Need Voltage Doubler Circuit

[Dave.H]

I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @ www.dse.com.au

Thanks
Dave

 

[Tim Wescott]

On Thu, 24 Apr 2008 07:31:54 -0700, Dave.H wrote:

I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to provide
the 45v B+, I've found circuits via Google, but they don't give the
values for 24 VAC. Also don't say what diode to use. I have about 50
1N4004's lying around I could use. Transformer I'm planning on using is
M6672 @ www.dse.com.au

Thanks
Dave

Voltage doublers give poor regulation, and regenerative sets want a good,
steady input voltage. You'll get better performance off of you pair of
wall warts, no matter how ugly things look.

--
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

 

[John Popelish]

Dave.H wrote:

I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @ www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp. A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply. Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end. The output is
taken across those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies. Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output. You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter. Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output. This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler. I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC. Your diodes and transformer will
allow about 300 to 400 mA load current. If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts. This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

 

[John Popelish]

Dave.H wrote:

On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:
(snip)
I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit. I think the
current draw of the radio is only about 8-10 mA. Transformer is M2860
@ www.dse.com.au.

Sounds like a plan. The light load voltage may be something
like 33 volts, so the peak would be 33*sqrt(2)=47, but the
bridge will waste about 2 volts. so that might do very
nicely. At that low current, the 2,200 uF capacitor will
produce a low ripple output. If the voltage comes out a
little high, add a resistor in series with the either the 30
volt secondary, or in series with the primary.

--
Regards,

John Popelish

 

[Mike Silva]

On Apr 24, 10:31 am, "Dave.H" <the19...@googlemail.com> wrote:

I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use.  I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

Here is a standard voltage doubler circuit:
http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

The 1N4004s will be fine. Use filter caps rated for 50V or more,
50-100uF, and consider adding a second stage of RC filtering by adding
a resistor (dropping, say, 10% of your voltage at your current draw)
followed by another filter cap (50-100uF again). For best results you
should also regulate the voltage going into the detector stage - a
simple resistor and zener will do. Many people also find that adding
bypass caps (0.01 to 0.001) across the diodes helps eliminate power-
supply radiated hum.

Mike

 

[Dave.H]

On Apr 25, 1:29 am, John Popelish <jpopel...@rica.net> wrote:

Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp. A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply. Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end. The output is
taken across those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies. Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output. You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter. Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output. This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler. I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC. Your diodes and transformer will
allow about 300 to 400 mA load current. If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts. This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

 

[Dave.H]

On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 1:29 am, John Popelish <jpopel...@rica.net> wrote:



Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp. A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply. Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end. The output is
taken across those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies. Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output. You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter. Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output. This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler. I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC. Your diodes and transformer will
allow about 300 to 400 mA load current. If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts. This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit. I think the
current draw of the radio is only about 8-10 mA. Transformer is M2860
@ www.dse.com.au.

 

[Mike Silva]

On Apr 24, 11:46 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:





On Apr 25, 1:29 am, John Popelish <jpopel...@rica.net> wrote:

Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use.  I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp.  A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply.  Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end.  The output is
taken across  those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies.  Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts.  Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output.  You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter.  Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output.  This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler.  I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC.  Your diodes and transformer will
allow about 300 to 400 mA load current.  If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts.  This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option.  Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit.  I think the
current draw of the radio is only about 8-10 mA.  Transformer is M2860
@www.dse.com.au.- Hide quoted text -

- Show quoted text -

Another option is to use the 30V transformer in the voltage doubler
circuit. Take your 40-45V off the lower filter cap (and add
regulation for the detector), and take 80V off the upper cap for the
power output stage that you talked about.

Mike

 

[Dave.H]

On Apr 25, 2:03 am, Mike Silva <snarflem...@yahoo.com> wrote:

On Apr 24, 11:46 am, "Dave.H" <the19...@googlemail.com> wrote:



On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 1:29 am, John Popelish <jpopel...@rica.net> wrote:

Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp. A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply. Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end. The output is
taken across those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies. Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output. You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter. Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output. This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler. I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC. Your diodes and transformer will
allow about 300 to 400 mA load current. If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts. This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit. I think the
current draw of the radio is only about 8-10 mA. Transformer is M2860
@www.dse.com.au.-Hide quoted text -

- Show quoted text -

Another option is to use the 30V transformer in the voltage doubler
circuit. Take your 40-45V off the lower filter cap (and add
regulation for the detector), and take 80V off the upper cap for the
power output stage that you talked about.

