Current sense on high side driver VNQ9025AJ

[DiogoPedrosa]

Hey,

I'm planning to use this high-side driver in a project distribution module, but I'm having trouble fully understanding how the current sensing works.

I've looked at the schematic, and I see a combination of two resistors, Rprot and Rsense. Is this supposed to act as a voltage/current divider?What is the purpose with these sensors? Also, I assume the CS (current sense) output from the driver is 5V, but in the datasheet, it's directly connected to the MCU without a voltage divider to 3.3V. As far as I know, STM MCUs can't handle 5V inputs. Is it necessary to add a voltage divider before connecting it to the ADC?

[MasterT · Accepted solution]

https://www.st.com/resource/en/datasheet/vnq9025aj.pdf 

Page 3:

>>CS Analog current sense output pin; it delivers a current proportional to the selected load current.

Page 11:

>> K2 IOUT/ISENSE = 5050

So, 30A produces ~6 mA at CS, setting R = 1 k, max voltage 6V.

The answer to why not divider is on page 26:

>>Calculation example:
For VCCpeak = -150 V; Ilatchup ≥ 20 mA; VOHµC ≥ 4.5 V
7.5 kΩ ≤ Rprot ≤ 140 kΩ.
Recommended values: Rprot = 15 kΩ

As you can see min. value is 7.5k, but it's not possible to set 7k since resistor 1k (dictated by K=5050) already out of range in case of high current switching.

And yes, 3.3V uCPU 'd limits R even to lower value, 500-560 Ohms or so.   Rprot = 7.5k serves as protection for 150 V (!), so 6-7 V that may appear at CS pin is last things to worry about.

 OTOH, 7-15k in series with adc input is not good, depends on sampling rate : once in a seconds 'd be fine, but high sampling rate demands OPA buffers