37

Below is the pinout for the receptacle: GND TX1+ TX1- Vbus CC1 D+ D- SBU1 Vbus RX2- RX2+ GND | | | | | | | | | | | | =+====+====+====+====+====+====+====+====+====+====+====+= | | | | | | | | | | | | GND RX1+ RX1- Vbus SBU2 D- D+ CC2 Vbus TX2- TX2+ GND You will note that all ...


28

No, the voltage does not distribute equally. The reverse leakage current for diodes is not a carefully controlled parameter, and can vary substantially from unit to unit, even from the same manufacturing batch. When placed in series, the diodes with the lowest leakage current will have the highest voltage across them, which will cause them to fail, which in ...


14

There are many types and subtypes of electrolytics, I will concentrate on those most common for hobbyists: the liquid electrolyte aluminum one. Most details can be read in the corresponding wikipedia article but in short: The cathode oxide layer can withstand up to 1.5V of reverse voltage it can receive permanent damage earlier, but 0.5V is deemed safe ...


11

and 2. As I recall from some professional audio gear manual (made in 1980s) that I had to mess with, miswiring +/- at the speakers is the common reason for big "trenches" in the spectrum that are not correctable by the equalizer. The reason is, the speakers' ranges overlap somewhat and in the overlapping band they may pretty much fight each other. ...


11

If you understand the body diode then you should see that if you apply significant negative voltage to the input the body diode needs to block or it will pass the incorrect polarity to the output, probably destroying the circuitry it is connected to. In normal operation (positive input) the body diode is shunted by the Rds(on) channel resistance of the ...


10

Given that the pins are open drain, the following circuit should work quite nicely: simulate this circuit – Schematic created using CircuitLab You'll get some wasted power as one of the resistors will be pulled straight to ground when an LED is turned on, however if you go for a low-current LED, you can increase the resistance to reduce the amount of ...


10

The biggest practical difference is that a Zener* diode is made to break down in a useful and predictable way, while a rectifying diode is made to predictably not break down. So if you buy a Zener diode that's designed to regulate to 12V, then you can figure that if you arrange for it to get the right current in the reverse direction, you'll get pretty close*...


9

The LED will definitely pass current in the reverse direction — nearly as much as in the forward direction. (Think of it as a 5-volt Zener diode.) This much current (and the heat that is produced) will definitely damage it, although probably not catastrophically. The simplest solution is to put an ordinary diode in parallel with the LED, pointing the ...


9

The MOSfet is not configured for reverse polarity protection. It is arranged so that it will turn OFF on over voltage. So it is over voltage protection. At voltage below 16V the MOSFET will be turned on by bias through R4. When the input voltage goes over ~17V the transistor Q2 turns on by current through the zener diode D3. Q3 takes away the bias voltage ...


9

MOSFETs have a neat party trick: they can conduct in both directions due to their symmetrical construction. 3-terminal FETs also have an issue: the body diode will conduct when the FET is connected and biased in ‘reverse’ (Vds negative for a pFET), even if the FET is ‘off’. The body diode is a side-effect of the internal source-substrate connection. At any ...


8

It should work. Here's another diagram of this approach. (fig.5 from Maxim App Note 636) When the battery has a proper polarity (as shown in the diagram): The body diode of the MOSFET is forward biased. VGS = -Vbatt+Vdiode < 0V, and the P-ch MOSFET is turned on. It is necessary for the body diode to be forward biased with battery in proper ...


7

Use a reasonably powerful audio amplifier, feed it with 20 Hz from a signal generator or something, and connect the magnet where the speaker would go. If this blows the output fuse of the amplifier or it catches fire or something, try again with more turns of a thinner wire when making your electromagnet.


7

Alas, no. If you give it a voltage under -0.3V it will probably be toast. So yes, you need additional protection against reverse polarity.


7

Did you see (and understand) Spehro Pefhany's comment? I take the liberty of turning it into an answer because that's also what I'd suggest. Add a diode in front of your circuit that shorts the voltage if the battery is reversed (=V2). In that case the (resettable) fuse will break the current flow. If the battery is not reversed (=V1) there is almost no ...


