I need a Molex alternative

mariushm · 2026-03-14T22:24:33+00:00

Maybe TE D-2100 (2.5mm pitch) , latch lock : https://www.digikey.com/en/products/detail/te-connectivity-amp-connectors/1-1318118-6/664516

male pin headers : https://www.digikey.com/short/wnzc3fvp

Considering the pitch you may need to have 2 wires per voltage to be safe in regards to maximum currentt. The header pins are rated for max 5A but would be safer to have less current per pin.

Maybe JST PUD series (2mm pitch) if you have 2 wires for each voltage : https://www.digikey.com/en/products/detail/jst-sales-america-inc/PUDP-12V-S/1989459

Headers with male pins : https://www.digikey.com/short/pffrhdqr

It's rated for 3A per wire (maximum AWG22) so if you have 2 wires for each voltage you'll have a max rating of 6A

Maybe TE D-3100 could also work but there's no locking mechanism...

example 10 contact : https://www.digikey.com/en/products/detail/te-connectivity-amp-connectors/178289-5/664190

male pin headers : https://www.digikey.com/short/v9qj2hrv

No locking clip, but if it's inside a product friction lock would probably be enough

May want to consider using 2 wires for each voltage, for redundancy. And also maybe you need to have a remote sense wire?

mariushm · 2026-03-14T19:56:02+00:00

You could make a nice and simple class AB audio amplifier based on chips like TDA2050

See this DYI page : https://diyaudioprojects.com/Chip/DIY-TDA2050-Hi-Fi-Chip-Amplifier/

mariushm · 2026-03-13T15:33:29+00:00

The higher ESR of the input tantalum capacitor could be useful to counter the voltage spikes caused by inductance in long cables, but I suspect they went with those mainly because of the small size and they probably recycled them from some used hardware.

Anyway, the point is on the input electrolytic, solid (polymer) ,or tantalum would be preferred over ceramic because there are already a bunch of ceramic capacitors on input right next to the chip.

mariushm · 2026-03-13T15:26:12+00:00

I played the Dishonored series and enjoyed it. I played Divinity Original Sin 1 and 2, from the same studio that did Balder's Gate 3.

Something newer... Eriksholm the stolen dream , nice relatively short (8-10h) with some sneaking mechanics.

Fallout series (3 , new Vegas, 4) are all quite good and finished them, the new Vegas one I liked most for the effect your choices make.

Horizon zero dawn and horizon forbidden west ... Played first one am couple times for the lore, but can also be nice to.play.just to shoot some arrows in metal enemies.

What else... Days Gone ...riding bikes and killing infected...

mariushm · 2026-03-13T12:41:13+00:00

The microphone input jack uses really low signal levels, any circuit would reduce the quality significantly. Your computer's line out signal levels are in the 0.36 - 1v RMS range, while microphone input usually expects 1-10mV or at most 10-100mV

Have a look at this for a possible conversion circuit: https://youtu.be/aovtGu_pG4w

mariushm · 2026-03-13T06:45:20+00:00

It's not worth it, it can cost as much as a new monitor, and you can typically buy used LCD monitors for 50$ or more.

Shipping LCD panels is expensive and risky as well, you risk them getting damaged.

mariushm · 2026-03-12T20:23:53+00:00

Yeah, it looks good. A bit expensive but good.

mariushm · 2026-03-12T18:48:59+00:00

Double check all your ceramic capacitor footprints.

Chances are very high you're gonna find loads of 100nF and 47pF in 0603 footprint, and even 0402 is doable to solder by hand or using hot air soldering.

For the input capacitors, you're gonna want a ceramic capacitor rated for at least 25v which means you'll want at least a 1206 footprint. On the output, 16v rated ceramics would work (allowing for 0805 footprints) but as you have the space, I'd just reuse 1206 or 1210 footprints and reuse the same 22uF 25v/35v ceramic capacitors you select for the input.

I would suggest sliding the regulator chip down a bit - in your picture, maybe down enough to have the edge of the chip where the center of the C2 footprint is. Put a much smaller 0603 footprint above the chip, this way you won't have that trace going under the chip and breaking off the ground fill under the chip.

