Personal computing discussed
Moderators: renee, mac_h8r1, Nemesis
JustAnEngineer wrote:http://www.weaknees.com/tivo-power-supply.php
notfred wrote:The black and gold ones are low ESR types and those are usually the ones that go rather than the standard electrolytics. Match voltage and uF (go bigger in voltage if you can't match exactly) and only replace low ESR with low ESR. If it's more than just 1 or 2, it may be worth getting them from Digikey rather than Frys.
sluggo wrote:I'd replace every electrolytic in the supply that's made by the failing part's maker. Just take pictures of the unit before you replace so you can be sure of matching the polarity on the new caps.
sluggo wrote:I'd replace every electrolytic in the supply that's made by the failing part's maker. Just take pictures of the unit before you replace so you can be sure of matching the polarity on the new caps.
Xylker wrote:It was good to know that a 50V cap would not hurt me.
just brew it! wrote:Before it blows the body off the end, possibly taking an eye out on the way past, and sprays the rest of the components with its insides.the voltage rating is just the maximum safe voltage the cap can be used at
notfred wrote:just brew it! wrote:Before it blows the body off the end, possibly taking an eye out on the way past, and sprays the rest of the components with its insides.the voltage rating is just the maximum safe voltage the cap can be used at
Xylker wrote:If you are interested, I can show more pics of the repair itself. But it is U-G-L-Y!
just brew it! wrote:Xylker wrote:It was good to know that a 50V cap would not hurt me.
Yeah, the thing people who haven't worked with component level electronics frequently don't realize up front is that cap voltage isn't like voltage on a power supply or battery. Having a higher voltage cap won't fry anything, because a cap only puts out as much voltage as is put into it; the voltage rating is just the maximum safe voltage the cap can be used at.
just brew it! wrote:@ludi -
Agree with #2 and #3, but #1 doesn't make sense. The lost capacity is only relative to the maximum energy the capacitor can theoretically store if charged to its maximum rated voltage, which does not affect the amount of energy stored at a given (lower) voltage. Energy stored in a capacitor is a function of the actual voltage across the capacitor and the capacitance.
ludi wrote:just brew it! wrote:@ludi -
Agree with #2 and #3, but #1 doesn't make sense. The lost capacity is only relative to the maximum energy the capacitor can theoretically store if charged to its maximum rated voltage, which does not affect the amount of energy stored at a given (lower) voltage. Energy stored in a capacitor is a function of the actual voltage across the capacitor and the capacitance.
Okay, after chewing on it a bit, I think I follow what you're saying but it's been a while since Physics I, and Wikipedia's description of the Farad is ambiguous. Supposing, for example, two simple capacitors rated 1000uF (0.001F) at 25V and 50V, respectively. Do both charge or discharge at the rate of 1V/s for 1mA of current, but the latter has the capacity to charge to (and discharge from) twice as much voltage?