7

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow. Not as much as normal, but some will still flow. And that means, if the lamp is turned on, dead ...


6

You only need a reverse polarity protection for your device if there is a possibility that the power input can get applied in the wrong direction. You would have to ask yourself how likely it would be that a reverse polarity condition could happen and then decide from that whether you need protection for that scenario. Most simple reverse polarity ...


6

Something like this should work. Note that the limit switches are different- one is normally closed and the other is normally open. If you have a relay with three form C contacts you can avoid using the diodes. simulate this circuit – Schematic created using CircuitLab The two contacts are used to reverse the motor in the usual way, but one ...


6

Here are the (simplified) rules by which a N-channel MOSFET obeys: If the voltage between gate and source is grater than the threshold, the current can flow between source and drain (in either direction). If the drain voltage is lower than the source voltage, the current flows from source to drain, even if the gate is not triggered (due to the body diode - ...


6

If one battery has less capacity than the others then it will drop to zero faster and may even reverse since the current through the load is going through the battery in reverse. In other words, it's possible that two good batteries have reverse-charged a weak one. This would be particularly likely if Mom replaced two of the three batteries and left an old ...


6

It depends on your application. The main issue with low-side protection is that you are disconnecting your ground reference. Many different systems work on the assumption that the 0V/Ground/Earth is shared between the devices. There can be many obvious and hidden ground connections. If by way of example you have a circuit that is connected to ground by ...


6

Tweeters can be damaged by low frequency signals, so a bypass capacitor is often used in series to act as a high-pass filter. Speaker drivers are really nothing more than an inductor placed next to a fixed/permanent magnet, so they don't have polarity in the same way a diode might. Given a signal, the magnetic field causes the driver to physically move. In ...


5

If you're making an avalanche relaxation oscillator, most transistors are not really designed or characterized for this so it should not be an enormous surprise that one maker's part will behave differently from another, or even other parts with the same part number from the same source. simulate this circuit – Schematic created using CircuitLab


5

EDIT: You have changed the schematic. Yes, in your schematic, at any point, out of the box or "aged" you can potentially be shorting the battery at every single switch over. It may be unlikely if the relays are exactly the same, but it is a very real risk, at any point in time. This is why the schematic you posted earlier, with the relays connected ...


5

I'm betting on this as the reason: In an industrial environment you will want the design to be somewhat ruggedized, and that's what this circuit does. It protects the input from Reverse-voltage (via BYM10-1000) Overvoltage (via 56\$\Omega\$ and Z2SMB36) High-voltage transients (via 56\$\Omega\$ and 2.2\$\mu\$F)


5

The p-mosfet trick won't work as both sides supply voltage. You could use a latching circuit (P-mosfet + npn bjt) but it would only work for the first battery you charge, unless you power cycle the circuit every time you change the battery. A further improvement would be to use a microcontroller to check the battery status before turning on a MOSFET, but at ...


5

Here's the relevant part of the system diagram from the datasheet you linked: A Schottky diode typically has a forward voltage drop of 0.2 to 0.3 V. So if 5 V is applied at "V+", then only about 4.8 V is available to supply the Richtek switching converter or any other circuits that might be connected to that same circuit node. The reason it's called a "...


5

Here is what the datasheet says in the ABSOLUTE MAXIMUM section about reverse voltage: This is completely clear and unambiguous. If you are still confused, look up "absolute maximum".


5

You're about 1/5th the way from the center on the left of the graph. You'll get a marginal reverse current, but it's so little it wouldn't matter to much electronics. Your battery would drain ever so slowly. Internal battery drain current would probably be higher. So basically your circuit does nothing.


5

Your circuit has issues and I would not try to "patch" these as there are simpler, more reliable solutions! I would use a poly fuse which is a self-recovering fuse. After it blows it will recover (become "good" again) after some time by itself. Of course you could also just use an ordinary glass fuse but make sure that the user can replace it without ...


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