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Back again, Was "Generate -100V"

D

Denis Gleeson

Jan 1, 1970
0
Dear all

Many thanks to all who contributed to my question about generating
-100V. Given
below for the readers information.

To test the possibilities with the LM3578 I was using a TIP30A as the
pass transistor.
I followed Figure 21 in the data sheet. (some one said its at
http://cache.national.com/ds/lm/lm1578A.pdf)
I used a 22uH 0.7A inductor because it was handy. I was also building
the thing on a plug in test board.

To follow up on some of the replies:

I said I didnt want to use a transformer and this is for a couple of
reasons that may not actually be valid. Im un-familiar with their
use(this is valid). I perceive them as being bulky and I need to make
a small PCB size for the product. I think they are expensive and if I
can get away without one Ill save money.

So if anyone can enlighten me and tell me where Ill get a suitable,
small and inexpensive transformer then Im interested.

Winfield Hill had some interesting things to say:
Note, to provide 100mA at 100V, you'll draw 833mA average from
your 12V supply. This translates to an inductor current ramp
with a 1.66A peak. Or actually 1.89A peak, taking into account
the 88% duty cycle with a 12:100V voltage ratio.

There's an efficiency issue with the LM3578 used as an inverting
regulator, due to its high Emitter Saturation Voltage, over 1.2V,
wasting 10% of your power right up front. So you'll need over 2A
peak capability. Oops! Sorry, the LM3578 is limited to 750mA.

I wonder if you would explain these figures further Winfield.
I cant see where the 833mA or 1.89A peak come from. Im not questioning
them I
just need to understand how they are derived.

I think the transistor I chose was ok but the inductor selection seems
to be a big issue for me. Anybody got any pointers on what size and
rating inductor I need.

Many thanks to all again.

Denis
 
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