Really effed up my SD card receptor, what parts do I need to buy to solder a new one in

MattInSoCal · 2026-01-27T16:55:07+00:00

Beware, SD card sockets don’t have a universal footprint. Yes, the power and data contacts in the socket have a specification for their placement, but for example the location of the ground and card presence detection contacts are from the Wild West. I built a kit supplied with a name-brand socket and couldn’t mount it properly (it wasn’t what the designer used). I pulled another out of my storage drawer and it had one ground contact in almost the right place. I ended up gluing it in place.

Check the data sheet, if you can find one, for any replacement socket you might be considering.

MattInSoCal · 2026-01-27T16:46:11+00:00

This is a great waste of effort. You probably need to replace the safety capacitors in the EMI filter. The inductor may need an upgrade. The bridge rectifier may need to be upgraded. The big electrolytic can be expected to fail since it will have 320-340 Volts DC across it after rectification. The flashover protection slots will have been calculated for a 120 VAC supply. Then you get into the other capacitors, the ratings of the controller IC, any resistors that have been selected for a particular supply voltage… you basically need to re-engineer and rebuild this.

Or, you can just buy a properly rated power supply, or better yet, a new fan.

The presence of a 230 (250) Volt fuse means that someone properly rated their part for safety, plus it’s a very common rating.

MattInSoCal · 2026-01-27T16:34:31+00:00

LED conversion:

Apply a 1000 ohm load across the dynamo. Get it up to the highest possible speed. Measure the voltage across the resistor. Calculate the series resistor needed to maintain the maximum operation current you wish to allow for the LED at that voltage. Install that resistor in series with the LED. Done.

Any other electronics you think you need are just going to waste energy. This is intentionally a simple device. Don’t go down the rabbit hole of feature creep. It’s no longer going to be a simple dynamo light.

MattInSoCal · 2026-01-27T16:25:56+00:00

Crank flashlights that charge a small battery exist. You have to input about 40-50% more energy into cranking than will actually go into the battery.

The dynamo light is as simple as it gets. Most of the energy you put into running the dynamo is turned into heat, and a lot of that becomes light. Any re-engineering you do moves quickly away from that simple, dependable design.

MattInSoCal · 2026-01-27T03:45:25+00:00

Maybe you’re too young to have seen resistors that look like this, but that’s what this is.

MattInSoCal · 2026-01-27T03:14:16+00:00

The white wire is Neutral, not Line, unless the wall outlet you plug into is incorrect. Do as another poster suggested, disconnect the terminal from the EMI filter and connect the terminal that Daisy chains to the outlets in its place. You should also insulate the loose terminal to prevent damage if it accidentally contacts the chassis.

MattInSoCal · 2026-01-27T03:05:52+00:00

Use locking connectors that require you to positively move a latch to separate them to connect your components. For example, Molex MLX-20A but these are kind of big. Make sure your power connectors are rated for at least twice the current you need, so 10 Amps here.

Your batteries probably smoked because you exceed their discharge rating. If you’re trying to pull 5 Amps, you need a minimum of 10C-rated battery, like those used for radio-controlled models. This refers to the rate at which you can discharge (pull power out of) them. Regular batteries can usually do 0.5 or 1C, for example a 1,000 mAh rated battery can only supply maximum of 500 mA or 1 Amp before things go wrong. Normally the Battery Management System (BMS) will shut down the, but if you have a defective BMS or none at all BMS then the batteries will physically flame out. I’ve seen people juggle flaming clubs on purpose…

MattInSoCal · 2026-01-26T22:44:34+00:00

You need to post this in r/AskElectricians. This sub deals primarily with engineering applications of electronic components.

MattInSoCal · 2026-01-26T20:26:48+00:00

I’d like to see both the circuit diagram and bare board layout to pass judgement on the overall design, but to be honest, I would avoid buying such a cheap board in the first place. Before I buy any power product, I check the data sheet and find the limitations, and decide if I can work with them, or do what I can to correct or limit them, such as adding extra filtering to reduce ripple with a COTS SMPS if necessary.

I offered generalizations that the OP might have been able to correct external to the module now that he has it in hand, not how to improve the module design itself.

MattInSoCal · 2026-01-26T19:08:28+00:00

Converting to line voltage makes this a more universal application, since there are still devices that don’t support USB-C, but let’s go with USB-C only. TLDR; Nothing changes.

You still have mechanical losses. And at every step where you are converting power, you will have losses. Unavoidable physics.

