Reducing Source Voltage To Power Supply

[whit3rd]

On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use.

Those six D cells give 9V fresh, over 6V until they're dead.
And the output impedance is low (similar to a regulated power supply).
So, look for a wall wart with regulated DC output in the 6 to 9V
range. It'll work.

 

[John Fields]

On Thu, 10 Jan 2008 14:58:18 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

On Jan 11, 6:26 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Jan 2008 08:43:50 -0800 (PST), "Dave.H"



the19...@googlemail.com> wrote:
On Jan 11, 3:37 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Jan 2008 05:51:50 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

---
You may want to consider something like one of these:

http://www.tamuracorp.com/clientuploads/pdfs/pg5.pdf

which you can get from Digi-Key.

In any case, how much current does the radio draw and what current
is your wall-wart rated for when it's outputting 9V?

--
JF

The wall wart is rated at 1 amp. Not sure what the current draw is.

---
OK, here's what you've got: (View in Courier)

FWB
+-----+
| +|----+--[7809]--+--->+9VDC
MAINS>----P||S--|~ | +| | |
R||E | | [BFC] | [LFC]
MAINS>----I||C--|~ | | | |
| -|----+-----+----+--->GND
+-----+

Since your wall-wart puts out 9VRMS when it's loaded with 1 ampere,
that's 12.73V peak. You'll lose about 1.4V of that across the
diodes in the bridge, leaving you with about 11.3V into the
regulator.

The 7809 has a worst-case dropout of 2.5V, so the bad news is that
for a 9V output you'll need 11.5V into it, and you've only got
11.3V. :-(

The good news is that your wall-wart's regulation is poor and with a
no-load output voltage of 12.5 it may be able to supply the voltage
needed by the 7805. Let's see...

For a 1A load we'll have an output, from the bridge, of:

Vout(1) = (VRMS * sqrt(2)) - 1.4V

= (9V * 1.414) - 1.4V = 11.3 volts

For no load we'll have:

Vout(2) = (12.5V * 1.414) - 1.4V = 16.27 volts

That means that from no load to full load we'll have a voltage
change of about 5 volts per ampere of current change.

So what?

Well, since the 7805 needs 11.5 V minimum into its input to provide
9V out, that means that (assuming the transformer's voltage VS
current change is linear) the load can never draw more than:

Vout(2) - Vdo 16.27V - 11.5V
Il(max) = --------------- = ---------------- = 0.954 ampere.
5 5V

Realistically, though, we haven't even considered ripple and we need
to do that in order to choose the value of the BFC.

In order to do that we need to know how much current your radio
draws, but you don't know how much that is.

No matter, we can work something out and then plug in your numbers
when you find out.

Let's say your radio's running on a watt. Then the current into it
will be:

P 1W
I = --- = ---- = 0.111A = 110mA
E 9V

And the voltage out of the bridge will be:

/ 5V * 0.111A \
Vout = Vout(2) - | ------------- |
\ 1A /

= 16.27V - 0.555V

~ 15.72V

Since the 7809's dropout voltage is 11.5V and we have 15.72V
available at the output of the bridge, the difference (4.22V) is the
ripple we're allowed.

The filter cap's capacitance, then, would be:

Idt 0.111A * 0.01s
C = ----- = ---------------- = 2.63E-4F = 263ľF
dV 4.22V

Since the bridge will be supplying current to the load as well as to
the cap when it's charging, the voltage available from the
transformer will drop somewhat during that time so it would be a
good idea to increase the value of the capacitor in order to
compensate for that.

Without going into it analytically, I'd guess that doubling the cap
would do it. Even better, throw 1000ľF in there; they're cheap!

Only one thing left to do and that's to check on whether you've got
enough headroom to get that 11.5V with low mains, and you now ought
to have enough information to be able to do that.

--
JF

Thanks, but it all sounds so complicated to me, I'm thinking it would
be easier to keep it on batteries

---
Oh, well...


--
JF

 

[Dave.H]

On Jan 11, 6:26 am, John Fields <jfie...@austininstruments.com> wrote:

On Thu, 10 Jan 2008 08:43:50 -0800 (PST), "Dave.H"



the19...@googlemail.com> wrote:
On Jan 11, 3:37 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Jan 2008 05:51:50 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

---
You may want to consider something like one of these:

http://www.tamuracorp.com/clientuploads/pdfs/pg5.pdf

which you can get from Digi-Key.

In any case, how much current does the radio draw and what current
is your wall-wart rated for when it's outputting 9V?

--
JF

The wall wart is rated at 1 amp. Not sure what the current draw is.

