An easy solution for that: Turn on the inverter (obviously with the input and output disconnected). It will come on for a moment, and then die. now the caps are discharged.
I reckon the lamp option is a good one since it indicates how much current it’s using!
I have the same scenario when switching off a switched mode PSU. On the 4A and bigger the mains cap takes some time to discharge so clutching the PCB with one hand to turn it over can be a shocking experience…
The only problem is those 60w lamps are starting to get very scarce. No shop sell them anymore and it’s not really something you can find at a junkyard.
4x 12V lamps from MIDAS is also an option.
What is the size of the capacitors inside the multi 3000va? Wild guess 50000µF
With a 230V 60W lamp, not so much. It doesn’t even glow at 50V.
Yeah, that would actually be a better option if you prefer a visual indication. When the lamps go dim, you know it’s charged.
No clue. I could open one up, but that’s too much effort today. A bit of googling seems to suggest that a 12V 2000VA has 4 x 15 000µF caps in it (or 60 000µF). So your guess is likely not far off.
Agreed! But there were many options of those incandescent lamps. You can still get oven lamps or halogen lamps at 220V. When you see one being thrown out grab it. You never know when you will need it…
Just thinking here…
So the capacitors are basically a dead short initially, going open circuit when charged.
So if you connect the battery via a, say 25ohm resistor, at 55V, you pass 2.2A initially, going to zero.
So, could you connect the battery via a push-button (capable of the 2.2A), a 25ohm resistor and an ACDC “48V” LED? When you push the button, the LED initially lights, then goes dark as the capacitor is charged. At which point you release the push button and close the main fuse / isolator.
Does that work? So the resistor should be capable of P = 2.2 * 2.2 * 25 = 121W, probably 100W since the time is so short?
Edit: This will be slowish, but the resistor could also be 10ohm, current 5.5A initially, which gives ~300W, but again, being so short you can undersize the resistor?
Edit 2: Then the LED also impedes the current. Does it burn out or only charge the caps over two days?
Edit 3: Sorry, thinking by typing this time… Obviously it just works without the LED, but is it possible to know when the precharge is finished, without (more) complicated voltage monitoring?
I can think of two options:
Put the LED on the Inverter DC bus (i.e. After the pre-charge resistor) with a reverse-biased zener in series and a resistor. Current will only start to flow in the LED, Zener, resistor circuit once the voltage is higher than the Zener voltage plus the LED volt drop. The resistor is just to limit the current. By selecting the Zener voltage, you can choose when the LED will start to glow.
Put the LED and it’s series resistor in parallel with the pre-charge resistor. Initially the voltage across the pre-charge resistor will be high enough that the LED will light up, but as the current decreases the voltage across the resistor (and therefore the LED) will decrease until the pre-charge is finished and the LED stays off.
You quite right. I use a 10ohm 25W resistor EVERY time I connect battery to inverter.
I = V/R
So current = 50/10 = 5A which is fine as it is only for charging the capacitors for a second or two !
Then quickly connect the battery to inverter directly and that is it !
OK, that seems easy. I wonder if anyone would be willing to test how well it works with the ACDC 48V LEDs… specifically if you can stop as soon as the LED drops out, or if the sparks are still there and you need to press the button longer (in which case the LED is moot).
Now who do we know with a bunch of these nice blue inverters on their desk?
(I’m only buying in Jan.)
Man if Izak would only go fetch his multimeter and measure the resistance of said globe we might actually be able to decide on suitable resister…
Haha ok. It measues 70Ω when cold. Calculation says it goes up to about ten times that once it gets hot.
I forgot to mention, if your inverter CAN power up on AC only (Axpert Kings cannot) then the caps will very likely be charged and not need the resistor method. Very easy to check. Apply AC to inverter and if inverter powers up, measure the inverter battery terminals. If you see close to the DC battery voltage eg 50V then you def don’t need the resistor method.
So I know this is overkill, but the LED idea keeps sticking in my mind…
I want it to light while you press the pre-charge button until the pre-charge is done, at which point you release the button. Wiring it across the pre-charge resistor should achieve this as @stanley said.
So it needs to work over a wide voltage range, otherwise it goes out too quickly. I found this circuit for a Wide Voltage Range LED using a FET, which goes up to 30V. FET datasheet.
Can someone recommend a different FET that would work from a similar low voltage up to 60V and still produce the 10-30mA current?
I think I would look into permanently fitting a large ferrite bead around the DC cabling to the inverter.
Inductive Reactance = 2(pi)F.L Capacitive Reactance = 1/(2(pi)F.C)
Normal running conditions F ( frequency) would be very low and so it won’t create losses.
But a rate of change of current would represent a high-frequency surge and it would represent an inline impedance.
Not fancy, won’t break, and always there even when you’re not.
I think this will do what you want:
The one I linked to does a constant 30mA and works up to 60V
A large ferrite bead wouldn’t add anywhere near enough inductance to limit the current when connecting the battery. The inductor required to limit the current so it doesn’t make a big spark would be very large.
I presume we are referring to a battery bank of 48V (or possibly 24V)
One needs a device that is rated at the maximum voltage so a 220V lamp will be slow.
Better to get a couple of 12V halogen downlighters. They draw 4A just to operate!
As the caps charge you can short circuit (or bypass) the lamps progressively to speed up the charging…