12 Volt Regulated Supply

D

Dave.H

Guest
I was looking at a 12 volt regulated power supply at Dick Smith
Electronics cat. number M9935 @ www.dse.com.au for a regen radio I'm
planning on building (http://www.schmarder.com/radios/tube/
1-12af6.htm) but the AU$24 is a bit too high for my liking. I was
thinking of using a center tapped transformer, 6.3-0-6.3V @ 500mA, in
series with a 1.3 watt zener diode, I need help with obtaining 12.6
volts from a center tapped transformer, I never had to do it before.
I thought that the two positive primary leads were connected, I don't
want to blow the transformer if this isn't the case.

Any help appreciated,

Thanks
Dave

Transformer is M2853, and zener diode is Z3543 @ www.dse.com.au
 
On May 1, 10:42 pm, "Dave.H" <the19...@googlemail.com> wrote:
I was looking at a 12 volt regulated power supply at Dick Smith
Electronics cat. number M9935 @www.dse.com.aufor a regen radio I'm
planning on building (http://www.schmarder.com/radios/tube/
1-12af6.htm) but the AU$24 is a bit too high for my liking. I was
thinking of using a center tapped transformer, 6.3-0-6.3V @ 500mA, in
series with a 1.3 watt zener diode, I need help with obtaining 12.6
volts from a center tapped transformer, I never had to do it before.
I thought that the two positive primary leads were connected, I don't
want to blow the transformer if this isn't the case.

Any help appreciated,

Thanks
Dave

Transformer is M2853, and zener diode is Z3543 @www.dse.com.au
Click to expand...
How would I install the zener diode? I thought they were used in
series with the voltage source much like a rectifier diode, but I just
read that they're used in parallel to the source. Excuse the dumb
question, I've only ever had experience with rectifier diodes like the
1N4007.

Thanks
Dave
 
On Thu, 1 May 2008 07:33:05 -0700 (PDT), "Dave.H"
<the1930s@googlemail.com> wrote:

On May 1, 10:42 pm, "Dave.H" <the19...@googlemail.com> wrote:
I was looking at a 12 volt regulated power supply at Dick Smith
Electronics cat. number M9935 @www.dse.com.aufor a regen radio I'm
planning on building (http://www.schmarder.com/radios/tube/
1-12af6.htm) but the AU$24 is a bit too high for my liking. I was
thinking of using a center tapped transformer, 6.3-0-6.3V @ 500mA, in
series with a 1.3 watt zener diode, I need help with obtaining 12.6
volts from a center tapped transformer, I never had to do it before.
I thought that the two positive primary leads were connected, I don't
want to blow the transformer if this isn't the case.

Any help appreciated,

Thanks
Dave

Transformer is M2853, and zener diode is Z3543 @www.dse.com.au

How would I install the zener diode? I thought they were used in
series with the voltage source much like a rectifier diode, but I just
read that they're used in parallel to the source. Excuse the dumb
question, I've only ever had experience with rectifier diodes like the
1N4007.
Click to expand...
---
For starters: (View in courier)

The way to get 12V out of the transformer is to ignore the center tap
and use the the two ends of the secondary as your 12V source.

There are two ways to do that, one called half wave rectification and
the other full wave rectification.

Half wave rectification works like this:


MAINS>--+ +--[DIODE>]--+---->17.1VDC
P||S |+
R||E--CT [BFC]
I||C |
MAINS>--+ +------------+---->GND


When the end of the secondary connected to the anode of the diode goes
positive, charge is allowed to flow through the diode and accumulate
in the BFC where, when the capacitor is full will be the peak value of
the AC waveform minus the voltage dropped across the diode.

That is:


VDC = (VRMS * sqrt(2)) - Vf

= (12.VRMS * 1.414) - 0.7V

~ 17.1V


Full wave rectification works like this:


MAINS>-+ +----+-[CR1>]---+---+---->16.4VDC
| | | | |+
P||S +-[<CR2]-+ | [BFC]
R||E-CT | | |
I||C +-[<CR3]-+-|---+---->GND
| | | |
MAINS>-+ +----+-[CR4>]---+

This time, when the top of the secondary goes positive, charge will
flow through CR1 into the BFC and then through CR3 back into the
negative end of the secondary.

