This is a YMMV / Caveat Emptor / DTTAH / ACNR type question and answer*:
ie the following may help but you MAY blow things up.
Is the buzzing because of the capacitor is bad or that a different component could be failing and having a secondary, audible effect on the capacitor?
Could be either. Capacitor is possible but I've not heard one do this. You may get a better idea by using a 'sound transfer unit' eg a wooden ruler or length of solid plastic bar or similar. ENSURE it is not electrically conductive to the extent that applying 600 VDC on one end and your ear on the other will not cause "problems" [tm]. You may wish to use a piece of rubber glove on one end as well BUT in practice even touching the capacitor outer SHOULD be safe.
Place one end of STU (sound transfer unit) against object to be tested and other end against ear or bone of head near ear. Sound can be localised and heard much better this way., Try on surrounding objects as well. Applying EHT to far end is to be frowned on - you should not have any EHT in a UPS.
I thought the AC capacitors didn't have electrolytes in them and hence can't understand why they would be buzzing loudly when they did not previously
AC capacitors will probably contain two x electrolytic capacitors of twice the target capacitance each, connected in opposed polarity.
Should not both these caps be interchangeable in this application?
There is a "reasonable chance" that the cheap fan cap WILL work OK here. Also, used or dead fans may be cheaper again.
I am thinking of plugging out the 20uF 250V pedestal fan capacitor and putting this capacitor into such a fan instead, to confirm if the cap itself is causing the noise - sounds like a good idea?
Yes. Sounds good. Always a risk that the fan will stress it worse than the UPS does but if it is good it should work OK.
I am not sure why the fans use 250V caps but my assumption is that these Chinese makers just stock up on one variety of the caps since most of the world are at or below 250V.
Many countries are 230 VAC. Some nominally 220 VAC. Voltages can be and are sometimes higher than nominal. 250 VAC is saying it will work anywhere.
Hence, for this experiment, using a 150V cap instead of the 250V one for the fan (in the US) should not damage either fan or cap?
Probably. Cap MUST be AC rated and suitable for whatever class of service it sees. X rates is phase-phase or phase-neutral or line-line. Y rated is phase-ground. Such ratings allow for surges, spikes etc. 250 VAC cap in 110 VAC system should be fine.
If it's confirmed that the cap's at fault, could I put in the 20uF 250V pedestal fan capacitor on the APC UPS board and expect it to have no detrimental effect on the UPS?
You can always EXPECT :-) ... .
But, yes, it will probably be OK BUT no guarantee.
My understanding is that this is a power cap and as long as the capacitance and voltage rating is good, other cap chars like ESR don't matter?
See above re X and Y rated. Also, some designs pass high currents at all times and some don't and a cheap cap may get sadder quicker.
*YMMV / Caveat Emptor / DTTAH / ACNR =
Your milage may vary
Let the buyer beware.
Don't try this at home (you can but ...)
All care, no responsibility
Given the specs say
Data retention current: 1.0 μA (MAX.) (V = 3 V, T = 25°C)
I see at least three problems:
The two 1 kΩ at the battery transistor will draw too much current when VCC is out. 40 kΩ each seems more appropriate to me. (2.2V / (1µA / 25)) ~= 88 kΩ; 40 kΩ should be good enough.
When the battery is half used its voltage will be below the minimum data retention voltage, plus you will have a (small) voltage drop at the transistor. This may or may not work, or, worst, it may work when you test it and stop working at a later time.
The 2N3904 will not cut until VCC is below 0.7~0.8 V. If VCC goes down slowly (electrolytic capacitors, ...) there will be a time during which both this transistor and the one feeding current from the battery will be active. In this situation the third transistor, the bottom one, will probably work in reverse and feed current from the battery to VCC (VCC is now 1 or 2 volts below Vbat). This will drain the battery through the 1 kΩ. In the worst scenario the leak will keep VCC above .7 V and, because it will never stop, it will drain the battery in a few days.
I'd suggest changing the two resistors at the base of the 2N3904 to a divisor that will cut it somewhere between 3 and 4 volts at VCC, possibly adding a third resistor from the emitter of the bottom transistor to introduce a Smith Trigger effect and avoid any possibility of oscillations during the transition.
Best Answer
A critical piece of information is missing here: battery voltage. Though it is probably 12 V lead-acid, you should state it clearly just to be sure.
Assuming this is correct, then you have approximately 173 Wh of capacity. Your router presents a 6 W load and your monitor a 24.7 W load, so a total of 30.7 W. Dividing one by the other, you would get about 5.6 hours of runtime, so your 10 hour figure is already off by a large amount.
Of course this is merely an upper bound on the runtime, under the assumption of 100% efficiency in energy conversion, new and fully charged batteries with the stated capacity (certainly not true for, say, many Chinese manufacturers of 18650 cells which claim absurd capacities like 10 Ah). There’s also an implicit assumption that the power converters involved (in the UPS and the wall-warts) do not consume energy for their operation.
Even assuming the stated capacity is right, note that you’ll need to convert the DC voltage of the batteries to AC, which is later converted to DC again at the wall-warts for the router and monitor. For instance, the 80 Plus standard for PC power supplies requires a minimum of 80% efficiency (and when they meet higher standards, they get awarded progressively higher certification levels like 80 Plus Bronze, Silver, Gold, Platinum and Titanium).
Lacking any other information (and, should you have it, please provide it here so we can make more accurate calculations), we could use 80% as a ballpark efficiency figure for both the UPS and the wall-warts. As such, your UPS would only deliver 80% of its capacity to the wall-warts, and the wall-warts in turn would only be able to use 80% of that energy, so you’re losing 80% * 80% = 64% to power conversion inefficiencies. Therefore, your 5.6 hour runtime would already be reduced to 3.6 hours. The actual figure would depend on the actual efficiencies of the UPS and wall-warts.
Also note that SMPSes generally run better when the load is not far off from its designed maximum load. In your case, a load of merely 30 W is being placed upon a converter capable of 1100 VA/600 W, so at best 5% of the maximum. You’ll hardly get peak efficiency at such a low load.
Even so, you’re reporting an actual runtime that’s nearly half of what was expected. So it seems likely that some capacity degradation on the battery has already taken place. If the batteries are a few years old already and you need the extra runtime, it may suit you to replace them with new ones, but if my assumptions are correct, I wouldn’t expect you to get much over 3 hours of runtime under your setup.
If you’re really interested, and depending on the available equipment and time to invest, there are many measurements you can make to improve these back-of-napkin calculations. With a Kill-A-Watt meter or similar, plus an ammeter, you can determine the actual efficiency of your wall-warts. You could take out the batteries from the UPS after charging and do a capacity test, say by using an appropriate 50 W 12 V load (approximately simulating the actual load due to inefficiencies), and time how long the battery lasts until the battery voltage drops to an end-of-discharge value (10.5 V seems like a reasonable value). Given the 173 Wh expected capacity, a 50 W load should lasts for about 3.5 hours. At least for lithium-ion batteries, if you’re getting less than 80% of the expected runtime, it’s advised that you replace the batteries.