Hello expert,
What is the output leakage current when these three IC is disable?
1.When TPS92520-Q1 is disabled with the UDIM PIN keep low level
2.When LM61460AFSQRJRRQ1 is disable with the EN PIN keep low level
3. When LMR50410YQDBVRQ1 is disable with the EN PIN keep low level
Hello Zhiyuan,
There is a path from V5D to CSN to GND. Across the HS driver there is some circuitry which goes from supply of the HS driver to the SW node. Then you can see there is a path from SW to CSN (output) via the inductor and the current sense resistor. There is a resistance from CSN to GND. When the channel is off (CHxEN = 0 or UDIMx = 0) then both HS and LS fets are off and we have ~2V on the output based on the resistor divider as shown in the diagram.
This ~2V on the output (CSN) is usually not an issue for LED conducting because the Vf is much higher 3-3.5V. Some LEDs (like amber) don't have that high a Vf and can conduct at much lower voltages. That means you need to put a pull down resistor from CSN to ground to lower the voltage. That current from 5V is set by the internal impedance. This is in the datasheet.
For example say you want to make sure the voltage is never above 1.5V then you would need to bypass leakage current of ~28uA. That means you need a resistor from CSN to GND that is 1.5V / 28 uA = 53.6k ohms. You can add as much as 18uA (absolute worst case addition) to ensure the worst case over process and temp. That would mean you need to have 1.5V / 46uA = 32.6 k ohms from CSN to GND.
To prevent the LEDs from glowing if they have a lower Vf then you will need a pull down resistor. You need to look at the characteristics of the LEDs, hopefully they have the Vf over temp and process so you can pick the desired voltage on CSN. Then select based on figure 6-3 curve in the datasheet.
-fhoude
