# Circuit to measure Voltage and Current (will it work?)

Hello,

I have a question, If my circuit will work:

1. Function: Gate is on GND Supplying a voltage betwenn 0V and 10V 10V Input is ~3.3V output If the voltage goes above 10V the diodes D2 and D3 clamp it to ~3,3V (This works as I understand)

2. Function Vgate = 3.3v Supplying a current between 0mA to 20mA the voltage drop at R3 should "Supply" the voltage divider again max output is 3.3V

Will the 2nd function work?

The Si2356 is a good choice for a 3.3 volt gate control. It will have an on-resistance of about 0.05 ohms and that is negligible compared to the 499 ohm resistor (R3) so 20 mA will produce 9.98 volts across R3. The impact of 150 k in parallel with R3 is also quite negligible (it'll drop to 497.3 ohms) so the real voltage developed for 20 mA is 9.95 volts.

When using it as a voltage potential divider there will be anything from 1 uA to 10 uA leakage into the "off" MOSFET but, in comparison with the current though the series combo of R1 and R2 isn't a big deal.

R1 and R2 convert 10 volts to 3.329 volts but beware of resistor tolerances. For instance, if using 1% resistors for R1 and R2, 10 volts in could mean the 3.329 volts is a bit higher at nearly 3.4 volts. I don't think it'll cause a problem other than the uncertainty of value. I'd use 0.1% resistors if it is critical for performance.

Your current source, I1, is probably upside down.

At 20 mA there will be 10 V across R3. If Q1 is off then the voltage at the top of R3 will rise to its maximum possible.

The 100k and the diodes will still protect the output.

I recommend that you turn off the grid when taking screenshots. This will improve legibility.

• But if Q1 is on, then there will be a voltage drop across R3 and Q1 which will "supply" the voltage divider, or? There will be no current at the input if the gate is at GND – Fatih Y Mar 1 '18 at 14:01
• A constant current source will raise the voltage to it's maximum to try to drive the set current, 20 mA in this case, through the circuit. If, for example, it is a 20 mA industrial transmitter powered from a 24 V supply then the voltage would rise to almost 24 V if the transistor is off. It's only when you turn on the transistor that the voltage would drop. – Transistor Mar 1 '18 at 14:12