Hi Guys,

In attached schematic,
if Vout is connected to MCU i.e STM8S (https://www.mouser.in/datasheet/2/389/dm00024550-1797235.pdf)
MCU operating voltage is 5V.

I want to do the circuit analysis with below 4 case:

1. Z1 = Open Z2 connected , Effect on MCU pin =?
2. Z1 = short Z2 conncted , Effect on MCU pin =?
3. Z2 = Open Z1 connected , Effect on MCU pin =?
4. Z2 = short Z1 Connected, Effect on MCU pin =?


Regards,
Shiv Mishra

 

Homework?

What's your best guess for those 4 situations?

You say a pin on an MCU. Is that MCU floating or are its ground pins grounded?

 

See Section 9.2 of the datasheet.



In this particular case, the voltage will be limited by the clipping action of the electrostatic discharge (ESD) diodes provided that your resistors are sufficiently high so as to limit the current into the diodes to a low enough level. It looks like you won't have any problems (see the table below) provided you NEVER SHORT Z1.

 

Cases 1, 2 and 4 should be obvious.

Case 3 depends on the current carrying capability of input protection diode, which I did not see in @DickCappels post. But, with 47K as Z2 it is probably OK.

Bob

 

Will microcontroller get damaged with excess voltage or excess current?

Yes.

 

This is battery voltage measurement circuit which is connected with uC ADC pin.

 

Cases 1, 2 and 4 should be obvious.

Case 3 depends on the current carrying capability of input protection diode, which I did not see in @DickCappels post. But, with 47K as Z2 it is probably OK.

Bob

It is in the second table: 20 ma.

With 8.4 volts through 47k (assume Z2 is open) to, let's say an input clamped to 3.0 volts with a diode is only (8.4V-(3V + 0.6V)) 47k, about 100 microamps. Nothing to worry about current wise. If I misread that and it is 84 volts you are sill at only 1.5 ma and still fine.

In the case of Z1 or Z2 being open you won't be able to read the battery voltage, unless in the case of Z2 being missing the battery voltage is within the ADC's range, that seems unlikely.

Just don't short Z1. If you did and see an unhappy result just remember: We said so.

 

Won't this clamp the input voltage at VDD + 0.6V, which is just a little above the spec. of VDD + 0.3V for all other pins but less than 6.5 V absolute max. So, probably OK/

 

It is in the second table: 20 ma.

With 8.4 volts through 47k (assume Z2 is open) to, let's say an input clamped to 3.0 volts with a diode is only (8.4V-(3V + 0.6V)) 47k, about 100 microamps. Nothing to worry about current wise. If I misread that and it is 84 volts you are sill at only 1.5 ma and still fine.

In the case of Z1 or Z2 being open you won't be able to read the battery voltage, unless in the case of Z2 being missing the battery voltage is within the ADC's range, that seems unlikely.

Just don't short Z1. If you did and see an unhappy result just remember: We said so.

Thread #8 told , it will clamp voltage only upto VDD+0.6V as per datasheet. Since in our case Voltage is 8.4V so it will clip the voltage above VDD+0.6V

Still i dont understand what if Z1 is shorted ?

In a nutshell ,ADC pin get damage with both excess of voltage and current.

Best Regards,

 

Still i dont understand what if Z1 is shorted ?

Hi,
If you apply 8.4V to MCU input, you will damage the MCU, possibly blow the MCU's internal clamp diodes
E

BTW: if the internal diodes fail as a short circuit the 8.4V could be applied to other devices on the 5V line.