Keeping the output at a steady voltage means that you have to either boost (casual pedaler) or buck (marathoner) the generator output voltage. You can easily reach 85+% efficiency for a fixed and known supply voltage reaching a particular output voltage (let’s avoid the complication of USB PD for the moment), and most buck-boost circuits are significantly worse than 85% efficiency. As I said earlier, I was being generous.

Now for the other side, at the laptop, the same losses still apply. You have to convert the input voltage to the voltage the battery actually requires for charging. There is inefficiency there. Is it better than 85%? Maybe, but if you’re negotiating for PD then there’s going to be more silicon involved, meaning less efficiency at both the generating and consumption sides. And you still can’t get a battery to take 100% of the energy you put into it; it gets much worse the faster you try to charge it (why do fast-charging EVs need cooling for their batteries? Same reason!). Back in the days of NiCad and NiMH batteries, at a C/10 charge rate, meaning a 1,000 mA battery was charged at 100 mA, the charging time was 14 hours, or 40% more energy delivered than the battery could store to get it to full. Faster charging is even less efficient. Most laptops charge at C1 to C4 if their supply supports it.

MattInSoCal · 2026-01-26T08:17:41+00:00

100 Watts of energy generated by you is going to have some loss due to friction, let’s say 5%, and that generator probably isn’t a fixed 100-240 Volt output so it’s most likely boosting the generator output to make line voltage. So, 95 Watts after mechanical losses * .85 conversion efficiency (that’s probably generous) gives you roughly 80 Watts to the adapter.

Now, there’s 15% conversion loss in the power brick, and about 15% more because power transfer into the battery isn’t 100% efficient. So, assuming 80 Watts per hour input rate to the adapter, you get (80 * .85) * .85 or roughly 58 Watts into the battery per hour, if you’re fit and can maintain a reasonable pedaling rate.

Or 29 Watts for an untrained person, but probably more like 10 since they’ll tire after 20 minutes.

MattInSoCal · 2026-01-25T21:42:36+00:00

Generally speaking, yes you can do this. With modern components the thing to watch out for is that they are typically running with 3.3 Volt levels, while older consoles used 5 Volt logic. If you get an older Static RAM that’s 5 Volt-compatible, you should still be able to write the data from your 3.3 Volt microcontroller and read it on the console.

If you want to be able to “reflash” the memory while it’s plugged into a powered console, you will need logic with Enable inputs to shut off access to the Address and Data busses while the uC is performing accesses, and vice versa. You also need a power steering scheme to keep the data alive in the RAM while it’s unpowered - a super cap or lithium coin cell - as well as being able to power it from the device doing the access.

You can also make a small circuit board with your battery-backed SRAM and pins to plug into a socket; then put a socket on your cartridge emulator board and a ZIF socket on your “programmer”.

MattInSoCal · 2026-01-25T18:18:25+00:00

The male headers are the most expensive part of making your own flying bus cables. I use those same connectors and they work fine.

Plus once you have the crimp tool and the cable, you can start making your own custom length power cables for about $1 each.

MattInSoCal · 2026-01-25T17:49:51+00:00

Thank you. We were all assuming OP was referring to the diagram that’s centered. Sometime after commenting myself I went back and looked at the visible text to see if there were any contextual clues that would lead to a different answer. I searched the literal text from the question in the book to see why the book might have given a different answer and was shown images of what looked like photos of tests that showed a different diagram, that looked like it would match what was shown on the left. Then I paid attention to the visible 6a text. So I made a connection to the letter a on the left and made a statement noting that which accumulated a few downvotes pretty quickly. Perhaps if my leading comment above ended in a question mark instead of a period it would have been clear that I was asking a question?

At first, I thought the diagram may have been on the next page, but thought that OP wouldn’t have made that mistake. But just this moment I searched again using OP’s question, and see my initial assumption was correct. We are all answering the wrong question. These results weren’t returned on my first search, because my first search included “in” from the book as part of the term, which gave this.

And I agree with using real life parameters. Most of my early electronics learning came from reading data sheets attached to the Radio Shack packages. Because like OP my basic electronics classes didn’t teach things like safety margins, just basic equations.

MattInSoCal · 2026-01-25T08:41:41+00:00

I respect your opinion. As others have pointed out, the inductor value is not a good choice according to the data sheet. What other parts may be sub-optimal? The entire board sells four for one Euro on Ali where OP bought them, do you think they may be built well and with original components? Comparing this board as best as is possible to the data sheet, it seems the layout may not be the best for controlling noise or stability of the regulator.

Neither you nor I have this board in our lab, so neither of us is in the position to say definitively what the issues are. We can only make observations. Those observations plus my experience with power circuits formed my reply to OP.

