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Circuit

How can I interface with a microcontroller to detect presence of a 48 V supply with a source impedance of 10 kΩ without pulling down the voltage by more than 0.1 V?

I need to step down the output voltage (48V) at the drain of the IRFZ44N power MOSFET to within 5V so that it can be read by an Arduino without loading and causing the voltage to drop below 48V.

The MOSFET is acting as a switch. In the end I need to be able to tell if the switch is on or off based on the output at the drain terminal of the mosfet. How do I go about this?

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    \$\begingroup\$ What’s the issue with just a voltage divider? \$\endgroup\$
    – winny
    Commented Jul 1, 2019 at 9:38
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    \$\begingroup\$ No, just two resistors dividing 48 V down to 5 V + some margin. What are your requirements on load on the 48 V and what's the imput impedance of your menasurement? \$\endgroup\$
    – winny
    Commented Jul 1, 2019 at 10:18
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    \$\begingroup\$ No, it has not. What is your source impedance/how many uA/mA of load current can you spare from your 48 V supply? What is the imput impedance of your measurement at 5 V? \$\endgroup\$
    – winny
    Commented Jul 1, 2019 at 10:27
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    \$\begingroup\$ Please disregard what you have drawn and take a step back. Your 48 V source as it is, what is the source impedance from it and/or how much current can you spare for your measurement? \$\endgroup\$
    – winny
    Commented Jul 1, 2019 at 11:02
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    \$\begingroup\$ "Say i use 1kohm and 10kohm, then, the output voltage will be 11kohm/(11kohm+10kohm(current limiting))* 48V= 25.1V" - but the load draws 0.5A at 48V, which is equivalent to 96 Ohms. What is the purpose of the 10k resistor and why does it have to be so large? \$\endgroup\$ Commented Jul 3, 2019 at 19:02

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I can spare in the order of a few Milli amps for measurement, as I do not want the 48v to drop

I'm measuring 5V using an arduino. So input impedance of an analog pin is 100Mohm

To be able to tell how much your 48 V will drop due to this, you must provide the source impedance of your 48 V rail.

Non the less, if you can spare a few mA of current for the measurement and have 100 Mohm input impedance downstream, I would recommend the following:

This will consume just under 2 mA and give you about 20 % margin until you saturate your measurement.

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT: With your updated information (for future reference here on EE.SE, please make sure not to sprinkle vital informaion in comments and include as much as you have in your original question), this above is my updated schematic. It will load your 48 V rail with about 45 uA, causing a <1% voltage drop across your 10 kohm source impedance.

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  • \$\begingroup\$ i have dited the question, circuit and title. Hope it is clear now \$\endgroup\$ Commented Jul 3, 2019 at 10:48

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