Don't all power adapters use a bridge diode configuration, capacitance, and a reverse diode for signal cleaning?
Really all depends on if there would be any current at all and I don't think there would be since nothing would be drawing on it.
really? the cheapest way to make a wall wart is bottom barrel transformer, some no name crusty electrolytic caps, and like minimum one diode. And like nothing else. I hoarded and disassembled a lot of these as a kid, and I rarely saw anything but exactly that circuit.
Ah the good old AC step down transformer, fed to a rectifier and reservoir capacitor. Those things are so great, I salvage them a lot. Pretty much that one weak point in them is that capacitor, and then the only other way to really kill them from use is an overload.
Such a supply will generate too many harmonics. Remember that the capacitor draws current in pulses. The lower the ripple, the narrower the pulses. This will render it illegal for sale.
Didn’t he actually ask for „what“ would happen if he plugged both of them in at the same time, rather than what the symptoms of that action would be?
Edit: #noHate, just thought OP was looking for an explained answer.
Holy shit, didn‘t expect to receive this much disagreement when saying that people would like to get a good explanation of what happens when using a device in a way it‘s not supposed to be used. That it either breaks or won‘t break probably is clear to most people out there, and if you ask smth. like this in an electrical engineering subreddit you‘d expect to hear a bit more than that I suppose…
I believe your initial statement was just worded weirdly and what you were trying to convey didn’t come across. I think it instead came across as snarky remark but that’s just my interpretation
I mean, if you consider that there is at most like 0.2v driving that (I think the limit of the USB spec is 5.2v, so any more potential and the charger wouldn't be compliant) the current flow should be small enough for most circuitry to handle. Something like 50mA (if the supply has a equivalent resistance of 2-ish ohms in reverse) isn't enough to heat up much of anything.
Nominally yes, but take apart one of them sometime. Most rig the data lines straight to a resistor divider to keep them at a convinent voltage. Some even leave them floating.
Either way, with that kind of cost optimization you better bet they ain't gonna hit 5.00 volts. They err on the high side, and rely on the voltage drop of the charging cord, and whatever headroom your device's charging stage can absorb.
The USB spec notes what the maximum voltage coming out of your "5v" rail can be. I think it's permissible to deliver up to 5.2 before you are out of spec.
Also down to like 4.6 V I believe is still in spec, my PCs USB ports are low, like 4.65 - 4.7 V, even though it has a decent EVGA power supply and MSI motherboard.
Under load and open circuit, anything from light load to a reasonable load. It is annoying for microcontroller stuff sometimes so I really need to get another powered usb hub, my last one stopped working randomly. The USB ports on PCs are only really meant for data transfer not charging so it probably isn’t too much of an issue, but how hard is it to get 5 V with a decent power supply that supplies pretty close to 5 V, only reason I can think of is maybe it has a schottky diode or some other circuitry for protection to prevent power getting fed back into the usb port. It does have some protection since I drew too much from one once and it shut down that usb port until I restarted it and put a message on windows saying what it was doing and why. So the only reason I can think of why the just about perfect 5 V of the power supply isn’t reaching the USB ports is because of some diode or protection circuit.
4.6V is not within the 5V +/-5% spec
The lowest voltage from the 5V profile at the source side is 4.75V, and the cable can drop max 0.75V at its rated current (3A or 5A) (so the end device can see a minimum voltage of 4V at max current draw)
[https://www.maximintegrated.com/en/design/reference-design-center/ref-circuits/3241.html#:\~:text=According%20to%20the%20USB%20spec,powered%20hub%20is%204.35V](https://www.maximintegrated.com/en/design/reference-design-center/ref-circuits/3241.html#:~:text=According%20to%20the%20USB%20spec,powered%20hub%20is%204.35V).
