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PTR08100W - Current drop while programming device

Christine Chuah
Prodigy 160 points
Other Parts Discussed in Thread: PTR08100W

 

Hi

We are encountering a weird behaviour on the PTR08100W power module and would like to seek for advise.

We have built the circuitry for 5V output with this power module using the following:

Cin --> 1 x 330uF(recommended electrolytic capacitor EEUFC1E331)

Cout --> 1 x 330uF(recommended electrolytic capacitor EEUFC1E331)

Rset --> 255ohm (2 x 510ohm resistor in parallel)

This 5V (5.22V) will be used as an input to three other linear regulators that each have an output current of up to 3A.

Our board works successfully with the Vin of the PTR08100W at ~5.0-5.3V (external power supply that can output up to 3A).

However, when the Vin of PTR08100W is increased above 5.8V, we observed that the current drawn from the external power supply will drop in the middle of our operation and our operation will fail. The current did not change when the Vin was at ~5.0-5.3V.

Attached is a .doc with the table of the experiment that we did.1070.PTR08100W_current_issue.doc

  • 0
    TI__Mastermind 38230 points

    Hi Christine :

    I reviewed the PTR08100 operating with Vout of 5.25V from a 5 Vout power source.

    The PTR08100 is a buck  power regulator. It must have a input voltage of at least Vo/0.83 or 6.32V  for   accurate and reliable operation. When Vout is 3.6V <Vo≤5.5V. the input voltage minimum is Vo/0.83 or greater .

    When the voltage is less than the  Vo/0.83 minimum ,the internal operation is not capable of any stable  voltage regulation. The minimum duty cycle for any regulation is Vin= Vout/0.83 .

    Any increase in input voltage above the  minimum Vo/0.83 with increase  the input current .The minimum input voltage is 6.32V . The input current for the 6.32V is 2.77A  when the output  load and voltage is 3Amp and 5.25V. The formula of input current is

     Efficiency=( Vout x Iout)/(Vin x Iin)  

      Input current = (Vout x Iout) /( Efficiency x Vin)   Input current = (5.25V x 3amps)/ 0.9 x6.32V)= 2.77 Amps.

     

    Best regards

    Tom