Voltage regulator

[Vladimir]

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english

 

[Mac]

On Wed, 02 Mar 2005 22:05:17 -0800, Vladimir wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english

What is K?

--Mac

 

[Ross Herbert]

On 2 Mar 2005 22:05:17 -0800, vkiryazev@yandex.ru (Vladimir) wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english

Need much more information....

I assume by Uin you are specifying a DC input voltage?

Do you want a PSU with linear regulator or a switch mode unit? How
much space do you have to fit the thing in?

Do you mean that both outputs are rated at 5.5A, ie. total Pout = 70W?

What do you mean by the term K?

You also need to specify what is the average normal load and any short
term peak load to be placed on the supply in order to be able select a
PSU with adequate output rating.

 

[Rob Gaddi]

Basic off-line converter. Try either http://www.powerint.com/ or
http://www.st.com/stonline/prodpres/discrete/vipower/vipower.htm for
places to start.

Vladimir wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english

 

[Rich Grise]

On Thu, 03 Mar 2005 08:39:44 +0000, Ross Herbert wrote:

On 2 Mar 2005 22:05:17 -0800, vkiryazev@yandex.ru (Vladimir) wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english

Need much more information....

I assume by Uin you are specifying a DC input voltage?

I "assume" the opposite, considering it's sped'd at 50 Hz. ;-)

Cheers!
Rich

 

[Fred Bartoli]

"Rich Grise" <richgrise@example.net> a écrit dans le message de
newsan.2005.03.03.17.59.23.837190@example.net...

On Thu, 03 Mar 2005 08:39:44 +0000, Ross Herbert wrote:

On 2 Mar 2005 22:05:17 -0800, vkiryazev@yandex.ru (Vladimir) wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english

Need much more information....

I assume by Uin you are specifying a DC input voltage?

I "assume" the opposite, considering it's sped'd at 50 Hz. ;-)

Isn't 50Hz almost DC?


--
Thanks,
Fred.

 

[Ken Smith]

In article <42275735$0$31418$636a15ce@news.free.fr>,
Fred Bartoli <fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote:
[...]

Isn't 50Hz almost DC?

There are as many decades below 1Hz as there are above it.
--
--
kensmith@rahul.net forging knowledge

 

[martin griffith]

On Thu, 3 Mar 2005 18:32:11 +0000 (UTC), in sci.electronics.design
kensmith@green.rahul.net (Ken Smith) wrote:

In article <42275735$0$31418$636a15ce@news.free.fr>,
Fred Bartoli <fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote:
[...]
Isn't 50Hz almost DC?

There are as many decades below 1Hz as there are above it.
--
I think that the lowest frequency is approx

1/(UTC - when did the Big Bang occur)

(+-3dB of course)


martin

"An eye for an eye makes the whole world blind"
Gandhi

 

[Geir Klemetsen]

"Mac" <foo@bar.net> skrev i melding
newsan.2005.03.03.08.36.48.324467@bar.net...

On Wed, 02 Mar 2005 22:05:17 -0800, Vladimir wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english


What is K?

Can it be Kelvin. Maybe he's thinking on working temperature. Btw: How many
degrees is that?

 

[Terry Given]

Geir Klemetsen wrote:

"Mac" <foo@bar.net> skrev i melding
newsan.2005.03.03.08.36.48.324467@bar.net...

On Wed, 02 Mar 2005 22:05:17 -0800, Vladimir wrote:


Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english


What is K?


Can it be Kelvin. Maybe he's thinking on working temperature. Btw: How many
degrees is that?

Good thinking. -123 degrees C. Now thats *cold*. Forget electrolytics.

Cheers
Terry

 

[Mac]

On Thu, 03 Mar 2005 21:52:09 +0100, Geir Klemetsen wrote:

"Mac" <foo@bar.net> skrev i melding
newsan.2005.03.03.08.36.48.324467@bar.net...
On Wed, 02 Mar 2005 22:05:17 -0800, Vladimir wrote:

Hi all!
I need a voltage regulator:Uin=220ą10%;50 Hz;Uout1=6.3 v;Uout2=-6.3 v;
Iout=5.5 A; K>=150;
Please, somebody hint me something about scheme. (Maybe you know
a link at web page with the design of such regulator).
TIA
PS Sorry for bad english


What is K?

