My teacher told me today that most modern welders use more power when just operating without welding; than actually when you use it to weld. Is this true, and how could that be?

This came up after we found a battery charger (very old) in the storage. I took it apart and it had a huge capacitor in it. I assumed it was for filtering, but my teacher said it was for power factor correctioning.

Can anyone please explain how PF correctioning works, and how important it is in modern electronics?

Thanks

 

I wouldn't believe what your teacher says about the welder.

Many loads are inductive, the mains ones being transformers and motors.

An inductor has an impedance, and , so you'd think that , well it does but what that doesn't account for is the power factor. Only resistors can dissipate power, in reality just multiplying the voltage by current gives VA, the apparent power.

Hang on a second, I forgot, it's much easier to link to Wikipedia rather than explain it myself.

http://en.wikipedia.org/wiki/Power_factor

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so the capacitor is used to shift the phase angles from the transformer? Is it for when the battery charger is just left on without charging a battery? All it basically is, is a huge transformer, some heavy rectifiers, two fuses, a mechanical timer, and a huge capacitor.

 

The capacitor is probably just for filtering ripple current. Why would would want power factor correcting in a battery charger is beyond me.

 

That's what I told my teacher and he practically embarrassed me in front of my class mates the way he was talking to me. If anyone has any experience or comments on why this would need power factor correction please post

Also I don't understand what he means by that welder thing. Doesn't make sense, I mean the no load voltage for most transformers is higher(current increases) is that what he meant? That would theorectically raise the current in most cases.

 

Your teacher is not correct. The welder is drawing useless (inductive) current when not welding, but it is not power. Power factor correction saves the power company from having to supply the useless (inductive) current but the power meter does not measure it, so it does not cost you. I have not measured the power used by my welder but I would bet that the power usage is lower when not welding!

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Power factor correction is done by means of capacitor banks at point of supply usually near the metering point, from a customer, to reduce the line current drawn from the supply.
These capacitor banks are switched on and off by means of a ct sensing circuit and set to keep the PF at around 0.95 - 0.97.
Some power boards measure kVArh and kVA as well and do apply extra charges for it hence a PF control will pay itself back in 1 to 2 years..

Charger
It depends where the capacitor of that charger was placed.
If across the incoming mains terminals 240 or 415 Volts (some chargers may be 2 phase supply) then it is for power factor correction.

If at the secondary side after the bridge rectifier then it would be for filtering so you have to check that out.
Normally the battery or accu will act like a type of capacitor and will smooth the ripple anyway.

Welder
Your teacher is wrong here!
An unloaded welder will draw magnetising current and some copper losses at a low PF say 0.6.
When in use the welder will use a lot more power depending upon the current setting and type of welding rods used.
In addition to the losses above, it will draw the welding current as well from the mains.
The powerfactor may improve a bit but will still be lagging at around 0.8 PF.

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There again you don't need a large smoothing capacitor in a barrtery charger.

Is this an electrolytic capacitor?

What's its voltage rating and capacitance?

If you can prove the teacher wrong then he'll be more embarrased than you can ever be!

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I "reverse engineered" the charger today and the capacitor is connected to the secondary. The secondary has two seperate coils that are connected to each other with four outputs. My teacher still won't except the denial on the welder thing, and what the capacitor is for?

So why would a capacitor be used for power factor correction on the secondary side instead of the primary side?


I still think it is for stablizing the power output for DC battery charging, I'll post a schematic of what the charger contains

 

In that case Sceadwian is right it's a smoothing capacitor and has nothing to do with power factor correction.

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Here's a schematic of the charger:



The motor is a mechanical timer which is quite interesting

I'm not quite sure if the transformer is constructed the way I pictured it... I have to take it apart tomarrow. In addition the value of the capacitor is arbitary as I didn't get it due to heavy rust on it

 

Thanks for the schema.

That Cap at the secondary is a small smoothing capacitor for the DC side of the charger.
It is probably there for reducing the ripple current a little when small accu's are charged up e.g. 12 Volts 2 Ah ( alarm type accu ).
There is no electronic circuitery in the charger which may require a smoother DC ?

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Thanks for the schema.

That Cap at the secondary is a small smoothing capacitor for the DC side of the charger.
It is probably there for reducing the ripple current a little when small accu's are charged up e.g. 12 Volts 2 Ah ( alarm type accu ).
There is no electronic circuitery in the charger which may require a smoother DC ?

The transformer is a typical centre tapped transformer, which in this case only need two rectifier diodes.

A single winding transformer has the usual 4 diode bridge.

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Nope, all it consists of is a large transfomer, a mechanical timer, two heavy power rectifiers, the capacitor, two fuses, and an ammeter. (This is a very old charger ;P) Simple but effective

 

There are several apparent errors in that schematic.

The capacitor is in a very strange place.
The battery and the meter are connected the wrong way round.

JimB

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