power supply mods

"H. Dixon"

** Beware - the " Sample Power supply 1 " schem will simply not work.

The other schems look OK.

..... Phil

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:

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

"John Popelish"

** Not true.

** 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

(snip)

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:

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

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:

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

--snip

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

H. Dixon

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

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

Thank you.

H. Dixon