5 ms·

What exactly does it kill? Does it fry the whole circuit or does it simply prevent any form of ethernet connectivity while plugged in?

by norswap 12y ago

wyager 12y ago
>What exactly does it kill? Everything. >Does it fry the whole circuit Yes.
- bitJericho 12y ago
  And if you're lucky it'll fry you too when you're handling the cable.
  - Roboprog 12y ago
    Me? Nah. BOFH says to luser "Can you hold this for a minute, here? We're looking into an intermittent connection problem. Yeah, right there, just a minute. Bzzzzzzt!"
function_seven 12y ago
It will fry the electronics. Ethernet is not intended to carry any significant current on the TX/RX pins. Applying 120 VAC to those pins will almost certainly cause the current to exceed the limits.
- oasisbob 12y ago
  It isn't strictly true that Ethernet separates data pins from power pins. Because the data transmission is differential, you can send power over the same cable. (The audio world has been using phantom power for years over XLR cables -- its not a new technique.) eg, Gigabit Ethernet has no unused pins, and can still support PoE.
  - function_seven 12y ago
    That's true, I should rephrase it as "Ethernet is not designed to carry any significant current. On any pins. At all. So don't connect your twisted-pair to your two-prong."
TimWolla 12y ago
I don't think any circuits will survive 110 / 230V.
- oofabz 12y ago
  There are circuits that can survive high voltages, but they are rare. You can buy power transistors (BJTs) which can handle 400V (VCE). The base can't take high voltages but it's simple to protect with a resistor. Here is a power supply I built that works directly with mains voltage, no transformer: https://frammish.org/realized.png https://frammish.org/realized.png https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-positive.jpg https://frammish.org/protoboard-positive.jpg
  - deutronium 12y ago
    Very interesting, do you mind explaining roughly how it works? And what the output voltage is.
    - oofabz 12y ago
      I'll try. This is a constant-current power supply, so the voltage depends on the resistance of the load. The LED on the left is the load. The circuit does not work well and I do not advise building it. It had extremely noisy output and tended to destroy the expensive LEDs I hooked up to it. The four diodes and large capacitor on the right are a full-bridge rectifier to provide 170V DC. The top left PNP transistors are a current mirror which reflects the current through the LED to the right. The 1k potentiometer on the base reduces the mirrored current in order for the logic to consume less power than the load. The top left transistor is mostly responsible for the total inefficiency of the power supply, because it adds 0.7V (VBE) on top of the voltage of the load. One of the large heatsinks is connected to it. The bottom NPN transistors are another current mirror but without a resistor. One would think you could just measure the current at the other end, so you get the opposite polarity from the start. But doing it this way makes it work properly at startup, when there is zero current. The two transistors on the right are a constant-current source where the current is defined by the two resistors. This current and the mirrored current from the LED are connected directly together so that they fight. The node they are fighting over is connected to the gate of the bottom-left MOSFET. The MOSFET is connected to the other heatsink. If the current through the LED is too low, the constant current source "wins", and pulls the gate voltage high, limited by the Zener diode to avoid overvolting the gate. If the current through the LED is too high, the mirrored current wins and pulls the gate low, shutting off current to the LED. When the current is shut off, the capacitor on the left continues feeding the LED until the current drops enough for the MOSFET to turn on again. This circuit also has a nasty tendency to get stuck in a steady state where the MOSFET is half on, causing an enormous amount of heat to be dissipated in the MOSFET and killing it. It might be possible to address this with an inductor to create oscillation and buffer current. But if I did it again, I would take another approach, and use an astable multivibrator to sample the LED current at discrete intervals instead of doing it continuously. This would allow time for the gate to settle either high or low so it doesn't get stuck in the middle.
kbrosnan 12y ago
It puts 120 AC volts into a system that is designed for 5 DC volts. Unless your ethernet port has a fuse between it and the circuitry it will burn the traces off the board, blow capacitors and other components and possibly start a fire.
asteli 12y ago
Regarding the plugging mains voltage into Ethernet /specifically/.... A good ethernet-enabled device should have isolation transformers (usually referred to as "magnetics"), either in the jack or as a separate component on the board. Long story short, it's very possible that if anything, you'd only destroy the magnetics, which while labor intensive to replace (lots of pins on a large component body!), aren't particularly expensive.
tbrownaw 12y ago
Any long-distance-capable wired networking will have electrical isolation requirements, so that connecting systems that have different ground voltages won't melt anything (this can apparently be a serious concern if you're networking outside a single building). So, the worst it should do is fry the network-facing side of the isolation part, and b0rk your network card. ...it looks like ethernet in particular mandates magnetic (rather than optical) isolation. So there's at least a possible mechanism for feeding too much AC power across the isolator, even if I'd expect the losses from trying to stuff such a low frequency (60 Hz) thru a component designed for MHz to make it rather harmless. I'd guess that feeding in much higher frequency power (say, around 1 MHz) would be much more likely to do interesting things than would the near-DC coming out of the wall.
- monochromatic 12y ago
  > So, the worst it should do is fry the network-facing side of the isolation part, and b0rk your network card. Yeah, or burn the building down.
- userbinator 12y ago
  Indeed, here is a typical schematic of an Ethernet interface: http://www.micro-examples.com/pics/094-EP2-ETHERNET-schematic.png http://www.micro-examples.com/pics/094-EP2-ETHERNET-schemati... So the part that's likely to get fried by excessive current is the isolating transformer, if mains is applied across either of the RX/TX pairs. If it's applied between one RX and one TX, then unless the insulation breaks down there is unlikely to be any current flow - and the isolation transformers are rated to 1.5kV as per the spec, so 120V or 240V won't do it. The 2kV cap for the TX ground reference also stops any current from flowing that way. I wonder if there's market for NICs with fuses or other protective circuitry on the inputs...

