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Sci.Electronics.Basics -> power supply mods
There are 9 messages in this thread.
You are currently looking at messages 1 to 9.
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Author: H. DixonDate: 21:42 03-08-08
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I've finished the standard "build yourself a adjustable power supply"
project and some small others. I was looking at this:
http://www.electronics-lab.com/projects/test/014/index.html
I'd like to try this and was wondering about actually adding it to the
power supply. The PS has two 24V center tapped transformers with the
primaries in parallel and the secondaries in series. Right now the
center taps are not used. Could I make use of the centers, couple of
resistors, add another bridge rectifier and supply the input for the
above voltmeter? It does require +-5 volts so I guess it would have to
resemble the schematic at the bottom where they use the 7805/7905 to
get the +-5V.
Just wondering if the above is a doable near future project.
TIA
H. Dixon
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Author: Phil AllisonDate: 22:11 03-08-08
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"H. Dixon"
>
> I've finished the standard "build yourself a adjustable power supply"
> project and some small others. I was looking at this:
> http://www.electronics-lab.com/projects/test/014/index.html
>
> I'd like to try this and was wondering about actually adding it to the
> power supply. The PS has two 24V center tapped transformers with the
> primaries in parallel and the secondaries in series. Right now the
> center taps are not used. Could I make use of the centers, couple of
> resistors, add another bridge rectifier and supply the input for the
> above voltmeter? It does require +-5 volts so I guess it would have to
> resemble the schematic at the bottom where they use the 7805/7905 to
> get the +-5V.
** Beware - the " Sample Power supply 1 " schem will simply not work.
The other schems look OK.
..... Phil
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Author: John PopelishDate: 22:12 03-08-08
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H. Dixon wrote:
> I've finished the standard "build yourself a adjustable power supply"
> project and some small others. I was looking at this:
> http://www.electronics-lab.com/projects/test/014/index.html
>
> I'd like to try this and was wondering about actually adding it to the
> power supply. The PS has two 24V center tapped transformers with the
> primaries in parallel and the secondaries in series. Right now the
> center taps are not used. Could I make use of the centers, couple of
> resistors, add another bridge rectifier and supply the input for the
> above voltmeter? It does require +-5 volts so I guess it would have to
> resemble the schematic at the bottom where they use the 7805/7905 to
> get the +-5V.
>
> Just wondering if the above is a doable near future project.
I'm pretty sure you will need an isolated supply for this
kit, in order to use it to measure a voltage referenced to
the negative side of your regulates supply.
The kit gives few detail,s, but the data sheet for the chip:
http://www.intersil.com/data/fn/fn3082.pdf
Gives more. It appears that the input common of the chip
(pin 32) is biased to some voltage between the supply rails.
So the supply voltages must center themselves around the
voltage being measured.
I suggest you get another small transformer, say, a 12 volt
center tapped .3 amp transformer and use a bridge rectifier
and the center tap to produce a raw + and - 7 volt supplies
that you regulate to + and - 5 volts @ 200mA with 7805 and
7905 regulators. Then you can connect the digital meter to
whatever you want to measure without worry that there will
be a supply conflict.
--
Regards,
John Popelish
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Author: Phil AllisonDate: 22:26 03-08-08
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"John Popelish"
> I'm pretty sure you will need an isolated supply for this kit,
** Not true.
> The kit gives few detail,s, but the data sheet for the chip:
> http://www.intersil.com/data/fn/fn3082.pdf
> Gives more. It appears that the input common of the chip (pin 32)
** The 7107 is the one used here ( LED drive) and that needs +/- 5 volts
supplies with the common on pin 21.
Usually, one just connects the minus input to the PSU common.
The 7106 has a floating ( ie battery ) supply.
..... Phil
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Author: John PopelishDate: 22:35 03-08-08
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Phil Allison wrote:
> "H. Dixon"
>> I've finished the standard "build yourself a adjustable power supply"
>> project and some small others. I was looking at this:
>> http://www.electronics-lab.com/projects/test/014/index.html
(snip)
> ** Beware - the " Sample Power supply 1 " schem will simply not work.
It might, if the positive and negative supply currents vary=20
only a small fraction of the total current if the LED=20
current passes through both regulators. But I can't see how=20
the LED current returns to the negative supply at all.
The data sheet:
http://www.intersil.com/data/fn/fn3082.pdf
shows the digital ground (the negative rail for the LED=20
switches) connecting to pin 27 but the schematic shows pin=20
27 connected only to a capacitor. And this is what is shown=20
on the data sheet as a typical ICL7107 design. WTF? How=20
does the current dumped by the segment drivers into digital=20
ground return to any supply?
Found it on page 5:
"=95 ICL7107 POWER SUPPLY: DUAL =B15.0V
V+ =3D +5V to GND
V- =3D -5V to GND
Digital Logic and LED driver supply V+ to GND"
But GND is pin 21, which is the common between the positive=20
and negative supplies, so the LED current is drawn only=20
through the +5 volt regulator, not the -5, so you are=20
exactly right. It can't work. I'm trying to keep up. ;-)
--=20
Regards,
John Popelish
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Author: Phil AllisonDate: 23:03 03-08-08
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"John Popelish"
Phil Allison wrote:
> ** Beware - the " Sample Power supply 1 " schem will simply not work.
It might, if the positive and negative supply currents vary
only a small fraction of the total current if the LED
current passes through both regulators. But I can't see how
the LED current returns to the negative supply at all.
