Stacking regulators in your case may not work. For the top regulator to
work its quiescent current must go through the load on the 5v rail. If
there is no load on the 5v output then it won't regulate. You can fix
this situation by putting a 470 ohm resistor permanently across the
lower regulator (this provides a continuous load of 11ma which is
greater than the top regulators 8ma maximum quiescent current so it will
regulate properly - see data sheet link below). This may be the best
solution if the wall wart is up to it.
But the wall wart could be a problem since it is unregulated and is
rated at 9v at some specified current which you haven't provided. All
regulators need some voltage 'overhead' to work. From the LM7805 datasheet
http://www.fairchildsemi.com/ds/LM/LM7805.pdf
The TYPICAL drop out voltage is 2 volts so if you want 9 volts out of
the regulator you need 11 volts going into it TYPICALLY - some 7805's
may need more. When you start putting a load on that wall wart the 14
volts output is going to fall and at 350ma it may be too low. My *guess*
is that unless the wall wart is rated at more than 9v @1Amp you have no
chance of making it work as you have suggested.
There are a few easy solutions to this: get a 9v regulated wall wart or
get a 12v unregulated wall wart but I suspect as much as possible you
want to roll-your-own. This regulator might solve your problem and still
use your wall wart:
http://www.ebay.com/itm/Boost-Buck-...375?pt=LH_DefaultDomain_0&hash=item43b703d86f
I haven't dealt with this supplier or used this circuit but it seems
reasonable and is from a top rated supplier on ebay so chances are good,
and if you are a hobbyist getting into electronics you will want to get
used to using ebay.
The voltage divider question:
The 78xx series is not a great candidate for this type of modification
because it's quiescent current is so high. It's not impossible but the
LM317 is much better for this (search
http://www.fairchildsemi.com for
the data sheet).
The simplest way to jack up the voltage output is to put a zener diode
in series with the gnd pin - for your example a 3.9v diode will do the
job nicely. The quiescent current is typically about 5ma and may be as
high as 8ma which is fine for the zener diode. Since the quiescent
current only changes by a maximum of .5ma this will have almost no
effect on the zener voltage and hence the output voltage
A brute force approach for the 7805 with voltage divider would go like
this:
We want to be able to plug any old 7805 regulator into the circuit and
be within 10% of the 9v value so we set the current through the voltage
divider to be much greater than the maximum quiescent current of the
regulator (8ma). To achieve this We make the design decision that the
divider current will be at least 80ma. Therefor the total resistance of
the divider must be no more than 112 ohms. The regulator gnd pin must be
raised by 4 volts so the bottom resistor (Rb) of the divider must be
4v/9v * 112 and the top resistor (Rt) equals the remainder. Rb = 49.8
ohm Rt= 62.2. The closest E24 values are 51 and 62 ohms (47 and 56 for
E12 values). Because of the amount of current involved you should check
the power dissipation in these resistors which is I*I*R which comes to
approximately 0.4 of a watt. In this example you need to use at least
1/2 watt resistors for Rt and Rb (if the circuit is to sit around
powered up for years and be reliable than 1 watt resistors would be a
better choice).
This design continually wastes 0.8 watt of power but has the advantage
that you could use it on a production line and get the right result
without adjustment but it is wasteful and inefficient.
An approach more suitable to a designer who will adjust his work is like
this:
The maximum change of the quiescent current in the regulator is 0.5ma
(DqI)so wee want to make that small (less than 10%) relative to the
current in the voltage divider. The top resistor (Rt) will always have 5
volts across it and for 5ma (DqI * 10) has a value of 1k ohm. The value
for Rb has to be selected by testing as quiescent current of the
regulator can vary from about 4ma to 8ma and that plus the current from
Rt must cause a 4v drop in Rb. So the value for Rb will be between about
360 to 440 ohms. In this case the power dissipated in Rt is just 25mW
and Rb is under 50mW. Common 1/4 watt resistors will be perfect and the
wasted power is under 0.1 watt. Just grabbing a 390 ohm resistor would
put the circuit within about a maximum of 11% of the desired output but
most people would trim that to be closer to the 9v.
The reason for not using Rb alone is because of how much the quiescent
current can change between devices (4 to 8ma) and how much it can change
in one device (0.5ma) depending on how it is used.
The zener diode works well with the 7805 but the resistor divider not so
well. The resistor divider works well with the LM317 because of its low
current through the adj pin but also for that reason the simple zener
approach does not work so well with the 317
HTH