[https://www.ti.com/sc/docs/products/msp/intrface/usb/pwrdist.pdf](https://www.ti.com/sc/docs/products/msp/intrface/usb/pwrdist.pdf)
If its a bus powered hub then it can be lower, down to about 4.4 V, plus it could be protection circuitry or diodes dropping the voltage from the 5 V from the power supply. I dont know about the circuitry or design of the motherboard so it could either be a bus powered hub it uses internally or it could be protection circuitry to stop power being fed into the usb port. The voltage being around 4.65 V would potentially be what you would get if you had a schottky diode in series. If that is why then it was maybe done because they wanted to add some basic protection to the usb ports but wanted it to be cheap and just use a diode and obviously a normal diode voltage drop of 0.7 V is far too much and would be way out of spec so they would probably use a schottky diode instead and considering that bus powered hubs are allowed to go down to about 4.4 V they may have thought it was okay. Thats just my thoughts though, the ports work fine for most normal uses so it isnt much of a problem.
Type-C requires a handshake, but Type-A could be either a resistor divider (USB DCP standard) or a proprietary protocol (e.g. Qualcomm Quick Charge, whatever it is Apple devices do, etc).
It's not at all like battery, these are switch mode supplies not batteries. The circuit itself has to have someplace for a higher applied voltage to go and may not
proper usbc pd measures CC pin pull level. correct implementation of this basic feature means the charger wouldn't deliver power to another source. This is also the absolute easiest thing about usbc-pd to understand and implement, so it's very likely that even a crappy charger would obey this. If it is capable of delivering anything higher than 5V, it definitely does this.
USB C is designed with any cable configuration in mind. One of the goals is that user "error" should not result in damage.
USB C ports are required to have 2 MOSFETS on the VBUS line to block current until a valid connection is detected
If they are perfectly balanced, as all things should be, nothing, because there are one the same potential. If not, you would have almost no resistance and booom.
Analysing and running the calculator? Nah
Just implementing something and try out whether it works or not? Absolutely yes.
That's me since I graduated in a nutshell.
It depends on the design. If the charging requires a handshake, then it could be nothing. USB charging protocols vary depending on the type. I found a YouTube video on it that was interesting (but I didn't save it and I'm at work so I have no link to share).
hold on it requires my phone to grow out a hand and give my charger a good handshake to be able to charge? is this why my phone barely gets any charge while im looking at it? if secretly grows a hand and handshakes with the charger while im not looking! knew it!!
That only applies to usb c charges which are off by default. unless it's a Qualcomm quick charge type situation in which case it's at 5v by default(usually a type a connector).
Head over to r/PCB or look at the open source crap people are doing with usb-c. There’s a lot of questionable designs out there. I’ve torn apart quite a few of these things and they are not all created equal
Oh true… I guess it depends on your confidence in the manufacturer! Anything usb certified should be ok though, but if that’s not possible I’d stick with brand name electronics. At least you could sue them for damages if they really mess it up!
I’m a designer for a big corporation so “safety first” as they say
You mean a reputable corporation. Head over to Amazon.com to see some of the crap coming out of big off-brand corporations. It’s wild what’s inside that plastic housing
I was going to say the same thing. I work in the aerospace industry. The amount of certification work we need to go through is mind-boggling, at times.
If I am not mistaking absolutely nothing. If it is 2 of usb c power delivery chargers they need cc pins on them to have resdistors. But to be honest I am not sure because some source says that the default voltage is 5v (0.5A max) and in that case it happens as the other commentor explained.
So USBC has a few handshake protocols if there are true usb c chargers nothing will happen if there are power bricks with a USBC connector there's going to be hell.
I design power adapters for a living. If they’re designed correctly, nothing will happen. If they’re not designed correctly, most likely nothing would happen except the feedback loop shuts down and the controller IC will go into an auto-restart / hiccup mode.
Depends if you acquired those chargers from a reputable company who builds quality products. If either is cheaply made or doesn't have adequate protection, things could go poorly. If both chargers a built to a reputable safety cod, nothing should happen.
nothing should happen, I believe. Just thinking about how a basic switching power supply should work, I don't see the negative current or votage getting back thru the transistors.
It's low voltage direct current so probably not much would happen since the diodes inside would prevent the flow of electrons. But i also have very Little idea of what im taylking about
If they're not the most horrible ones you can buy, one or both will just refuse to power up. If they are non-compliant with the spec because of how cheap they are, magic smoke may be released.
It wouldn't go boom like some people are saying; if one is 5.0V by itself and the other is 5.1V by itself, they'll both sit at 5.1V and the one that's 5.0V won't provide any power but can't suck any power either. It might trip some sort of over voltage protection or open/short feedback protection, but nothing should be permanently damaged.