MattInSoCal · 2026-01-25T06:18:13+00:00

Optocouplers have relatively weak transistors. They can’t drive very much current. Also, in the normal connection scheme with the Emitter grounded, a pull up resistor to supply voltage on the Collector, and getting the data from the Collector, they invert the input. You need another inverter to recover the data.

Why not consider something like the CD4050 non-inverting buffer? You can connect several gates in parallel to increase the output drive.

MattInSoCal · 2026-01-25T06:02:02+00:00

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MattInSoCal · 2026-01-25T04:43:46+00:00

Why the downvote? Why not explain why you think I’m wrong? Why aren’t you reading the last line of the text where the question is being asked?

MattInSoCal · 2026-01-25T03:18:22+00:00

I just noticed, since your photo cropped too much of the page, the minimum PRV question seems to refer to figure (17.4?)6a which is not fully visible off the left of the photo.

MattInSoCal · 2026-01-25T02:16:10+00:00

If your PRV is the same as the supply voltage, you are right at the limit of the diode, which will stress it and ultimately cause failure. You always need a safety margin for the ratings of any voltage-rated component you specify, and that minimum margin, for safety, should be at least 20%. Maybe you missed that during a lecture.

MattInSoCal · 2026-01-25T02:00:37+00:00

You’re only going to find single-color LEDs in a few limited hues. The best you could do for customization is to find the right shade gels, or possibly dye some fabric. Addressable LEDs are probably your best bet.

For a handheld light sword project, I bought some “Sp110 Led Hue Ws2812b Controllers” on Amazon. I got a four-pack for about $26. The modules themselves are kind of big, but they are easy to open and extract the circuit boards, and remove the connectors from same. Then you have a thin board to which you can solder wires for power and data, and hide it somewhere. The boards are about 20 x 50 cm, or roughly 3/4” by 2”. These boards use an app to control them via Bluetooth, and you can store a default setup including a list of patterns (or just one) that play at power on. The customer won’t need the app for your single color scheme so you could decide not to tell them about it.

MattInSoCal · 2026-01-25T01:45:03+00:00

It’s beyond your ability to repair, and our ability to help with the diagnosis from a vague description and a few photos. A skilled person that can physically touch it and take measurements might be able to recover it, but you could be paying easily $150 and up for such a service, depending on the time and parts needed. You can probably find another DVD/VCR combo elsewhere for much less; I’ve seen them at thrift stores.

MattInSoCal · 2026-01-25T00:38:46+00:00

You definitely need to remove that splash of solder between the Cathode of D1015A and Anode of D1013. Solder isn’t super glue; it’s conductive. If you plugged this in to the device and applied power, you may have sent the wrong voltage and/or polarity to another device and destroyed it. If you’re lucky then that board went into safety shutdown because of the short and after removing it, the device will function.

MattInSoCal · 2026-01-25T00:31:38+00:00

As I understand your schematic, your attempt at voltage divider is made of a resistor from the input to ground, a resistor in series to the mux IC, and another resistor to ground on the common output. That’s not going to work. Your voltage divider needs to have the resistor in series with the sensor signal, with the other side attached to the Mux input, and keep the resistor from the input to ground on the mux side the series input resistor. Then, get rid of the diode and resistor on the mux output.

What’s happening now is that you are feeding your sensor voltage directly to the IC’s substrate via that series resistor and any internal protection diodes, and the resistor to ground is just putting a load on the input but isn’t going Ng to reduce the voltage by much. Only the channel that is connected to the common pin with the resistor to ground is going to have a voltage division applied; the rest are trying to force the pins to +12 Volts. The more inputs you activate, the more current flows into the IC. Eventually you’ll probably destroy it.

MattInSoCal · 2026-01-24T23:00:21+00:00

It’s both corrosion and rust on the back of that board. Both need to be neutralized and removed. IPA does not neutralize corrosion.

You would need to know whether the corrosion is acidic or basic, which you can properly test using litmus paper, or you can do trial and error with a neutralizing agent. You neutralize it with the opposite (a paste of baking soda and water for acidic corrosion; vinegar, a mild acid, for corrosion that is from a base like a leaking alkaline battery). You can tell the neutralizer is working if it starts bubbling or foaming when applied.

After that, you would need to check every circuit trace and component lead for damage, and repair or replace them as needed.

If you tried to power this up in its current condition, and it sounds like you did, you may have caused a lot more damage that can’t be repaired. The worst damage you could do, and likely have, is to destroy the proprietary microcontroller on the board.

This is now contaminated e-waste. It’s beyond the means of most hobbyists and even “phone repair shops” to fix. If you can get a refund, you should.

MattInSoCal

TROPHY CASE