# how to improve voltage reguration.

Discussion in 'Electronic Basics' started by KM, Oct 19, 2004.

1. ### KMGuest

I have use a LC3803 Linear regurator to design a flyback regurator. A
tronsformer is wind with single primary and double secondary coil (one
for 3volts and one 7volts) the FB was return from 3 volt output. It is
found that the output of 7 volts when unloaded is floating high to 13v
when only 3volts output are loaded. It is found that the overshoot is
causing the raise. What is the possible solution to this?

2. ### John PopelishGuest

Are the 3 and 7 volt outputs both positive?
Do the 3 and 7 volt outputs share a common return?
What are the rated full load currents for each output?

3. ### Terry GivenGuest

basically your problem is due to the lack of coupling between the
regulated and unregulated outputs. Firstly, ensure tight coupling in
your transformer. Secondly, "stack" the windings - say you have 3T and
7T each for the 3V and 7V supplies. rather than having a 3T and a 7T
winding, use a 3T winding & a 4T winding - in other words, tap a single
winding.

Thirdly, the lack of load on the 7V supply is also a problem. Depending
on the 7V load, a simple solution is a series R such that the RC time
constant is longer than the "overshoot" (which is probably only 1us or
so). Or slap a minimum load on the 7V supply.

Fourthly, you could change your feedback circuit - instead of a voltage
divider connected to the 3V rail only, attach another resistor from the
divider to the 7V rail. select the resistors such that when 3V = 3 and
7V = 7 the voltage divider sits at the correct voltage (whatever that
happens to be). This method basically averages out the errors - instead
of one well-regulated output and one badly regulated output, you get two
moderately well regulated outputs.

cheers
Terry

4. ### Rich GriseGuest

What's a linear regurgitator?

Cheers!
Rich
(no offense intended, KM - I'm making a joke about "Engrish" at your
expense. )

5. ### KMGuest

No, there are individual return seperately
the rated full load are 2.2A for 3 Volts and 1.2 A for 7volts respectively.

6. ### N. ThorntonGuest

Sounds like a good description: feed the current in, then when you
open the switching tr, the inductor regurgitates the energy.

NT

7. ### John PopelishGuest

That rules out sharing currents in the 3 volt winding.
I think I would try winding part of the 7 volt winding bifilar with
the 3 volt winding and putting the rest on top of that. And you will
probably need to add some minimum load to the 7 volt output so that it

8. ### ClarenceGuest

How about a "Snubber" on the unloaded winding to control overshoot? I can not
imagine a supply design without some form of rise-time load for a lightly
loaded output. Ripple control is most important at low currents.

9. ### John PopelishGuest

Isn't that what a small resistive load across the filter cap amounts
to?

10. ### Terry GivenGuest

depends what you mean by small, but yeah. likewise for a resistor in
series with the "low" power winding (in this case that wont work as 1.2A
aint small)

cheers
Terry

11. ### ClarenceGuest

No. It is a value of a resistor which is a load only when the rise time (Dv/Dt)
of the overshoot is fast enough to pass through a Cap in series with the
resistor. An Ac terminator is one way to look at it.

I usually use something like a .1mfd in series with a 100 ohm resistor. Only
conducts the ac component. It was very common when a Vibrator was the AC
source.

12. ### Rich GriseGuest

Would it be way persnickety to do 2 turns of the 7, then 3 bifilar, then
the other 2 of the 7? i.e., not only embed it sideways, but lengthwise,
so to speak. (transformer design is black magic, AFAIK.)

Thanks,
Rich