Things You'll Need
Instructions
Examine the formula used to calculate the desired output voltage. It is Vout = Vref(1 + R2/R1) +IqR2. R1 is the resistor to be placed at the regulator̵7;s output pin, while R2 is the resistor to be placed at the regulator̵7;s ground or common pin. Iq is the quiescent current output from the ground pin through R2.
Calculate the output voltage for values of R1 = 470, and R2 = 220. Use the LM7805, which has a reference value of 5 volts. If Iq is 3 mA, or milliamps, the equation is Vout = 5(1 + 220/470) + 220*0.003 = 7.96 volts.
Add the chip to the breadboard so it is parallel and has all three of its pins inserted into different columns. Attach a jumper wire between the input pin 1 and a breadboard row or column that is the voltage source for the circuit.
Place one end of the 0.22 capacitor at pin 1, and connect the other end to a row or column that is the breadboard̵7;s ground. Insert the 470 resistor between common pin 2 and output pin 3. Connect one end of the 220 resistor to pin 2 as well, and connect its other end to ground.
Attach the 0.1 capacitor between pin 3 and ground. Attach the 330 resistor, which functions as the load, the same way.
Adjust the power supply so that it outputs 9 volts. Connect its leads to the circuit, then turn it on. Measure the voltage across the load resistor by switching the multimeter on and placing its leads on each side of the 330 resistor. A representative output reading is 7.8 volts.