lamp circuit w/ capacitor bank? 3

[PaulKraemer]

Hi,

I am trying to get a basic understanding of the circuit in the attachment I uploaded. This circuit is used to power a lamp module in a UV curing system. The primary of the lamp ballast transformer is supplied with 480 VAC. I believe this is transformed to approximately 900-1000 VAC across terminals C and T on the ballast transformer secondary. The C terminal of the ballast secondary passes through a capacitor bank to one side of the lamp. The T terminal on the ballast secondary appears to go directly to the other side of the lamp. Between the capacitor bank and the lamp, I believe the potential transformer is used to transform the high voltage down to a lower voltage that is used by the sensor board.

What I don't understand is what the capacitor bank is doing. If I have 900-1000 VAC across terminals C and T (before the capacitor bank), would I expect the same voltage across terminals CL and L after the capacitor bank, I know that capacitors stores energy in an electric field, so I assume this capacitor bank is somehow conditioning the output to the lamp, but I'd like to get a conceptual understanding of what exactly it is doing.

If anyone can clear this up for me, I would greatly appreciate it.

Thanks and best regards,
Paul

 

[itsmoked]

It appears the capacitors are there to limit the current. In an AC circuit they are another form of resistor. By switching those DK contacts the number of capacitors in series with the lamp is changed. The controller board is on that side of the caps so it can monitor what voltage is being applied to the lamp and control how many caps are in circuit. The lamps have wildly varying conductance that changes with temperature. Cold they don't conduct well, hot too well. The caps allow the controller to manage that.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[PaulKraemer]

Thank you Keith - that is a big help.

If I were to measure the AC voltage between terminals CL and L (across the potential transformer primary and also the lamp), would you expect this voltage to vary based on the how many caps are in the circuit? It totally makes sense to me how this could control/limit the current through the lamp, but if there is a variable voltage across the potential transformer primary, wouldn't this mean that there would also be a variable voltage supplied to the sensor board?

Thanks again,
Paul

 

[Compositepro]

The potential transformer must be sending a signal to the sensor board, not supplying power.

 

[itsmoked]

If I were to measure the AC voltage between terminals CL and L (across the potential transformer primary and also the lamp), would you expect this voltage to vary based on the how many caps are in the circuit?

Yes. Because the DKs are discrete steps you will see changes of, guessing, a couple of volts.
Do be careful as that voltage is likely higher than most Digital Multimeters. It would be safer to measure the secondary controller side and do NOT mix that up.

It totally makes sense to me how this could control/limit the current through the lamp, but if there is a variable voltage across the potential transformer primary, wouldn't this mean that there would also be a variable voltage supplied to the sensor board?

Most definitely. That's OK though as there are ways to deal with that. Notice one cap is always there in-circuit so the controller is always powered even with no tube.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[ScottyUK]

Just something to be aware of because you may see some peculiar results if you measure lamp current and voltage while you are playing with this circuit: discharge lamps have a negative resistance characteristic. That is to say, the greater the lamp current, the lower the volt-drop across the gas column.

Not sure what purpose contact 1DK serves, unless there's a minor drawing error.

 

[Compositepro]

ScottyUK, that is strange. My guess would be that the line depicting a wire that bypasses contact 1DK should not be there. Otherwise there is no way to turn the lamp off.

 

[ScottyUK]

Well, 1LC would drop the primary out so it could definitely be turned off. I agree with you though - the shorting link is an error otherwise what purpose does 1DK serve?

 

[itsmoked]

It's the vertical link that shouldn't be there. There must always be a capacitor connected or the controller can never start controlling.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[PaulKraemer]

Thank you Keith, SkottyUK, and CompositePro for helping me try to understand this circuit. I agree that there must be an error in the drawing. The machine is powered at the moment, but as soon as I can turn it off, I should be able to verify how the capacitor bank and the contactors 1DK, 2DK, and 3DK are actually wired.

The machine I am trying to troubleshoot actually has four identical (but independent) lamp circuits. Three of them work fine, but on the fourth one, the 1 amp circuit breaker (1CB) trips immediately when the lamp ballast is powered. I have been in touch with the manufacturer and they have been given me some steps to determine what is causing this. They say the most likely possibilities are:

(1) Faulty lamp module
(2) High impedence short between the coil and the contacts on mercury relays 1DK, 2DK, or 3DK
(3) Capacitor shorting to ground
(4) Faulty lamp ballast transformer

I will perform all the tests they suggest and try to isolate the problem, but I have never seen this type of protective circuit before (circuit breaker between ground and what appears to be a center-tap on the transformer secondary). I am not sure exactly how all of the above (1-4) would result in excessive current between the center-tap and ground, but I can perform the tests they have suggested safely even if I don't fully understand the theory behind the protective circuit.

Thank you all for your help with this.

Best regards,
Paul

 

[Skogsgurra]

A final note that indicates that the schematic shouldn't be trusted is that the transformer primary terminal 3 and 4 are not connected.

BTW, It seems that this is one of the resonant, starter-less, ballasts (with high leakage transformer) that was discussed one or two years ago.

Can't locate that thread. But it sure is there. Somewhere.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[itsmoked]

Nice catch Gunnar..

I'd ask them for a new "correct" schematic.

Never heard of:

High impedance short between the coil and the contacts on mercury relays

Seems absurd.

They left out: A capacitor that has shorted - NOT to ground.


And indeed what is that CB to ground?!?


Keith Cress
kcress -

http://www.flaminsystems.com

 

[Compositepro]

Just a guess, but that connection may create a ground reference that:

1) Keeps the voltage to ground at 1/2 of the lamp voltage, to reduce stress on the wire insulation.
2) The circuit breaker will trip on a HV ground fault.
3) Gas discharge lamps often require a trigger to start the lamp, and this can be the voltage from the lamp electrode to a closely placed metallic reflector at ground potential. This distance is far shorter than the distance between the two lamp electrodes.

The voltage to the lamp will be very high for starting but once the lamp is lit the voltage will be low due to the capacitor impedance being much greater than the lamp. Thus the potential transformer for feeding a voltage signal to the sensor board. The currant and voltage to the lamp give the power and also an indication of lamp condition.