Mike

Is this how I would hook it up?
http://s222.photobucket.com/albums/dd237/ozguy89/?action=view&current=FullWavePowerSupply.gif
Sorry for the large image file, this image was emailed to me to help
in building a power supply for a battery transistor radio.

 

[Dave.H]

On Apr 25, 2:03 am, Mike Silva <snarflem...@yahoo.com> wrote:

On Apr 24, 11:46 am, "Dave.H" <the19...@googlemail.com> wrote:



On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 1:29 am, John Popelish <jpopel...@rica.net> wrote:

Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp. A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply. Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end. The output is
taken across those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies. Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output. You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter. Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output. This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler. I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC. Your diodes and transformer will
allow about 300 to 400 mA load current. If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts. This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit. I think the
current draw of the radio is only about 8-10 mA. Transformer is M2860
@www.dse.com.au.-Hide quoted text -

- Show quoted text -

Another option is to use the 30V transformer in the voltage doubler
circuit. Take your 40-45V off the lower filter cap (and add
regulation for the detector), and take 80V off the upper cap for the
power output stage that you talked about.

Mike

What value resistor would I need to run a 2 volt LED off the 30 volts?

 

[Dave.H]

On Apr 25, 3:25 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 2:03 am, Mike Silva <snarflem...@yahoo.com> wrote:



On Apr 24, 11:46 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 1:29 am, John Popelish <jpopel...@rica.net> wrote:

Dave.H wrote:
I have a regen radio powered by two 23 volt wallwarts in series, but
would like to use a 24 VAC transformer with a voltage doubler to
provide the 45v B+, I've found circuits via Google, but they don't
give the values for 24 VAC. Also don't say what diode to use. I have
about 50 1N4004's lying around I could use. Transformer I'm planning
on using is M6672 @www.dse.com.au

The diodes should work fine as long as the DC current load
is less than about a half amp. A simple doubles connects
two capacitors (negative side of one and positive side of
the other) to one side of the 24 supply and two diodes
(cathode or banded end of one and anode of the other) to the
other side of the 24 volt supply. Then connect the free
anode to the free negative capacitor end and the free
cathode to the free positive capacitor end. The output is
taken across those last two connections, so that the two
capacitors are acting as a pair of stacked half wave
supplies. Under no load, this will produce at least
24*sqrt(2) *2 minus two diode drops or about 66 volts. Keep
in mind that the 24 volts AC output may rise quite a bit
under no load, since it is rated for full load.

Any load will sag some the voltage, and add twice line
frequency ripple to the output. You may want to add an RC
low pass filter stage with the series R selected to lower
the voltage to closer to your desired 45 volts, and the
capacitor from that output to the other side of the supply
as an additional ripple filter. Or you may just lower the
voltage under load by adding some resistance in series with
the 24 volt AC output. This will not reduce the ripple as
much, but will make the 24 volt transformer run a little
cooler. I just looked up the transformer and it has several
taps, so you can try different combinations to get very
close to the desired output voltage.

For the two doubler capacitors, I would use something like
5,000 to 10,000 microfarads per amp of output load and rated
for at least 16 volts DC. Your diodes and transformer will
allow about 300 to 400 mA load current. If your load
current is at or below 400 mA, that implies something around
2,200 to 4,700 uF 16 or 25 volt capacitors for the doubler.

Of course, if you use the full 30 volt winding, you can get
pretty close to 45 volts output with a bridge rectifier made
with 4 of your diodes and a single capacitor, say, 2,200 to
4,700 uF, rated for 50 volts. This will give you a voltage
around 40 to 45 volts, depending on load current that can be
up to about 0.7 amps.

--
Regards,

John Popelish

I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit. I think the
current draw of the radio is only about 8-10 mA. Transformer is M2860
@www.dse.com.au.-Hidequoted text -

- Show quoted text -

Another option is to use the 30V transformer in the voltage doubler
circuit. Take your 40-45V off the lower filter cap (and add
regulation for the detector), and take 80V off the upper cap for the
power output stage that you talked about.

Mike

What value resistor would I need to run a 2 volt LED off the 30 volts?

Would it be possible to run the tube filament (2 volts, 60 mA) off a
6.3 volt, 500 mA transformer, using something like resistors? I would
like to eliminate using "D" cells.
Thanks,
Dave

 

[Mike Silva]

On Apr 24, 1:20 pm, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 2:03 am, Mike Silva <snarflem...@yahoo.com> wrote:

Another option is to use the 30V transformer in the voltage doubler
circuit.  Take your 40-45V off the lower filter cap (and add
regulation for the detector), and take 80V off the upper cap for the
power output stage that you talked about.

