| ??? 05/02/11 05:09 Read: times |
#182149 - Did you actually read? Responding to: ???'s previous message |
Richard Erlacher said:
IIRC, the point of the expression is just exactly that ... If you increase the output current, the ripple is increased. The point of that is that one must know what the tolerable ripple at the input is. That means that one has to know what the unregulated input looks like at the desired load. It's a somewhat confusing thing to consider, but, given all the correct information, it produces a reasonable estimate of the minimal input capacitance.
It also tells you that if you can tolerate larger input ripple, a smaller input cap will do the job adequately. What it doesn't address is the heat (power) dissipated in the regulator if you have so much "overhead". But did you really read what I wrote? Really? It does say that if you need a larger current, you can get away with a smaller capacitor. That is a broken formula since a larger current generates a larger ripple that needs a larger capacitor to overcome. The capacitor should be scaled with current/ripple (larger for higher current, smaller for larger accepted ripple) and not scaled with 1/(current*ripple). My rewritten formula that added parentheses or multiplied the capacitance with the current instead of dividing with the current does scale the capacitor with the required current. |
| Topic | Author | Date |
| LDO/Capacitor | 01/01/70 00:00 | |
| It seems... | 01/01/70 00:00 | |
| there's an old "standard" ... | 01/01/70 00:00 | |
| Any other LDO | 01/01/70 00:00 | |
| LP38502, LP38692, ... | 01/01/70 00:00 | |
| Thanks | 01/01/70 00:00 | |
| Formula and parentheses | 01/01/70 00:00 | |
| You're right, but that wasn't the point | 01/01/70 00:00 | |
| Did you actually read? | 01/01/70 00:00 | |
| I did read it, but incorrectly | 01/01/70 00:00 | |
Easy to go wrong with multiple divisions | 01/01/70 00:00 |