When the polarity reverses and the bottom of the secondary goes
positive, charge will flow through CR4 into the BFC and then back into
the negative end of the secondary through CR2.

The full wave rectifier has the advantage of using a capacitor with
half the capacitance required for the same ripple out of a half-wave
rectifier, plus some others, so lets assume that's the supply you'll
build.

OK, now for the Zener.

A Zener diode belongs to a class of devices called "shunt regulators"
because they shunt current that a load doesn't need to ground while
maintaining the voltage across the load constant even as the supply
voltage changes and the load current varies.

In your case since you have a 16.4V supply and you want to use that
12V Zener, we look at the data sheet for a 1N4742 and we find that IZt
is 21mA and IZmax is 76mA.

IZt is the Zener test current and is the current through the Zener
which will cause the voltage dropped it fall within the device's
specifications, and IZmax is the maximum current which can be pushed
through the diode without damaging it if the ambient temperature is
25C or lower.

Now, if we look at the regulator circuit we'll have:


16.4V>--[R]--+---------+
|+ |
[ZENER] [LOAD]
| |
GND>---------+---------+

Let's assume that your load draws 50mA and that we'll allow IZt to
flow through the Zener.

Then we'll have, for the value of R:

Vsupply - Vz 16.4V - 12V
R = -------------- = ---------------- = 61.9 ohms
IZt + Il 0.021A + 0.05A

which is pretty close to 62 ohms, a standard 5% value.


The power the resistor will dissipate will be:


P = (Vsupply - Vz) * (IZt + Il)

= (16.4V - 12V) *(0.021A + 0.05A)

= 0.312 watts

So a half-watt resistor would be fine.

Your Zener circuit now looks like this:

6.3VAC 16.4VDC 12VDC
/ / /
MAINS>-+ +----+-[CR1>]---+---+-[62R]-+--------+
| | | | |+ |K |
P||S +-[<CR2]-+ | [BFC] [1N4742] [LOAD]
R||E-CT | | | | |
I||C +-[<CR3]-+-|---+-------+--------+
| | | |
MAINS>-+ +----+-[CR4>]---+
\
6.3VAC


But...

There's a very much better way to go, and that's to use a series
regulator like a 7812 in place of a Zener, like this:


6.3VAC 16.4VDC 12VDC
/ / /
MAINS>-+ +----+-[CR1>]---+---+-[7812]--+
| | | | |+ | |
P||S +-[<CR2]-+ | [BFC] | [LOAD]
R||E-CT | | | | |
I||C +-[<CR3]-+-|---+----+----+
| | | |
MAINS>-+ +----+-[CR4>]---+
\
6.3VAC

Not only will this give you better regulation with line and load
changes, it'll allow you to use the full output capability of the
transformer (half an ampere) with a suitable heatsink on the 7812.

So, the last thing to figure out is how big the BFC has to be.

We start by looking at the droput voltage of the 7805, (that's the
voltage difference between the output and the input which will assure
the output stays in regulation) and it's 2.5V, so that means we can
never let the input fall below:


Vin = Vout + Vdo = 12V + 2.5V = 14.5 volts


Since the BFC is being charged with sinusoidal pulses and being
discharged by the load when the pulse voltage isn't high enough to
supply the load current, ripple will appear across the capacitor and
we have to make sure that ripple never falls below 14.5V or the 7812
will go out of regulation. to do that we can say:


Idt
C = -----
dV

where C is the capacitance in farads,
I is the load current in amperes,
dT is the ripple period in seconds, and
dV is the allowable ripple in volts.


Then:


Idt 0.5A * 0.01s
C = ----- = ------------- = 0.002F = 2000ľF
dV 2.5V


Not bad, but that's cutting it pretty thin so let's lower the ripple
to one volt and see what we get.


