Hello

I am new to the electronic world, and this is my first post here.

I have a power regulated circuit (TPS55065-SO Switch mode Regulator 1.5~40V input.
it wont start until the Voltage input become 5.03V, but the person who made it say it should start with minimum Voltage of 3.6V

i have changed the inductor (L1) to different values (22UF, 47uf & 100uf) and it did not work. i mean we still need to input 5.03V to power it on.

Also if we power the circuit with 5.03V it will still be on even if we drop the input power to 2V

We would really appreciate any suggestion and help.

Here is the schematic and the other photo show the output voltage of the regulator on the actual board.

Thanks a lot

 

Perhaps the IC has a low-voltage lockout function which operates at power-up?

 

I am not sure if i get what you mean, but the datasheet of the IC say it can operate on low voltage
Unregulated input voltage, V(driver)-2: –0.5 V to 40 V
Unregulated inputs, V(AIN), V(ENABLE)-2:–0.5 V to 40 V

Here is the datasheet link: http://pdf1.alldatasheet.com/datasheet-pdf/view/465218/TI1/TPS55065-Q1.html

I think i found something in the datashhet, it say in the ELECTRICAL CHARACTERISTICS page
V(driver) Unregulated input voltage: 1.5v to 40v
V(driver) Start-up condition voltage: 5V (TEST CONDITIONS: IO = 500 mA)

could this be it,i mean it must get 5v to start working, then it can operate on 1.5~40V

 

You need a capacitor on pin 9.
Page 12:
"Extension of the Input Voltage Range on V(driver)
To ensure a stable 5-V output voltage with the output load in the specified range, the V(driver) supply must be greater than or equal to 5 V for greater than 1 ms (typical). After a period of 1 ms (typical), the logic may be supplied by the Vout regulator and the V(driver) supply may be capable of operating down to 1.5V."

The chip is doing exactly what it is supposed to do.

Here is the datasheet:

 

could this be it,i mean it must get 5v to start working, then it can operate on 1.5~40V

That was my thinking. #12 has found the relevant info confirming it.

 

Thank you #12 & Alec_t

You need a capacitor on pin 9.
Page 12:
"Extension of the Input Voltage Range on V(driver)
To ensure a stable 5-V output voltage with the output load in the specified range, the V(driver) supply must be greater than or equal to 5 V for greater than 1 ms (typical). After a period of 1 ms (typical), the logic may be supplied by the Vout regulator and the V(driver) supply may be capable of operating down to 1.5V."

The chip is doing exactly what it is supposed to do.

Here is the datasheet:

i checked the board and i found out that pin 9 is not connected to anything.
can you tell me what we will get if we add a capacitor on pin 9!!

i mean will it change any thing?? as its already working when it gets 5.03 for about a sec, then it can be still on when the voltage is dropped to 2v

Thanks again for your help

 

5VG is a supply voltage (page 8). I guess if you're not going to use it for anything, you don't need a filter on it. At first, I thought it had some effect on inrush current limit (page 2), but I think I was wrong.

 

thank you again guys