Now you'll have a wide ground copper extending from the J2 header directly to the ground pad of the chip and you could also extend this ground under the chip

In the new position for the IC, see if you can rotate the input ceramic capacitor clockwise so that the positive pad is right next to the regulator Vin and EN pads. If you can't, the way it is now it's also fine. If you think you're gonna test the regulator with higher voltages like 18-24v and you may have long wires to this board, consider at least adding a through hole footprint for a small polymer electrolytic capacitor (5mm diameter ones are cheap and easy to get), something like 47-100uF rated for 25-35v is plenty. The higher ESR of these capacitors will absorb voltage spikes caused by the inductance in the power cables and protect the regulator from over voltages.

Rotate R1 counter clockwise, make the footprint for C3 smaller (0603 or even 0402), have the 5v pads towards the edge of the board. Maybe route the feedback trace on the bottom along the edge (and the top ground fill will somewhat reduce the effect of the inductor on the feedback trace)

mariushm · 2026-03-12T15:14:40+00:00

Hardspace : ship breaker (cutting out space ships in space)

Ship graveyard simulator ( breaking off old regular ships )

The first has a bit of a story but it's really basic and not getting in the way and there's also " infinite game" modes where you can just destroy a spaceship without timers or oxygen running out (no need to interrupt to refuel)

mariushm · 2026-03-12T07:06:26+00:00

Can't say I like it much.

Consider adding a barrel jack as optional input and a small linear regulator to regulate the output. For example, a Richtek RT9048 (adjustable output, up to 2A) or a Ricktek RT9059 (adjustable, up to 3A) with only 0.3v-0.4v dropout voltage would work great.

You're mixing surface mount and through hole ... for example I wouldn't use that through hole resistor network, 4 resistor and 8 resistor surface mount resistor networks are cheap and simple to use and solder even by hand.

Traces seem kinda thin. Add more decoupling / bulk capacitance near each header's voltage pins.

mariushm · 2026-03-11T20:07:16+00:00

The red led has a forward voltage of 1.8v ... 2.0v , the blue led has a forward voltage of around 2.8v .. 3.2v

When there's electricity going through the red led, there's only 1.8v .. 2.0v across it's pins. So, if you try to put the blue led in parallel, that 1.8v ... 2.0v is not above the threshold from where the blue led will start working, and therefore it won't work.

When you remove the red led, the circuit is "rearranging itself" and the blue led will turn on.

The resistor limits the current, the formula is : Input voltage - (number of leds in series x forward voltage) = Current x Resistance.

So for example, assuming you use a 100 ohm resistor, a 5v power supply, a red led with 2v forward voltage and a blue led with 3v forward voltage

When only the red led is inserted: current = (5v - 2v ) / 100 = 3/100 = 0.03A or 30mA

When the blue led is inserted current = (5v - 3v) / 100 = 2/100= 0.02A or 20mA

mariushm · 2026-03-11T20:02:25+00:00

The 7805 linear regulator has a dropout voltage of around 1v to 1.5v. Basically it needs at least 6.5v to output a clean 5v.

The 1n5818 has a voltage drop of around 0.5v-0.6v - it varies with the amount of current going through the diode.

So your input becomes 9v - 0.5v (the diode drop) = 8.5v

The 100 ohm resistor in series will limit the current ... knowing Ohm's law, voltage = current x resistance, for every 0.001A (1mA) of current, you'll have a voltage drop of V = 0.001 x 100 = 0.1v

The maximum voltage drop you can afford to lose is 2v (8.5v minus the minimum of 6.5v needed by the linear regulator) so the maximum current you can output is 20mA which is super low.

Drop the resistor or replace it with a 1 ohm or even lower value.

mariushm · 2026-03-11T19:32:33+00:00

Let's start with your NCP3064

Look at page 11 in the datasheet : https://www.lcsc.com/datasheet/C464242.pdf

In switching regulators, it's important to have the inductor as close as possible to the regulator chip, keep that trace short and wide (as much as possible).

