i'm trying to build an AC to DC power supply via bridge rectifier. the filtered voltage that was appearing(before i connect the zener and the series resistor) was around 4.6V. i want to use a 1N4728A Zener diode as voltage regulator, the datasheet tells me it has a Zener voltage of 3.3V, but when i build it in a Circuit Simulator its only giving me 3.2V, maybe i did something wrong with the calculations



here's what i did to calculate for the Series Resistor(R3) for the Zener
1n4728A Zener datasheet: Zener voltage(Vz) = 3.3V and knee current(Ik) = 1mA
load resistor = 1k
filtered voltage = 4.6V

since the voltage that will appear in the load is parallel to the Zener which is 3.3V then the voltage that will drop in the Series Resistor is 4.6-3.3 = 1.3V.
before attaching the Zener diode and Series Resistor the current that was passing in the load resistor was 4.6V / 1000Ohm = 4.6mA.

once i attach the Zener diode the voltage that will appear in the load is the same(3.3V) then i can calculate the amount of current that will pass on the load once i attach the Zener diode. 3.3V / 1000Ohm = 3.3mA.
and the amount of current that the Zener Needs(Knee Current) was 1mA

then by KCL I(series resistor) = I(zener) + I(load)
I(series resistor) = 1mA + 3.3mA = 4.3mA

and by ohm's law i'll need R(series resistor) = 1.3v / 4.3mA = 302 Ohm.

the voltage appearing is only 3.2V instead of 3.3V

 

You can increase the voltage if you pass more current through the zener diode. The downside of this solution is reduced efficiency. Real zener diodes have a spread of zener diode voltages (~ 5-10%). Look at the data, at what current the voltage is normalized, and set the current of the zener diode.

 

3.2V/3.3V is about a 3% error, pretty good for a 5% zener.