0
\$\begingroup\$

current protection attempt

Is this current protection technique going to work? I came up with this because I want to make a current protection circuit and I do not have a shunt resistor at hand. I think the circuit should work in the following way: Initially, at power up, the voltage across the diode D1 is almost zero, so the output of the op amp OA1 is less than the non-inverting input of OA2 (formed with a voltage divider from a 9v power source), hence the output of OA2 will close the N mosfet Q1. This in consequence will close the P channel mosfet Q2. The 10 V zener diode Z1 will have a voltage drop of 10 v and the rest of 30 v will create a current of I_z =30/500 = 0.06A which is assumed to be lower than the maximum rated current of Z1. Therefore the mosfet Q2 is conducting.

However, if the current through R_load is increasing too much, say to create a voltage drop greater than 1v over the diode D1, (this means less than 50 A through diode, according to this), then the output of OA1 will be greater than that of the voltage divider, and the output of OA2 then will be low, opening Q1, which in turn will open Q2.

Is this how it will work? Feel free to comment on this design ... Which are the possible problems?

\$\endgroup\$
1
\$\begingroup\$

The issue with using diodes is the fairly steep V/I curve, which is temperature dependent. If you look at the curve on the datasheet you linked, the current associated with a 0.8V drop changes from 6A to 11A between 25C and 125C, and with the forward drop you will be dissipating a fair amount of heat, so the trip point is going to be less well controlled than with a shunt resistor. enter image description here

\$\endgroup\$
  • \$\begingroup\$ Then, if I replace the diode with a shunt resistor and change the values of the resistors R4,R5,R6,R7, is this a good design overall? \$\endgroup\$ – C Marius Oct 16 '18 at 17:59
  • \$\begingroup\$ Everything else looks OK. For the shunt resistor, there are ones specifically designed for current sense duty, that have a low thermal coefficient. \$\endgroup\$ – Phil G Oct 16 '18 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.