HID Ballast Circuit - Kirchoffs Law

[Shiju]

Hi All,

its understood that HId lamps operate at 110V while its running, eventhough the i/p supply voltage is 230V...........the remaining voltage being dropped across the ballast.
Now what confuses , is the current drawn by the lamp and the input current.
for example.. if the lamp wattage is 1000W, input current to the circuit will be 1000/230V (neglecting losses,p.f etc).the current drwn by the lamp should be 1000/110v (if im not mistaken)..........
but it makes no sense as one end of the lamp remains connected to the neutral,........
Going back to basics what Kirchoffs law says is current entering in the circuit should be equal to current leaving the circuit.........

 

[cbarn24050]

Hi, yes in a series curcuit the current is the same in all components. Your problem is that you decided to ignore pf.

 

[steve066]

The ballast probably contains a transformer. The transformer steps down the voltage to the 110V the lamp needs, and at the same time, it steps up the current.

Kirchoffs law doesn't apply because the transformer makes two separate isolated circuits, thus the current can be different.

Steve

 

[cbarn24050]

Hi Steve, you might be right if a ballast was a transformer, but it isn't, it's a choke.

 

[Shiju]

Hi Again,

Yes, Cbarn the ballast is just a choke in this case, the ordinary reactor type not the Autotransformer type.
If PF is considered as Steve suggests
For a PF of 0.85, Lamp wattage = 1000W, Ballast loss = 50W

Input Current, Ip = (1000 + 50)/(230*0.85) = 5 .37 A

As HID lamps operate at 110V whole they are working,

Lamp Current, Ic = 1000/ 110 = 9 A

Which is contradicting as the circuit is a series one……

 

[cbarn24050]

Hi Shiju, where are you getting the pf value from? it should be just under 0.5 from the figures you supplied, or is there a pf capacitor that you haven't taken into account.

 

[Shiju]

Cbarn, its the PF capacitor i have accounted...
pls. note these arent measured values...im going by the theory....

 

[cbarn24050]

From your answer I'm not sure wether you now understand or not. You still havent explained where you got your pf of .85 from.

 

[steve066]

Looking back at your original post, you said:

"if the lamp wattage is 1000W, input current to the circuit will be 1000/230V "

This is incorrect. Nothing in the ballast or lamp operates at 1000W and 230V. You are taking a wattage from the lamp and the voltage from the input to the ballast. You can't use the voltage at one part of a circuit and the power at another part to find current.


Steve

 

[tinfoil]

You have to be careful with the use of the term 'power factor' when referring to lighting, as it is sometimes used to describe ballast 'conversion efficiency' rather than the phase angle between current and voltage!

 

[Skogsgurra]

To make things even more confusing:

There are three power factors: the Displacement Power Factor, the Distortion Power Factor and the Total Power Factor.

The former is defined to be cos(phi1) where phi1 is the angular displacement between voltage fundamental and current fundamental.

The total power factor is defined as P/S where P is power and S is appearent power. The symbol for it is often the greek letter Lamda (looks like an inversed "y", which can be seen on ballasts.

If power factor is being used for "conversion efficiency", then that particular manufacturer does not know what he is doing.

 

[tinfoil]

Good overview of ballasts:

http://www.advancetransformer.com/literature/pdf/FLB_Pocket_Guide.pdf