Can it be Kelvin. Maybe he's thinking on working temperature. Btw: How many
degrees is that?

Well, 273 K is 0 C, IIRC. So, if it is Kelvin, then at least we won't have
to worry about the transformer getting too hot.

--Mac

 

[Robert Baer]

Vladimir wrote:

I have designed the circuit of a voltage regulator. It is here:
www.chessrating.narod.ru/voltreg.gif
I have omited an adjustment of output voltage. Both the ICs is the
positive linear regulators.
Could you please look on it and discover any mistakes?
Please take into account my parameters:
Vin=220vą10%, freq=50 Hz,
Vout1=6.3 v,Vout2=-6.3 v,I(load)=5.5 A, stabilization factor>=150.
Thanks in advance.
I persume that you know that there aer some limitations on the old

3-terminal IC voltage regulators.
1) Needs about 2V or so Vin-Vout to operate.
2) Your bottom regulator is seeing twice the input voltage compared to
what the top one sees; make sure the worst case input voltage is no more
than 35V (if i remember correctly).
3) Worst of all, none are speced to pass more than one amp. However that
can be fixed via the use of a power pass transistor connected to the
regulator. The method may be found in older data sheets; all of the
really cool stuff seems to have been dropped form the data sheets over
the years (since the 80's).

 

[Clarence_A]

"Vladimir" <vkiryazev@yandex.ru> wrote in message
news:cf3329a5.0503250751.1fd19466@posting.google.com...

I have designed the circuit of a voltage regulator. It is here:
www.chessrating.narod.ru/voltreg.gif
I have omited an adjustment of output voltage. Both the ICs is
the
positive linear regulators.
Could you please look on it and discover any mistakes?
Please take into account my parameters:
Vin=220vą10%, freq=50 Hz,
Vout1=6.3 v,Vout2=-6.3 v,I(load)=5.5 A, stabilization
factor>=150.
Thanks in advance.

This is a JOKE? Right?

The input is shorted!

 

[John Woodgate]

I read in sci.electronics.design that Clarence_A <no@No.com> wrote (in
<9Y61e.2762$zl.1706@newssvr13.news.prodigy.com&gt about 'Voltage
regulator', on Sat, 26 Mar 2005:

This is a JOKE? Right?

The input is shorted!

A lot of people make that mistake. You CAN'T run two bridge rectifiers
with opposite polarity outputs from one transformer secondary winding.
It result in two parallel, back-to-back diodes directly across the
winding, BANG!!!

If you have only one winding available, it's quite OK to use half-wave
rectification, with doubled-value filter capacitors. There is no net DC
in the winding.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[John Woodgate]

I read in sci.electronics.design that Vladimir <vkiryazev@yandex.ru>
wrote (in <cf3329a5.0504030504.6ea2568c@posting.google.com&gt about
'Voltage regulator', on Sun, 3 Apr 2005:

Hello! Could you please look on this scheme:
www.chessrating.narod.ru/voltreg2.gif
Is it proper way to connect resistor divider to the positive regulator?
(Positive regulator is used for negative output.)
Thanks in advance!

No. The resistor on the right is short-circuited.

All you need is a perfectly ordinary circuit. Then you earth the +
output terminal. Use Courier font:

____
+ o--------| |-----+---+-o +
|____| | |
| [R1] Earth
Input |_______|
| Output
[R2]
|
- o-----------+-------------o -
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Roger Johansson]

Vladimir wrote:

Hello! Could you please look on this scheme:
www.chessrating.narod.ru/voltreg2.gif
Is it proper way to connect resistor divider to the positive
regulator? (Positive regulator is used for negative output.)

It doesn't look right. The second resistor is connected between ground
and ground, and is useless.

I am too lazy to look up the circuit you should use, but I can tell you
how to do it.

Use the standard circuit for the positive regulator you use.

Assuming that the input power supply is floating, a transformer
connection to the mains, you can then safely ground the positive output
from the regulator and use the ground=negative from that regulator
circuit as your negative output.

That negative output will be the same as the negative end of the input
power supply.

If you want to combine this negative power supply with a positive one,
remember to connect their common ground to the positive output of this
negative supply, and no other point.

In short, build the circuit just like you would build a positive power
supply, then use its outputs as a negative power supply.


--
Roger J.