Mike

Is this how I would hook it up?http://s222.photobucket.com/albums/dd237/ozguy89/?action=view¤t...
Sorry for the large image file, this image was emailed to me to help
in building a power supply for a battery transistor radio.- Hide quoted text -

- Show quoted text -

Not for my suggestion. I'm suggesting, if you also want a higher
voltage for a power output stage, that you use a voltage doubler like
so:http://www.amazon.com/Losing-Faith-Preacher-Atheist/dp/187773313X

The bottom of the two capacitors is your 0V or ground reference. The
top of the two capacitors will give you 80-90V with a 30V
transformer. The junction of the two capacitors will be at half that
voltage or 40-45V. Thus you have low voltage (suggest additional
filtering or regulation) for your detector and audio voltage
amplifier, and higher voltage for your power stage. Each capacitor
needs to be rated 50V or higher, since each one receives the full
secondary peak voltage.

Also, you asked about AC on the filaments. That's likely to lead to
hum. A good idea would be to turn the 6.3VAC into well-filtered DC
and drop that to 1.4V with a resistor. The resistor will help protect
your filaments from turn-on surge, which is why it's a better approach
than building a 1.4V regulated source.

Mike

 

[Mike Silva]

On Apr 24, 2:19 pm, Mike Silva <snarflem...@yahoo.com> wrote:

Not for my suggestion.  I'm suggesting, if you also want a higher
voltage for a power output stage, that you use a voltage doubler like
so:http://www.amazon.com/Losing-Faith-Preacher-Atheist/dp/187773313X

Umm, wrong link. Sheesh! :-O

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

 

[John Popelish]

Dave.H wrote:

On Apr 25, 3:25 am, "Dave.H" <the19...@googlemail.com> wrote:

What value resistor would I need to run a 2 volt LED off the 30 volts?

If the 30 volts is DC , you just apply Ohm's law using the
desired current and the amount of voltage the resistor must
waste. For instance, if you had 30 volts DC and you wanted
a resistor that would pass 10 mA (1/100 amp) to the LED
while using up 28 of the 30 volts (leaving 2 volts for the
LED), you plug those numbers into E=I*R and rearrange to
solve for R. 28 V=(1/100 A)*R, or R=28*100=2800ohms. You
could use thew standard values of 2700 or 3000 for a little
more or less than 10 mA. But you would actually want to
wait until the supply was built and measure the DC voltage
before calculating the resistor.

Would it be possible to run the tube filament (2 volts, 60 mA) off a
6.3 volt, 500 mA transformer, using something like resistors? I would
like to eliminate using "D" cells.

It would be possible and a lot more efficient than using the
30 volt supply for the 2 volt filament. Same formula.
AC voltages are measured in RMS volts, because that gives
the equivalent DC voltage as far as heating effect on
resistors goes. So you need to have a resistor that wastes
6.3-2 =4.3 volts while passing 0.06 amps. So 4.3=0.06*R, or
R=4.3/0.06 = 71.67 ohms. Standard values close to that are
68 and 75. If the 2 volt filament can run off a D cell (1.5
volts) you can probably go with the higher resistance,
especially if the transformer actually outputs a little more
than 6.3 volts with this light load.

You also want a resistor that stands the heat that produces,
so multiply amps through times volts across that resistor to
solve for the watts it is producing.
4.3V*0.06A=0.258 watts.

A little too much for a 1/4 watt resistor to handle, so a
1/2 watt or larger is needed.

But check if you have a 6.3 volt center tapped transformer.
You can get 6.3/2 or 3.15 volt from one end to the center
terminal and lower the power you have to waste in the
resistor.

--
Regards,

John Popelish

 

[Dave.H]

On Apr 25, 12:01 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

"Tim Wescott"



Voltage doublers give poor regulation,

** The standard voltage doubler has exactly the *same* ( load ) percentage
regulation as the same transformer would provide with a bridge rectifier and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

and regenerative sets want a good, steady input voltage.

** Only way to get that is with a regulated supply.

You'll get better performance off of you pair of
wall warts, no matter how ugly things look.

** Not true.

Line voltage and load variations will affect them just like any unregulated
supply.

The actual percentage load regulation depends on the size of the transformer
and how much load is applied - ie light loads = good regulation.

