Hello TI team :)
Here are my application details:
- Vin= 13.8V
- Vout= 12.4V
- Load= 5A
- Inductor= 4.7uH
- RT= 249kOhm
- Rsns= 10mOhm
- Radj= 2kOhm
- PFET delay delta= 12ns
I wanted to check for current runaway using equation 13 in the datasheet: V(FD) + V(ESR) > [V(IN) * t(ON)] / t(OFF)
For this I need to know t(ON) and t(OFF). During current limit, I was calculating them this way:
- t(ON)
- Since the feedback voltage will be 0V in the case of a short, I imagine t(ON) is as long as it takes for the inductor current to ramp up to the current limit threshold.
- V(L) = L * di/dt
- di = I(ADJ) * R(ADJ) / R(SNS) = 40uA * 2kOhm / 10mOhm = 8A
- V(L) = V(IN) - V(OUT) = 13.8V - 0.0V = 13.8V
- dt = L * di/V(L) = 4.7uH * 8A / 13.8V = 2.72us
- t(OFF)
- Using equation 10 in the datasheet
- t(OFF) = {4.1u * [[V(IN) / 31] + 0.15]} / {[V(FB) * 0.93] + 0.28V} = 4.1u * 0.595 / 0.28 = 8.71us
Using equation 13:
- [V(IN) * t(ON)] / t(OFF) = 13.8V * 2.72us / 8.71us = 4.31V
Of course it is unrealistic to have the diode forward voltage and the inductor voltage drop exceed 4.31V.
Am I doing something wrong in my calculations of t(ON) and t(OFF)? My application doesn't seem that unique, so I would be surprised if the LM25085 in this configuration was not protected against output shorts.
Thank you!!
