I am currently building a data logging device for monitoring battery packs which contain 8 Lithium cells. The device will periodically log the voltages of all 8 cells, as well as log temperatures collected from 3 thermistors. My device has an STM32L series microcontroller.

As of now I am trying to figure out what the input circuit for the ADC should be. Due to the high power nature of the system, and it containing two large industrial motors which are subject to varying loading, I am assuming the battery pack would be subjected to high voltage and current spikes. I will be taking taps from each cell and connecting it to my device.

In terms of protection, I have done some research and found I should probably be looking into TVS and clamping diodes as protection means. The input range that needs to be measured is 0-32V. I have guesstimated the following circuit:

enter image description here

The TVS diode is SMAJ33A (http://www.littelfuse.com/~/media/electronics/datasheets/tvs_diodes/littelfuse_tvs_diode_smaj_datasheet.pdf.pdf) with a standoff voltage of 33V, and maximum clamping voltage of 53.3V @ 7.5A (Ipp). I am not sure how to select the TVS diode (or if I even need one!), so I would really appreciate any advice on this. For the clamping diode part, I am following ADC input protection?. I have selected Bat54S dual Schottky barrier diodes. This is to hopefully provide over voltage protection.

Do you think my choice of circuit protection is sufficient for this application? Any advice would be really appreciated. Thanks!

  • \$\begingroup\$ Be sure your protection diodes don't leak \$\endgroup\$
    – user76844
    Commented Sep 29, 2018 at 11:52
  • \$\begingroup\$ The leakage current is 1uA @ the standoff voltage (33V) according to the TVS datasheet. Is this enough to affect ADC readings? \$\endgroup\$
    – Russell
    Commented Sep 29, 2018 at 12:33
  • \$\begingroup\$ I do agree with other contributors, remove diodes, but remove TVS also, it can only fail if over stressed but no great benefit. Only make sure R1 can withstand expected surges (i.e do not choose a 0603 or similar) and move C1 to the opamp input, filters the sooner the better for EMI and also helps shunting surges. \$\endgroup\$
    – carloc
    Commented Sep 29, 2018 at 15:26
  • \$\begingroup\$ Ok, thank you. If I choose an 0805 rated at 0.5W, this should be sufficient hopefully? \$\endgroup\$
    – Russell
    Commented Sep 30, 2018 at 3:10
  • \$\begingroup\$ Check resistor max voltage and surge capability with your expected surge. Sidenotes 1.Resistors are often poorly specified AFA their surge capability, you can go for a specific Surge Characterized resistor (all vendors have specific products) or make tests and educated guesses on "normal" ones. The first ones are not significantly more expensive than normal ones but it may prove hard to assure assemblers buy exactly the part number you choosed. It depends a lot on what's behind. 2.Very often the hardest part is guess what expected surge may be, in doubt stick on some appropriate IEC standard. \$\endgroup\$
    – carloc
    Commented Sep 30, 2018 at 4:55

1 Answer 1


The most important component that needs to be there is R1 and it is 100 k ohm so you're good to go. Any IC including the AD820 will have ESD protection diodes at it's input already. You added additional BAT54 diodes in parallel to them, that will not hurt but they're not really needed either.

Those internal (on chip) ESD diodes will protect the chip sufficiently as long as you limit the current, which you do with R1.

The TVS D2 will only offer additional protection as long as there is something (like series resistance) present to limit the current. If the input voltage is enough to make D2 conduct but the current itsn't somehow limited, D2 will self distruct.

So: all in all your schematic looks fine to me, you could remove some components if you want to save cost but it will also work as it is now.

  • \$\begingroup\$ Thanks very much for your advice! I will remove the BAT54 diodes as from what you have explained, there is really no reason for them. For the TVS diode D2 from what I understand now, it will be useless if I don't add a series resistor before it? if I added, say a 1k resistor, would this be sufficient? \$\endgroup\$
    – Russell
    Commented Sep 29, 2018 at 12:27
  • \$\begingroup\$ The TVS is only useful if you want it to take care of low-current spikes etc. But as I mentioned, the protection by R1 and the build-in ESD protection of the AD chip is sufficient, you don't really need that TVS. A TVS could be useful if your design required R1 to have a much lower value like 1 kohm. But the 100 kohm you have basically takes care that the current can never exceed damaging values. \$\endgroup\$ Commented Sep 29, 2018 at 13:41
  • \$\begingroup\$ Ok, I think I understand now. So in fact there are already clipping/TVS diodes built into the AD chip, however we can only inject a very small current into the inputs, so my large 100K resistor is putting a large enough current limit in place? Having done a bit more research, it looks like industry standard is IEC 61000-4-2. If I take the 8kV standard, then current is limited to (8kV/100000) = 80mA (looks like its still too high for this), but is this the correct line of thinking? \$\endgroup\$
    – Russell
    Commented Sep 30, 2018 at 3:16
  • \$\begingroup\$ Also, I have checked the STM32 datasheet and it looks like it has ESD protection on all pins too. Also I have checked in the STM32 datasheet and it says "Electrostatic discharge (ESD) (positive and negative) is applied to all device pins until a functional disturbance occurs. This test is compliant with the IEC 61000-4-2 standard.". Does this mean that it would be ESD safe if I was to remove the OpAmp buffer and RC filter, and just have in effect the 100K resistor between the exposed user input and the ADC signal pin? \$\endgroup\$
    – Russell
    Commented Sep 30, 2018 at 3:20
  • \$\begingroup\$ Yes, ALL ICs have ESD protection, they need it. Without it the inner circuits would be damaged too easily. Yes your 80 mA calculation is correct. But such an 8 kV ESD pulse is very short. When an ESD pulse strikes a chip's pin directly, up to 4 A can flow. Yes that is 4 A. But the on-chip ESD protection is designed to handle that. So really, an 80 mA pulse is nothing. \$\endgroup\$ Commented Sep 30, 2018 at 10:07

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