LM3478: Operation of current sense

Hello Chiranjeevi M R,

One question upfront: Why would you prefer the very old LM3478 over one of our modern parts like the LM5155/6 or even the LM5158 with integrated FET?

> The option to set current limit by choosing Rsn works like if voltage across Rsn (which is given as input to Isn) crosses threshold (~190mV), the MOSFET will be turned off till the next cycle.
Correct. During one cycle, the current through the inductor / FET / sense resistor will rise until a "certain" threshold will be reached.
At that point the FET will be turned off until the end of the cycle.

> The short circuit sensing happens if there is ~300+ mV at Isn pin. But when will this happen?
> As soon as the pin voltage reaches 190mV, the MOSFET is turned OFF right?
You are right that during normal conditions the overcurrent comparator will never trip.
This can only happen in case of a severe permanent short.

> And I couldn't completely understand the plot attached with this post.
I am sorry, I cannot see this plot.
Adding an image with drag and drop or copy and paste does not work in this forum.
Please use the Insert-> video/image/file feature to add a picture.

> The duty cycle is set using feedback resistors, right?
Indirectly yes, but not only by these resistors.
Ignoring the voltage drops, the duty cycle id defined with this equation which only depends on the ratio of input voltage and output voltage: D = 1 - (Vin / Vout)
The feedback resistors define which output voltage shall be reached.
But when the input voltage varies, the duty cycle will change accordingly.

> If no load is connected and current ripple is small, then current through MOSFET will be low.
Please do not confuse the AVERAGE load current and the PEAK inductor current.
At a light load, the current through the MOSFET will still grow with the same slope (defined by the inductor size) as it does in case of a big load.
It will turn off earlier in case of a light load and the falling slope of the inductor current will be longer and less steep.

> If it is low enough to make current sense voltage less than 0.1V, does the duty cycle reach 100%?
> If so, how is the output voltage set to desired value?

The current sense signal is NOT compared against a fixed threshold.
It is basically compared against the sum of the voltage feedback signal and the slope compensation ramp.
The slope compensation can be configured via an external resistor, if necessary.

At light loads the on time of the FET will become shorter, not longer.

If the energy that is transmitted to the output is bigger than what is getting consumed, the Overvoltage Protection will kick in:

7.3.1 Overvoltage Protection
If at anytime the voltage at the feedback pin rises to VFB+ VOVP, OVP is triggered.
OVP will cause the drive pin to go low, forcing the power MOSFET off. With the MOSFET off, the output voltage will drop.
The LM3478 will begin switching again when the feedback voltage reaches VFB + (VOVP - VOVP(HYS)).
See Electrical Characteristics for limits on VFB, VOVP and VOVP(HYS).

In the other direction, if VOUT is much higher than VIN and in case of a heavy load and if the current sense resistor is not calculated properly,
the controller could run into a situation where it actually reaches 100% duty cycle which might even destroy the FET or the inductor.
Therefore, please take care when calculating the sense resistor.

Best regards
Harry