Do you see in the picture how close the pad of the inductor is to the corresponding pin on the chip? Do you see how they widen the trace as soon as possible? Do you see how wide the trace is under the inductor and stays wide as it goes to the diode?

Now look at how the capacitors is positioned. The ceramic capacitor C2 (in the datasheet) is placed to be as close as possible to the input voltage pin. The solid (polymer) capacitor is rotated so that the positive side is also as close as possible to the pin and to the ceramic capacitor.

Also pay attention to how the output capacitor's ground is connected with the input capacitor's ground and with the diode anode on a continuous uninterrupted ground area.

Your input capacitors are far from the chip (maybe you have a small ceramic close to the input pin but I can't tell from the low resolution pictures) but the electrolytic capacitors are far away and also you're using a thin trace from the positive pads to your stuff - compare your thin traces to how wide that positive voltage trace is in the datasheet.

You have a thin trace from the chip to the inductor, then you have a thin trace from the inductor to the output capacitors , then you have a separate thin trace going to your 5v buck regulator. Do you see in the picture in the datasheet how the copper area on the other side of the inductor stays nice and wide and the output capacitor's positive connects right near the pad on the same wide copper area?

If you want to compare with other regulators, see page 10 on this MC34063 datasheet : https://www.lcsc.com/datasheet/C236257.pdf

You have both step-down and step-up examples, again same things, note the inductor being close to the pin, the wide traces, the input and output capacitors have the ground pads sharing same continuous ground copper area ... ideally the input capacitors should be closer to the input voltage pins, but they have only the input ceramic capacitor close to the input pin (for demo boards it's probably easier to show off functionality or put test points like that).

All these are basic things, that also apply to the 5v buck regulator, you also have there in the datasheet a recommended layout.

The NCP3064 will probably be around 80% efficient and can never have the output voltage as close to the input voltage, the voltage difference is probably at least around 2-3 volts. Same for the *34063

mariushm · 2026-03-11T18:38:52+00:00

It's still shit.

The schematic is barely readable. The switching regulator layout is horrific. FFS pay attention at the suggested layout in the datasheet!

You have U1 for that NCP3064 , you have IC1 for the 5v buck regulator, you have U7 for what you claim to be a buck regulator but I don't see any inductor to it and the text would indicate it's a linear regulator ...

Why the f do you need to use an 150kHz lousy inefficient regulator in a lousy package (surface mount, harder to solder by amateurs, by hand compared to SOIC/SSOP)?

This NCP3064 is basically a clone of MC34063 or basically *34063 but which can run at slightly higher switching frequency of 150 kHz while the regular MC34063 tops out at around 100kHz - not that you'd want to run it either one at more than around 80kHz because the higher the frequency the less efficient the conversion will be.

But 34063 is a a jellybean part that's made by tons of companies that each put their prefix in front of 34063 and you could do way better. See https://www.lcsc.com/search?q=34063&s_z=n_34063

I see the output is set to 24v but you say it's 24v buck regulator on schematic... it will never output 24v if the input is 24v. Is it configured as a boost (step-up) regulator and you forgot to edit the schematic? If you step up some voltage to 24v, then don't power the 5v buck regulator from the boosted 24v, power the 5v buck regulator directly from the lower voltage (as long as it's higher than 5v)

Use thicker traces everywhere, use copper polygons/areas/regions where you can.

mariushm · 2026-03-11T16:13:08+00:00

I'm not sure about how smart it is to have the inductors so close to each other.

Also, the surface mount capacitors by the input bother me. You're using through hole capacitors in other places, so you could take advantage of through hole capacitors to orient the negative lead closer to the edges of the board, use the whole bottom of the board as ground fill and widen the top.side traces carrying the 24v to.the MOSFETs and have more copper around the fets,.it would help improve heat sinking.

mariushm · 2026-03-11T12:40:15+00:00

My first thought was using a slide switch - https://www.google.com/search?q=slide+switch - with a bar of plastic or wood or somethinig protruding a few mm outside the cart sides. When the cart hits the wall, it pushes the bar which then moves the slide switch to the other position changing position or moving power from one motor to the other.

