fig
Village Idiot
It's not like you tinker with just anything you happen to pick up.....right?
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Modifying my cheap Chinese power supplies...
- Thread starter Big Monk
- Start date
vigilante398
Authorized Vendor
I think I still have one that hasn't been gutted. I wasn't using any of the other voltages, so I designed a drop-in board that had 10x 9V regulators so I could get an isolated 1.5A @ 9V out of every tap, so one of mine has that inside. Because tubes.
Big Monk
Well-known member
Interesting that you bring this up.
4-5 months back I bought a Mosky Iso-10 trying to have an all-in-one supply and solve some issues from digital noise from the Strymon pedals I own:
Here I'm thinking "Wow! 7 300 mA/9v, 1 500 mA 9v and 2 adjustable 500 mA outs! Let's try it!" Never mind that the total output capacity, judging by the labels at least is 3,600 mA with a 2,500 mA wall-wart!
So I get it hooked up and the noise is gone. So it's definitely isolated. But I keep feeling a lot of heat on the chassis and a smell, not a burning smell per se, but the smell when electrical stuff is getting warm/kinda hot.
Keeps happening so I returned it. Obviously, in this case, the Mosky seemed to have overstated the current capacities of the taps where the O'traki supplies seem to understate them.
I ended up hooking all the same pedals up to the O'traki before returning the Mosky and the noise was back but no overheating issues. Got a replacement for the Mosky and the same thing. Ended up returning that as well.
fig
Village Idiot
We're all tinkerers. That's how we ended up here, LOL
Big Monk
Well-known member
I did this for the Christmas village about 5 years ago. Took an old Dewalt bit case and made a 6 output power supply with limiting resistors on Perfboard so that I did not have to turn the battery boxes for small street lamps, etc. off/on anymore.
finebyfine
Well-known member
man i just realized i should buy one like this and gut it for a power supply build. mine like this is probably less than an enclosure that would end up being an awkward size for a power supply anyway
vigilante398
Authorized Vendor
So based on the datasheet the output of that regulator is
Vout = 1.23 (1 + 6530/1000)
Vout = 9.26V
So that regulator is supplying the 9V, and it looks like there's a zener diode on the far right side of the board (conveniently close to the 12V output) that I'll bet they're using for the 12V. Zener regulation works fine for low current, but I definitely wouldn't push the output on that one much.
Vout = 1.23 (1 + 6530/1000)
Vout = 9.26V
So that regulator is supplying the 9V, and it looks like there's a zener diode on the far right side of the board (conveniently close to the 12V output) that I'll bet they're using for the 12V. Zener regulation works fine for low current, but I definitely wouldn't push the output on that one much.
Big Monk
Well-known member
The only effect I use 12 vDC for is my old Boss CS-2. I'd like that to be adjustable in a perfect world.
Ideally on this smaller supply I'd want all the 9 vDC taps to be 300 mA and the 12 vDC tap to be adjustable from 9-18 vDC at 100 mA.
vigilante398
Authorized Vendor
Well we know the regulator is capable of 3A, so I would say you should be able to get 300mA per output no problem if you replace or jumper the ferrite beads for the 100mA ports (the three that say "010" on them) as long as your wall wart can handle it. Unfortunately the only way to get that 12V tap to be adjustable is to remove the zener and add a new regulator, like an LM317 with a pot for adjustment. You'll need to figure out where to put it though. To get up to 18V you'll either need a boost regulator or you'll need to start using an 18V wall wart instead of the 15V.
fig
Village Idiot
Question sir: If you have in mind what ideally and perfect are, and should this endeavor not pan out, have you considered designing and building one custom to that criteria?
Big Monk
Well-known member
So I think I'm going to do the following to make the little power supplies a little more versatile:
1.) Run the supply off a larger 18 vDC/2A wall-wart. This should output 18 vDC from the 15 vDC port
2.) Change the Zener diode to accommodate this change and still output 12 vDC from the 12 vDC port.
3.) Rework the resistances off the regulator to still give me 9 vDC with an 18 vDC input
4.) Change the 100 mA ferrite beads for 300 mA.
Besides the $10 wall-wart, this should be about $1 in parts.
1.) Run the supply off a larger 18 vDC/2A wall-wart. This should output 18 vDC from the 15 vDC port
2.) Change the Zener diode to accommodate this change and still output 12 vDC from the 12 vDC port.
3.) Rework the resistances off the regulator to still give me 9 vDC with an 18 vDC input
4.) Change the 100 mA ferrite beads for 300 mA.
Besides the $10 wall-wart, this should be about $1 in parts.
vigilante398
Authorized Vendor
2. A 12V zener will give you 12V regardless of input, but it will decrease the amount of current you get out of it because you're dumping more voltage in the regulation. So you don't need to swap the zener, but you may want to get a higher power 12V zener if you're sticking with the zener.
3. You don't need to make any changes to the regulator. the output voltage is hard set independent of the input, so you'll get 9V regardless of the input. It will actually pull a little less current current from the 18V because of the efficiency of the regulator.
Harry Klippton
Well-known member
Big Monk
Well-known member
Very nice.
When you say "pull a little less current", do you mean less available current for that tap(s)? Or less loss as a result of the higher voltage?
I'm thinking the minimization of I^2*R loss in my industry, i.e. sending power on 345 kV lines results in less loss as opposed to 13.2 kV, etc.
vigilante398
Authorized Vendor
Nope, the same amount of current will be available on the regulator output, but the amount of current the 9V regulator pulls from the 18V input is decreased.
V_in * I_in = V_out * I_out + efficiency loss
so
18V * Input current = 9V * output current + efficiency loss
When input voltage goes up from 15V to 18V, your regulator isn't pulling as hard on the input.
vigilante398
Authorized Vendor
Here's an example:
Let's say you have a 300mA load on your 9V regulator. The datasheet doesn't have the efficiency vs output load graph like I wanted to see, but let's assume 85% efficiency, which is a reasonable approximation for this thing.
With the stock power supply:
Vin = 15V
Vout = 9V
Iout = 0.3A
15V * I_in = 9 * 0.3 + loss
15V * I_in = 2.7 + loss
Loss with 85% efficiency of a 2.7W load is going to be 47.6mW, which gives us
15V * I_in = 3.176
I_in = 212mA
So let's look at the same thing with an 18V supply:
18 * I_in = 3.176
I_in = 176mA
Thanks for coming to my TED talk
Let's say you have a 300mA load on your 9V regulator. The datasheet doesn't have the efficiency vs output load graph like I wanted to see, but let's assume 85% efficiency, which is a reasonable approximation for this thing.
With the stock power supply:
Vin = 15V
Vout = 9V
Iout = 0.3A
15V * I_in = 9 * 0.3 + loss
15V * I_in = 2.7 + loss
Loss with 85% efficiency of a 2.7W load is going to be 47.6mW, which gives us
15V * I_in = 3.176
I_in = 212mA
So let's look at the same thing with an 18V supply:
18 * I_in = 3.176
I_in = 176mA
Thanks for coming to my TED talk
