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Sounds OK in principle but without the schematic for your LTC3780 board I can't be sure.
I'm happy that your supply was working OK but that "Indestructible" has some really awful design and construction practice - you shouldn't just plonk one earth wire on top of another while adding to the solder blob and gently caressing it with the iron (he's doing that so as to not melt the whole blob and have the first wires fall off). It's bad because the interface between the different layers of partially melted solder won't be reliable. The joint should be mechanically stable before you simultaneously apply the iron and the solder. Also, and especially for home built or prototypes, you should protect the mains wiring with heatshrink sleeving, or in some other way, so that when you are poking about inside trying to get it to work you don't accidentally touch the mains. I use complete mains inlet, filter, switch devices which avoid the need for so much mains wiring - they cost a bit more but are much safer and better.
The picture shows how I get 12V at 5A into a box:
I've said this before, but it may save others from problems, any supply based on the LTC3780 will be noisy, it can be very efficient but it's really hard to get rid of the noise spikes which may affect other circuits.
I am using a server power supply unit to get the 12V output. I will reconnect all the wires with terminals to each other to make a good connection.
I've replaced the left two potentiometer with external potentiometer. The middle one is a 200k Ohm potentiometer for the current limiting function of the board, but it looks like the value of the potentiometer is twice as much as it should be.
I think my board can deliver up to 20A, which is not really what I want. (my current and voltage detection screen is only able to display 10A)
Since the (wiper) voltage of a potentiometer can't suddenly be negative when there isn't any negative voltage applied, and the output current limit is to much when completely turned to the right, I should remove the last half of the range.
That means that the range I want is 100k Ohm, and after re-thinking I won't need that 100k Ohm resistor in series with the 100k Ohm potentiometer, since the current limiter will always start half way, thus at 10A.
I want to replace the "Max I Adj" with a 100k Ohm potentiometer, since the 200k Ohm gives twice as much current as I can handle.
Is my thinking done right? Or am I missing something?
1 of 1 people found this helpful
Working on the assumption that you board is like the one described in this page:
Then I think your first idea was best: change VR2 to 100k and put 100k between it and R7.
Should reduce the maximum current to about 6.5A.
It's working like that schematic.
But the thing is, the maximum current with the 200k Ohm potentiometer isn't 10A but 20A.
Is that second resistor really necessary?
If I'm correct it will increase the minimum current limit by half the range of the 200k Ohm potentiometer.
2 of 2 people found this helpful
It's not obvious just what the maximum current will be , I estimate that you need 1.6V at D2 anode so if the sense resistor really is 0.007R then you get 0.07V on the amplifier input at 10A so you need a gain of 1.6/0.07 to get current limiting = 22.8, so VR2 should be (1.6M/(22.8-1) )-1 = 72.39k - use the 100k pot with 270k in parallel with the outer two terminals.
I think the best bet would be to make a model and simulate the circuit. You can download LT Spice for free and it has a model for the LTC3780 and you can easily find some op amps in LTSPice standard library that are close to the ones on your board (it won't matter if they are not exactly the same).
Then you can be sure how well it will work.
I am using my self-built variable lab bench power supply for a few months now, and it's working perfectly!
The one thing that I noticed (and has damaged one channel of the psu) is that the current limit of 10A is reached with the current limiting potentiometer at half of it's range. According to this instructables (with the video) I am using a 200k Ohm potentiometer to set the current limiter.
As you might wonder right now, what happens if I set the pot to it's max setting. That will make the current exceed the 10A limit easily.
I am thinking about using a 100k Ohm potentiometer with a 100k Ohm resistor in series so that I only have 'half' the range of the 200k potentiometer.
Will that make the current output limit go back to it's designed 10A max?
I have burned two diodes during the 'let's see what it can handle' session last week, so I don't want that to happen again.