---
OK, here's what you've got: (View in Courier)

FWB
+-----+
| +|----+--[7809]--+--->+9VDC
MAINS>----P||S--|~ | +| | |
R||E | | [BFC] | [LFC]
MAINS>----I||C--|~ | | | |
| -|----+-----+----+--->GND
+-----+

Since your wall-wart puts out 9VRMS when it's loaded with 1 ampere,
that's 12.73V peak. You'll lose about 1.4V of that across the
diodes in the bridge, leaving you with about 11.3V into the
regulator.

The 7809 has a worst-case dropout of 2.5V, so the bad news is that
for a 9V output you'll need 11.5V into it, and you've only got
11.3V. :-(

The good news is that your wall-wart's regulation is poor and with a
no-load output voltage of 12.5 it may be able to supply the voltage
needed by the 7805. Let's see...

For a 1A load we'll have an output, from the bridge, of:

Vout(1) = (VRMS * sqrt(2)) - 1.4V

= (9V * 1.414) - 1.4V = 11.3 volts

For no load we'll have:

Vout(2) = (12.5V * 1.414) - 1.4V = 16.27 volts

That means that from no load to full load we'll have a voltage
change of about 5 volts per ampere of current change.

So what?

Well, since the 7805 needs 11.5 V minimum into its input to provide
9V out, that means that (assuming the transformer's voltage VS
current change is linear) the load can never draw more than:

Vout(2) - Vdo 16.27V - 11.5V
Il(max) = --------------- = ---------------- = 0.954 ampere.
5 5V

Realistically, though, we haven't even considered ripple and we need
to do that in order to choose the value of the BFC.

In order to do that we need to know how much current your radio
draws, but you don't know how much that is.

No matter, we can work something out and then plug in your numbers
when you find out.

Let's say your radio's running on a watt. Then the current into it
will be:

P 1W
I = --- = ---- = 0.111A = 110mA
E 9V

And the voltage out of the bridge will be:

/ 5V * 0.111A \
Vout = Vout(2) - | ------------- |
\ 1A /

= 16.27V - 0.555V

~ 15.72V

Since the 7809's dropout voltage is 11.5V and we have 15.72V
available at the output of the bridge, the difference (4.22V) is the
ripple we're allowed.

The filter cap's capacitance, then, would be:

Idt 0.111A * 0.01s
C = ----- = ---------------- = 2.63E-4F = 263ľF
dV 4.22V

Since the bridge will be supplying current to the load as well as to
the cap when it's charging, the voltage available from the
transformer will drop somewhat during that time so it would be a
good idea to increase the value of the capacitor in order to
compensate for that.

Without going into it analytically, I'd guess that doubling the cap
would do it. Even better, throw 1000ľF in there; they're cheap!

Only one thing left to do and that's to check on whether you've got
enough headroom to get that 11.5V with low mains, and you now ought
to have enough information to be able to do that.

--
JF

Thanks, but it all sounds so complicated to me, I'm thinking it would
be easier to keep it on batteries

 

[gearhead]

On Jan 10, 8:29 am, "Dave.H" <the19...@googlemail.com> wrote:

On Jan 11, 3:01 am, gearhead <nos...@billburg.com> wrote:





On Jan 10, 7:37 am, gearhead <nos...@billburg.com> wrote:

On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use.  I have no
problem using a wall wart, but I can't find one of the correct
voltage.  I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC.  Is there a way to halve this voltage with resistors
without generating too much heat?

I think your best bet is to use a LM7809 voltage regulator after the
filter cap.

I just checked Radio Shack's web site.  They don't carry the 7809, but
you can use the adjustable LM317T and set the output voltage to 9.
Here's the datasheethttp://www.fairchildsemi.com/ds/LM/LM317.pdf
Of course, using the 7809 would be a bit simpler but you would have to
order it and have it shipped from an online supplier like Digikey.com
or a miscellany house like allelectronics.com.  To avoid shipping
charges on a part that costs about a dollar, it makes sense to stop in
at RS for the LM317T.  Radio Shack's web site will even tell you
whether the Radio Shack nearest you has the part in stock.

