Skip to content

Power & Energy

How a low-voltage cutoff protects a lithium cell

A low-voltage cutoff switches a load off before it drains a lithium cell too far. A comparator watches a fixed fraction of the cell's voltage against a steady reference. While the cell is healthy its output holds a MOSFET on, and the moment the cell reaches the trip point, about 3.0 V, the output drops, the load goes off, and the cell keeps the rest of its charge.

Why stop at 3 V?

Below about 2.5 to 3 V a lithium cell can be damaged for good: its copper current collector starts to dissolve, and a cell drained that deep may never charge safely again. So every lithium pack has an under-voltage cutoff, usually inside its protection chip.

The circuit

  • Reference: a 2.0 V Zener fed from the cell through 4.7 kΩ, on the comparator's − input. It stays at 2.0 V whether the cell is at 4.2 V or 3.0 V.
  • Divider: R1 = 10 kΩ over R2 = 20 kΩ gives the + input two thirds of the cell.
  • Switch: the comparator's output drives the gate of an N-channel MOSFET on the load's low side.

The trip point

trip = Vref × (R1 + R2) ÷ R2 = 2.0 × 30 ÷ 20 = 3.0 V

Two thirds of 3.0 V is 2.0 V, so that's where + falls below − and the output drops. For a different trip, solve for R2: with R1 = 10 kΩ, 4.0 V needs 10 kΩ and about 3.3 V needs 15 kΩ.

The Zener needs a little current to hold its voltage: (3.0 − 2.0) V ÷ 4.7 kΩ ≈ 0.2 mA, enough for a steady reference and far too little to matter to the cell. Fed so little, a real Zener sits a touch under its rated voltage, so this one trips at about 2.95 V.

Live experiment

Pick the trip point

A cell starting at 3.6 V runs a lamp through a MOSFET, and a comparator watches R1 (10 kΩ) over R2 against a 2.0 V Zener. Pick R2 and run 7 s: none never trips, 10 kΩ trips at 4.0 V before it starts, 15 kΩ at 3.3 V, and 20 kΩ at about 2.95 V.

Try it yourself: free lesson

“Never run it flat” is lesson 4 of 6 in Solar Sound, Power & Energy World 3. Free with an account. Lessons unlock in order, so you’ll do the 17 before it first.

Add hysteresis

A cell's voltage recovers a little when its load goes off. It creeps back over the trip point, the load comes on, the cell sags, and the load flickers. Real cutoffs add hysteresis: a little positive feedback from the comparator's output to its + input, so the turn-on point sits a few hundred millivolts above the turn-off point. It's a Schmitt trigger.

Ready-made chips

The protection board on a phone, laptop or power-bank cell does this along with over-charge and short-circuit cutoffs, usually with one small chip and a pair of MOSFETs. For a battery-powered project, a voltage supervisor chip or a load switch with an enable pin can replace the discrete parts, but the idea is the same comparison against a reference.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Never run it flat” is lesson 4 of 6 in Solar Sound, Power & Energy World 3. Free with an account. Lessons unlock in order, so you’ll do the 17 before it first.

Or start from the very first lesson, no account needed

Practice questions

Three quick questions from the lessons. Pick an answer to check it.

  1. Question 1 of 3Why switch the lamp off when the cell reaches 3.0 V?
    Show the answer

    Answer: Below about 2.5–3 V a lithium cell can be damaged for good. Run too low, a lithium cell's copper starts to dissolve and it may never charge safely again. Every lithium pack has a cut-off.

    • Why not “The lamp would burn out”? A falling cell makes a lamp dimmer, not hotter. The cutoff is there to protect the cell.
    • Why not “To make the lamp brighter”? Switching the lamp off makes it dark. The point is to keep the cell's last charge in it.
    • Why not “The comparator needs exactly 3 V”? The comparator works at any voltage in its range. 3.0 V is chosen for the cell, and the divider sets it.
  2. Question 2 of 3While U1's + is above its −, its output is…
    Show the answer

    Answer: High: M1 on, lamp on. A comparator's output goes high when + is above −, and low when it's below. Nothing in between for long.

    • Why not “Low: M1 off, lamp off”? Low is for + below −. That's when the cell has reached the trip point.
    • Why not “Negative”? Powered from the cell and ground, the output can't go below 0 V.
    • Why not “Half way”? A comparator has huge gain, so it snaps to one rail or the other, almost never in between.
  3. Question 3 of 3Swap U1's inputs: the divider on −, the reference on +. The lamp…
    Show the answer

    Answer: Stays off while the cell is healthy. The comparator now says high only when the cell is low: exactly backwards.

    • Why not “Never turns off”? It's off from the start, while the cell is healthy, and only comes on once the cell falls past the trip point.
    • Why not “Works the same”? Swapping + and − inverts the answer: high now means the cell is below the trip point.
    • Why not “Flickers”? A comparator snaps cleanly either way. Swapping the inputs makes it backwards, not unsteady.
Keep practising: free lesson

“Never run it flat” is lesson 4 of 6 in Solar Sound, Power & Energy World 3. Free with an account. Lessons unlock in order, so you’ll do the 17 before it first.

Questions people ask

What is a low-voltage cutoff?

A circuit that disconnects the load when a battery's voltage falls to a set level, so the battery isn't over-discharged. For lithium-ion cells the level is usually between 2.5 and 3.0 V per cell.

What voltage should a lithium battery cut off at?

Most lithium-ion cells are rated down to about 2.5 to 3.0 V per cell, and the datasheet gives the exact figure. Cutting off at 3.0 to 3.3 V gives up a little capacity but is kinder to the cell, since little energy is left below about 3.3 V anyway.

How do I work out the divider for a cutoff circuit?

Trip voltage = reference × (R1 + R2) ÷ R2, with R1 on top. Rearranged, R2 = R1 × reference ÷ (trip − reference). For a 2.0 V reference, R1 = 10 kΩ and a 3.0 V trip, R2 = 10 kΩ × 2 ÷ 1 = 20 kΩ.

Every topic