What exactly does it kill? Does it fry the whole circuit or does it simply prevent any form of ethernet connectivity while plugged in?

by norswap 12y ago

What exactly does it kill? Does it fry the whole circuit or does it simply prevent any form of ethernet connectivity while plugged in?

wyager 12y ago

>What exactly does it kill? Everything. >Does it fry the whole circuit Yes.

bitJericho 12y ago
And if you're lucky it'll fry you too when you're handling the cable.
- Roboprog 12y ago
  Me? Nah. BOFH says to luser "Can you hold this for a minute, here? We're looking into an intermittent connection problem. Yeah, right there, just a minute. Bzzzzzzt!"

function_seven 12y ago

It will fry the electronics. Ethernet is not intended to carry any significant current on the TX/RX pins. Applying 120 VAC to those pins will almost certainly cause the current to exceed the limits.

oasisbob 12y ago
It isn't strictly true that Ethernet separates data pins from power pins. Because the data transmission is differential, you can send power over the same cable. (The audio world has been using phantom power for years over XLR cables -- its not a new technique.) eg, Gigabit Ethernet has no unused pins, and can still support PoE.
- function_seven 12y ago
  That's true, I should rephrase it as "Ethernet is not designed to carry any significant current. On any pins. At all. So don't connect your twisted-pair to your two-prong."

TimWolla 12y ago

I don't think any circuits will survive 110 / 230V.

