6 volt LT at 2 amps

[Keithstv]

That circuit as it stands would only be OK up to 100mA.  The capacitor charges at the peak of the AC input voltage, charging to approximately 11.2V (1.414 x 9V - 1.5V. Peak of the AC voltage - 2 diode forward voltage drops). The capacitor will then be discharged by the load current. A rough approximation of the volt drop is given by  (Load current x 10mS)/ capacitor value. 10mS is approximately the time between the points on the AC input voltage that the capacitor is charged. This volt drop is the ripple voltage seen across the capacitor. In this case at a load current of 100mA the volt drop will be 100mA x 10mS / 470uF or approximately 2.1V. Subtract this from the peak voltage across the capacitor and you have the minimum input voltage to the regulator, in this case 9.1V.  With the 78 series regulators the minimum  input voltage is 3V greater than the output voltage. In this case this is 9V. So the regulator would be OK up to about 100mA. Any greater current and the output would dip as the voltage across the capacitor drops below 9V. Increasing the capacitor to 4700uF would reduce the ripple voltage to about 0.2V at a load current of 100mA or approximately 2V at a load current of 1A.

Looking at the data sheet for the LM338 the minimum input to output voltage at 2A is approximately 2.3V. This would need a minimum input voltage of 8.3V. Following the calculations above this would require a capacitor of 6800uF to work. With a 470uF capacitor there would be lot of ripple voltage on the output.

Note that the LM338 also requires two external resistors to set the output voltage and a substantial heatsink.

Hope that helps.

Keith