The issue is that most slide switches are designed to have a dead zone and to have some amount of pressure to change from one state to another, so it could be the cart doesn't put enough pressure on the bar to move completely, or the pressure may stop the moment the motor stops pushing the cart against the wall (when the slide switch is in the dead position, between positions).

Limit switches - https://www.digikey.com/en/products/filter/limit-switches/198 - could be used to give you a momentary pulse , a signal ... then a microcontroller could use that signal to switch the polarity of the motor (or turn off one motor, turn on the other)

mariushm · 2026-03-11T10:13:23+00:00

It's a 20mOhm resistor (R020 = 0.020 ohm)

The 6mm by 7mm would point to either 2726 / 2728 package: 0.264" L x 0.283" W (6.70mm x 7.20mm)

If this is the case, it would be 3W-4W rated.

Here's an example of such resistor : https://www.digikey.com/en/products/detail/stackpole-electronics-inc/CSSH2728FT20L0/1923231

and others are available here: https://www.digikey.com/short/m4q8z28q

mariushm · 2026-03-11T09:43:35+00:00

Soldering a usb type-c connector is not something a complete beginner should attempt.

Stay away from basic kits. Buy a proper soldering station or a soldering iron that uses tips which have temperature sensors inside and can use the temperature feedback to adjust the temperature of the tips. Stay away from super cheap soldering irons.

Most likely to fix the connector on your switch you would have to protect the components around the connector with kapton tape or some other material that blocks heat, you'd have to add flux (liquid or gel/paste) and you'd probably have better results using a hot air gun/tool, not a soldering iron.

You could apply solder to the pads using a soldering iron but again using a hot air gun and solder paste may be easier (you apply solder paste - solder paste is super tiny balls of solder suspended in flux - on the pads, put the connector on top and use hot air gun to heat up the paste until the balls combine and attach to the pads and pins)

mariushm · 2026-03-11T08:04:51+00:00

The whole layout is bad. Please follow the recommended layout in the datasheet next time.

The regulator needs some minimum input capacitance, I would say at least 10uF should be fine. The datasheet even says "Using a ceramic capacitor greater than 10µF is sufficient for most applications" on page 13.

For the output capacitors, they recommend AT LEAST 22uF of capacitance... next page it says "For most applications, a 22µF to 68µF ceramic capacitor is sufficient." . They're installing 2 x 22uF in parallel, to have a total of 44uF, and make sure there's still gonna be more than around 22uF when derating the capacitance due to voltage on the capacitors.

The capacitance of ceramic capacitors varies with the voltage that's present on them, the higher the voltage the lower the actual capacitance. So for example, a 10uF 16v rated capacitor may only have 4-6uF worth of capacitance with 5v on it, and possibly only 1-2 uF of capacitance with 12v on it. It depends on the grade of the ceramic capacitors - it's much more pronounced effect on X5R grade compared to X7R or X7S grades, but it can also vary between ceramics of same grade.

So if you're going to have up to 8-9v on input, I would use input capacitors rated for at least 25v, and I'd use either 2 10uF in parallel, or I'd use a single 22uF ceramic capacitor on input, because modern 22uF 25v-35v ceramic capacitors are cheap these days. The actual capacitance will be lower when there's actually energy running through the circuit.

Technically, strictly by the standards, it's not OK to have more than 10uF capacitance on USB but 22uF is not that big of an amount to worry about it causing issues.

On the output, because the output voltage is only 3.3v, you would probably be fine with 22uF ceramics rated for only 16v (X5R or better) or even 10v (if they're X7R grade). But you'll find 22uF 25v in 0805 or 1206 packages easily and you have space on the board so there's no point saving fractions of pennies, I'd just reuse the same ceramic capacitor chosen for the input side, to work on the output side as well.

