensive!

~~~~~~~~~~~~~ You wouldn't laugh at my igloo if you knew how cold my beer is!

I've added the capacitors due to 'not' having the power to run the circuit. So I considered them as a 'step up' or leveler. I'm now understanding that I need to balance the capacitors and hope the conversation will go deeper in configuration and determination of resistance/proceedure required.

I always felt that the first capacitor recieves charge, then the next and so on, till the circuit requires the current built up. I had a feeling that each capacitor was recieving the same power and the stronger ones were absorbing first, then resisting and allowing the weaker to catch up. I didn't think of this as a problem as much as I thought they are balancing themselves. It all happens so fast!

My first thought with balancing went to 'blocking diodes'... no reverse direction, yet still allows the weaker inline to catch up.

Any way I twist this around in my head, balancing makes sense.

quote:Originally posted by kcarring

Ron_o brings out a very important point above, regarding the voltage divider across caps...

When you build a capacitive transformer, if for example it is a step down, the proper term for that is in fact "voltage divider"...

and, if you are doing this from mains power...

The explosive results he speaks of, can be pretty exciting.

Balancing is pretty crucial when playing with those circuits, which, can be, by the way, huge money savers, as discussed by Goerge Wiseman at Eagle Research.

Large Transformers are really pretty expensive!

~~~~~~~~~~~~~ You wouldn't laugh at my igloo if you knew how cold my beer is!

Old in age, not in mind, so 'Teach me something new'!

Resistors ( to balance the voltage ) are only necessary across capacitors when capacitors are wired in series.

if a bunch of capacitors are connected in parallel ( as in your solar charger ) it is better to use capacitors of equal uF value , this is to balance the energy approx equally between each cap. This balanced charge is also very important if a cap dump circuit is being used.

BTW photoflash caps work the best in this application

Bleed resistors are used to "bleed" power from capacitors when the supply voltage is removed.

In the case of charge/voltage balancing resistors as when capacitors are wired in series , these resistors serve a similar function when power is removed as the "bleed" resistor as mentioned above.

Hi Ron talking of cap dump My set up is using 2 10 watts solar polycrystalline panels connected in series connected to a sg oscillator ckt and charging two sla 12v 7 ah batteries What size of cap is needed for dumping into the 2 batteries ? My solar out put range from 32 to 36 v dc .... when connected to the circuit is 13 vdc and 2 output range from 12.99 / 12.77 I want to optimized the voltage juice from my solar set up Do I need a cap in the input as well??? thanks totoalas

I would suggest a starting value of 47000uF and this should be connected to the solar panels via a 6 amp blocking diode this in turn would then be connected to your sg oscillator.

A "cap dump" circuit is basically a capacitor that is charged, being ( at some defined voltage ) discharged into a battery This requires a simple voltage comparator circuit ( usually a thyristor , zener diode and a couple of leds ) , the SG oscillator would not be required.

Have just had a look at your Excel spreadsheet , looks ok to me ( i don't use spreadsheets , i tend to write my own software to do the calculations , saves me having to learn how to use them !, lol . So i'm not the best person to test it)

ron

I will do a couple more example calculations for you to run through .

Apprieciate that Ron! I'm not so savy on writing programs, so the spreadsheets. We all use what we can! LOL! As long as we all get there.

quote:Originally posted by ron_o

Hi Olddawgsrule

Have just had a look at your Excel spreadsheet , looks ok to me ( i don't use spreadsheets , i tend to write my own software to do the calculations , saves me having to learn how to use them !, lol . So i'm not the best person to test it)

ron

I will do a couple more example calculations for you to run through .

Old in age, not in mind, so 'Teach me something new'!

I have not reached voltage (before dump) that you're talking of. Typically mine dump before they reach 21v. With that said, I did buy 35v capacitors. Probably a little low for voltage from a safety stand point.

But... they are matching units and the price was right...

If you have not purchased yet... Ron mentioned; "photoflash caps work the best in this application".

This is something I'm going to look into deeper.

I'll let Ron correct me if incorrect. I purposely purchased capacitors of a higher rate than I thought the running max voltage would be. If you're expecting higher, than mine are under rated for your use.

quote:Originally posted by totoalas

Hi Ron talking of cap dump My set up is using 2 10 watts solar polycrystalline panels connected in series connected to a sg oscillator ckt and charging two sla 12v 7 ah batteries What size of cap is needed for dumping into the 2 batteries ? My solar out put range from 32 to 36 v dc .... when connected to the circuit is 13 vdc and 2 output range from 12.99 / 12.77 I want to optimized the voltage juice from my solar set up Do I need a cap in the input as well??? thanks totoalas

Old in age, not in mind, so 'Teach me something new'!