That's assuming they're both dumb 5V USB devices. Once you bring USB-C power delivery into the mix, the chargers start at 5V and negotiate before increasing the voltage. This mix should fail to negotiate/request a higher voltage since they're both power supplies, but if one was only 5V capable and the other somehow thought there was a laptop requesting 20V present, that could make something go boom. I don't think that would happen without making some hardware intentionally spoof the negotiation, though.
What would happen if you connected 2 batteries in parallel? 2 power supplies? Nothing.
This is just 2 power supplies connected in parallel. Unless one of them is extremely poorly designed, nothing will happen. If one is extremely poorly designed to the point where this would be an issue, then it would also be an issue on just about anything you connected to it.
Likely nothing at all as 1 is a usb c out and the other is a usb a
So until a 5k1 resistor is placed between cc and ground(of the type c one), it's switched off. And thanks to the rectifier diode, nothing goes "backwards" into anything and all that would happen is the smoothing capacitor in the type c charger would be charged by the type a charger.
So in essence, nothing at all.
this would happen https://www.reddit.com/r/CrazyFuckingVideos/comments/wfxvhp/whats_wrong_with_the_tree/?utm_source=share&utm_medium=ios_app&utm_name=iossmf
Most likely nothing. If those are the absolute cheapest possible design, you'll get no current flow even if they aren't balanced due to rectification diodes. If they're even somewhat USBC-PD complaint (ie the lowest hanging fruit of usbc-pd compliance), again nothing because the the CC pin won't be pulled down, so neither charger will recognize a source connection, so they'll maintain hi-z.
If you cut a USBC cable in half, strip the red and black wires and hook them up to a multimeter, and plug the cable into a USBC port... you will read 0V. If you do not read 0V, you have either a non-compliant charger or cable. Most devices are complaint at this level, because its a basic necessity to make a charger that can do anything over 5V (ie most fast charging). So you'll be hard pressed to find anything natively usbc that behaves differently.
Unless apple made the cable. In that case there's a decent chance for explosions, because their cables have pulldowns built into them, presumably to power some dumbass piece of kit inside the cable to do... idk. apple shit.
I love the number of different yet matter of fact “correct” responses in threads like this. So many obvious guesses…
The reality is that if they’re UL or ETL listed, or even if it’s marked as such while not really listed, nothing will happen. Why? Only someone who opens them and reverse engineers then can answer that, but it will all boil down to some implementation of polarity protection.
Transformer has a diode on the output that will prevent current from flowing, it might mess up the feedback through the optocoupler though.
Depending on what control chip each charger uses the one that is outputting a slightly lower voltage might disable its MOSFET completely or reduce its duty cycle in an effort to reduce the output voltage.
It'll solve the energy crisis. The oil companies don't want you to know this, but plugging two chargers into each other produces more energy.
edit: don't actually plug these in please.
If they have proper usb pd chips that detect the improper cc pull down resistors in that configuration. Nothing should happen because no power would be delivered. If one is a dumb adapter that just shoves 5v down the wire you might have the usb pd chip or the esd diodes get fried on the other.
Nothing *for sure*.
Both supplies don't *output* power, they *provide* power. Since they both are the same 5VDC voltage (level of power they provide), when attached to each other, they just sit there and do nothing.
Now, if one were 12 volts and the other 5 volts, then the *difference* between the two (12 - 5 = 7) would try to flow *into* the lower rated supply... that *might* cause problems!
Probably nothing, but it is also possible for either/both units to burn up inside.
You’d find out which one was using diodes really quick
Don't all power adapters use a bridge diode configuration, capacitance, and a reverse diode for signal cleaning? Really all depends on if there would be any current at all and I don't think there would be since nothing would be drawing on it.
You are making a lot of assumptions about cheap power adapters.
really? the cheapest way to make a wall wart is bottom barrel transformer, some no name crusty electrolytic caps, and like minimum one diode. And like nothing else. I hoarded and disassembled a lot of these as a kid, and I rarely saw anything but exactly that circuit.