I don't live in the US, I live in Australia, and it just so happens
that Dick Smith Electronics sells the 7809 regulator (Z6550).  I would
be interested in using this and need a diagram (simple how to wire it
up. The power supply schematic has the transformer connected to the
bridge rectifier, connected to the DC output with a cap or two across
the output.- Hide quoted text -

- Show quoted text -

Okay, a basic flow chart goes like this:

xformer-->bridge rectifier-->filter cap-->regulator-->radio
If you have any confusion about how to connect the bridge and the cap,
just ask.
How big a filter cap you need depends on the radio's current draw. If
you have a junk box, just grab a cap with at least 1000 uF, the more
the better. Up to a point. You could need as much as 10,000 uF, but
it sounds like the radio is pretty small so I really doubt that.
Now, on to the regulator.
The 7809 regulator has three terminals: in, ground, and out.
Ground is common; the rectifier's output, the cap and the radio all
share the same ground, or negative.
You connect the rectifier's positive output to the regulator's input
terminal.
Connect the regulator's output terminal to what used to be the radio
batteries' positive terminal.
Three-terminal regulators are not hard to use, even for a beginner.
Good idea to read the datasheet anytime you are using an unfamiliar
device:
http://www.st.com/stonline/products/literature/ds/2144/l7806ab.pdf

 

[Michael A. Terrell]

"Dave.H" wrote:

On Jan 11, 3:01 am, gearhead <nos...@billburg.com> wrote:
On Jan 10, 7:37 am, gearhead <nos...@billburg.com> wrote:

On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

I think your best bet is to use a LM7809 voltage regulator after the
filter cap.

I just checked Radio Shack's web site. They don't carry the 7809, but
you can use the adjustable LM317T and set the output voltage to 9.
Here's the datasheethttp://www.fairchildsemi.com/ds/LM/LM317.pdf
Of course, using the 7809 would be a bit simpler but you would have to
order it and have it shipped from an online supplier like Digikey.com
or a miscellany house like allelectronics.com. To avoid shipping
charges on a part that costs about a dollar, it makes sense to stop in
at RS for the LM317T. Radio Shack's web site will even tell you
whether the Radio Shack nearest you has the part in stock.

I don't live in the US, I live in Australia, and it just so happens
that Dick Smith Electronics sells the 7809 regulator (Z6550). I would
be interested in using this and need a diagram (simple how to wire it
up. The power supply schematic has the transformer connected to the
bridge rectifier, connected to the DC output with a cap or two across
the output.

http://www.fairchildsemi.com/ds/LM/LM7809.pdf figure 7, page 22.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[John Fields]

On Thu, 10 Jan 2008 08:43:50 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

On Jan 11, 3:37 am, John Fields <jfie...@austininstruments.com> wrote:
On Thu, 10 Jan 2008 05:51:50 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

---
You may want to consider something like one of these:

http://www.tamuracorp.com/clientuploads/pdfs/pg5.pdf

which you can get from Digi-Key.

In any case, how much current does the radio draw and what current
is your wall-wart rated for when it's outputting 9V?

--
JF

The wall wart is rated at 1 amp. Not sure what the current draw is.

---
OK, here's what you've got: (View in Courier)

FWB
+-----+
| +|----+--[7809]--+--->+9VDC
MAINS>----P||S--|~ | +| | |
R||E | | [BFC] | [LFC]
MAINS>----I||C--|~ | | | |
| -|----+-----+----+--->GND
+-----+

Since your wall-wart puts out 9VRMS when it's loaded with 1 ampere,
that's 12.73V peak. You'll lose about 1.4V of that across the
diodes in the bridge, leaving you with about 11.3V into the
regulator.

The 7809 has a worst-case dropout of 2.5V, so the bad news is that
for a 9V output you'll need 11.5V into it, and you've only got
11.3V. :-(

The good news is that your wall-wart's regulation is poor and with a
no-load output voltage of 12.5 it may be able to supply the voltage
needed by the 7805. Let's see...

For a 1A load we'll have an output, from the bridge, of:


Vout(1) = (VRMS * sqrt(2)) - 1.4V

= (9V * 1.414) - 1.4V = 11.3 volts


For no load we'll have:


Vout(2) = (12.5V * 1.414) - 1.4V = 16.27 volts


That means that from no load to full load we'll have a voltage
change of about 5 volts per ampere of current change.

So what?

Well, since the 7805 needs 11.5 V minimum into its input to provide
9V out, that means that (assuming the transformer's voltage VS
current change is linear) the load can never draw more than:


Vout(2) - Vdo 16.27V - 11.5V
Il(max) = --------------- = ---------------- = 0.954 ampere.
5 5V


Realistically, though, we haven't even considered ripple and we need
to do that in order to choose the value of the BFC.

In order to do that we need to know how much current your radio
draws, but you don't know how much that is.

No matter, we can work something out and then plug in your numbers
when you find out.