The electrolyte capacitors I'm using are considered 'bleed'... The 'photo-flash' style would be considered...??

Two things hit me right off the bat.

If I'm looking for longer duration, then bleed sounds right. But, if quick, which I assume photo-flash would be, then repetitive quick charge and release could be better.

Many assumptions being taken here...

I need to get a few of these and see what they do!

Thank you for the suggestion! Or should I say BTW... LOL... I am paying attention and appreciating every minute of this!

quote:Originally posted by ron_o

Resistors ( to balance the voltage ) are only necessary across capacitors when capacitors are wired in series.

if a bunch of capacitors are connected in parallel ( as in your solar charger ) it is better to use capacitors of equal uF value , this is to balance the energy approx equally between each cap. This balanced charge is also very important if a cap dump circuit is being used.

BTW photoflash caps work the best in this application

Bleed resistors are used to "bleed" power from capacitors when the supply voltage is removed.

In the case of charge/voltage balancing resistors as when capacitors are wired in series , these resistors serve a similar function when power is removed as the "bleed" resistor as mentioned above.

ron

Old in age, not in mind, so 'Teach me something new'!

We may be getting our "wires crossed" here , so here we go :-

A charged capacitor can retain it's charge for quite some time and if for example in the good old days of valves ( tubes ) it was common to have capacitors that were charged to several hundred volts ( or more ! ) obviously if you came into contact with one of these you would get quite a shock. To reduce this shock hazzard a high value resistor ( about 470,000 ohms ) was connected across the terminals of the capacitor , so that when power was removed from the circuit the capacitor would slowly discharge through the resistor. Hence the term "bleed" resistor ( slowly "bleed" the power away )

Certain types of capacitor if allowed to stand even after being fully discharged can self charge back up ( this is due to retained ( stress energy ) in the dilectric being released , again the "bleed" resistor stops the charge build up.

caps.... due to the method of manufacture of electrolytic caps not only do they have capacitance but they also have inductance and internal resistance. The internal inductance and resistance limit the rate at which a capacitor can discharge.

Photoflash caps are designed/manufactured to minimumise inductance and resistance and therefore can discharge at a higher rate , hence being more suitable for cap dump battery charging circuit applications.

With regard to working voltage selection for caps , a good rule of thumb is 1.5 volts times the maximum expected working voltage , so your use of 35 volt working caps is spot on.

What's to stop you using 100 or 200 volt caps , you know that your circuit will be safe then ? Answer is caps rely upon the voltage across the plates to maintain the di electric, if you go way outside the normal working voltage of your application the cap suffers ( typically much lower capacitance than what it should be )

BTW When i learn't to program computers , the few commercial programs that were about "cost the earth" and so you had to learn to write your own and it's just "stuck" with me. Now of course their are a wide range of programs available , but i just hate having to learn to use them so where ever possible i write my own code.

As to; With regard to working voltage selection for caps , a good rule of thumb is 1.5 volts times the maximum expected working voltage , so your use of 35 volt working caps is spot on.

That was yet another assumption of not going to high in voltage, but yet trying to remain safe. I'll call it 'dumb luck' for what I knew then about capacitors...

I have several 300-450v capacitors from my salavging, but just didn't seem right to be using them. Now I know why!

Old in age, not in mind, so 'Teach me something new'!

Capacitors , a practical guide ( in several postings # 5 )

Please Note a "typo" error was made in one of the examples :-

V(charge) was stated to be 12 volts , it should have read 14 volts , the correct value of 14 volts had been used in the calculation.

My thanks to fellow IAEC member , Olddawgsrule for pointing out this error

The following examples should be taken as guides / approx only ( the equations used are correct , capacitors are not perfect ! )

example 1

Let us assume that we have a small electric motor that draws a constant load of 2.4 watts when it is powered by a DC supply who's voltage ranges from 14 to 10 volts

What size capacitor would run the above motor for a period of 5 minutes ( Cap start voltage 14 volts , switchoff voltage 10 volts )

Since the motor draws a constant load of 2.4 watts over the stated voltage range we are going to use the CONSTANT POWER DISCHARGE equation.