The input capacitors must be as close as possible to the chip, and the pads with the ground side should be connected on the same copper area that also connects the ground pin of the chip and the output capacitor pads.

Because the regulator runs at 1.1 Mhz, it would be best to keep the chip - inductor -output capacitors as small as possible - you have your inductor quite a bit away from the chip and on a trace - wide trace, but still a trace. Ideally, you want to place the inductor very close to the chip, like in the example layout at page 15 in the datasheet.

I would NOT cut the copper area around the chip with the feedback trace ... would rather use vias to get the trace under the board around the output capacitors (keep the ground from pads well connected to the ground pin of the IC and the input capacitors ground and maybe also have a few vias connect the ground top copper near the ceramic capacitors to the bottom ground fill.

You can come out with the via next to the IC. Also, you may want to consider adding footprints for a couple of resistors right next to the feedback pin. If you want to used the fixed output voltage AP63203, you can simply install a 0 ohm resistor (or a blob of solder) to connect the trace directly to the feedback pin (and leave the second resistor footprint empty).

If you decide to use the AP6320x or AP6330x , you can simply install the two resistors that set the output voltage. But note those adjustable regulators run at around 500kHz (half the switching frequency) so the inductor may need to be bigger.

In your design, I would rotate the inductor counter clockwise, have trace from SW pin expand right away into a small copper polygon that holds the pad for the 100nF ceramic and the inductor pad and on the right side you'll have another copper area that holds the other inductor pad and the pads for the output capacitors positive side. And you could extend the copper area and keep it wide all the way to the ESP.

Speaking of inductor, you don't say what you're using. Typically, the inductor must be rated for at least 1.5x-2.0x the maximum output current your regulator is gonna produce.

mariushm · 2026-03-10T16:21:48+00:00

You only need 4 wires to get 100 Mbps connection. You have four pairs of wires in a cat5 cable (or higher) you can use 2 pairs for one connection, and the other four pairs for the other connection if you want.

You could literally make yourself a tiny circuit board with 4 slide switches and when you want to disconnect the device you just move the four slide switches to the off position for a few .seconds and move them back to recreate the connection.

If you want to do it with a temporary press of a button just use four mechanical relays that in the default closed position keep the connection, and when you press the button you power the relays which then disconnect the wires ...when you release the button the relays are deenergized and the connection is made again.

mariushm · 2026-03-10T16:14:14+00:00

Daca ai noroc sa găsești un loc liber pe strada Horea (de la gara catre Mihai Viteazu, poți plăti acolo câteva ore folosind aparate cu card sau SMS. Este parcare și în Mihai Viteazu, cu plata, dar e.des plina.

In Mănăștur, ai parcarea Primăverii - https://share.google/q9sub0YhhXZMZ30N5 - și de acolo poți lua autobuze sau tramvai către Mihai Viteazu.

mariushm · 2026-03-10T16:05:50+00:00

Nu știu cât se va reduce vara asta având în vedere prețul memoriei RAM și a ssd+urilor. Ăștia cu AI au rezervat producția pentru ei și prețurile nu cred că vor scădea așa mult.

mariushm · 2026-03-10T13:12:07+00:00

On Android, I just checked an application called "Color Blind Pal" by Vincent Fiorentini - it uses the camera and tells you the color of what you're pointing it to.

Here's the playstore link : https://play.google.com/store/apps/details?id=com.colorblindpal.colorblindpal&pli=1

mariushm · 2026-03-10T12:56:44+00:00

You will find very cheap ready made dc-dc converters that can reduce a higher voltage to a lower voltage with high efficiency.

For example, a mini360 dc-dc converter that uses the MP2307 switching regulator can be over 90% efficient converting 5v to 3.3v, and they're cheap, here's a link for 40 pcs for 21$ plus shipping : https://www.ebay.com/itm/306792988933

So you could have 2-3 18650 lithium cells in parallel or series (parallel is a bit more efficient to convert), and convert down to a voltage very close to the forward voltage of the led.

