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I have a low-power fuel cell which I am connecting to a capacitor to charge a capacitor to power a small motor. The cell itself does not have enough power to power the motor. After the capacitor is full, I can discharge it to power the motor. However, I am doing this manually. enter image description here

How can I do this automatically without using a microcontroller?

Is there a switch that tells when the capacitor is fully charged and opens the path for the current to the motor and goes back to blocking current when voltage is not sufficient? (Note:The motor requires 2V which my cell can provide)

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  • \$\begingroup\$ Whats the period it takes for the capacitor to fill ? \$\endgroup\$
    – tobalt
    Commented Mar 31, 2023 at 18:56
  • \$\begingroup\$ I add 1K resistor and it takes 30 minutes \$\endgroup\$ Commented Mar 31, 2023 at 19:00
  • \$\begingroup\$ A comparator and a transistor? \$\endgroup\$
    – winny
    Commented Mar 31, 2023 at 19:11
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    \$\begingroup\$ FYI: this kind of thing is called "energy harvesting" and maybe if you search that keyword you will find something. Usually it is a tiny solar panel, not a fuel cell \$\endgroup\$ Commented Mar 31, 2023 at 19:14
  • \$\begingroup\$ by the way I am curious where someone can buy a low-power fuel cell? \$\endgroup\$ Commented Mar 31, 2023 at 19:14

2 Answers 2

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Once triggered, this circuit keeps M2 conducting down to about 400 mV at the input capacitor C1 and then turns off for the next charging cycle.
With R5 and R7 the trigger voltage can be adjusted to the treshold voltage of the MOSFETs. In an ideal Vgs FET selection they are not needed.
The trigger voltage of this circuit is not very precise and temperature dependent, but the benefit is a very low current consumption.
C2 and C3 work as a kind of charge pump and keep the FET Vgs high enough to let them conduct while the supply falls far below the threshold.

One of the FETs begins to conduct at rising V_CAP and the capacitive coupling to the other FET lifts its Vgs over the theshold. This positive feedback turns both FETs full on very fast.

The way back to the off state begins when the remaining charge of C2 or C3 can no longer keep the connected FET conducting.

The component values used in this schematic may be far away from the needed optimum in the real application.

schematic

simulate this circuit – Schematic created using CircuitLab

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Not having full your full requirements, it's hard to do a precise design.

But basically, you need 2 things :

  1. an electronic "switch" : that would be some kind of transistor (probably either bipolar or mosfet)

  2. a circuit telling you when you get enough voltage in the capacitor and when you have to stop emptying it because the voltage is too low (no point to keep using up energy once the voltage is too low to keep the motor spining). So I would say some kind of comparator with hysteresis.

So some idea would be the following :

schematic

simulate this circuit – Schematic created using CircuitLab

COMP1 is a comparator (if you don't know them, its a bit like an Op-Amp without negative feedback, but optimized for it). Depending on the model, you might need to add a pull-up resistor to the output.

Note that all resistors will need to have huge resistance, or you will consume too much power. I guess that you will use no resistor bellow 100k.

D1 and R1 provide a stable supply voltage to the comparator (if D1 is a 1.8V zener, then the supply voltage is 1.8V). Make sure to use a big resistor for R1 in order to avoid waisting your precious energy. Note that this is an OK solution only because the comparator draws very very little current.

The comparator COMP1 and R2 to R5 form an non inverting comparator with histeresis. This means that when the voltage of the capacitor rises above Vth+, then the output goes high. If it falls bellow a different threshold Vth-, the output goes low. Vth+>Vth-. For example, if you choose your resistors for Vth+=3V and Vth-=2V, then the motor will turn on when the capacitor reaches 3V, and go out when it emptied at 2V (without hysteresis, you have a single threshold, so the motor would oscillate quickly between on and off).

The N-MOS M1 will act as a closed switch if the voltage at its input (gate) is greater than some threshold. Note that 1.8V is quite low for a MOSFET, so you have to carefully pick one that turns on at such low voltage. Alternatively, you can use a bipolar transistor, but it will waist some energy.

A few extra notes :

  • you probably want a free-wheeling diode to protect the mosfet when you switch off
  • I'm not really sure how this will behave if you get less voltage than the minimal working voltage of the comparator. So maybe still get a switch just after the capacitor, excepted that now, you only switch it on once at the beginning, and then everything keeps going by itself

Let me know if you need help to dig further (in that case, I will need full specification of your needs : voltage at which to switch on/off the motor, current produced by the cell, current consumed by the motor, and some idea about your level in electronics to adapt my explanations)

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  • \$\begingroup\$ Comp has its inputs shorted and its max supply of 1.8 V is too low to enable most MOSFETs. \$\endgroup\$
    – tobalt
    Commented Mar 31, 2023 at 19:47
  • \$\begingroup\$ good point for the short in the schematics, I corrected it. For the mosfet, I mentionned that to get it working at 1.8V, one needs to carfully chose the mosfet. A random N-MOS will probably not do the job. But there are still 600 references with Vth<1.5V in stock at digikey (about 10%). \$\endgroup\$
    – Sandro
    Commented Mar 31, 2023 at 19:55
  • \$\begingroup\$ For max supply, it all depends on the voltages OP wants. If a cut of voltage of for example 2.5V is fine, then we can increase the supply voltage to maybe 2.3V. One might also replace the mosfet by a bipolar transistor + resistor, but I don't like wasting energy on low power applications. \$\endgroup\$
    – Sandro
    Commented Mar 31, 2023 at 19:57

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