power supply mods

[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.

Just wondering if the above is a doable near future project.


TIA


H. Dixon

 

[Phil Allison]

"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

 

[Phil Allison]

"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

 

[Phil Allison]

"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

 

[H. Dixon]

--snip

Found it on page 5:
"• ICL7107 POWER SUPPLY: DUAL ±5.0V
  V+ = +5V to GND
  V- = -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.  ;-)

These display segments are common anode types - is that maybe why?

H. Dixon

 

[H. Dixon]

The problem involves the balance of current through the two
regulators.  The simple 18 volt battery supply with two
regulators shown will work only of the two regulators see
almost identical load currents.  This is because that is the
only way they will divide the 18 volt battery voltage in
half.  If 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.  so 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.  That is why I
recommended building a supply derived from a 12 volt center
tapped secondary transformer.  The center tap acts as that
mid point that separates the positive supply current from
the negative supply current.

Ok I "think" I understand. I'll go the separate transformer route.

Thank you.


H. Dixon