> I'm hoping someone can give me some help understanding 
> this circuit- i'm
> just a hobbyist with no training in electronics and i'm 
> struggling with this
> one.

There are far easier circuits for a beginner to start on. 
While this looks very simple a confusing factor is that the 
AC is being half wave rectified (only positive half cycles 
pass through the diode) and so the DC voltages you are 
measuring are driven by half cycle sine waves with a half 
cycle off time between each one. This means that if you 
measure with either an AC or a DC meter you will get a 
reading BUT it will not represent an "RMS" or true value. 
The DC meter is confused by the AC component and the AC 
meter is not calibrated to measure such an odd waveform. If 
your volt  meter was a "true RMS"  one then the answers 
would be correct BUT could not easily be calculated.

An easy way to get results which are much more intuitive is 
to add a smoothing capacitor from ground (negative terminal) 
to the junction of D1/R1/R2. (In practice this may 
necessitate increasing R1 substantially - see below). The 
voltages will now assume a smoother DC value and ohms law 
will start to look like it works.

Let's look at what you'd get in such a case. For starters 
I'll assume that V after the diode is about 21V (it would be 
higher) to make the results comparable to what you see now.

V_D1 cathode = 21V.

Vbattery = 10.5 V = per cell voltage of 10.5/8 =~ 1.3V which 
is in the range of what you may see for charging.

V across 27R = 21V-10.5V = 9.5V
I_ 27R = V/R = 9.5/27 +1 350 mA
!!!!!!!!!!!!!!!!!
This is not what you see in practice.
The half wave and loading makes I much lower than would be 
expected.

A you see V_27R = 2.95V =~ 3V you would expect a current of 
I=V/R = 3/27 = 110 mA.
This is about what you see (!).

What this tells us is that more needs to be known to 
understand this circuit. The lower than expected voltage at 
D1 cathode may be due to loading of the half wave supply but 
MAY be because the transformer is designed to current limit. 
This is done with some small power supplies either due to 
economy or as a means of protection or regulation.

It is entirely possible that the circuit values were arrived 
at by "playing". They may have altered R1 to get the desired 
value of current.

Note that as Vin >> Vbattery this approximates a constant 
current source. As Vbattery rises with charge state the 
current only reduces slightly as Vin is high enough to 
provide substantial current. The effect is masked by the 
half wave high ripple waveform.

R2 and the LED simply use the voltage across R1 as a LED 
power supply. R2 sets LED current.

*IF* there was a large enough smoothing cap after D1 then.

VDC ~= 1.4 x Vin AC.

U charge = V/r = (VDC-Vbattery)/R1

V_R1 = VDC-VBattery

VR2 = VR1 - VLED.
VLED for red  LED ~= 1.2V.

I LED = V_R2/R2 = (VR1-1.2)/R2

As long as R2 >> R1 then R2 does not muchly affect charge 
current.

More if wanted, but try that first.


        Russell

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