Hello, I am troubleshooting a circuit that uses the IPS511S high side power mosfet switch. While testing this circuit I have noticed that the load resistance can’t be too high in order for the part to completely turn off when the ‘IN’ pin is held low. If the load resistance is open circuit, Vout will be ~8.3V when ‘IN’ is at ground and Vcc is 24V. If instead I use a 100k ohm load resistor, Vout will be 0V as expected. In both cases raising ‘IN’ to 5V will cause Vout to be 24V. Below is a table I made to summarize these results. I don’t understand this behavior and appreciate any help.
Thank you
Sean
Rload = open circuit
IN = ground, Vcc = 24V, Vout = 8.3V (this is not expected output behavior)
Rload = 100k ohms
IN = ground, Vcc = 24V, Vout = 0V (this is expected output behavior)
Rload = 100k ohms or Rload = open circuit
IN = +5V, Vcc = 24V, Vout = 24V (this is expected output behavior)
Hello, I am not sure about this one. An Engineer will take a look and offer some assistance.
I suspect that your load is not truly an open circuit, but rather the equivalent resistance of a multimeter or other similar instrument being used to obtain the voltage measurement. Such resistances are typically in megaohm territory.
Semiconductor switches all leak to some degree, and it doesn’t take much current to develop ~8.3V across a megaohm or ten…
From the information given, the described behavior would appear to be quite normal.
Thank you for mentioning the semiconductor leakage current. This is a very good point. You are correct that my “open-circuit” test scenario would in reality be in the megaohm range as you stated. So, like you said, the part’s leakage current spec could easily generate ~8.3V across several megaohm impedance. Thank you for helping me understand the root cause of this behavior, which I now understand is normal.
