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Powering Fluorescent Ballast IC

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Fluorescence

Electrical
Oct 19, 2008
42
Hello All,

I am making a dimmable fluorescent ballast for fluorescent light using IR21593 Fluorescent driver IC......(datasheet)


Pagfe 11 of this application note shows a full circuit set-up involving the IR21593....


I wish to power this IC from a 14 Volt rail which i have, but i am wondering if this is actually possible, since it looks like it can only be powered by a charge pump running off the half bridge?

....The datasheet (figure 5, page 15) shows this IC being "self" powered by a charge pump involving the diodes and capacitors in figure 5.

-I do not wish to do this as i have a 14 Volt rail which i can use.....however, it looks to me as if i possibly have no choice -it looks like the IR21593 can only be powered by this charge pumup arrangement?

-This IC also contains a high side driver and the energy for this is gotten from capacitor C3 (figure 5, page 15).....i am wondering how the charge would get into this capacitor if i didn't use the charge pump (i.e. if i just used my 14V rail straight into Vcc)?

Any considerations greatly appreciated.
 
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Seems simple enough. Looks to me that you can eliminate D1, D2 and C2 and just feed 14V into pin 13. Leave D3 and C3 connected and power for the high side driver will come from your 14V supply.

 
Hello , thankyou for your reply.

I hear you say that it can be powered from my 14 Volt rail and that is a relief to me.

Also, considereing the power supply for the high-side gate drive (D3 and C7 on pins 13,14,15 of IR21592 of page 11 of the following app note:


)

....I cannot see how any current can get through diode D3....since it will always be reverse biased. Therefore, the high side drive would have no volts to drive the upper Mosfet?

Sorry to keep asking about this.
 
Think about what happens when M3 turns on.

pin 15 = 0V
pin 13 = 14V
the capacitor C7 charges
 
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