C(farads) = 2 x t(seconds) x P(watts) / ( V(start) ^2 - V(over) ^2 )

C(farads) = 2 x 60 x 5 x 2.4 / ( 14^2 - 10^2 )

C(farads) = 600 x 2.4 / ( 196 - 100 )

C(farads) = 1440 / 96

C( farads ) = 15

So in the "ideal" world a 15 Farad capacitor , charged to 14 volts and then discharged through the above motor until the voltage on the cap was 10 volts would keep the motor running for 5 minutes ( 300 seconds )

The above equation can of course be re-arranged to tell us how long a particular value of capacitor would power the motor

t(seconds) = C(farads) x ( V(start)^2 - V(over)^2 ) / 2 x P(watts)

Using the values from the previous example BUT using a 47000uF cap ( 0.047F )

t(seconds) = 0.047 x ( 14^2 - 10^2) / 2 x 2.4

t(seconds) = 0.047 x ( 196 - 100 ) / 4.8

t(seconds) = 0.047 x 96 / 4.8 = 0.94 seconds

CONVERTING CAPACITANCE TO NOMINAL BATTERY CAPACITANCE ( AMP HRS )

Ah = 0.5 x C(farads) x V(charge)^2 / 3600 x Nominal Battery Voltage

let V(charge) = 14 volts and Nominal Battery Voltage = 12 volts

An LED is driven by a constant current source ( this means that no matter what input voltage is used , providing it meets the minimum working voltage and is no more than the maximum circuit design voltage , that the current drawn/supplied will be constant )

Assume that the charge voltage is 14 volts ( V(start)) and the cutoff voltage ( V(over)) is 10 volts and that the total constant current drawn by the LED and its constant current source is 60 milliamps ( 0.06 amps )

Lets say we need it to run for 4 hours

C(farads) = 0.06 x 4 x 60 x 60 / ( ( 14 - 10 ) C(farads) = 0.24 x 3600 / 4 C(farads) = 0.24 x 900 = 216 Farads

Please Note..... this is not the best way to power a LED , i have used it just to explain the equation and how to use it. If for example the LED was driven at a 50 % duty cycle the average current would be half and so the capacitors value could also be halved

Not a problem , the next section will be looking at time constants , quantity of charge at a given time ( charge/discharge ). This will then be put to good use to show how to calculate capacitor capacitance (Farads) for a cap discharge battery charger.

Capacitors , a practical guide ( in several postings # 7 )

The equation for capacitor discharge is :- Vc = Vo x e^(-t/RC)

where

Vo = capacitor fully charged voltage

Vc = voltage "removed" from capacitor

t = time to discharge capacitor from Vo to Vo - Vc volts ( ie voltage remaining in capacitor ) note that t is a negative number

R = resistance in Ohms through which the capacitor discharges

C = capacitance in Farads

NOTE :- R x C is known as the time constant

Example

A 0.132 Farad capacitor is charged to 35 volts , it is then discharged for a period of 0.5 seconds through a 8 ohm resistor . How many volts have been discharged and how many volts are remaining in the capacitor ?

Vc = 35 x e^(-0.5 / 8 x 0.132)

Vc = 35 x e^(-0.47348)

Vc = 35 x 0.62283 = 21.79905 volts ( "removed" from capacitor )

Voltage remaing in capacitor = 35 - 21.79905 = 13.20095 volts

Capacitors , a practical guide ( in several postings # 8 )

In part 7 the capacitor discharge equation was introduced , this will now be re arranged to be more useful in the design of capacitor pulse battery charging.

Here comes the math :-

Vc = Vo x e^(-t / RC )

Vc / Vo = e^(-t / RC )

In( Vc / Vo ) = ( -t/ RC ) ( Note In = log to base e , also known as "natural" log )

In( Vc / Vo ) / -t = 1 / RC

-t / In( Vc / Vo ) = RC

-t / R x In( Vc / Vo ) = C ( in Farads )

Example

What value capacitor ( in Farads ) when charged to 35 volts and discharged to 14.5 volts through a 5 ohms resistor in a period of 0.35 seconds would be required ?

Note voltage "removed" from capacitor = 35 - 14.5 = 20.5 volts

Hi about Caps What is the difference in Start Eun and KVAR capacitors Im planning to install on anything that rotate in my house lol and want to gave at least near unity in my aircon washing machine dryers etc etc can you please show us the way..... thanks

These as the names suggest are used to start and run single phase AC induction motors / compressors etc. The Start capacitor is used to provide the initial starting torque to get the motor / compressor running, at some point ( approx 75 % of full speed ) the start capacitor is switched out of circuit and replaced by the Run capacitor.

