I have a problem about making an negative regulator, which have parameters following
+ Vin = 40 V
+ Vout = -12V
+Iout = 1A
I need schematics not use transformer. Help me please
Thank you every one seeing the thread!!

 

i have used following schematic, but Io very small. approximatly about 10 ma

 

Everyone, please help me now
I really need it

 

Hello,

Take a look at figure 7 on page 10 of the attached PDF for an idea.

Bertus

 

Cant be done without an inductor.

 

have you got schematic "Buck boost converter using tl494"

 

Cant be done without an inductor.

ofcourse, inductor is really needed. but i have not schematic yet. can you help me

 

Mr Kid,

You need to learn how to use a web search to find technical literature on electronic parts.

Have you looked at the TI web page on the TL494? I found it instantly using Google.

 

Sign up here and put in your numbers.

http://www.ti.com/lsds/ti/analog/we...ntr_en&HQS=sva-web-webdesigncntr-vanity-lp-en

 

Everyone, please help me now
I really need it

Gee, only 22 minutes after posting a question you are already following it up with a demand that everyone drop everything they are doing and immediately give you what you want.

Wish I could say that was a record, but I seem to recall someone else doing it in under ten minutes a year or so ago.

 

Cant be done without an inductor.

Is that true? I haven't looked at it, but it seems like a flying capacitor circuit could accomplish it (though not as well, I would imagine).

 

Pretty hard to get -12V at 1A with only a capacitor

 

Pretty hard to get -12V at 1A with only a capacitor

Is it? I doesn't seem too unreasonable -- but I've never built one so perhaps I am missing something.

I was actually thinking you were saying that an inductor was required in order to get a negative voltage from a positive DC supply at all and so I was only thinking in terms of getting a negative voltage. Of course the OP's second post demonstrates that a flying capacitor approach will produce the negative voltage (albeit it with the puny drive capabilities of a 555), so I should have figured that you were talking in terms of the current requirements.

It seems that the key thing is that your flying capacitor is only changing by the ripple voltage on each cycle, call that Vr, and so the amount of charge transferred each cycle from input to output will be C·Vr and at a switching frequency F the current will therefore be:

I = C·Vr·F

If we spec a 100 mV ripple voltage and a 100 kHz switching frequency, then we are only talking about a 100 uF flying capacitor. So none of that seems unreasonable.

The driver has to be able to charge the capacitor in half a cycle, so if we use five time constants (we can probably get by with just three and if we increase the driver voltage a bit can even get by with two or less, but let's use five) then the output impedance of the driver needs to be

5RC = (1/F)/2

R = (1/CF)/10 = (Vr/I)/10

So for 100 mV of ripple at 1 A, you need an output impedance of 10 mΩ. I don't know how realistic that is, though if the constraints can be relaxed then you can probably get by with 0.1Ω or so. Nor do have I a feel for how low impedance you need for the driver (and inductor, for that matter) in an inductive circuit.

But I do agree that an inductive approach seems for more practical in most situations.