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I built this circuit for Vdd of a 8-bit MCU (3.3V nominal) and the input rating is 12-24V.

But when 12V input is used, MCU Vdd pin only receives around 1V, I guess Q1 isn't turning on.

The circuit is modified from Maxim Integrates AN-760. 760's intended for 5V IC Application, so I modified D1 to a lower voltage Zener.

Maybe I am doing this the wrong way, some insights would be helpful.

schematic

simulate this circuit – Schematic created using CircuitLab

Travis Su
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    The circuit is working as intended. It's a protection circuit, not a voltage regulator. It's supposed to disconnect the load (turn off Q1) if the input voltage is high enough to cause the Zener diode to conduct significant current. Otherwise, Q1 is turned on with a very low voltage drop. – Dave Tweed Sep 14 '22 at 11:46
  • @DaveTweed So there's a possibility that the D1's Voltage is too low at 3.9V since the MCU could be able up to 4V of absolute maximum rating, I should select a D1 that's range in 4-4.5V? – Travis Su Sep 14 '22 at 12:26
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    You said that you're supplying 12V, and you want 3.3V to the MCU. That means that you need a voltage regulator, not this circuit. I'll say it again: This circuit is not a voltage regulator. – Dave Tweed Sep 14 '22 at 12:54
  • @DaveTweed Voltage Regulator don't usually offer protection isn't it? – Travis Su Sep 14 '22 at 14:24
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    Well, yes, voltage regulators do "protect" a load against input voltage variations. This circuit is not a voltage regulator. – Dave Tweed Sep 14 '22 at 15:11

1 Answers1

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Figure 1. The circuit from the application note.

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Figure 2. The OP's circuit.

Here are the most obvious errors in your circuit.

  1. Q1 is back-to-front. Supply should be connected to the emitter.
  2. Q2 is upside-down. The emitter should be connected to the incoming supply.
  3. You've missed this connection so D1, D2, Q2, R2 and R3 are doing nothing for Q1.

Update since the schematic has been fixed:

From the How it works section,

The transistor Q2 acts as the control element for Q1 and will turn on (turning Q1 off) when the voltage at the power-supply input is equal to the sum of the zener voltage, due to diode D1 and Q2's own VBE voltage at a collector current of about 650uA. Q2 and D1 together produce a typical trip voltage of 5.85V at 25°C. Approximately 0.53V of this is due to the VBE voltage of Q2. The remaining 5.32V is produced across D1. Note that the zener diode D1, although a nominal 5.6V device, is being operated at a very low reverse current, about 200uA as defined by the VBE of Q2 together with the 2.7k resistor.

This is telling you that the circuit is designed to switch off the output if the input rises above 5.85 V to protect a circuit with a nominal 5 V power requirement. You have modified the circuit so that it will switch off when the input voltage rises above 3.9 V + 0.53 V.

As has been pointed out more than once in the comments, This circuit is not a voltage regulator. It is an over-voltage protection circuit.

But when 12 V input is used, MCU Vdd pin only receives around 1 V.

Since 12 V > 3.9 V + 0.53 V the protection turns off the output. The circuit is working as intended by the original designer.

Transistor
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