On page 8 of the LM2621 datasheet it suggests bootstrapping when the output is between 2.5V and 5.0V by connecting Vdd to Vout. In each of the application examples this is done through a 500Ω resistor (R1). What are the criteria for selecting a value for R1? On the bench (4V in, 5V out, 200mA) the 500Ω resistor produces >200mV of output ripple, and shorting it reduces ripple to acceptable levels.
Hi Chris,
The output ripple of the converter shouldn’t be correlated with the R1 resistor. The maximum VDD operating current is 110uA and really shouldn’t affect the output voltage ripple either. Perhaps your output capacitor is too small and when you connect VDD directly, the extra capacitance that was on VDD is now applied to the output thus reducing the ripple. What size is your output capacitor?
Regards,
Dan
Daniel Tennant What size is your output capacitor?
What size is your output capacitor?
Currently using a single 100µF tantalum. We've tried as much as 500µF and as little as 22µF with little effect. I should note that this ripple is at the hysteresis frequency. Circuit is essentially identical to Example 1 in the datasheet.
Hello Chris,
R1 is used to filter the supply input (VDD) of the part in combination with C3. The 500ohm value may have been used in order to reduce the capacitance of C3 therefore if you could use a higher value of C3 and a lower value of R1 (anyway I would suggest to keep R1 lower than 1Kohm).
For what concern the output ripple I am not sure that what causing the improvement of the ripple is that R1 is shorted, most likely is the fact that you are adding 100nF at the output hence the self resonance of that cap may contribute to reduce the ripple.
My suggestion is to leave the circuit like it is now and to try to add another 100nF capacitor at the output.
Furthermore, could you please send me some switching waveforms in order to figure out if the part is working properly?
Giuseppe