..... Phil

Actually the wallwarts are regulated, I think they're M9926 @ www.dse.com.au

 

[Dave.H]

On Apr 25, 4:19 am, Mike Silva <snarflem...@yahoo.com> wrote:

On Apr 24, 1:20 pm, "Dave.H" <the19...@googlemail.com> wrote:

On Apr 25, 2:03 am, Mike Silva <snarflem...@yahoo.com> wrote:

Another option is to use the 30V transformer in the voltage doubler
circuit. Take your 40-45V off the lower filter cap (and add
regulation for the detector), and take 80V off the upper cap for the
power output stage that you talked about.

Mike

Is this how I would hook it up?http://s222.photobucket.com/albums/dd237/ozguy89/?action=view¤t...
Sorry for the large image file, this image was emailed to me to help
in building a power supply for a battery transistor radio.- Hide quoted text -

- Show quoted text -

Not for my suggestion. I'm suggesting, if you also want a higher
voltage for a power output stage, that you use a voltage doubler like
so:http://www.amazon.com/Losing-Faith-Preacher-Atheist/dp/187773313X

The bottom of the two capacitors is your 0V or ground reference. The
top of the two capacitors will give you 80-90V with a 30V
transformer. The junction of the two capacitors will be at half that
voltage or 40-45V. Thus you have low voltage (suggest additional
filtering or regulation) for your detector and audio voltage
amplifier, and higher voltage for your power stage. Each capacitor
needs to be rated 50V or higher, since each one receives the full
secondary peak voltage.

Also, you asked about AC on the filaments. That's likely to lead to
hum. A good idea would be to turn the 6.3VAC into well-filtered DC
and drop that to 1.4V with a resistor. The resistor will help protect
your filaments from turn-on surge, which is why it's a better approach
than building a 1.4V regulated source.

Mike

I see that I replied to your post by mistake. I meant to reply to
John Popelish 's message about using a bridge rectifier and cap to
boost the 30 Volts AC to 45 volts DC. Seems much more simpler to me.

Dave

 

[Mike Silva]

On Apr 24, 10:01 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

** The standard voltage doubler has exactly the *same* ( load ) percentage
regulation as the same transformer would provide with a bridge rectifier and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

Are we sure about that? Only one-half of the capacitor stack is
charged each half cycle, but both capacitors are discharging each half
cycle.

Which brings up a circuit I ran across years ago and was able to find
again yesterday. This one claims to charge both capacitors each half
cycle, making it a true full-wave circuit. I haven't tried it or
analysed it - I just put it out for consideration.
http://www.kwarc.org/bulletin/99-04/tech_corner.htm

Mike

 

[Mike Silva]

On Apr 25, 7:59 am, "Phil Allison" <philalli...@tpg.com.au> wrote:

I knew you'd have one of your outbursts on this, but I thought the
question interesting enough to endure it. I hope others will weigh
in.

 

[Dave.H]

On Apr 25, 2:04 am, John Popelish <jpopel...@rica.net> wrote:

Dave.H wrote:
On Apr 25, 1:38 am, "Dave.H" <the19...@googlemail.com> wrote:
(snip)
I might go with that bridge rectifier option. Sounds simpler, and I
believe I have a 2,200 MFD, 50 volt cap lying around.

I found a 30 volt, 500 mA transformer I'll use with the bridge
rectifier, Smaller and cheaper than the one amp unit. I think the
current draw of the radio is only about 8-10 mA. Transformer is M2860
@www.dse.com.au.

Sounds like a plan. The light load voltage may be something
like 33 volts, so the peak would be 33*sqrt(2)=47, but the
bridge will waste about 2 volts. so that might do very
nicely. At that low current, the 2,200 uF capacitor will
produce a low ripple output. If the voltage comes out a
little high, add a resistor in series with the either the 30
volt secondary, or in series with the primary.

--
Regards,

John Popelish

Should I regulate the output? How about a voltage regulator IC like
the LM317?

 

[Rich Grise]

On Fri, 25 Apr 2008 04:45:37 -0700, Mike Silva wrote:

On Apr 24, 10:01 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:
** The standard voltage doubler has exactly the *same* ( load ) percentage
regulation as the same transformer would provide with a bridge rectifier and
capacitor filter.

http://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/voldoub.html

Are we sure about that? Only one-half of the capacitor stack is
charged each half cycle, but both capacitors are discharging each half
cycle.

Which brings up a circuit I ran across years ago and was able to find
again yesterday. This one claims to charge both capacitors each half
cycle, making it a true full-wave circuit. I haven't tried it or
analysed it - I just put it out for consideration.
http://www.kwarc.org/bulletin/99-04/tech_corner.htm

I don't know why anyone could need anything that's not here:
http://www.tpub.com/neets/book7/27m.htm

Cheers!
Rich