Run capacitors are most usually Polypropylene film types ( remain in circuit until motor switched off )

Start capacitors are usually non-polarised Aluminium Electrolytic types ( can only remain in circuit a short while )

KVAR type capacitors are made with a high grade metallized polypropylene film with "self heaing " properties in the event of dielectric breakdown. KVAR capacitors are used to correct power factor ( ratio of working power to total power ). Typically power factor correction is only used in 3 phase applications.

In single phase applications the capacitor that actually runs the motor is chosen to produce maximum efficiency .

hI rON Thanks for the link ... have read some long time ago Just to share my adventure with kvar I assembled two run caps in parallel and plugged it on same outlet with my ac unit .... I manually switch it on when I switch on my Ac Without load it draws 5 amperes ( Caps) My bill last summer 700 per month for 3 months w/o caps 6 pm to 6 am on a 500 watt 50 hz window ac unit Tested for 2 months on same time frame with caps - my bill 400 Then i brought to Philippines and tested with a 5 yr old 500 w a/c unit my analog meter slowed down when the caps are switched on.... The 20 yr old meter exploded 4 months after the test ( not related and maybe to old age) the new meter is digital and need to test again but this time with correct KVAR capacitors Im planning to make another try using remote control units ( for garage door opener) to energized contactor with discharged resistor / pig tail wires for our a/c and washing machine..... My objective to make it near as possible to unity . and see..... for the KVAR sizing - using Lifasa table..... Our electric utility rates is the number one in Asia for the most expensive power rates for the past 10 years ..... We need to do something with normal 36 to 38 deg C weather and washing machines the two basic use for this application Any ideas are welcome totoalas

Obviously your experiments confirm that power factor correction does work. The power loss / waste problem is mainly due to power losses in the supply cables to the equipment/load. To cure/fix these losses it is necessary to connect the power factor correction capacitor as close as possible to the load ( you should aim for 95% of demand unity as a maximum )

Thanks ron For the crash course in Capacitor 101 lol Since the KVAR Ive seen available in the market is 5 KVAR and the calculation does not meet the available value Series or parallel combination will do ???? Any recommendation on handheld kvar instrument ( cheap one hehehehehhh) as I will check it out in Zuhai China maybe UNI T which is a reliable brand will do.... what are the things to look for a meter on a single phase application ...... thanks again ........

Sorry about the length of that article , but it does the job far more eliquently than i could.

I don't know of a manufacturer of cheap power factor meter. Power factor as you can imagine is quite a major issue with big energy users and so finding a meter that can measure low correction factor requirements is your main requirement.

If you point me in the direction of a manufacturer i will have a look at the models.

RonO I think your quizing me here! In the below Guide I see what I believe to be a mistake. Hopefully I pass the quiz...

quote:Originally posted by ron_o

Capacitors , a practical guide ( in several postings # 5 )

CONVERTING CAPACITANCE TO NOMINAL BATTERY CAPACITANCE ( AMP HRS )

Ah = 0.5 x C(farads) x V(charge)^2 / 3600 x Nominal Battery Voltage

let V(charge) = 12 volts and Nominal Battery Voltage = 12 volts

Ah = 0.5 x 15 farads x 14^2 / 3600 x 12 volts

Ah = 7.5 x 196 / 43200

Ah = 1470 / 43200 = 0.034 Ah

ron

Shouldn't this line; Ah = 0.5 x 15 farads x 14^2 / 3600 x 12 volts Read; Ah= 0.5 x 15 farad x 12? Being 12 volts is the V(charge)? Then resulting in .025Ah?

Old in age, not in mind, so 'Teach me something new'!

I think i know why you thought the equation was wrong , topic 5 was dealing with capacitors as batteries . Now since we should not discharge a battery below its nominal battery voltage ( in this case 12 volts ) we only really have the full use of a battery between it's fully charged voltage and its nominal voltage. This is why in the capacitor example only the energy ( expressed as amp/hrs ) available from 14 volts down to 12 volts is calculated. Obviously , if you wished to you could calculate to any "discharge" voltage.

Has i said previously , these examples are written "on the fly" ( brain to keyboard ) i know ( or at least i think i know, lol ) what i'm doing / heading but obviously i'm not fully explaining whats going on , again sorry. I shall be more careful in future.