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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
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