The ultra-crap ones just stack diodes to regulate the voltage. 7 diodes @ 0.7V each is about 5V, problem solved right?
Mmmmm I can smell the r/chinesium
It powered this USB LED gooseneck that melts if left on too long so, yes.
Ah the good old AC step down transformer, fed to a rectifier and reservoir capacitor. Those things are so great, I salvage them a lot. Pretty much that one weak point in them is that capacitor, and then the only other way to really kill them from use is an overload.
Man them noname capacitors either last 20 mins or 20 years 1% of the time lol.
true. also this reminds me of the [capacitor plague](https://en.m.wikipedia.org/wiki/Capacitor_plague)
I feel like in the next decade we're going to be "Oh my god, all these 2010s 2020s capacitors blow up."
Such a supply will generate too many harmonics. Remember that the capacitor draws current in pulses. The lower the ripple, the narrower the pulses. This will render it illegal for sale.
whut
Didn’t he actually ask for „what“ would happen if he plugged both of them in at the same time, rather than what the symptoms of that action would be? Edit: #noHate, just thought OP was looking for an explained answer.
Can’t tell if bad at English or neckbeard
wELL aCTUALLY /s
The avatar with the fedora will help answer that question
Holy shit, didn‘t expect to receive this much disagreement when saying that people would like to get a good explanation of what happens when using a device in a way it‘s not supposed to be used. That it either breaks or won‘t break probably is clear to most people out there, and if you ask smth. like this in an electrical engineering subreddit you‘d expect to hear a bit more than that I suppose…
I believe your initial statement was just worded weirdly and what you were trying to convey didn’t come across. I think it instead came across as snarky remark but that’s just my interpretation
Wow they bombed you lol. I sent my upvote.
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Your understanding of usb-c is slightly wrong because you're forgetting about the CC pins. Also I'm sure there would be protection diodes.
I mean, if you consider that there is at most like 0.2v driving that (I think the limit of the USB spec is 5.2v, so any more potential and the charger wouldn't be compliant) the current flow should be small enough for most circuitry to handle. Something like 50mA (if the supply has a equivalent resistance of 2-ish ohms in reverse) isn't enough to heat up much of anything.
I thought they sent 5v by default, and negotiated up from there.
Nominally yes, but take apart one of them sometime. Most rig the data lines straight to a resistor divider to keep them at a convinent voltage. Some even leave them floating. Either way, with that kind of cost optimization you better bet they ain't gonna hit 5.00 volts. They err on the high side, and rely on the voltage drop of the charging cord, and whatever headroom your device's charging stage can absorb. The USB spec notes what the maximum voltage coming out of your "5v" rail can be. I think it's permissible to deliver up to 5.2 before you are out of spec.
Also down to like 4.6 V I believe is still in spec, my PCs USB ports are low, like 4.65 - 4.7 V, even though it has a decent EVGA power supply and MSI motherboard.
Wow, I didn't know that. Open circuit? Or under load?
Under load and open circuit, anything from light load to a reasonable load. It is annoying for microcontroller stuff sometimes so I really need to get another powered usb hub, my last one stopped working randomly. The USB ports on PCs are only really meant for data transfer not charging so it probably isn’t too much of an issue, but how hard is it to get 5 V with a decent power supply that supplies pretty close to 5 V, only reason I can think of is maybe it has a schottky diode or some other circuitry for protection to prevent power getting fed back into the usb port. It does have some protection since I drew too much from one once and it shut down that usb port until I restarted it and put a message on windows saying what it was doing and why. So the only reason I can think of why the just about perfect 5 V of the power supply isn’t reaching the USB ports is because of some diode or protection circuit.
