We are trying do design a circuit which works by harvesting energy from a split core current transformer. Most of the electronic works correctly, we use a 1.5F 5.5V SuperCap to provide enough power for an STM32F411 micro-controller and an LoRa transceiver. Most of the time the micro-controller is in deep-sleep, allowing the cupercap to charge enough for sending phases.

What we want now is to turn on power to the micro-controller only when voltage at the supercap is over the desired voltage (say 3.6V). The hard part is the we don't have stable "high" voltage at the beginning, when the supercap is empty, to provide power to active circuits.

My idea is to use a Zener diode to control a Mosfet like this. Do you think it would work? If I use a high resistance value before the Zener, will it be enough to avoid drawing too much current.

  • 3
    \$\begingroup\$ STM32F411 has a brown-out detector with a programmable threshold from ~2.2 to ~2.9 V? Can you use this and possibly also have the uC monitor the voltage to be sure it's enough for other components before enabling them? \$\endgroup\$
    – The Photon
    Jun 12, 2019 at 16:34
  • \$\begingroup\$ There is also a "programmable voltage detector" with threshold programmable up to ~3.0 V. This can either be monitored programmatically or set to generate an interrupt. \$\endgroup\$
    – The Photon
    Jun 12, 2019 at 16:39
  • 1
    \$\begingroup\$ Look for LDO's with UVLO ( under-voltage lockout) Any design like this may cycle on-off ) in marginal power situations. \$\endgroup\$ Jun 12, 2019 at 16:52
  • 1
    \$\begingroup\$ The LMV393 comparator has a supply voltage of 2.7V. You could power the chip off of the cap, and then have a voltage divider to the positive input set for 3.6V. Just need a low voltage reference (I'm thinking something like the LM431) for the negative input to compare against. \$\endgroup\$
    – Stiddily
    Jun 12, 2019 at 17:23
  • \$\begingroup\$ manbe something likr MAX803 - how mich current can you spend on the monitoring circuit? \$\endgroup\$ Jun 12, 2019 at 23:56

1 Answer 1


Hold OFF MCU power till minimum threshold reached on PS capacitor.

It's a challenge to make predictable power switches for an application such as this where the voltage rises very slowly.
Brownout detectors within MCUs are designed to hold the device reset when the power drops to a critical level. However what happens at very low voltages is unpredictable and for most devices (including MCUs) not specified.

As an example, consider the TI TLV803S Voltage Supervisor which is very typical of the external (to an MCU) reset devices available. For detecting a power fail point this sort of supervisor is ideal, BUT operation at voltages starting from zero and rising is complicated. This particular device is nice because they actually characterized the operation at very low voltage:

enter image description here

Notice above that the device is not working as one might expect until the VDD reaches about 1.1V. It's clear from the graph and statements in the datasheet that both the output voltage and the current that can be sunk on the reset line would be a major cause for concern in any design.

I've used a TLV803S in the circuit below that will be predictable starting from 0V and up. The circuit switches on the load at about 2.95V and off at about 2.93V. You could select another variant to get a different voltage.


simulate this circuit – Schematic created using CircuitLab

In the schematic above from 0 - 0.75V current is leakage values, and the load is OFF.
From about 0.75 - 1.1V the current through Q1 (the mirror of Q2) keeps M1 in cutoff so the Load is OFF.
Above 1.1V the TLV803S is operating in normal mode and M1 is held OFF, M2 is clamped by the Q1 current source so the load is OFF.

When the VDD voltage reaches 2.95V, the TLV803S raises the reset line and M1 turns ON, M2 turns ON and the load is powered. The switch consumes about 1.5mA.

If VDD falls below 2.93V, then the load is turned OFF. At this voltage level the switch consumes 670uA.

Note: I'm sure you understand, but it's worth pointing out that using a CT in an application like this can be dangerous. The CT can produce very high output voltages if not loaded correctly. Your circuit should look like the following, and if there is a risk that the CT can be disconnected from your PCB while still having a mains cable through it then you should add a high value burden resistor or two Zener diodes directly across the ouptut to limit the voltage.


simulate this circuit


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