How To Convert A 9V Battery To 3.3V DC

To reduce a 9V battery to 3.3 volts, use a zener diode, such as a 1N746 or a 1N4728A. Choose the appropriate one based on how much power it can dissipate.

A 1N4728A has a 3.3-volt and a 1 W power rating. It can supply, on average, a stable 3.3 volts to a circuit or another component. The maximum current Izm is approximately 1 W/3.3 V = 303 mA. This means you must use a series resistor to ensure that the value of the current flowing through the diode never exceeds this amount, else it will become damaged or destroyed.

One option is a 330-ohm resistor. Using Ohm's Law, the maximum zener current I is (Vin – Vout)/R = (9 V – 3.3 V)/330 ohm = .0172 A = 17 mA. This is within the diode's power rating, since P = IV = (17 mA)(3.3 V) = 57 mW. Note that the current is also well within the diode's calculated Izm rating above. These calculations do not factor in the zener resistance, which is important for precision measurements.

Step 1

Attach the positive side of the 9 V battery to one side of the resistor. If you are using a battery holder, this is the side with the red lead.

Step 2

Connect the other end of the resistor to the cathode side of the zener diode, so that it will be reverse-biased. This is the side indicated by a mark.

Step 3

Wire the remaining diode terminal to the negative side of the battery. If you are using a battery holder, this is the side with the black lead.

Step 4

Place the multimeter on a DC voltage setting. Measure the voltage across the diode by placing a multimeter lead on each terminal. It should read approximately 3.3 volts. Note that the voltage between the battery and ground remains at 9 V.