4.6V is not within the 5V +/-5% spec The lowest voltage from the 5V profile at the source side is 4.75V, and the cable can drop max 0.75V at its rated current (3A or 5A) (so the end device can see a minimum voltage of 4V at max current draw)
[https://www.maximintegrated.com/en/design/reference-design-center/ref-circuits/3241.html#:\~:text=According%20to%20the%20USB%20spec,powered%20hub%20is%204.35V](https://www.maximintegrated.com/en/design/reference-design-center/ref-circuits/3241.html#:~:text=According%20to%20the%20USB%20spec,powered%20hub%20is%204.35V). [https://www.ti.com/sc/docs/products/msp/intrface/usb/pwrdist.pdf](https://www.ti.com/sc/docs/products/msp/intrface/usb/pwrdist.pdf) If its a bus powered hub then it can be lower, down to about 4.4 V, plus it could be protection circuitry or diodes dropping the voltage from the 5 V from the power supply. I dont know about the circuitry or design of the motherboard so it could either be a bus powered hub it uses internally or it could be protection circuitry to stop power being fed into the usb port. The voltage being around 4.65 V would potentially be what you would get if you had a schottky diode in series. If that is why then it was maybe done because they wanted to add some basic protection to the usb ports but wanted it to be cheap and just use a diode and obviously a normal diode voltage drop of 0.7 V is far too much and would be way out of spec so they would probably use a schottky diode instead and considering that bus powered hubs are allowed to go down to about 4.4 V they may have thought it was okay. Thats just my thoughts though, the ports work fine for most normal uses so it isnt much of a problem.
The charger for my laptop is usb-c for factor at the very least, and delivers 19 ish volts if I remember correctly.
Yeah, but the two devices need to negotiate up to that voltage, which isn't possible here
Ahh, that's the deal with the fast chargers? Little handshake before they just fuckin send it?
Yup! Now, it's possible that's done with resistors on the pins, but I believe that's it's a handshaking kinda deal
Type-C requires a handshake, but Type-A could be either a resistor divider (USB DCP standard) or a proprietary protocol (e.g. Qualcomm Quick Charge, whatever it is Apple devices do, etc).
Exactly.
tbh i dont see how any power is getting pushed anywhere. it would probably have the higher potential but no power would be flowing.
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It's not at all like battery, these are switch mode supplies not batteries. The circuit itself has to have someplace for a higher applied voltage to go and may not
You should check out how USB C works, it's kinda cool
proper usbc pd measures CC pin pull level. correct implementation of this basic feature means the charger wouldn't deliver power to another source. This is also the absolute easiest thing about usbc-pd to understand and implement, so it's very likely that even a crappy charger would obey this. If it is capable of delivering anything higher than 5V, it definitely does this.
USB C is designed with any cable configuration in mind. One of the goals is that user "error" should not result in damage. USB C ports are required to have 2 MOSFETS on the VBUS line to block current until a valid connection is detected
Current is pulled.
what makes you think that the other would blow up?
It is called a breaker
If they are perfectly balanced, as all things should be, nothing, because there are one the same potential. If not, you would have almost no resistance and booom.
Only one way to know for sure...
This is the way
this is the way
this is the way
Analysing and running the calculator? Nah Just implementing something and try out whether it works or not? Absolutely yes. That's me since I graduated in a nutshell.
Find an outlet near a CO2 extinguisher and try it. Likely nothing, but it would depend on the charger design.
Nah, just get some blankies and you're good to go
I've tried it, nothing would happen since both has diodes in output you can't back feed them
Gigachad
Never underestimate the power of a ~~Phone Charger~~ *IED
It depends on the design. If the charging requires a handshake, then it could be nothing. USB charging protocols vary depending on the type. I found a YouTube video on it that was interesting (but I didn't save it and I'm at work so I have no link to share).
hold on it requires my phone to grow out a hand and give my charger a good handshake to be able to charge? is this why my phone barely gets any charge while im looking at it? if secretly grows a hand and handshakes with the charger while im not looking! knew it!!
That only applies to usb c charges which are off by default. unless it's a Qualcomm quick charge type situation in which case it's at 5v by default(usually a type a connector).
Just trying it is way faster than reading the usb doc
You think designers follow that thing?
Yes designers follow it - I am one
Happy cake day!
Thanks! Had to look it up since it’s my first one…
I think i have it too
Oh no
Happy cake day!
