Questions about simple power supply

I need to put together a simply power supply to convert 28V AC into 12V DC.
The regulator part is no problem, I've done that before. However, I was
wondering if I could get a little help on the "converting AC to DC" part.
Obviously I need a rectifier, could anyone recommend one to use for 1A max
load. And the biggest problem I always have is figuring out capacitor
values. What kind of cap and what value should I use to smooth out the DC
before I feed it to the regulator (LM 340T12)? Is there a formula for this
cap value? Is anything else required?

Thanks,
Harry

 

A bridge rectifier made of 4 1N4002 (or higher voltage) diodes is good
for about 1.5 amps (.75 amp per diode). With this high of AC voltage
( 28*1.414= 39 volts) you have so much excess that ripple is of little
concern, so 1000 uf may be enough to keep the minimum above the 14 or
15 needed to keep the regulator functioning. You might even consider
adding a resistor between the rectifier and storage capacitor or in
series with the primary to keep the regulator and transformer cooler.
The ideal solution would be to have a 28 volt center tapped secondary,
so you could produce 14*1.414= 19 volts with only two diodes, and use
about 4700 uf for the filter. This would make the regulator run much
(1/4 the rise) cooler.

 

The capacitor is about 8000uF per amp for 1 volt ripple, so
if you start with 28 and lose 1.5 to the rectifiers and
another 3 to the regulator, and the output is 12 and the
current is 1 amp, then you have an extra 11.5 left over
for ripple, which means the capacitor can be 11.5 times
smaller than 8000uF, or 696 uF.

-Bill

 

Unfortunately the transformer isn't center tapped and it's already part of
the project (it's got two secondaries) so I'm stuck working with the 28V.

However, I forgot to multiply my 28AC by that magic number of 1.414 (I'm
still fairly new to electronics), which means I'm gonna be over the 35 volt
maximum that the LM340T12 can handle as input volts. Could I use a resistor
to reduce the volts down to at least 35V, if so, would I put in after the
cap, just before the regulator. Also, since the amps would change how would
I figure out the value of it?

One last question, what's the difference between a rectifier, bridge
rectifier, and full rectifier. Are these all terms for the same thing or
are there different types of rectifiers?

Thank you for the response,
Harry

 

I would put it before the cap, to reduce the ripple as well as the
voltage. But just adding a resistor will not lower the voltage till
current is passed, so if the regulator is not loaded, the voltage will
still be too high. So you will also need some shunt component to draw
a minimum current through the resistor, like a zener diode. Though,
you could put a big zener in series to drop a fairly fixed voltage
over a wider range of current. Your main problem is going to be all
the heat produced by all this voltage drop.

What is the expected range of the regulator's load current?

A rectifier is a diode. A bridge rectifier is 4 diodes connected so
you apply AC to two nodes, and get DC out of two other nodes. A half
bridge is two diodes connected either cathodes together or anodes
together.
See:
http://www.tpub.com/neets/book7/27c.htm

 

Thank you for all the answers.

I don't expect too much current to be drawn by my circuit. Here's a link to
what I'm trying to supply power for:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page2.htm

Just scroll down to the "Electronic Thermostat and Relay Circuit". I'm
assuming the main draw will be from the relay and that ain't much. I just
always want to make sure I have more than enough juice, so that's why I'm
thinking of a 1A regulator.

I'm thinking of maybe using a LM317HV as the regulator instead. It accepts
input volts of up to 60V. However, it's not preset like my previous choice
of regulator, so it will need extra parts, but since my previous choice
would need extra parts too, I might as well just go with something that's
made for higher volts.

Thanks,
Harry

 

If you power the relay directly at the rectifier storage capacitor,
with a resistor in series selected to produce the correct voltage
across the coil when it is energized, you can reduce the regulated
current so low (probably below .05 ampere) that a simple zener
regulator and series dropping resistor is all the regulation you will
need.

For instance, a 12 volt, 1 watt zener and a series resistor of about
470 ohms, 2 watts will provide a 12 volt supply with enough capability
to power everything but the relay. Very simple and reliable.

One other change I would make would be to power the comparators
directly off the 12 volt line, not the 5.1 volt line. That way you
could raise the 220 ohm series resistor to about 470 ohms and lower
the current consumption a bit more. The comparators are able to
function quite will with an unregulated supply, so a double regulated
supply is not very useful.

 

If you are stuck with a 28 Volt secondary you could put a 24
regulator in front of the 12 Volt regulator.

The circuit would be: Bridge rectifier - Filter Capacitor - LM7824
- LM7812 - Load

Add some 0.1uF capacitors where they would normally be in a
regulator circuit.

The heat load and voltage drops will be split across two devices.

The voltage drops will not be current dependant.

Two 12 Volt regulators could be used if the output voltage of the
first is raised to 24 volts by an appropriate voltage divider. See a
data sheet on how to do this.

The regulators can be attached to the same heat sink with out
insulators.

Rob.

 

One last question, for the 1000uf Cap, what kind should I use (ie: ceramic,
etc.)?

Thanks,
Harry

 

One quick question, what kind of cap should I be using for this (ie:
ceramic, etc.)?

Thanks,
Harry

 

50 volt electrolytic.

 

Elementry Watson :
Rectifier capacitors of this size are always electrolitic.
Well, this is my experience anyway.
Rex

 

btw
you wanted a formula for the ripple, well

for full wave rectifier (bridge) there is something called ripple factor = γ

defined as

rms ripple volts / d.c. volts

equals: 1 / 4 f C RL

but as others have said, since you are regulating it, you will not see it
anyway, hence you can do with a much smaller capacitace.

 

sorry, mistake

γ = 1 / 4 √3 f C RL
F.W.

What you want is to make gamma approach zero

You can see that

increasing f or C or RL are all desired. In practice RL is your load

so you can't change it, f is 50 whichever way you look at it, so you are

left with C. The larger C is the less ripple you will have.



btw

If you are interested in how the formula is arrived at let me know,

though opening an all-rounder electronics book at the chapter of

power supplies should reveal all.





btw
you wanted a formula for the ripple, well

for full wave rectifier (bridge) there is something called ripple factor = γ

defined as

rms ripple volts / d.c. volts

equals: 1 / 4 f C RL

but as others have said, since you are regulating it, you will not see it
anyway, hence you can do with a much smaller capacitace.

 

Portion removed

Portion removed

Just to be picky

The LM 317 will only accept an input of 60 volts if you stay within
the following limit taken from the data sheet.

"Input-Output Voltage Differential +40V, -0.3V"

For a 60 volt input the minimum output voltage would be 20 volts.
Most regulators follow a similar limiting factor

The 317 will still be OK for this application.

Rob.

 

It will be an electrolytic cap, but use a value larger than
the minimum, say 2200uF.

-Bill

 

Does it matter if it's an aluminum electrolytic or tantalum electrolytic?

Thanks,
Harry

 

Does it matter if it's an aluminum or tantalum electrolytic cap?

Thanks,
Harry

 

For 1000 uF 50 volt, it _will_ be aluminum - tantalums appear to be
limited to smaller values - under 100 uF at 50 V, somewhat higher for
low voltages.

You won't find ceramic capacitors much over 1 uF.

 

Aluminum is the obvious choice based on cost, size and availability.
Tantalums in supply filters also have the nasty habit of developing
internal shorts when the voltage is applied too fast. And since the
voltage rise is limited only by the impedance of the power
transformer, the voltage will pop up in much less than a half cycle.