Hi All,

I'm looking at the VN340SP-33-E datasheet (http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00060183.pdf). Now, I'm a little confused as to how this IC is controlled. It sounds to me like it is voltage controlled (because of the logic level table & because of the MOSFETs).

However, in the diagram on the datasheet, they use a current limiting resistor Rin. Why is that resistor being used and how would I calculate the R value?


Thanks.

 

"Each I/O is pulled down when
overtemperature condition of the relative channel is verified."

The control inputs are bidirectional. Rin decouples the uC control output and the uC's fault detection input. So if the relays input is pulled low via the overtemp protection it doesn't affect the uC's control output.

The inputs have high impedance, specified in the datasheet is 25uA for 30V input voltage,i.e. you can use quite high resistor values. Should work with anything up to 100k (at 5V). The lower limit is determined by the max output current of the component that provides the control signal.

 

This is an outstanding part by the way. Tough as nails!!

 

Thanks for the explanation praondevou.

So if the input is pulled low, the mcu essentially sees a resistor and then ground? In my case, I am using an optocoupler and 12V as my input logic signal. If the optocoupler is on (12V supplied) and there is no resistor, if the input is automatically pulled low, will this cause a short? And if the resistor is there, then it will just draw current?

I actually have an existing design that I was looking at. The reason I asked about the resistor is because in the design, a 1k resistor is being used. In the datasheet it says that the current per input cannot be greater than 10mA. I guess in normal operation the resistor value does not matter, but if the input is pulled to ground and I have a voltage of 12V for the input, would this cause problems since it exceeds the 10mA specified? I've attached a picture of the design.

Thanks!

 

Yes for 12V I would increase the 1k resistors.

 

Thanks praondevou.

I took a bit more time to analyze this. If my understanding of the resistor works as I previously described, the circuit I posted would have a current of 8.804mA going into an input if overtemperature occurs and the input pin is pulled to ground.

I'm looking at designs from another engineer and trying to make sense of the circuits (I'm not that experienced). Thanks for helping me clear this up.

 

If my understanding of the resistor works as I previously described, the circuit I posted would have a current of 8.804mA going into an input if overtemperature occurs and the input pin is pulled to ground.

I agree. It's probably even less because the switch inside the VN340 that pulls the input to ground will have some voltage drop too. It is not mentioned in the datasheet. (only the voltage of the DIAG pin)