justDIY
Active Member
I would like to build some hv9910 based constant-current regulators for using in my experiments with LEDs. I would like to build one converter, that can handle several different output options, by changing the current programming resistor(s).
I have a basic knowledge of the forumla involved with sizing the components, and have built an excel spreadsheet to help keep it all straight. I have had prototypes working on the breadboard of both the buck and buck-boost configurations.
it would seem from my experience, the less output current required of the converter, the more inductance is required, in an almost linear relationship.
for example:
given an input voltage of 24v and an estimated output voltage of 36v, I come up with these inductace values:
0.550 A requires roughly 880 uH
1.100 A requires roughly 440 uH
now my question is, if I build the converter to handle 0.55 A, but later change the programming resistor to regulate for 1.1 A, what effect will the larger than required inductor have? Will its extra inductance simply be wasted capacity, or will it have some detremental effect on the circuit?
I have a basic knowledge of the forumla involved with sizing the components, and have built an excel spreadsheet to help keep it all straight. I have had prototypes working on the breadboard of both the buck and buck-boost configurations.
it would seem from my experience, the less output current required of the converter, the more inductance is required, in an almost linear relationship.
for example:
given an input voltage of 24v and an estimated output voltage of 36v, I come up with these inductace values:
0.550 A requires roughly 880 uH
1.100 A requires roughly 440 uH
now my question is, if I build the converter to handle 0.55 A, but later change the programming resistor to regulate for 1.1 A, what effect will the larger than required inductor have? Will its extra inductance simply be wasted capacity, or will it have some detremental effect on the circuit?