oofabz 12y ago
There are circuits that can survive high voltages, but they are rare. You can buy power transistors (BJTs) which can handle 400V (VCE). The base can't take high voltages but it's simple to protect with a resistor. Here is a power supply I built that works directly with mains voltage, no transformer: https://frammish.org/realized.png https://frammish.org/realized.png https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-negative.jpg https://frammish.org/protoboard-positive.jpg https://frammish.org/protoboard-positive.jpg
- deutronium 12y ago
  Very interesting, do you mind explaining roughly how it works? And what the output voltage is.
  - oofabz 12y ago
    I'll try. This is a constant-current power supply, so the voltage depends on the resistance of the load. The LED on the left is the load. The circuit does not work well and I do not advise building it. It had extremely noisy output and tended to destroy the expensive LEDs I hooked up to it. The four diodes and large capacitor on the right are a full-bridge rectifier to provide 170V DC. The top left PNP transistors are a current mirror which reflects the current through the LED to the right. The 1k potentiometer on the base reduces the mirrored current in order for the logic to consume less power than the load. The top left transistor is mostly responsible for the total inefficiency of the power supply, because it adds 0.7V (VBE) on top of the voltage of the load. One of the large heatsinks is connected to it. The bottom NPN transistors are another current mirror but without a resistor. One would think you could just measure the current at the other end, so you get the opposite polarity from the start. But doing it this way makes it work properly at startup, when there is zero current. The two transistors on the right are a constant-current source where the current is defined by the two resistors. This current and the mirrored current from the LED are connected directly together so that they fight. The node they are fighting over is connected to the gate of the bottom-left MOSFET. The MOSFET is connected to the other heatsink. If the current through the LED is too low, the constant current source "wins", and pulls the gate voltage high, limited by the Zener diode to avoid overvolting the gate. If the current through the LED is too high, the mirrored current wins and pulls the gate low, shutting off current to the LED. When the current is shut off, the capacitor on the left continues feeding the LED until the current drops enough for the MOSFET to turn on again. This circuit also has a nasty tendency to get stuck in a steady state where the MOSFET is half on, causing an enormous amount of heat to be dissipated in the MOSFET and killing it. It might be possible to address this with an inductor to create oscillation and buffer current. But if I did it again, I would take another approach, and use an astable multivibrator to sample the LED current at discrete intervals instead of doing it continuously. This would allow time for the gate to settle either high or low so it doesn't get stuck in the middle.

kbrosnan 12y ago

It puts 120 AC volts into a system that is designed for 5 DC volts. Unless your ethernet port has a fuse between it and the circuitry it will burn the traces off the board, blow capacitors and other components and possibly start a fire.

asteli 12y ago

Regarding the plugging mains voltage into Ethernet /specifically/.... A good ethernet-enabled device should have isolation transformers (usually referred to as "magnetics"), either in the jack or as a separate component on the board. Long story short, it's very possible that if anything, you'd only destroy the magnetics, which while labor intensive to replace (lots of pins on a large component body!), aren't particularly expensive.

tbrownaw 12y ago

Any long-distance-capable wired networking will have electrical isolation requirements, so that connecting systems that have different ground voltages won't melt anything (this can apparently be a serious concern if you're networking outside a single building). So, the worst it should do is fry the network-facing side of the isolation part, and b0rk your network card. ...it looks like ethernet in particular mandates magnetic (rather than optical) isolation. So there's at least a possible mechanism for feeding too much AC power across the isolator, even if I'd expect the losses from trying to stuff such a low frequency (60 Hz) thru a component designed for MHz to make it rather harmless. I'd guess that feeding in much higher frequency power (say, around 1 MHz) would be much more likely to do interesting things than would the near-DC coming out of the wall.

monochromatic 12y ago
> So, the worst it should do is fry the network-facing side of the isolation part, and b0rk your network card. Yeah, or burn the building down.
userbinator 12y ago
Indeed, here is a typical schematic of an Ethernet interface: http://www.micro-examples.com/pics/094-EP2-ETHERNET-schematic.png http://www.micro-examples.com/pics/094-EP2-ETHERNET-schemati... So the part that's likely to get fried by excessive current is the isolating transformer, if mains is applied across either of the RX/TX pairs. If it's applied between one RX and one TX, then unless the insulation breaks down there is unlikely to be any current flow - and the isolation transformers are rated to 1.5kV as per the spec, so 120V or 240V won't do it. The 2kV cap for the TX ground reference also stops any current from flowing that way. I wonder if there's market for NICs with fuses or other protective circuitry on the inputs...