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schematic

simulate this circuit – Schematic created using CircuitLab

I am not able to get proper 3.3volt output at the output capacitor. Since 3.3 volt zener diode have been used, there is a drop in the voltage along zener diode when 3.3v is flowing.

What should I do?

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  • \$\begingroup\$ I suspect it would help if you included a BJT in your circuit to enhance the current compliance. This might require changing the zener, though. Or else using a 2-bjt circuit to lower, then raise back up, the Vbe difference to compensate things a bit. I think you may need to write more about why you want something like this. \$\endgroup\$
    – jonk
    Oct 25, 2019 at 7:02
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    \$\begingroup\$ The resistor divider produces 1/11*36 = 3.272 Volts. Your 3.3Volt zener just does not get enough voltage to conduct. \$\endgroup\$
    – Oldfart
    Oct 25, 2019 at 7:04
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    \$\begingroup\$ you can simply remove R2 to address the problem pointed by @Oldfart. You also have to make sure that your load does not drain more than 3.3mA, otherwise you have to lower R1's value \$\endgroup\$
    – joribama
    Oct 25, 2019 at 7:18
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    \$\begingroup\$ Welcome to EE.SE! Please note that voltage isn't flowing. Current is flowing. \$\endgroup\$
    – winny
    Oct 25, 2019 at 7:38
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    \$\begingroup\$ If you are using this circuit as reference voltage for MCU, then it's a really bad idea. \$\endgroup\$
    – EEMuks
    Oct 25, 2019 at 8:09

1 Answer 1

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If you remove R2 in your circuit, you should be able to measure the requested 3.3V reference voltage at the output of the capacitor.

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