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33

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 ...


22

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 ...


12

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 ...


9

A -0.3 volts will not hurt any capacitor, even one rated for 5 VDC. Also Schottky diodes are available in very high current modules if needed. A negative 3 volts may hurt them by confusing the chemicals in the dielectric, much the same as reverse charging a battery. But there the similarity ends. Capacitors can be drained to zero volts for a long time, yet ...


9

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 ...


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

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

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 ...


6

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.


6

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 ...


6

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 ...


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

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

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

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 - ...


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 ...


5

The polarity protection works correctly as explained in Mosfet in reverse polarity protection. The rest is the Typical Applications given by Microchip in the MCP16301/H datasheet. So, I don't see any issues there. I don't know if you have considered the inrush current when applying 30V while C2 initially forms a short: it should not exceed max Pulsed Body-...


4

Electrolytic capacitors can withstand for short instants a reverse voltage for a limited number of cycles. In detail, aluminum electrolytic capacitors with non-solid electrolyte can withstand a reverse voltage of about 1 V to 1.5 V. Solid tantalum capacitors can also withstand reverse voltages for short periods. The most common guidelines for tantalum ...


4

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 the ...


4

What you didn't see is that with 20V applied minus the volt drop of the 1N5819 (leaving maybe 19.3 volts), there will be 19.3 volts applied across two series zeners that don't want a terminal voltage greater than 16.6 volts. This will equal smoke. The other thing is the an LED might have a forward volt drop of 1.5 volts before it starts glowing and, with ...


4

You'll want a FET with low Rdson at the 2.7-3V Vgs so that you'll have minimal power losses when the battery is correctly inserted. The appnote you link to actually has some useful examples in table 1, e.g. it gives ILRML6401 as having 85 mΩ at 2.7V. That is worse than what the mfg datasheet of that FET promises (in the graphs below from the datasheet), but ...


4

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 ...


4

Sounds to me like you need an 'ideal' bridge rectifier. Fortunately, these are available from Linear Technology. Check out LT4320: http://www.linear.com/product/LT4320


4

There are various solutions to this problem. They range from diy circuits to IC's built for this specific purpose. The most basic circuit uses series diodes with every supply rail. This does mean you lose power in the diodes as they have a voltage drop. It might not be a problem for you if you are dealing with very low-power devices that draw almost no ...


4

For such a low allowable drop, you need a zero voltage drop diode. It can be done with a P-Channel enhancement mode MOSFET. E.g. a IRFD9120 is sufficient for up to 1A drain current. (Use a multimeter and check how much current your camera draws. Check it with display on and zoom moving.) simulate this circuit – Schematic created using CircuitLab


4

You can treat the voltage as a height. https://www.youtube.com/watch?v=ZSrzvoJcaJ8&g To measure the voltage we need two points in the space. One of this point is treated as a reference point. We have a very similar situation when we try to measure a height of an object. We need a reference point. The most common reference point is "above mean sea level"...


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