I can also vary the base voltage and control the amount of current through the LED or other load devices. However, for some loads, even this circuit might be an improvement over a simple resistor.

This parallel circuit contains a resistor and a lamp. A current of 5 A flows through the cell. The current splits at the junction. 3 A flows through the resistor and 2 A flows through the bulb.

Using properties of a series circuit, the sum of the voltages across the LED and resistor R must equal the supply voltage. \({V_s} = {V_{LED}} + {V_R}\) Now substitute the supply voltage (\(6V ...

The circuit in question ... A transistor is also wired up, switching a 2.2 K resistor in parallel with the LEDs. When turned on by the PIC, this transistor causes roughly a 10 mA current to ...