Meopta Recifier KINO 100XM

[MarinBNM]

Hello;

 

I've got question about this rectifier.

 

This is schematic of it:

 

http://gallery.filmvorfuehrer.de/view_photo.php?set_albumName=Open-Air-Cinema-Biograd-na-Moru%2C-Croatia&id=meopta_kino_100xm

 

So, question is what V4 diode is for? Is it just for safety measures if someone switch poles on XBO so lamp won't ignit or some other reasons? Like maybe got something with that 100V DC comming from other transformerr via 300Ohm resistor which is used for lamp striking??

 

 

 

Also something about capacitors C1 and C2...C1 is conected directly and C2 is connected to DC via R3 (for current limiting thrue C2 isn't it?), but why C1 isn't concted via some resistot like C2??? Is this resistor realy need to be here???

 

 

 

Also what is R2 for (intended to don't leave DC without load when XBO iisn't ON) but how when K switch depress when we switch ON rectifier???

 

 

Thanks for answers!!!!!!

 

 

P.S.

If schematic isn't good let me know then i get better resulution, or if somebody needs it for him own!

[stefan2]

Marin,

 

 

V 4 ist a so called Booster diode, to raise, together with an auxiliary rectifiier the no load voltage over 100 V. In this case the main transformer can be built for a significantly lower voltage (near to the operating voltage), and does not have to built up 80 Volts. Capacitor C1 is a filter cap, after L1 as primary filter element, that charges up to let's say 40 Volts. C 3 carges to 110 Volts, V 4 obstructs. If R3 would not be present, the lamp would draw excessive current during start, that might severely demage the electrodes of the lamp. The higher no load voltage is needed to assist in arc striking, but inrush current must be limited C1 does not require the resistor, as is is just charged to operating voltage, therefore does not affect the insrush current at startup.

 

R2 is to decharge the capacitors when switched off. The contact K is part of the main contactor that controls the threephase input.

 

SNIP

 

 

This rectifier is virtually indestructible, V1 and V 4 are of the street railway type, the transformer is well sized, but it does have some obstacles.

On the main rectifier as well as on V4, about 3 Volts are lost, resulting of approx. 300 Watt heat, which is pulled out by means of the fan, another 26 Watts consumption. THe transducer circuit takes 5 Amps at 40 Volts, 200 Watt of idle loss. The Booster circuit looses over resistor R1 during lit condition of the bulb, 120 V - 40 V is 80 Volt at 0.5 Amp, 40 Watt of loss.

 

Altogether around 550 Watts of losses, for a 2,5 kW bulb... thats a total efficiency of around 75%, with power factor around 0,4 compared to 95 to 98 % efficiency in modern switch mode supplies, at power factor 0,99.

 

Unless electricity is nearly free, it pays within a short ammount of time to replace with switchmodes.

 

 

Stefan

[MarinBNM]

V 4 ist a so called Booster diode, to raise, together with an auxiliary rectifiier the no load voltage over 100 V

 

Thanks Stefan for answer!

 

But if we remove V4 we also gonna get 100V from auxillary rectifier?

 

I do some experimets with this kind of rectifier, only transformer is singlephase, and I conect 100V DC directly to main DC feed to lamp via 300Ohm resistot and relay contact which cutt-off 100V after lamp strike and everything work OK. What you think about this, is V4 really nesasery?

[stefan2]

If V 4 is omitted, the lower impoedance of the main rectifier will short circuit the booster, like in striked arc condition. The voltage will reach 40 Volts. One way is to short out V4 with a contactor after lamp is lit. Saves about 100 Watts in power consumption.

 

Hope this litle picture helps, omitting everything unneccessairy.

 

Stefan

[MarinBNM]

That idea to short diode after powering lamp is good to save energy! :D

 

I think on You're schematic that you didn't put booster voltage to negative, you put that wire with diode on plus, didn't then other wire from transformer must come to - pole of DC?

 

This is mine schematic, it realy work I try it wright now I'm striking 900W lamp without any problems!!!

 

What You think about it?

 

[MarinBNM]

OK, problem why lamp couldn't ignit with diode was in first smoothing capacitor (newer used, but obviously junk :lol: )

 

Now everything working wright like on this schematic, only C1 capacitor and cables to him heats too much, but I think that OK because this is single phase rectifier, so has much ripple voltage then three-phasse.

 

[THX]

Hello,

 

by the way: The 110V stand-by Voltage-Supply is able to drive a 500W Xenon bulb with approx. 5A ;) You only have to strike the lamp manually and for quite a long time (1sec or so). But never try this ;) I don't think that these Selen-Rectifiers like the current of 5A very much ;)

[stefan2]

OK, problem why lamp couldn't ignit with diode was in first smoothing capacitor (newer used, but obviously junk :lol: )

 

Now everything working wright like on this schematic, only C1 capacitor and cables to him heats too much, but I think that OK because this is single phase rectifier, so has much ripple voltage then three-phasse.

 

 

Marin. my way of connecting the booster gives a boosted voltage higher than yours, as the no load voltage of the rectifier is added to the booster. Therefore you might use a smaller transformer (2 x 24 Volts in series=48V) will normally do. 70 Volt secondary is not that common to get.

 

Assuming No Load around 35 Volts, 48 * 1.4142 = 65 Volts plus 35 V is 100V boosted voltage.

 

THreephase theretically has 3.8% ripple, if fully balanced, unfiltered single phase 48% ripple. So with singlephase inductor filters MUST be utilized, as current ripple shall be minimal. Capacitors do not help to reduce current ripple. Inductors are adding much to weight and discomfort in carrying around.

 

Stefan

[MarinBNM]

Stefan, thanks for help, but I have 100V from auxillary tranformer, so i haven't any problem about no-load volatage now! but you're idea iv very good because we can use smaller secondary volate for booster voltage!!