Idt 0.5A * 0.01s
C = ----- = ------------- = 0.005F = 5000ľF
dV 1V


Still not bad, and that'll make up for the capacitor's tolerance on
the low side.

So your final power supply will look like this:


MAINS>-+ +----+-[1N4001>]---+----+---[7812]--+---->+12VDC
| | | | |+ | |
P||S +-[<1N4001]-+ | [5100ľF] | [100nF]
R||E-CT | | | | |
I||C +-[<1N4001]-+-|----+-----+-----+---->GND
| | | |
MAINS>-+ +----+-[1N4001>]---+


The 100nF cap on the output keeps everthing stable, and both caps
should be mounted as close to the 7812 as possible.

Also, there's nothing critcal about the BFC; a general purpose
aluminum electrolyticwill do fine and if you cant find 5100ľF the
next largest size rated at around 25V will be even better.

JF
 
On Thu, 1 May 2008 11:17:47 -0700 (PDT), "Dave.H"
<the1930s@googlemail.com> wrote:

On May 2, 3:13 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 1 May 2008 07:33:05 -0700 (PDT), "Dave.H"

So your final power supply will look like this:

MAINS>-+ +----+-[1N4001>]---+----+---[7812]--+---->+12VDC
| | | | |+ | |
P||S +-[<1N4001]-+ | [5100ľF] | [100nF]
R||E-CT | | | | |
I||C +-[<1N4001]-+-|----+-----+-----+---->GND
| | | |
MAINS>-+ +----+-[1N4001>]---+

The 100nF cap on the output keeps everthing stable, and both caps
should be mounted as close to the 7812 as possible.

Also, there's nothing critcal about the BFC; a general purpose
aluminum electrolyticwill do fine and if you cant find 5100ľF the
next largest size rated at around 25V will be even better.

JF

I like that 7812 idea. The tube (12AF6) draws 150 mA, then there's
the current draw of the 12 volt B+, a zener won't handle all that. A
5600 MFD 40 volt capacitor is AU$7.30, I don't like paying that for
low quality Asian capacitors, so I'm paralleling 3 2200 MFD 25 volt
caps in series to give a total 6600 MFD. Each of these caps cost $0.97
Click to expand...
---
You can't "parallel caps in series".


According to the tube specs the plate current will be about 1mA, so
the total draw, with 175mA of filament current, will be 176 mA. With
that current, the capacitance needed to get 1 volt of ripple will be:


Idt 0.176A * 0.01s
C = ----- = ---------------- = 1.76E-3F = 1760ľF
dV 1V

If your 2200ľF caps have a tolerance of +/-10% then the lowest their
capacitance will be is 1980ľF, so you'll only need to use one.

Matter of fact, even if they're +/-20% that comes out to 1760ľF, which
is right on the money, so you could still get away with just one!


JF
 
On Thu, 01 May 2008 14:41:44 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 1 May 2008 11:17:47 -0700 (PDT), "Dave.H"
the1930s@googlemail.com> wrote:

On May 2, 3:13 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 1 May 2008 07:33:05 -0700 (PDT), "Dave.H"

So your final power supply will look like this:

MAINS>-+ +----+-[1N4001>]---+----+---[7812]--+---->+12VDC
| | | | |+ | |
P||S +-[<1N4001]-+ | [5100ľF] | [100nF]
R||E-CT | | | | |
I||C +-[<1N4001]-+-|----+-----+-----+---->GND
| | | |
MAINS>-+ +----+-[1N4001>]---+

The 100nF cap on the output keeps everthing stable, and both caps
should be mounted as close to the 7812 as possible.

Also, there's nothing critcal about the BFC; a general purpose
aluminum electrolyticwill do fine and if you cant find 5100ľF the
next largest size rated at around 25V will be even better.