Let's say your radio's running on a watt. Then the current into it
will be:


P 1W
I = --- = ---- = 0.111A = 110mA
E 9V

And the voltage out of the bridge will be:

/ 5V * 0.111A \
Vout = Vout(2) - | ------------- |
\ 1A /

= 16.27V - 0.555V


~ 15.72V


Since the 7809's dropout voltage is 11.5V and we have 15.72V
available at the output of the bridge, the difference (4.22V) is the
ripple we're allowed.


The filter cap's capacitance, then, would be:


Idt 0.111A * 0.01s
C = ----- = ---------------- = 2.63E-4F = 263ľF
dV 4.22V


Since the bridge will be supplying current to the load as well as to
the cap when it's charging, the voltage available from the
transformer will drop somewhat during that time so it would be a
good idea to increase the value of the capacitor in order to
compensate for that.

Without going into it analytically, I'd guess that doubling the cap
would do it. Even better, throw 1000ľF in there; they're cheap!

Only one thing left to do and that's to check on whether you've got
enough headroom to get that 11.5V with low mains, and you now ought
to have enough information to be able to do that.



--
JF

 

[Peter Bennett]

On Thu, 10 Jan 2008 08:41:57 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

I'm a bit confused as to what the LM7809 regulators actually are.
I've heard they're integrated circuits, but isn't the TO-220 case for
power transistors or am I wrong?

Yes, the 78xx family of voltage regulators are integrated circuits -
but they only have three external connections (input, output, and
ground), so a three terminal TO-220 package is quite suitable. They
may also have to dissipate significant power, so they need to be in a
heat-sinkable case.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Dave.H]

On Jan 11, 3:37 am, John Fields <jfie...@austininstruments.com> wrote:

On Thu, 10 Jan 2008 05:51:50 -0800 (PST), "Dave.H"

the19...@googlemail.com> wrote:
I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

---
You may want to consider something like one of these:

http://www.tamuracorp.com/clientuploads/pdfs/pg5.pdf

which you can get from Digi-Key.

In any case, how much current does the radio draw and what current
is your wall-wart rated for when it's outputting 9V?

--
JF

The wall wart is rated at 1 amp. Not sure what the current draw is.

 

[Dave.H]

On Jan 11, 3:37 am, "Bob Monsen" <rcmon...@gmail.com> wrote:

"Dave.H" <the19...@googlemail.com> wrote in message

news:66b3847d-eb48-4300-a4e0-b399edcf2d0e@l32g2000hse.googlegroups.com...

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

An LM317 is an alternative to the LM7809 others have been recommending. It
is a variable regulator, in which the voltage is set by a voltage divider on
the output. Radio shack does carry these.

BTW, your 9VAC transformer is just right for your device, which is 9VDC.
Once you regulate it down, you'll be right on target, and the voltage won't
dip (like it would with a 6VAC transformer).

Regards,
Bob Monsen

I'm a bit confused as to what the LM7809 regulators actually are.
I've heard they're integrated circuits, but isn't the TO-220 case for
power transistors or am I wrong?

 

[Dave.H]

On Jan 11, 3:01 am, gearhead <nos...@billburg.com> wrote:

On Jan 10, 7:37 am, gearhead <nos...@billburg.com> wrote:

On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

I think your best bet is to use a LM7809 voltage regulator after the
filter cap.

I just checked Radio Shack's web site. They don't carry the 7809, but
you can use the adjustable LM317T and set the output voltage to 9.
Here's the datasheethttp://www.fairchildsemi.com/ds/LM/LM317.pdf
Of course, using the 7809 would be a bit simpler but you would have to
order it and have it shipped from an online supplier like Digikey.com
or a miscellany house like allelectronics.com. To avoid shipping
charges on a part that costs about a dollar, it makes sense to stop in
at RS for the LM317T. Radio Shack's web site will even tell you
whether the Radio Shack nearest you has the part in stock.

I don't live in the US, I live in Australia, and it just so happens
that Dick Smith Electronics sells the 7809 regulator (Z6550). I would
be interested in using this and need a diagram (simple how to wire it
up. The power supply schematic has the transformer connected to the
bridge rectifier, connected to the DC output with a cap or two across
the output.

 

[gearhead]

On Jan 10, 7:37 am, gearhead <nos...@billburg.com> wrote:

On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use.  I have no
problem using a wall wart, but I can't find one of the correct
voltage.  I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC.  Is there a way to halve this voltage with resistors
without generating too much heat?


I think your best bet is to use a LM7809 voltage regulator after the
filter cap.

