Smart diode for two primary cells in parallel? - Page 1
Forum section: electronics / projects

author: ricko_uk
author_rank: Super Contributor
timestamp: February 27, 2022, 07:30:40 pm
content: Hi,I have an application where two primary 3.6V C-size lithium thionyl chloride batteries are connected in parallel on the PCB itself (i.e. through the PCB track and NOT inside a battery pack). This is the battery datasheet:https://docs.rs-online.com/e5b6/0900766b800cc0ca.pdfAssuming neither one of the batteries can be installed in inverted polarity (i.e.they are mechanically always inserted the right way around) do you need a smart diode on each of them to prevent them form discharging them into each other and/or for safety and/or for any other reason?Thank you

author: Gyro
author_rank: Super Contributor
timestamp: February 27, 2022, 07:38:10 pm
content: Are they the only power source? If they are a backup supply for something mains powered for instance, you would need something to prevent any accidental charging current.EDIT: If it is just the two cells, then, as long as they are always replaced as a pair, I don't think you would have a problem.

author: ricko_uk
author_rank: Super Contributor
timestamp: February 27, 2022, 07:46:16 pm
content: Thank you Gyro,there is another power source (also 3.6V) from another rail but they are separated by a TPS2116 Powerpath Mux with priority switchover so that is the other source falls below a certain value then these two batteries are switched automatically as the source.You can see the block diagram of the TPS2116 on page 10 of the datasheet here:https://www.ti.com/lit/ds/symlink/tps2116.pdf?ts=1626672560005Thank you

author: Gyro
author_rank: Super Contributor
timestamp: February 27, 2022, 07:53:23 pm
content: The datasheet includes reverse current blocking with max 150nA leakage current @105'C, so I think you would be fine (apart from device failure, but a diode can fail too).

author: ricko_uk
author_rank: Super Contributor
timestamp: February 27, 2022, 08:36:16 pm
content: Yes that is protected that way, but the client is adamant to have two smart diodes each in series with each of the two batteries I am talking about (not the other alternative source).He says to avoid any one of the two batteries discharging into the other one (again NOT into the second/alternate source of power).Is there any reason for those two series smart diodes?Thank you

author: David Hess
author_rank: Super Contributor
timestamp: February 28, 2022, 01:34:43 am
content: Low drain applications like memory backup would normally also include a resistor of like 10 kilohms in series with the battery which would further limit the current if they were connected in parallel.

author: AndersJ
author_rank: Frequent Contributor
timestamp: February 28, 2022, 01:41:18 am
content: What is the purpose of the 10K resistor?

author: Gyro
author_rank: Super Contributor
timestamp: February 28, 2022, 08:46:56 am
content: It's normally put there to provide some protection against accidental shorts, or accidental reverse charging of the cell by the main power source. Probably not relevant in this case because the OP is using two C sized Lithium primary cells, suggesting much higher current drain.Quote from: ricko_uk on February 27, 2022, 08:36:16 pmAs discussed previously, I agree with you that, as long as it is mechanically impossible to reverse insert a cell, and they are both replaced at the same time, the diodes should be unnecessary. However, if it's possible for one cell is end of life and the other new one, then he may have a point.If it's a matter of keeping the customer happy, there's no harm in putting in a dual P-Channel MOSFET with low threshold voltage as a pair of ideal diodes. It's an SO8 or smaller single package solution. It might be worth it for a (relatively) happy customer and a peaceful life!

author: Kleinstein
author_rank: Super Contributor
timestamp: February 28, 2022, 09:56:49 am
content: With Modern electronics one may not need the full voltage. So there may be the option to use just 2 normal diodes. With the usually low current needed the forward voltage of a 1N4001 is not that large. The slight loss in capacitance due to an earlier shunt down may be offset by reduced standby current. Simple diodes tend to be more reliable than a more complicated active ciruit.The PMOSfet as a diode works for reverse polarity protection, but not really to avoid current transfer between the 2 batteries.