JF

I like that 7812 idea. The tube (12AF6) draws 150 mA, then there's
the current draw of the 12 volt B+, a zener won't handle all that. A
5600 MFD 40 volt capacitor is AU$7.30, I don't like paying that for
low quality Asian capacitors, so I'm paralleling 3 2200 MFD 25 volt
caps in series to give a total 6600 MFD. Each of these caps cost $0.97

---
You can't "parallel caps in series".


According to the tube specs the plate current will be about 1mA, so
the total draw, with 175mA of filament current, will be 176 mA. With
that current, the capacitance needed to get 1 volt of ripple will be:


Idt 0.176A * 0.01s
C = ----- = ---------------- = 1.76E-3F = 1760ľF
dV 1V

If your 2200ľF caps have a tolerance of +/-10% then the lowest their
capacitance will be is 1980ľF, so you'll only need to use one.

Matter of fact, even if they're +/-20% that comes out to 1760ľF, which
is right on the money, so you could still get away with just one!
Click to expand...
---
Oops...


Idt 0.150A * 0.01s
C = ----- = ---------------- = 1.50E-3F = 1500ľF
dV 1V


Just one will work...

JF
 
On May 2, 3:13 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 1 May 2008 07:33:05 -0700 (PDT), "Dave.H"



the19...@googlemail.com> wrote:
On May 1, 10:42 pm, "Dave.H" <the19...@googlemail.com> wrote:
I was looking at a 12 volt regulated power supply at Dick Smith
Electronics cat. number M9935 @www.dse.com.aufora regen radio I'm
planning on building (http://www.schmarder.com/radios/tube/
1-12af6.htm) but the AU$24 is a bit too high for my liking. I was
thinking of using a center tapped transformer, 6.3-0-6.3V @ 500mA, in
series with a 1.3 watt zener diode, I need help with obtaining 12.6
volts from a center tapped transformer, I never had to do it before.
I thought that the two positive primary leads were connected, I don't
want to blow the transformer if this isn't the case.

Any help appreciated,

Thanks
Dave

Transformer is M2853, and zener diode is Z3543 @www.dse.com.au

How would I install the zener diode? I thought they were used in
series with the voltage source much like a rectifier diode, but I just
read that they're used in parallel to the source. Excuse the dumb
question, I've only ever had experience with rectifier diodes like the
1N4007.

---
For starters: (View in courier)

The way to get 12V out of the transformer is to ignore the center tap
and use the the two ends of the secondary as your 12V source.

There are two ways to do that, one called half wave rectification and
the other full wave rectification.

Half wave rectification works like this:

MAINS>--+ +--[DIODE>]--+---->17.1VDC
P||S |+
R||E--CT [BFC]
I||C |
MAINS>--+ +------------+---->GND

When the end of the secondary connected to the anode of the diode goes
positive, charge is allowed to flow through the diode and accumulate
in the BFC where, when the capacitor is full will be the peak value of
the AC waveform minus the voltage dropped across the diode.

That is:

VDC = (VRMS * sqrt(2)) - Vf

= (12.VRMS * 1.414) - 0.7V

~ 17.1V

Full wave rectification works like this:

MAINS>-+ +----+-[CR1>]---+---+---->16.4VDC
| | | | |+
P||S +-[<CR2]-+ | [BFC]
R||E-CT | | |
I||C +-[<CR3]-+-|---+---->GND
| | | |
MAINS>-+ +----+-[CR4>]---+

This time, when the top of the secondary goes positive, charge will
flow through CR1 into the BFC and then through CR3 back into the
negative end of the secondary.

When the polarity reverses and the bottom of the secondary goes
positive, charge will flow through CR4 into the BFC and then back into
the negative end of the secondary through CR2.

The full wave rectifier has the advantage of using a capacitor with
half the capacitance required for the same ripple out of a half-wave
rectifier, plus some others, so lets assume that's the supply you'll
build.

OK, now for the Zener.

A Zener diode belongs to a class of devices called "shunt regulators"
because they shunt current that a load doesn't need to ground while
maintaining the voltage across the load constant even as the supply
voltage changes and the load current varies.