I just checked Radio Shack's web site. They don't carry the 7809, but
you can use the adjustable LM317T and set the output voltage to 9.
Here's the datasheet
http://www.fairchildsemi.com/ds/LM/LM317.pdf
Of course, using the 7809 would be a bit simpler but you would have to
order it and have it shipped from an online supplier like Digikey.com
or a miscellany house like allelectronics.com. To avoid shipping
charges on a part that costs about a dollar, it makes sense to stop in
at RS for the LM317T. Radio Shack's web site will even tell you
whether the Radio Shack nearest you has the part in stock.

 

[John Fields]

On Thu, 10 Jan 2008 05:51:50 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

---
You may want to consider something like one of these:

http://www.tamuracorp.com/clientuploads/pdfs/pg5.pdf

which you can get from Digi-Key.


In any case, how much current does the radio draw and what current
is your wall-wart rated for when it's outputting 9V?


--
JF

 

[Bob Monsen]

"Dave.H" <the1930s@googlemail.com> wrote in message
news:66b3847d-eb48-4300-a4e0-b399edcf2d0e@l32g2000hse.googlegroups.com...

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

An LM317 is an alternative to the LM7809 others have been recommending. It
is a variable regulator, in which the voltage is set by a voltage divider on
the output. Radio shack does carry these.

BTW, your 9VAC transformer is just right for your device, which is 9VDC.
Once you regulate it down, you'll be right on target, and the voltage won't
dip (like it would with a 6VAC transformer).

Regards,
Bob Monsen

 

[amdx]

"Dave.H" <the1930s@googlemail.com> wrote in message
news:66b3847d-eb48-4300-a4e0-b399edcf2d0e@l32g2000hse.googlegroups.com...

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

To eliminate the heat problem, you can use a non-polarised capacitor
between the transformer secondary and the rectifier bridge. This will drop
some voltage similar to a resistor. I have always picked the value
by trial and error, someone may speakup with the math required to calculate
the proper value. I have found the speaker crossover caps to
be useful in this application. WAG, 10uf to 100uf just give you a start.
Also note the output voltage of your 9VAC transformer will drop down
from your measured 12.5-13 VAC when you load it to its rated current.
Mike

 

[amdx]

"gearhead" <nospam@billburg.com> wrote in message
news:bda6fdd9-a46f-44af-8f06-17d760261b51@c23g2000hsa.googlegroups.com...
On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?

Trying to use a resistive voltage divider to drive an active load is
not likely to work out well, because then the load itself is part of
the voltage divider. When the load increases and decreases, the
voltage changes.
In order for a resistive voltage divider to supply a somewhat steady
voltage while supporting an active (varying) load, the impedance of
the divider has to be very low, which means you would have to use
chunky, low-value power resistors and waste lots of current. Just a
bad way to do it.

I think your best bet is to use a LM7809 voltage regulator after the
filter cap.

 

[gearhead]

On Jan 10, 5:51 am, "Dave.H" <the19...@googlemail.com> wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use.  I have no
problem using a wall wart, but I can't find one of the correct
voltage.  I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC.  Is there a way to halve this voltage with resistors
without generating too much heat?

Trying to use a resistive voltage divider to drive an active load is
not likely to work out well, because then the load itself is part of
the voltage divider. When the load increases and decreases, the
voltage changes.
In order for a resistive voltage divider to supply a somewhat steady
voltage while supporting an active (varying) load, the impedance of
the divider has to be very low, which means you would have to use
chunky, low-value power resistors and waste lots of current. Just a
bad way to do it.

I think your best bet is to use a LM7809 voltage regulator after the
filter cap.

 

[Tim Wescott]

On Thu, 10 Jan 2008 05:51:50 -0800, Dave.H wrote:

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc, but
there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct voltage.
I have one rated at 9 VAC, outputting in the range of 12.5-13 VAC. Is
there a way to halve this voltage with resistors without generating too
much heat?

You can do it with a resistive divider, but what you really want to do is
use a voltage regulator.

Radio Shack has strayed far from their roots, but if you're in the US you
may want to go to the components section of one and see if they still
have the little "Experimenter's Notebooks" -- you want the one that
covers linear power supplies. Look in there for directions on making a
regulated 6V supply, and while you're in the store pick up the parts you
need.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

 

[Dave.H]

I'm building a power supply for an 8 transistor radio, that originally
used 6 "D" cells, involving a rectifier bridge, filtering caps etc,
but there's on problem,which is I can't seem to find an AC transformer
outputting 6 volts that fits in the case I want to use. I have no
problem using a wall wart, but I can't find one of the correct
voltage. I have one rated at 9 VAC, outputting in the range of
12.5-13 VAC. Is there a way to halve this voltage with resistors
without generating too much heat?