author: Gyro
author_rank: Super Contributor
timestamp: February 28, 2022, 12:57:16 pm
content: QuoteAh, good point. I was thinking of the gross reverse polarity situation, I was forgetting that MOSFET ideal diodes don't work at small reverse bias.This all probably depends on the actual current draw from the batteries (it would be helpful to know), a pair of C sized Lithium thionyl chloride batteries are good for a fair amount of current. The voltage drop of 1N4001s might become significant, highish voltage rated Schottkys ought to have minimal reverse leakage though.

author: richard.cs
author_rank: Super Contributor
timestamp: February 28, 2022, 01:05:14 pm
content: Are both cells in an identical thermal environment? If not temperature cycling can drive current flow between the cells which eventually leads to capacity loss. In practice I have only seen this happen in a device which paralleled a secondary cell, with an entirely different voltage(temperature) coefficient), with a thionyl chloride cell.Certainly batteries of thionyl chloride cells in parallel exist without diodes, I have some factory-made shrink-wrapped Panasonic ones. Some manufacturers give information on paralleling, from memory they call for the cells being from the same production batch.

author: ricko_uk
author_rank: Super Contributor
timestamp: February 28, 2022, 07:34:23 pm
content: Thank youDespite the very low current (200mA peak and 20mA typical) the ideal diode the client wants to use is SM74611KTTR (and 8A part!!). Datasheet here:https://www.ti.com/lit/ds/symlink/sm74611.pdf?ts=1646076468236&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FSM74611%253Fqgpn%253Dsm74611- Would that part have any issues at low currents?Quote from: Gyro on February 28, 2022, 12:57:16 pm- What do you mean "MOSFET ideal diodes don't work at small reverse bias"? Do you mean if the two batteries have only few hundred mV difference in voltage? And why is that? Would it work if one battery was at say 1V and the other at 3.6V?- Are you suggesting to implement a discrete solution or use an ideal diode in a single package?Thank you

author: Gyro
author_rank: Super Contributor
timestamp: February 28, 2022, 08:11:05 pm
content: The MOSFET ideal diode (there are probably much better terms) works by turning the MOSFET on, in parallel with its normal parasitic diode (which has the normal >0.7V voltage drop) to avoid this drop. The MOSFET is used with Drain and Source reversed compared to normal. This has no adverse effect (apart from conduction being in the same direction as the parasitic diode). To turn on a P-channel MOSFET, the gate must be negative with respect to the Drain/Source. This works fine for reverse polarity protection, where, if the polarity is wrong, the gate is taken positive instead, turning off the MOSFET (the parasitic diode is also reverse biased).The problem is that, for small differential positive voltage differences, the gates are still negative, so both MOSFETS remain turned on. It is possible to do, but would require sense resistors and opamps to drive the gates. This would be tricky when the currents are so low.So the MOSFET ideal diode solution will work fine to protect against one (or both) cells being reverse connected, but not for avoiding small current current transfers between batteries of slightly different voltages. I don't know if this is sufficient for the customer.EDIT: I'm not sure from the TI datasheet how much voltage their "Smart Bypass Diode" needs to turn on / off. It seems to require sufficient to activate its internal charge-pump power supply. This is almost certain to be more than the few mV difference between two similar cells. I don't think they will do the job.EDIT1: They appear to work by turning the MOSFET off for brief intervals when forward biassed in order to charge up their internal supply (using the parasitic diode 0.7V drop). So they would both turn off when the cells are not in use. I can't see any specification for minimum forward current, although there must be some kind of lower limit to still allow their charge-pumps to operate. [EDIT: From the test circuit in Fig 8, the lower limit appears to be1A.]

author: Kleinstein
author_rank: Super Contributor
timestamp: February 28, 2022, 10:25:04 pm
content: The active ideal diode circuit should work, but it usually has additional supply current even if idle.What does not work is using just 2 MOSFETs and the voltage from the output side to directly drive the gate. So the simple circuit sometimes used for reverse polatiy protection with a MOSFET instead of a diode.200 mA peak and 20 mA "normal" is a rather high current for a LI primary cell. This would likely include some SMPS and thus voltage drop would matter.At this high current the active ideal diode circuit is likely the way to go.The TI part is a bit odd one, specially made for PV without an extra supply. For a high power part the leakage current (up to 1 µA) may be significant and the start up for the charge pump may result in some initial voltage drop. So the worst case drop can be quite large, though one for a short time.