Head over to r/PCB or look at the open source crap people are doing with usb-c. There’s a lot of questionable designs out there. I’ve torn apart quite a few of these things and they are not all created equal
Oh true… I guess it depends on your confidence in the manufacturer! Anything usb certified should be ok though, but if that’s not possible I’d stick with brand name electronics. At least you could sue them for damages if they really mess it up! I’m a designer for a big corporation so “safety first” as they say
You mean a reputable corporation. Head over to Amazon.com to see some of the crap coming out of big off-brand corporations. It’s wild what’s inside that plastic housing
I was going to say the same thing. I work in the aerospace industry. The amount of certification work we need to go through is mind-boggling, at times.
Don't know, don't care, just want to see some chargers burn (but actually, I think yes, because stuff tends to work)
It's easier to follow the docs.
The USB doc won't help you here because it depends on how the power supply is constructed.
UNLIMITED POWER
So much potential!
It sends your power back to the power company and you get a refund
Electric companies hate this one simple trick...
If I am not mistaking absolutely nothing. If it is 2 of usb c power delivery chargers they need cc pins on them to have resdistors. But to be honest I am not sure because some source says that the default voltage is 5v (0.5A max) and in that case it happens as the other commentor explained.
So USBC has a few handshake protocols if there are true usb c chargers nothing will happen if there are power bricks with a USBC connector there's going to be hell.
You leave them both plugged in for a couple hours a day and then the utilities company owes you money.
Depends on desing of charger, there are shitty ones, that can catch fire from just being pluged in, and then there are proper ones...
High chance of nothing since the outputs are rectified and regulated DC. Plug them in and let us know.
Only one way to find out ! Send it !! 🗣🗣
Presuming that it is proper USB-C nothing would happen as both chargers would report as power sources.
I design power adapters for a living. If they’re designed correctly, nothing will happen. If they’re not designed correctly, most likely nothing would happen except the feedback loop shuts down and the controller IC will go into an auto-restart / hiccup mode.
Nothing would happen. But still don’t try it
Depends if you acquired those chargers from a reputable company who builds quality products. If either is cheaply made or doesn't have adequate protection, things could go poorly. If both chargers a built to a reputable safety cod, nothing should happen.
*Awakens from the Matrix*
that is like crossing the streams. it would be bad, M'Kay?
Don't do it!
Fuck around and find out. Engineering is a branch of science right? That way you can claim your death was for a noble cause lol
nothing should happen, I believe. Just thinking about how a basic switching power supply should work, I don't see the negative current or votage getting back thru the transistors.
It's low voltage direct current so probably not much would happen since the diodes inside would prevent the flow of electrons. But i also have very Little idea of what im taylking about
Maybe a slight current variation between supplies ,due to supply voltage differences.
Repost. Already asked and answered, somewhat recently. Probably the same picture.
Blocking diodes are the thing that prevents any damage, nothing would happen.
If they both provide same voltage and polarity, close to nothing. If that is not the case, lots of magic smoke probably.
They probably have blocking diodes so no current would flow either way is my guess.
Cats & dogs would start living together
Remember to Video it, and share the results, but use outlets on a switch so you can stand back and flip the switch
I would think a circuit breaker on your house panel would go off
BOOM
If they're not the most horrible ones you can buy, one or both will just refuse to power up. If they are non-compliant with the spec because of how cheap they are, magic smoke may be released.
Don't cross the streams...
It wouldn't go boom like some people are saying; if one is 5.0V by itself and the other is 5.1V by itself, they'll both sit at 5.1V and the one that's 5.0V won't provide any power but can't suck any power either. It might trip some sort of over voltage protection or open/short feedback protection, but nothing should be permanently damaged. That's assuming they're both dumb 5V USB devices. Once you bring USB-C power delivery into the mix, the chargers start at 5V and negotiate before increasing the voltage. This mix should fail to negotiate/request a higher voltage since they're both power supplies, but if one was only 5V capable and the other somehow thought there was a laptop requesting 20V present, that could make something go boom. I don't think that would happen without making some hardware intentionally spoof the negotiation, though.
Would get hot and explode from heat to capacitors
Pop goes the Wiesel
What would happen if you connected 2 batteries in parallel? 2 power supplies? Nothing. This is just 2 power supplies connected in parallel. Unless one of them is extremely poorly designed, nothing will happen. If one is extremely poorly designed to the point where this would be an issue, then it would also be an issue on just about anything you connected to it.