In your case since you have a 16.4V supply and you want to use that
12V Zener, we look at the data sheet for a 1N4742 and we find that IZt
is 21mA and IZmax is 76mA.

IZt is the Zener test current and is the current through the Zener
which will cause the voltage dropped it fall within the device's
specifications, and IZmax is the maximum current which can be pushed
through the diode without damaging it if the ambient temperature is
25C or lower.

Now, if we look at the regulator circuit we'll have:

16.4V>--[R]--+---------+
|+ |
[ZENER] [LOAD]
| |
GND>---------+---------+

Let's assume that your load draws 50mA and that we'll allow IZt to
flow through the Zener.

Then we'll have, for the value of R:

Vsupply - Vz 16.4V - 12V
R = -------------- = ---------------- = 61.9 ohms
IZt + Il 0.021A + 0.05A

which is pretty close to 62 ohms, a standard 5% value.

The power the resistor will dissipate will be:

P = (Vsupply - Vz) * (IZt + Il)

= (16.4V - 12V) *(0.021A + 0.05A)

= 0.312 watts

So a half-watt resistor would be fine.

Your Zener circuit now looks like this:

6.3VAC 16.4VDC 12VDC
/ / /
MAINS>-+ +----+-[CR1>]---+---+-[62R]-+--------+
| | | | |+ |K |
P||S +-[<CR2]-+ | [BFC] [1N4742] [LOAD]
R||E-CT | | | | |
I||C +-[<CR3]-+-|---+-------+--------+
| | | |
MAINS>-+ +----+-[CR4>]---+
\
6.3VAC

But...

There's a very much better way to go, and that's to use a series
regulator like a 7812 in place of a Zener, like this:

6.3VAC 16.4VDC 12VDC
/ / /
MAINS>-+ +----+-[CR1>]---+---+-[7812]--+
| | | | |+ | |
P||S +-[<CR2]-+ | [BFC] | [LOAD]
R||E-CT | | | | |
I||C +-[<CR3]-+-|---+----+----+
| | | |
MAINS>-+ +----+-[CR4>]---+
\
6.3VAC

Not only will this give you better regulation with line and load
changes, it'll allow you to use the full output capability of the
transformer (half an ampere) with a suitable heatsink on the 7812.

So, the last thing to figure out is how big the BFC has to be.

We start by looking at the droput voltage of the 7805, (that's the
voltage difference between the output and the input which will assure
the output stays in regulation) and it's 2.5V, so that means we can
never let the input fall below:

Vin = Vout + Vdo = 12V + 2.5V = 14.5 volts

Since the BFC is being charged with sinusoidal pulses and being
discharged by the load when the pulse voltage isn't high enough to
supply the load current, ripple will appear across the capacitor and
we have to make sure that ripple never falls below 14.5V or the 7812
will go out of regulation. to do that we can say:

Idt
C = -----
dV

where C is the capacitance in farads,
I is the load current in amperes,
dT is the ripple period in seconds, and
dV is the allowable ripple in volts.

Then:

Idt 0.5A * 0.01s
C = ----- = ------------- = 0.002F = 2000ľF
dV 2.5V

Not bad, but that's cutting it pretty thin so let's lower the ripple
to one volt and see what we get.

Idt 0.5A * 0.01s
C = ----- = ------------- = 0.005F = 5000ľF
dV 1V

Still not bad, and that'll make up for the capacitor's tolerance on
the low side.

So your final power supply will look like this:

MAINS>-+ +----+-[1N4001>]---+----+---[7812]--+---->+12VDC
| | | | |+ | |
P||S +-[<1N4001]-+ | [5100ľF] | [100nF]
R||E-CT | | | | |
I||C +-[<1N4001]-+-|----+-----+-----+---->GND
| | | |
MAINS>-+ +----+-[1N4001>]---+

The 100nF cap on the output keeps everthing stable, and both caps
should be mounted as close to the 7812 as possible.

Also, there's nothing critcal about the BFC; a general purpose
aluminum electrolyticwill do fine and if you cant find 5100ľF the
next largest size rated at around 25V will be even better.

