I try to build a low voltage drop regulator to bring the voltage down from 12 V to 5 V (ignore everything right of the capacitor C3). I have a 12 V voltage supply and I want to feed my OpAmp with a voltage below 5 V. To do this I put 3 bzx84c3v3 zener diodes in series (U3, U4, U5), everyone of them has a breakdown voltage of 3.3 V. So theoretically, the voltage should drop from 12 V to 2.1 V, but it drops to only 5.56 V. Putting more zener diodes doesn't really help. Does any one have an idea why the voltage doesn't drop accordingly to the breakdown voltage?

why no voltage drop

  • \$\begingroup\$ Why not use an integrated LVDO regulator? Cheaper, simpler, more accurate, etc. than attempting to do this yourself. \$\endgroup\$
    – Reinderien
    Dec 8, 2020 at 17:07
  • \$\begingroup\$ @Reinderien It's homework, I needed to build a LDO myself. \$\endgroup\$
    – Maxim
    Dec 8, 2020 at 17:20
  • \$\begingroup\$ Apart from the specific issue this attempted subtraction of a fixed amount from the input voltage seems an odd design. Typically a regulator compares the output to a reference while trying to be immune to the source voltage variation. \$\endgroup\$ Dec 8, 2020 at 19:36

1 Answer 1


If you observe the datasheet, you would find the appropriate working conditions where zener diode provides 3.3V drop across it on reverse breakdown. Just observe the below image from the datasheet. Bottom curve shows for 3.3V.

enter image description here

Closely looking at the given values, maximum voltage drop it would observe is 2.9V at 1mA current flow.

enter image description here

And the op-amp that you used, generally have input currents in the range of µA. That would mean the voltage across zener diode goes even low. From the result you posted. 5.56/3 = 1.85V drop is there.

For proper 3.3V. appropriate amount of current should flow through the diode.

Try looking at datasheet for in-depth reference.

Solution: Try putting pull-down resistor at opamp input(Proper value has to be selected).


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