The data sheet:
http://www.intersil.com/data/fn/fn3082.pdf
shows the digital ground (the negative rail for the LED
switches) connecting to pin 27 but the schematic shows pin
27 connected only to a capacitor. And this is what is shown
on the data sheet as a typical ICL7107 design. WTF?
** Looks like misprint in figure 8, pin 27 should be labelled as pin 21.
The -5 volt supply needs only small current, hence the simple inverter schem
to derive it from the +5 volt one.
...... Phil
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Author: H. DixonDate: 10:21 04-08-08
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--snip
> Found it on page 5:
> "=95 ICL7107 POWER SUPPLY: DUAL =B15.0V
> =A0 V+ =3D +5V to GND
> =A0 V- =3D -5V to GND
> =A0 Digital Logic and LED driver supply V+ to GND"
>
> But GND is pin 21, which is the common between the positive
> and negative supplies, so the LED current is drawn only
> through the +5 volt regulator, not the -5, so you are
> exactly right. =A0It can't work. =A0I'm trying to keep up. =A0;-)
>
> --
> Regards,
>
> John Popelish
These display segments are common anode types - is that maybe why?
H. Dixon
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Author: John PopelishDate: 12:15 04-08-08
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H. Dixon wrote:
> --snip
>=20
>> Found it on page 5:
>> "=95 ICL7107 POWER SUPPLY: DUAL =B15.0V
>> V+ =3D +5V to GND
>> V- =3D -5V to GND
>> Digital Logic and LED driver supply V+ to GND"
>>
>> But GND is pin 21, which is the common between the positive
>> and negative supplies, so the LED current is drawn only
>> through the +5 volt regulator, not the -5, so you are
>> exactly right. It can't work. I'm trying to keep up. ;-)
>>
>> --
>> Regards,
>>
>> John Popelish
>=20
> These display segments are common anode types - is that maybe why?
The problem involves the balance of current through the two=20
regulators. The simple 18 volt battery supply with two=20
regulators shown will work only of the two regulators see=20
almost identical load currents. This is because that is the=20
only way they will divide the 18 volt battery voltage in=20
half. If either draws more current, its half of the battery=20
voltage will collapse.
The meter circuit powers all the LEDs from the positive=20
supply (common anode), but returns that current to the=20
supply not through the -5 volt connection (that would=20
balance the regulator currents) but through the common=20
connection between the two regulators. so all that current=20
has to return to the battery through a 1k resistor, upstream=20
of the regulators, and that is not possible for the ~200=20
milliamperes and a 1k resistor, because that would require a=20
200 volt drop across the resistor, and only 18 is available.
If you replace the 18 volt battery with two 9 volt batteries=20
(or supplies) stacked in series, with the mid point of that=20
stack connected to the output common, then the LED current=20
comes from the plus 9 supply and returns to it through its=20
common connection, while the minus 9 volt supply carries=20
only the much smaller non LED chip current. That is why I=20
recommended building a supply derived from a 12 volt center=20
tapped secondary transformer. The center tap acts as that=20
mid point that separates the positive supply current from=20
the negative supply current.
--=20
Regards,
John Popelish
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Author: H. DixonDate: 11:27 05-08-08
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On Aug 4, 12:15=A0pm, John Popelish <jpopel...@rica.net> wrote:
> H. Dixon wrote:
> > --snip
>
> >> Found it on page 5:
> >> "=95 ICL7107 POWER SUPPLY: DUAL =B15.0V
> >> =A0 V+ =3D +5V to GND
> >> =A0 V- =3D -5V to GND
> >> =A0 Digital Logic and LED driver supply V+ to GND"
>
> >> But GND is pin 21, which is the common between the positive
> >> and negative supplies, so the LED current is drawn only
> >> through the +5 volt regulator, not the -5, so you are
> >> exactly right. =A0It can't work. =A0I'm trying to keep up. =A0;-)
>
> >> --
> >> Regards,
>
> >> John Popelish
>
> > These display segments are common anode types - is that maybe why?
>
> The problem involves the balance of current through the two
> regulators. =A0The simple 18 volt battery supply with two
> regulators shown will work only of the two regulators see
> almost identical load currents. =A0This is because that is the
> only way they will divide the 18 volt battery voltage in
> half. =A0If either draws more current, its half of the battery
> voltage will collapse.
>
> The meter circuit powers all the LEDs from the positive
> supply (common anode), but returns that current to the
> supply not through the -5 volt connection (that would
> balance the regulator currents) but through the common
> connection between the two regulators. =A0so all that current
> has to return to the battery through a 1k resistor, upstream
> of the regulators, and that is not possible for the ~200
> milliamperes and a 1k resistor, because that would require a
> 200 volt drop across the resistor, and only 18 is available.
>
> If you replace the 18 volt battery with two 9 volt batteries
> (or supplies) stacked in series, with the mid point of that
> stack connected to the output common, then the LED current
> comes from the plus 9 supply and returns to it through its
> common connection, while the minus 9 volt supply carries
> only the much smaller non LED chip current. =A0That is why I
> recommended building a supply derived from a 12 volt center
> tapped secondary transformer. =A0The center tap acts as that
> mid point that separates the positive supply current from
> the negative supply current.
>
> --
> Regards,
>
> John Popelish
Ok I "think" I understand. I'll go the separate transformer route.
Thank you.
H. Dixon
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