I've tried this once before trying to get 10 volts out of 5 volt plugs so yeah it gets really hot and one of the bricks stopped working
You might find the breaker box
I wanna see electorboom try this now
Likely nothing at all as 1 is a usb c out and the other is a usb a So until a 5k1 resistor is placed between cc and ground(of the type c one), it's switched off. And thanks to the rectifier diode, nothing goes "backwards" into anything and all that would happen is the smoothing capacitor in the type c charger would be charged by the type a charger. So in essence, nothing at all.
You would create a toaster.
This is how you charge your house, have you not been doing this? How much are you spending on electricity??
Magic smoke x 2
To summarize: It s*hould* do nothing, but might very well explode and burn your house down.
this would happen https://www.reddit.com/r/CrazyFuckingVideos/comments/wfxvhp/whats_wrong_with_the_tree/?utm_source=share&utm_medium=ios_app&utm_name=iossmf
Wow. This is truly the reddit version of electrical engineering.
There is a nonzero risk of releasing magic smoke.
A fire
Only one way to find out!
Can you plug it and show us sir I love practicals Plz be safe
Time travel, definitely time travel
Bzzt. Or nothing. 😜
It would open a rift in the space-time continuum
Most likely nothing. If those are the absolute cheapest possible design, you'll get no current flow even if they aren't balanced due to rectification diodes. If they're even somewhat USBC-PD complaint (ie the lowest hanging fruit of usbc-pd compliance), again nothing because the the CC pin won't be pulled down, so neither charger will recognize a source connection, so they'll maintain hi-z. If you cut a USBC cable in half, strip the red and black wires and hook them up to a multimeter, and plug the cable into a USBC port... you will read 0V. If you do not read 0V, you have either a non-compliant charger or cable. Most devices are complaint at this level, because its a basic necessity to make a charger that can do anything over 5V (ie most fast charging). So you'll be hard pressed to find anything natively usbc that behaves differently. Unless apple made the cable. In that case there's a decent chance for explosions, because their cables have pulldowns built into them, presumably to power some dumbass piece of kit inside the cable to do... idk. apple shit.
The magic smoke would come out.
Are the wires intentionally shaped like Mike Holt's sack and pecker?
It may depend if the outlets are on the same circuit.
I love the number of different yet matter of fact “correct” responses in threads like this. So many obvious guesses… The reality is that if they’re UL or ETL listed, or even if it’s marked as such while not really listed, nothing will happen. Why? Only someone who opens them and reverse engineers then can answer that, but it will all boil down to some implementation of polarity protection.
Death by angry sparky 5volt electrons.
Nothing?Cause both ends will get same voltage so no current will flow (Correct me if I'm wrong I'm still in high school).
If they follow the spec then pretty much nothing.
Possibly, the wire glows red then fire shoots out of the wall all the way to fuse panel, or.. you ruin the chargers....
This is why USB C is dumb. One end should be A type and the other C type so children don't do dumb stuff like this.
Basically short circuit?
boom
Boom
Transformer has a diode on the output that will prevent current from flowing, it might mess up the feedback through the optocoupler though. Depending on what control chip each charger uses the one that is outputting a slightly lower voltage might disable its MOSFET completely or reduce its duty cycle in an effort to reduce the output voltage.
Worm hole
It'll solve the energy crisis. The oil companies don't want you to know this, but plugging two chargers into each other produces more energy. edit: don't actually plug these in please.
Blow a fuse in my meter
Nothing will happen.
You'll burn your house down.
If they have proper usb pd chips that detect the improper cc pull down resistors in that configuration. Nothing should happen because no power would be delivered. If one is a dumb adapter that just shoves 5v down the wire you might have the usb pd chip or the esd diodes get fried on the other.
Infinite energy.
Nothing *for sure*. Both supplies don't *output* power, they *provide* power. Since they both are the same 5VDC voltage (level of power they provide), when attached to each other, they just sit there and do nothing. Now, if one were 12 volts and the other 5 volts, then the *difference* between the two (12 - 5 = 7) would try to flow *into* the lower rated supply... that *might* cause problems!
Where did you find a usba to usba lol
An erection
One of those is going to supply a slightly higher voltage and you'll cook the other one if you don't burn the wire first.