buckdc/dc converter
In the above figure \$Cin\$ is the bulk capacitor and \$Rstarup1\$ and \$Rstartup2\$ are the startup resistors (Startup circuit).For the given circuit I am using UCC28C40.In it's datasheet it is said that \$Cin\$ is used to maintain an acceptable minimum bulk voltage level (Vbulk(min)).
But what is purpose of setting bulk voltage to minimum value? What will happen if these components (Cin and startup resistors) are not included?
This paper describes the typical inductance and SRF of an electrolytic capacitor.
One way to really know what's going on is to hook the capacitor up to an impedance analyzer.
The difference in ESR between 100kHz and 150kHz is small (\$22 m\Omega\$ vs. \$25m\Omega\$). The minimum impedance is at around 70kHz.
Remember, for most forward-type switching power supply designs (i.e. buck) the size and number of output capacitors is much more influenced by the ESR you need to keep the output ripple within specification, not so much by the capacitance you need to maintain regulation when the power train isn't delivering energy - generally you wind up with much more C than you need need to get the ESR that you want.
Also, even when you exceed the SRF and the ESL becomes dominant, the part is still a capacitor - just with some inductance in series. The ESR (caused by ESL) has to become extremely huge before the part ceases functioning altogether as a capacitor, which will happen at extremely high frequencies (where the ESR approaches the load resistance). This article explains the concept very nicely.
(It's kind of like thinking about a gyrator circuit - it simulates inductance, but isn't really an inductor as it doesn't store energy in a magnetic field.)
Trust me, lots of power supplies operating at or above 100kHz are using electrolytic capacitors as output filters very successfully.
Looks like everyone who said it was the loop compensation was right! But I've led everyone on a bit of a wild goose chase - the problem was a mistake on the schematic that must have occurred when it was being modified from one iteration to the next: R6 is meant to connect to COMP, but is connected to FB! So, it was indeed the loop compensation not working properly!
Best Answer
Without the start-up circuit the chip will never receive any initial power to begin driving the MOSFET and hence it won't receive full normal power (needed) from the auxiliary winding on the transformer.
The start-up resistors are only present to ensure that the chip gets enough voltage to start driving the MOSFET - then it can receive normal operating power through the extra winding.
A bulk capacitor is needed on virtually any power supply regulator or indeed any chip. It provides a local circulatory path for high currents taken by those chips and, from the perspective of the power feed makes life easier. Ultimately it reduces EMI emissions and also susceptibility - it is a fundamental component in virtually every circuit ever developed.
Absolutely!! When the MOSFET turns on it draws a significant surge of current from the power feed lines and, if those lines have inductance and resistance (which they all do have) then the surge will cause a fall in line voltage for the period of that surge current. The capacitor is there to supply that surge current and reduce the effect of surge volt drop down the wires.