JF
Click to expand...
I like that 7812 idea. The tube (12AF6) draws 150 mA, then there's
the current draw of the 12 volt B+, a zener won't handle all that. A
5600 MFD 40 volt capacitor is AU$7.30, I don't like paying that for
low quality Asian capacitors, so I'm paralleling 3 2200 MFD 25 volt
caps in series to give a total 6600 MFD. Each of these caps cost $0.97

Dave
 
On May 2, 5:41 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 1 May 2008 11:17:47 -0700 (PDT), "Dave.H"



the19...@googlemail.com> wrote:
On May 2, 3:13 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 1 May 2008 07:33:05 -0700 (PDT), "Dave.H"
So your final power supply will look like this:

MAINS>-+ +----+-[1N4001>]---+----+---[7812]--+---->+12VDC
| | | | |+ | |
P||S +-[<1N4001]-+ | [5100ľF] | [100nF]
R||E-CT | | | | |
I||C +-[<1N4001]-+-|----+-----+-----+---->GND
| | | |
MAINS>-+ +----+-[1N4001>]---+

The 100nF cap on the output keeps everthing stable, and both caps
should be mounted as close to the 7812 as possible.

Also, there's nothing critcal about the BFC; a general purpose
aluminum electrolyticwill do fine and if you cant find 5100ľF the
next largest size rated at around 25V will be even better.

JF

I like that 7812 idea. The tube (12AF6) draws 150 mA, then there's
the current draw of the 12 volt B+, a zener won't handle all that. A
5600 MFD 40 volt capacitor is AU$7.30, I don't like paying that for
low quality Asian capacitors, so I'm paralleling 3 2200 MFD 25 volt
caps in series to give a total 6600 MFD. Each of these caps cost $0.97

---
You can't "parallel caps in series".

According to the tube specs the plate current will be about 1mA, so
the total draw, with 175mA of filament current, will be 176 mA. With
that current, the capacitance needed to get 1 volt of ripple will be:

Idt 0.176A * 0.01s
C = ----- = ---------------- = 1.76E-3F = 1760ľF
dV 1V

If your 2200ľF caps have a tolerance of +/-10% then the lowest their
capacitance will be is 1980ľF, so you'll only need to use one.

Matter of fact, even if they're +/-20% that comes out to 1760ľF, which
is right on the money, so you could still get away with just one!

JF
Click to expand...
Oops. Didn't mean to say that that they were in series. I'm only
paralleling them. I already have one 2200 uF 50 volt cap I'll use.
 
On Fri, 2 May 2008 19:54:11 +1000, "David L. Jones"
<altzone@gmail.com> wrote:

"Dave.H" <the1930s@googlemail.com> wrote in message
news:9d258a2d-a0da-4809-af3b-98a947ece05d@k1g2000prb.googlegroups.com...
I was looking at a 12 volt regulated power supply at Dick Smith
Electronics cat. number M9935 @ www.dse.com.au for a regen radio I'm
planning on building (http://www.schmarder.com/radios/tube/
1-12af6.htm) but the AU$24 is a bit too high for my liking. I was
thinking of using a center tapped transformer, 6.3-0-6.3V @ 500mA, in
series with a 1.3 watt zener diode, I need help with obtaining 12.6
volts from a center tapped transformer, I never had to do it before.
I thought that the two positive primary leads were connected, I don't
want to blow the transformer if this isn't the case.

Any help appreciated,

Thanks
Dave

Transformer is M2853, and zener diode is Z3543 @ www.dse.com.au

How about $7, delivered to your door:
http://cgi.ebay.com.au/Switching-Power-Supply-12V-1A-for-CCTV-camera-AU-P24_W0QQitemZ370046177343QQihZ024QQcategoryZ20589QQssPageNameZWDVWQQrdZ1QQcmdZViewItem

Dave.

Simply brilliant and only $7 AD. w/postage.
Click to expand...

I may get one just to see what is inside.
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