 

[Robert Monsen]

Vladimir wrote:

Hello!
Suppose, i have a required voltage on the output of linear
regulator(Vout=6.3),
but output current is too high (Iout=7.5). I need Iout=5.5 A.
How I can work it out? I think about a follow way. (It is quite
possible that it is wrong.)

I should increase output voltage, then choose two appropriate
resistors:

|----[ R1 ]----o V=6.3; I=5.5A
V>6.3; Iout=7.5 |
----------------|
(Output of regulator) |
|----[ R2 ]----|[ground]

Could you please advise me something about it.

Linear regulators usually regulate voltage, not current. The current is
then defined by the load, not the regulator.

6.3/5.5 = 1.14 ohms.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Mark Jones]

Robert Monsen wrote:

Vladimir wrote:

Hello!
Suppose, i have a required voltage on the output of linear
regulator(Vout=6.3),
but output current is too high (Iout=7.5). I need Iout=5.5 A.
How I can work it out? I think about a follow way. (It is quite
possible that it is wrong.)

I should increase output voltage, then choose two appropriate
resistors:

|----[ R1 ]----o V=6.3; I=5.5A V>6.3;
Iout=7.5 | ----------------|
(Output of regulator) |
|----[ R2 ]----|[ground]
Could you please advise me something about it.


Linear regulators usually regulate voltage, not current. The current is
then defined by the load, not the regulator.

6.3/5.5 = 1.14 ohms.

If there was no way to modify the linear regulator, you could limit the current
with a separate current limiter. This wastes power though. Check out some of
these ideas:

http://www.discovercircuits.com/C/cur-lim.htm

 

[Rich Grise]

On Tue, 05 Apr 2005 03:46:56 -0700, Vladimir wrote:

bigcat@meeow.co.uk wrote in message news:<1112670756.951875.257730@l41g2000cwc.googlegroups.com>...
Vladimir wrote:
Hello!
Suppose, i have a required voltage on the output of linear
regulator(Vout=6.3),
but output current is too high (Iout=7.5). I need Iout=5.5 A.

First i am sorry for my bad English.

No problem. It is better than the English of some English-speakers.

I have this regulator:
_________
| Volt. | Output (Vout=+6.3; Iout=7.5A)
____|Regulator|_________
| |
|_________|
(Adj)|
|
I can adjust output voltage using resistor divider (from 1.25v to 30v).
But i need the other current value. Iout must be equal to 5.5 amperes.
Can it be implemented, but without much complexity?

I think you misunderstand about voltage, current, and resistance. The
voltage is regulated at the set point, yes.

However: the load (what you're supplying power to) will only draw as
much current as it needs. The 7.5A spec on the existing regulator means
merely that it has the _ability_ to provide that much current. If the
load presents a resistance of, say, two ohms, then at 6.3 volts, it
will only draw 3.15 amps.

Think of a water faucet. When it's off, you still have full water pressure
(which will show on a gauge), but the amount of flow depends on how far
the valve is open. Even if the pipe is very big, and has the ability
to provide, say, one liter per second, if the valve is open only a
little bit, it might provide, say, half a liter per minute, but the
available pressure will be the same (except for losses in the system,
which are called "internal resistance").

So the supply itself can not control both voltage and current - the
load will only take what it needs.

Hope This Helps!
Rich

 

[Roger Johansson]

Vladimir wrote:

First i am sorry for my bad English.

That is not the problem, your english is good enough.

But you do not explain a lot of things.

Is this a practical real problem or a theoretical problem?

Give us the data on the regulator/power supply.
What are you using this circuit for?
What is the load, what is this power supply used to drive?

I have this regulator:
_________
| Volt. | Output (Vout=+6.3; Iout=7.5A)
____|Regulator|_________
| |
|_________|
(Adj)|
|
I can adjust output voltage using resistor divider (from 1.25v to
30v). But i need the other current value. Iout must be equal to 5.5
amperes. Can it be implemented, but without much complexity?

The method of using resistors to set the output voltage is only
acceptable if you need very low current, but you seem to need a lot of
current.

Can you change the regulator instead, to give the voltage you need?
Have you built the regulator yourself, do you have the schematic for it?
Is it the same regulator circuit as the one we talked about a few days
ago?

Is the same regulator/power supply used for some other load at the same
time as you need this reduced voltage?

What is the real practical problem you are trying to solve?
Are you trying to control a motor, or a lamp, or what?


--
Roger J.