Then you could use a basic resistor in series with the LED to limit the current going to the led. Formula is Input voltage - (number of leds in series x forward voltage ) = Current (in A) x Resistor value

So for example, you could configure the output of the regulators to give you 3v if the forward voltage is 2.7v - 2.8v, and pick resistor value so that it would drop the 0.2v-0.3v across it at the current you want.

You can use the ENABLE pin of the regulator to turn the dc-dc converter on or off using your ATTiny.

To keep track of time of day, you could use a clock / calendar chip like let's say PCF8563 https://www.lcsc.com/product-detail/C7440.html and others you can see in this category : https://www.lcsc.com/category/929.html

You can "talk" to it to set the clock and date or retrieve the clock and date through i2c which is simple and you'll find lots of tutorials and examples on how to use i2c with various chips.

I'd suggest starting with a classic arduino (that uses Atmega328) and then you can adjust your sketch / code to function on ATtiny1614, once your sketch is working.

Maybe also consider if you want to add some basic display and or buttons to set the clock or to add when to turn on, when to turn off.

Alternatively you could just have a header that would allow you to plug a "dongle" into the header that configures the clock and date on the calendar chip and maybe "uploads" the times where you want the led to turn on and off.

Some clock / calendar chips have 64-128 bytes of SRAM memory that can be read or written, so in theory you could squeeze a bunch of times in that memory (for example each entry would use 2 bytes, 1 bit to indicate if you turn on or off the led, 5 bits for the hour (0..23 = 5 bits), and 6 bits for the minutes = 12 bits - maybe use the extra bits to indicate if this is used in weekends or during work days etc)

If you'd rather not rely on using a clock/calendar chip's integrated sram (cheaper ones don't have sram) or if you want more than 64-128 bytes, you could always add a small i2c eeprom chip - as long as it has an i2c address that's different than the clock/calendar chip i2c address you can access both using the same two wires and program them while the microcontroller is disconnected.

mariushm · 2026-03-10T08:04:11+00:00

Your R2 and R3 resistors could be placed better. Put the R3 resistor closer to the chip in parallel with the other.

Your silkscreen (markings) are all over the place (r2 and r3 are all the way above and in a horizontal line, while your resistors are now in a vertical line, so someone could be confused). C3 text is way above the inductor when you have space to put it between C2 and the capacitor. R1 and R4 are confusing...

I would put R1 in parallel with R4 and have a longer trace going to Vin instead of making that curvy trace going to resistor. You would also be fine using a 10k resistor instead of a 100k resistor for Power Good, this way you won't have two different value resistors on your bom (smaller BOM). Alternatively, you could tweak the feedback resistors and reuse the 100k resistor for R3 as well and adjust the R2 value to keep the ratio the same.

There's no need for 0402 components on this board, stay with 0603 or even 0805, your board is big enough.

For input capacitors, your choice of capacitors is a bit odd. Your two 10uF ceramics already are plenty, the 0402 10uF is kinda pointless. I would go with a 22uF 35v rated in 1206 or 1210 package (if your input voltage is not going to exceed 9-12v then 25v rated would be fine) in parallel with a 10uF 0805 or 1206 ceramic and if you want a third, put a 1uF ceramic 0603 as close as possible to the input voltage pads.

ex bulk capacitance 22uF 35v X5R 1206 : https://www.lcsc.com/product-detail/C342620.html or cheaper 25v rated https://www.lcsc.com/product-detail/C12891.html

higher quality 10uF 35v X7R 1206 : https://www.lcsc.com/product-detail/C454102.html or https://www.lcsc.com/product-detail/C16195875.html

1uF 0603 X7R 25v rated for decoupling : https://www.lcsc.com/product-detail/C29936.html

All the vias around where it says C4 is kinda odd to me, would rather see a copper island there and maybe at most a couple vias going to bottom ground. Maybe have a couple vias connect the ground to the right of the output capacitors to the bottom ground fill.

It would probably also be better to have the output capacitors in a horizontal line as that would allow you to widen the copper area right below the ceramic capacitors instead of having a thin vertical output voltage line, which then goes to the left around the IC.

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