OK,

As you say you can ignore R2 and the LED1 for the moment they have 
nothing to do with 'regulating' the voltage. As you state this is a 
half wave rectified circuit.

The rough formula used for this is - Output voltage = input x 0.45 
(this is only true if there is no capacitor after the diode).

So you have (21vac - 0.6) x 0.45 = 9.18vdc. The 0.6 is the drop in 
voltage across the diode which is due to it's semiconductor 
properties. The max current that this will pass is  9.18v/27 = 340mA.

Now you have measured less current, that will be because there is a 
resistance in the battery pack itself, and it is only drawing 103mA.

This looks to be correct as you have measured a voltage drop of 2.96v 
across R1 so current being drawn by your battery pack is 2.96v/27ohm = 
109mA.

The LED indicator circuit is drawing about 11mA, so moving your meter 
to the input side of the circuit you should see about 115mA for the 
circuit. The current drawn will depend upon the state of charge and 
internal resistance of the battery pack.

Colin



:: I've measured some voltages whilst everything was connected: the
:: secondary
:: of the transformer is about 21 VAC. There is a drop of 2.96 VDC
:: across the
:: 27R, and a voltage of 10.5VDC across the battery. At the time, the
:: current
:: was 103 mA.
--
cdb, colin@btech-online.co.uk on 29/05/2008

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