Need help reducing voltage

If you are trying to . . . and expecting for a 3VDC being at the other lead of that dropping resistor, that is not going to happen until you connect on that 3 Volt at 3 amp load which would actually compute as needing the use of a 1.8 ohm @ 27 watt power resistor for that level of voltage drop.

With your present miscalculation in the use of such a high value of . . . 620 ohm . . .resistor you should
experience that 9 volts of expected drop, down to 3 VDC, but with it ONLY having the capability of supplying a feeble . . .and thats heavy on the feeble . . . . . .14 or so milliamperes of current at that 3V level.
As you try to pull more current, that voltage plummets.

Sooooooo what current consumption level of are you needing at the 3V end of the resistor ?

 

The voltage "drop" across the resistor will vary with the current going through it, according to Ohm's Law.
E = I x R
The voltage across a resistance equals the value of the resistance times the current through it. So if you have a 620 ohm resistor, and the current being drawn by whatever is connected to the "output" side of the resistor is 10 mA (0.01 A), then the voltage drop across the resistor will be 6.2 V. If the source is 12.0 V, then the load will see a voltage of 5.8 V. But if the load current increases to 15 mA, then the voltage across the load will decrease to 2.7 V.

This is why the only way to get a stable voltage reduction that does not vary with the load current is to use a regulating device or circuit.

AND, the reason you do not see any effect from the 620 ohm resistor is because you probably are measuring it with a meter that has a very high input impedance. Using the same equation, calculate the voltage drop across the resistor when the load current through it is 1 uA (1 microamp, or 1 one-millionth of an amp).

ak