Can batteries sink reverse current? - Page 1
Forum section: electronics / projects

author: NDani
author_rank: Contributor
timestamp: February 21, 2025, 12:46:14 pm
content: Hi all,I am designing a circuit which operates from a 3 V coin cell battery but accepts voltages up to 5 V from an external signal source. The input of my device is a logic gate. Most logic gates, such as the SN74HC04 will tolerate input voltages higher than the supply voltage as long as input clamp current limitations are observed. For the SN74HC04 (and similar logic devices) the maximum input clamp current is 20 mA, which, if clamping occurs, flows through the upper internal protection diode. To limit this current, a resistor in series with the gate input is usually used, as far as I know.The situation is illustrated by the attached image. In the example, the series resistor is 150 Ohm, which limits the clamp current to about 9 mA, assuming a 0.6 V drop across the diode. (I don't want to use a higher resistor value, as that would limit the speed of the circuit.)Here is my question. I assume the reverse current would flow through the battery. Is that acceptable? Can batteries sink such reverse currents without being damaged? Or should something be changed to the circuit to prevent that? I am more interested in the general concept than ideas specific to the attached circuit, but my end goal is to make a similar circuit function reliably.Thanks!

author: tszaboo
author_rank: Super Contributor
timestamp: February 21, 2025, 12:55:55 pm
content: Not a good idea. A battery, elem, will not like that.It's an easy fix, you can clamp the input with a (lov voltage) zener. Or divide it down. Use a comparator if necessary. Or make the input with an optocoupler.

author: NDani
author_rank: Contributor
timestamp: February 21, 2025, 01:49:43 pm
content: Thank you! Clamping with a Zener is a good idea, it didn't occur to me before. A simple voltage divider would not do the job right now because the input signal amplitude may be higher or lower than 5 V. But the Zener should fix that.

author: radiolistener
author_rank: Super Contributor
timestamp: February 21, 2025, 02:18:12 pm
content: for very small current its not a problem you can use it as a standby charger for the battery. But if current is significant it can lead to battery overcharge and it can lead to damage and fire. But significant current also can damage your IC, because internal protection diodes don't expect high current. This is why external clamp diodes are installed - to protect IC, it's more cheap to replace external clamp diodes than expensive IC. This is even more important for ADC input, because IC cost can be up to 100-200 USD, so it's a good idea to install external clamp diodes to protect expensive IC, even if such diode is more expensive than cheap one due to better RF specification (capacitance, etc), it still 100 times cheaper than expensive IC.

author: amyk
author_rank: Super Contributor
timestamp: February 22, 2025, 12:43:01 am
content: Quote from: NDani on February 21, 2025, 12:46:14 pmThat's called "charging", and obviously depends on the type of battery

author: tszaboo
author_rank: Super Contributor
timestamp: February 22, 2025, 07:58:51 am
content: Quote from: amyk on February 22, 2025, 12:43:01 amYou don't change 3V primary coin cells.

author: Gyro
author_rank: Super Contributor
timestamp: February 22, 2025, 10:57:37 am
content: I was looking at the datasheet for a fairly large CR2450 type Lithium button cell for another thread yesterday. It states 1uA maximum for reverse charge, basically just enough to cope with the worst case leakage of a Schottky reverse current protection diode and no more.

author: Peabody
author_rank: Super Contributor
timestamp: February 22, 2025, 03:43:04 pm
content: There is an LIR2450, but it's a regular lithium rechargeable, nominally 3.7V, and you charge it with CC, then CV at 4.2V, then termination. Opinions differ on whether you could apply, say, 3.9V as a continuous trickle charge without harming the cell. Big Clive says you can even apply 4.2V continuously with no problem. He's an engineer, and I'm not. But I've never seen a lithium charging IC that didn't terminate completely when the battery has been charged. Anyway, I would think the LIR2450 could absorb significant protection current provided it wasn't already fully charged, and some current, briefly, even if it was.As current flows through the protection diodes, initially it just replaces what the battery would have been supplying to the load, and that should do no harm. But once the diode current exceeds the load current, current would start to flow back into the battery, and/or the voltage would start to rise.It seems to me that this question points out the fact that typical protection designs with diodes going from the pin to Vcc have their limits, and are not the same as a zener or transorb to ground.Maybe it's just hard and expensive to do zeners on a wafer.

author: tszaboo
author_rank: Super Contributor
timestamp: February 22, 2025, 04:08:00 pm
content: Quote from: Peabody on February 22, 2025, 03:43:04 pmIt's the kind of discussion whether or not we can jump out from a moving car.You step on the break, the car stops, you get out.

author: NDani
author_rank: Contributor
timestamp: February 22, 2025, 05:26:44 pm
content: Thanks for your answers! The battery is a primary cell and I have no intention to charge it. As many of you pointed out, it is now clear that reverse current into the battery should be avoided. In my application, a parallel Zener to ground seems to be the easiest fix.

author: wraper
author_rank: Unknown
timestamp: February 22, 2025, 05:50:37 pm
content: IC won't like it as well. You should not pass current through protection diodes, unless it's very small and you really want to save parts on input clamping/voltage divider. Parts that are OK with input voltage higher than VDD have a different kind of input protection.

author: Zero999
author_rank: Super Contributor
timestamp: February 22, 2025, 07:20:03 pm
content: Quote from: NDani on February 21, 2025, 12:46:14 pmA high value resistor will not slow it down, if you put a capacitor in parallel, to bypass it at AC. A 2M2 series resistor will limit the current to <1µA. The input capacitance of a 74HC series gate is 5pF, so go for 10pF in parallel and it will form a capacitive divider at AC, which will convert 5V to 31/3V. Note that this is layout dependant. If you've got long, wide traces above a ground plane to the gate, then it will add to the capacitance, so the bypass capacitor will need to be larger.

author: aeg
author_rank: Regular Contributor
timestamp: February 22, 2025, 11:55:37 pm
content: Just use the right part - 74AHC04, not 74HC04. ESD diode stays off.