Hi, TI team:
could you tell me the CS pin(10 pin) output voltage and current when in the fault mode? our Vvs = 11.5V.
I find the value is 4.5-6.5V in the datasheet, it is right?
Please check the CS pin divide resistor value, I used 510R and 1K, so when in the fault mode, CS pin voltage is 3V ~ 4.5V? my MCU pin is 3.3V, so I worry the 4.5V will damage MCU, do I need change the divide resistor to 510R and 510R? the CS pin voltage is 2.25V ~ 3.25V, it will not damage MCU pin?
Hey,
Thank you for your patience. Yes, the divider will have to be adjusted for 3.3V.
For example,
From the datasheet the RCS is picked based on the linear range of the VCS (4V).
VCS FLT = 6.5V -> 3.3V (to protect MCU)
VCS linear = 4V ->2.3V (1V band for FLT cases)
Since the fault high is lowered the linear range has to be lowered as well. 3.3V will be the fault high the RCS can be selected for a range of 2.3V.
-> ((2.3V x 300) / 1) = 690
Since the FLT high voltage from IC (6.5V), RCS resistance (690), and the final voltage (3.3V) is determined we can solve for R1 in the divider.
VOUT = VIN (R2 / (R1 + R2)) -> 3.3V = 6.5 (690 / (X + 690)
R1 = 669
BR,
Alan
Hi, Alan:
I have other question: I set the current limited value is 1.33A, if the circuit shorted, the current will be 5A. could you tell me TPS4H160BQPWPRQ1 will provide 5A, only send the fault signal to MCU? or TPS4H160BQPWPRQ1 can not provide 5A, even the circuit shorted.
We need to do the big current test, and our design should have protection, if TPS4H160BQPWPRQ1 current only is set by resistor - Rcl, so we can pass this test.
If the output shorted, what will the TPS4H160BQPWPRQ1 do? still provide 1.33A, and send fault signal to MCU? or will provide bigger current, but the output voltage will be decreased?
Hey,
For better protection from a hard short-to-GND condition (when the INx pins are enabled, a short to GND occurs
suddenly), the device implements a fast-trip protection to turn off the related channel before the current-limit
closed loop is set up. The fast-trip response time is less than 1 μs, typically. With this fast response, the device
can achieve better inrush current-suppression performance.
For a visual representation the datasheet has this diagram:
BR,
Alan
Hi, Alan:
so when shorted happened, the chip will turn off the output, right?
and could you tell me how to calculate the CS pin voltage? it is depend on R5511/5506?
my SW team asked me the MCU_HSS_CS voltage when chip works normally and abnormally, they will set the threshold value
Hey,
Yes, for better protection from a hard short-to-GND condition (when the INx pins are enabled, a short to GND occurs suddenly), the device implements a fast-trip protection to turn off the related channel before the current-limit closed loop is set up. The fast-trip response time is less than 1 μs, typically.
5501 and 5511 will create a voltage divider on the spec above so using the divider equation R(CS) = 5506 is chosen.
BR,
Alan
