4
\$\begingroup\$

I want to create a passive circuit which will cut off a the power supply from a LiPo battery when the voltage of the battery drops to/below 3.2V.

I followed the video: https://youtu.be/NecbHbGAkkc?si=HhB0D730yQR2n9oM.

Timestamp: 9:00 mark.

The reference circuit in the video

So using the equation shown, for a cutoff of \$3.2V\$, \$V_{out} = V_{ref}(\frac{2.8k}{10k} + 1)\$, then \$V_{ref} = 2.5V\$.

So just for clarity:

\$V_{out} = 2.5 \times (\frac{2800}{10000} + 1) = 3.2V\$

The equation is not clear in the video, but as he says using \$R1\$ as \$2k\$ and \$R2\$ as \$1k\$ causes a \$V_{cutoff}\$ of 7.5V.

So here is my implemented \$V_{cutoff}\$ circuit:

enter image description here

There are some other circuits in the chain:

The circuit roughly follows Battery Connector -> Reverse Polarity Protection -> \$V_{cutoff}\$ -> output LED

+Batt_Cutoff implies its the positive input to the cutoff circuitry.

Battery Connector Battery Connector

Reverse Polarity Protection RPP

PCB Layout

Below is my implemented circuit. PCB layout

When hooked up to a LiPo, the voltage continued to discharge below 3.2V. When I have connected the circuit to a bench power supply, I found the indicator LED to turn off at around 2.6V.

What could be the issues for this/what to test for? Have I implemented the resistor value equation correctly? Could the properties of the MOSFETS be changing the properties of my circuit?

New contributor
Harveyn4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
\$\endgroup\$
6
  • \$\begingroup\$ Hi. Can you show exactly how the circuits are connected in the three schematic you show? \$\endgroup\$ Commented Oct 29 at 14:29
  • \$\begingroup\$ Using a comparator for detecting the voltage being too low on a battery. Ditto this similar idea but at a lower trigger voltage. Both have hysteresis. \$\endgroup\$ Commented Oct 29 at 14:34
  • 1
    \$\begingroup\$ in addition to the important points in the answers - maybe consider a TLV431 instead of the TL431 ... esp a lower minimum-current variation of it ; also the 1.25V vs 2.5V should allow larger resistor on top while keeping same Z into the input, and thus less current used in the shunt/sense resistors \$\endgroup\$ Commented 2 days ago
  • 4
    \$\begingroup\$ @Harveyn4444 Then I think it is pretty clear that Q2 should have the drain and source swapped. It will always conduct through the body diode \$\endgroup\$ Commented 2 days ago
  • 3
    \$\begingroup\$ YOUR MOSFET DRAIN AND SOURCE ARE SWAPPED - it actsas a bad diode. (As noted by Tyassin and Bravale). Mentioned again as this is THE critical error. \$\endgroup\$ Commented 2 days ago

5 Answers 5

Reset to default
13
\$\begingroup\$

I think the main problem is the P-MOSFET Q2, drain-source are swapped, the body diode is forward biased. In the picture from the video is correctly connected.

Suggestion: you can improve power consumption when battery is UV condition, using a low quiescent current comparator like LT6700. Another improvement is you can use higher value for resistors than TL431. It is a window comparator with internal 400mV reference, and you can simulate in LTSpice free. With LT6700-1 you can add hysteresis using the non inverting comparator and feeding back to inverting comparator input. This is an example with low cut at 3.2 V and switching on at 3.44 V. The hysteresis will depends on R2 value.

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ I think you mean "forward biased"; "directly polarized" sounds like a mistranslation. \$\endgroup\$ Commented 2 days ago
  • \$\begingroup\$ @Hearth : edited, I knew it was not the right expression, but did not come to mind. \$\endgroup\$ Commented 2 days ago
6
\$\begingroup\$

Your biggest mistake is the R4 value. You need to reduce the R4 resistor value to approximately (3.2V - 2V)/1mA = 1.2kΩ. Thus, you can use a 1kΩ resistor. Because the minimum TL431 cathode current for regulation is 0.4mA typical (value), with a maximum of 0.7 mA.

Also, add some hysteresis by adding a 680k resistor connected between the MOS drain and TL431 Vref pin (2)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • \$\begingroup\$ Thank you very much, thank you for explaining the R4 value calculation. I have one question about 3.2 - 2V, why 2V? Is this related to the Vref which is 2.5V? Do you have any good links for the hysteresis R value as I'm not too familiar with that topic. \$\endgroup\$ Commented Oct 29 at 15:23
  • 2
    \$\begingroup\$ This 2V is the TL431 "saturation voltage"- the minimum voltage drop that we can see between cathode and anode of a TL431. Did you swap the drain with the source? \$\endgroup\$ Commented 2 days ago
4
\$\begingroup\$

The tl431 isn’t true comparator. Although it can be used as crude one it brings some complications like some current always flow through it, even the Ref pin is below 2v5.

Also you didn’t involve a hysteresis so the mosfet will chatter when battery is at cutt-off voltage.

\$\endgroup\$
1
\$\begingroup\$

Here's the LTspice sim of G36's circuit with a low Vgs(th) P-MOSFET, and Rhys selected to give about 40mV of hysteresis:

enter image description here

\$\endgroup\$
1
\$\begingroup\$

I want to create a passive circuit which will cut off a the power supply from a lipo battery when the voltage of the battery drops to/below 3.2V.

Ahh, another context-less YouTube video advising the wrong thing :/ The video uses, or at least means to use, the LiPo protection as a prompt to explore TL431, a pedagogical aid if you will. It is there to teach you how to use TL431 as a comparator. You cannot actually use such a circuit to protect any batteries though!

Specifically, you do not want to connect a shunt regulator (TL431) and a low impedance (~kΩ) reference divider in parallel with a LiPo battery! The current the reference divider and the TL431 draws will over-discharge the battery, and will act exactly against your goal of protecting the battery.

Whatever protection circuit you will end up using must pass at most a current equivalent to the self-discharge current of the battery. That's exactly what dedicated LiPo protection chips do. Use a LiPo protection chip!

If you don't want to use a protection chip, you can use a modern "nanocurrent" CMOS comparator with built-in voltage reference. The resistors used for the feedback network will need to have suitably high values - in the megaohms range. The whole thing should draw <1μA of idle current, at least in that ballpark.

\$\endgroup\$
2
  • \$\begingroup\$ If you want a low battery discharge the R divider and comparator can be placed after mosfet, so they are disconnected also. Once the charger is plugged again after cutt-off ocured, it can set the mosfet on with some circuitry/logic, so the R divider and comparator start to measure again. \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ A momentary switch connected from the MOSFET gate to ground will start the circuit. \$\endgroup\$ Commented 20 hours ago

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.