Using a shunt on Live side to measure current

hi leo

perhaps i'm not making myself clear...my shunt is positioned on the LIVE wire of the AC power supply cable and is connected to an evm6736.

So looking at the schematic below nothing is connected to CUR1+ or CUR1- and my shunt is connected to CUR2+ and CUR2- and I have removed jumper JP2.  The potential of CUR2+ and CUR2- is at line voltage which is 220V rms. That means the potential at IN+ and IN- is also in that same ballpark so i do not understand why the IN+ & IN- pins (plus the 430) aren't cooked....i realise this is a fundamental electrical issue which is plainly escaping me.

thanks

Hi Tobias & Leo

Yes the shunt is connected in series with the load...as per the diagram below.

I believe that nothing gets cooked because whilst the connections at the live vable are at a potential of 220V that is only 220V with respect to ground.....IF THEY WERE CONNECTED TO GROUND...but as nothing along the way to the SD24 is connected to ground that potential is of academic interest.

But that still begs the question what happens internally at the pin? Surely something internal there has a ground connection???
And how does that correlate with the datasheet saying max voltage at a pin must be less than 3.3V....so a 'potential' of 200V is ok but a 'voltage' is not?

Hi Moshe,

Yes, the maximum voltage at a device pin is limited to the high level of your supply voltage. So about your question of voltage and potential, yes you can connect a potential of 200 V but than your GND of the device needs to be at 197 V. Due to the risk of an electrical shock I really advise not to do this!

In our case, we are using for current sensing a current transducer, please refer to the attached picture. This means that the high potential of the power line is galvanically isolated from our board supply and the voltage at the current input shouldn’t exceed 1 V or to be more detailed 920 mV.

Best regards,

Tobias

Hi Tobias

Thanks for that.

Now consider I2+ & I2- in your diagram...there is no galvanic isolation there, so what happens within the SD24 pin that those connect to (both I2+ & I2- are at a potential of 220V).
And also referring to your diagram, VCC and VSS would be drawn from the utility cables..ergo GND is at 220V also....so one way or another ground HAS to be at 220V for the EVM to work....hmmm, i think that answers my question above and explains everything!!!

Carrying on along that theme...if VSS is NOT at 220V (ie connected to utility cable) and a shunt is used it will cook the MSP430...yes?

thanks!

Hi Tobias

This is really getting confusing.

Please ignore the current transformer in the diagram for the purposes of this discussion.

Also, please note that i am connecting the shunt on the LIVE cable....so please ignore the 'N'  & 'L' marking and use the (L) & (N) marking. 

Therefore I2+ & I2- ARE at 220V....so NOW what happens at the MSP pin?

cheers

moshe 

As the ground plane, Vss is also at a potential of 220V with respect to ground (only).

I don’t have the EVM, but following the documentation it looks like the connections and labels are as on this picture, you can verify if I’m right;

Maybe you can now better explain how you have everything connected.

My connections using your diagram:

Shunt is on L(N) not N(L). All other connections are the same.

I think you tries to make me (and other) crazy!

Here is the Shunt on N(L);

moshe jacobson said:

same side as the shunt) which is on N(L)

And here is the Shunt on L(N);

moshe jacobson said:

Shunt is on L(N)

???