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TPS61169: If Shutdown Mode is On then also IR LED will ON

Part Number: TPS61169

Hello Team ,

We are using TPS61169 in our board for IR LED Booster.we are facing one issue with this booster like when shutdown mode is on then also this booster will generating 5V as output 


IR LED -----4 IR LED with Series connection (1.75V forward Voltage ) 


Test Case---1

CTRL : Pull Down  

VIN : 5V

Vout or VCC_IRLED : 5V

Observation : All IR LED will ON with 50% intensity 

Test Case---2

CTRL : Pull Up 

VIN : 5V

Vout or VCC_IRLED : 5.89V

Observation : All IR LED will ON with 100% intensity 

For more Information use below image : 

  • Hello Team , 

    We need a solution to trun-off the all IR LED in shutdown mode.

    Thanks ,


  • Hello Team ,

    For No load condition - IR LED voltage drop is low and it is 1.25V so = 1.25 X 4 = 5V and that is still equal to the input voltage so that occurs the LEDs little ON in the Shutdown condition 

    Please Provide any solution to turn off the IR LED in shutdown mode.



  • Hi Shubham,

    Since TPS61169 is a asynchronous device and there is no method to prevent current flow from VIN to load. Adding more IR LED is a effective way to turn off LED in shutdown mode.



  • Hello Robin ,

    We are using TPS61169 is below condition :

    No of IR LED : 2

    IR LED Forward Voltage : Min 1.2V and MAX 1.7V

    Current Setting : 300mA  

    Booster Input Voltage : 3V3

    A. When Driver is enable we measured below voltage and current :

    Booster Input Current : 372mA

    Booster Output Voltage : 3V

    Booster Output current : 304mA

    Two LED is working but in that case booster works in VIN>VOUT condition .so let me known have this condition will effect the booster operation and how we will avoid this condition .

    B. When Driver is disable we measured below voltage and current :

    Booster Input Current : 126mA

    Booster Output Voltage : 2.78V

    Booster Output current : 125mA

    When driver is disable then also LED is ON and 125mA current will flow.we need a solution that will off the LED in driver disable condition .

    Please provide the some solution two solve above two issue 

    we are waiting for your response .

    Thanks ,






  • Hi Shubham,

    We get your point and reply you later.



  • Hi Shubham,

    For your questions:

    A. The real load consists of diode and LED, so real output voltage is 2*Vf+0.7V. According to your question B's description, the Vf>1.4V when 300mA, which means the real output is more than 3.3V and the booster can work normally but internal switch operates at a ultra low duty cycle. 

    B. When disable, 2*Vfmin+0.7V=3.1V<3.3V, a proper solution is utilize two diodes in series.