# Tag Info

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First you need to find a relay that rated 24 V/80 A (or higher current) and then wiring should be like this: simulate this circuit – Schematic created using CircuitLab When V1 is connected it will energize the relay and your "equipment" get its power from V1 and when disconnected relay will switch to V2. You probably have some capacitors at the ...

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If you have a common ground, you could put the switch between the battery grounds and the breadboard grounds. If not, you probably need a 3-pole switch.

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A Christmas string LED is not the best choice. Digikey has thousands of different LEDs and switches. Learn to use the filters. Avoid surface mount parts (harder for a beginner). For LEDs, you probably want through hole parts. A common size is about 0.2" dia. For the switch, you probably want slide, panel mount. Here is one possibility: https://www.digikey....

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If you are worried about shocks, then you need a GFCI protecting the circuit. A fuse is no guarantee of protection as 30mA can kill a human, and you won't get a fuse to blow at that current. But the main thing is to design the enclosure properly, and use appropriately rated switches for the voltage. If the switches aren't fully insulated, then ground the ...

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Well if you are lucky and what your switch is doing is grounding a pull-up resistor, you can get by with connecting the high V side to ground with your cable. If it's completing the circuit I don't know if you can do much

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the threshold voltage of the LED can be exploited to cause the middle led to require both supplies to be on. simulate this circuit – Schematic created using CircuitLab

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I think the capacitor inside the power adaptor has some contributions on delay of switching. Try modifying the adaptor either by removing or replacing the capacitor with lower uf.

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Use an AND gate! simulate this circuit – Schematic created using CircuitLab

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This will do what you are asking for: simulate this circuit – Schematic created using CircuitLab

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simulate this circuit – Schematic created using CircuitLab

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See the green box and the purple box below: - When the start button activates, the item I have labeled "green M" is a coil that activates a contact (purple M) and so latches the circuit because the start button switch can be released and "purple M" keeps those nodes shorted.

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This switch has a combination of tabs for a crimp connector, as well as square pins. Using a "crimp connector" is a three step operation. First, strip off about 5-10mm or 1/4inch of the insulation from the end of the wire. Second, insert the exposed wire into the end of the "crimp connector", just in the half covered by the plastic sleeve. Inside, the ...

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Why not crimp it? Do you have a crimping tool? It can be done perfekly if you have one. Else, you could still do it with a pair of nose pliers. Advantage is that you can remove the connections later. Why not solder it You cannot remove the connections easily (desoldering will do). You can put in a sleeve into the cable, solder and then secure the ...

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There are a couple of potential harms: The inrush to the capacitor when the PMOSFET is turned ON. Some calculation is needed to be sure that the peak current is not too high. I guess that's no problem, but it's better to be sure. Does the power-on reset circuit work if there's some voltage left? A resistor could discharge it surely. I do not know if any ...

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You need to provide voltage to turn on the LEDs. You need to have a series resistor to control the current through the LEDs. Get three 1.5V cells in series to get 4.5 V. Have a series resistor of atleast 100 ohms. Here is the way to identify the LED positive and negative. Here is the required minimum voltage to turn on the LED. Remember to put resistor ...

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A LED will light up at its forward voltage, which is 2.1-3V depending on its color. Also keep in mind that LEDs are sensitive to polarity: simply swapping the two pins of the LED in your circuit may work. However, please consider some generic comments: you won't be able to light up the 120V night light when connecting batteries to its plug, since it needs ...

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A typical motor control switch/relay circuit: simulate this circuit – Schematic created using CircuitLab

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If the switches flip simultaneously, you've got two 1 pF caps in series. Calculate that capacitance. Then you have 500 fF in parallel with that total capacitance. Again calculate the total capacitance. Since you're not given the inductance or length of the connecting wires, you must assume it to be negligible. No calculus needed. Simply consider the ...

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Although a very "academic" exercise, assuming "1 μΩ" SPDT switches of "0 pF" switching in zero time... Assuming a clock at 5e8 rate driving both switches at 50% duty cycle; from 0 to 1ns $dV=\dfrac{Ic*dt}{C_1} = \dfrac{1mA*1ns}{1pF}=10V$ and at 1ns C1=1pF,10V switches in series with C2=1pF,0V after t=1ns switch {C1+C2=0.5pF,10V}//{C2=0.5pF,...

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The MUX chip must have larger supply voltages than the signal voltages. If you route RS232 with the RS232 MUX which is about +/- 12V signal to any of the nine output connectors, there it will get connected to CAN MUX. Therefore the CAN MUX must also have +/- 12V supplies so it does not clamp the signal. You also can't connect CAN and RS232 simultaneously to ...

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If the required signal rate is slow slow enough, use the 1mA to charge a capacitor and when needed dump the capacitor through a latching relay's appropriate coil.

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If you are able to convert this 10V/1mA (1) to a bit more mA @ 1V or 2V you may want to consider using a "photovoltaic relay" or "solid state mosfet relay": where pins 3 and 4 are the isolated terminals A and B. You should check the maximum required "trigger forward current" or "the minimum control current" or whatever description that tells you which ...

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Something like this should work. The standby current draw is less than a microamp. Depending on the battery size, the battery might last almost as long as without a load. "Jonk" loves these kinds of problems, he will likely have something better. simulate this circuit – Schematic created using CircuitLab

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The crosstalk, even in a 50 ohm system, may be too high for a SPDT in one IC. I suggest you implement "T" switches: two SPST in series, with the center_node either "shorted to Ground" or floating.

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The lever switches the contacts as a 2P2T but the stationary contacts are dark and not visible but likely oxidized from arc heat. The typical universal brake motor has high current from BEMF. Sometimes repairable surface if sufficient material exists. It may be an alloy of copper. Better to replace worn parts.

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As noted by others you should try changing the brushes. I had a washing machine with a universal motor stop working. (No notice, just went from working normally to stopped mid-wash.) The brush and commutator surfaces showed no signs of damage. The only visual indication was that the brushes were much shorter than a new pair. Fitting new brushes restored ...

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You need the transistor to be a saturated switch. The datasheet for your transistor and nearly every other transistor shows that its base current should be 1/10th its collector current for it to saturate well. Then the base current must be 300mA/10= 30mA. Can the ESP01 output 30mA at a reasonable voltage? The saturation voltage loss is shown as a max of 0....

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I wouldn't attempt the diode solution. The commenters are correct in that wiring the two power supplies together like that is likely to cause problems; I've myself seen a PC that won't shut down properly if back-fed 5V through USB from a bad hub. What I would suggest is connect both the grounds together, but connect only one of the power supplies. That ...

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You can use a latch circuit and MCU together to work out solution on your own. Once, button is pressed, the MCU will be powered MCU will ASAP latch the power latch (Power Enable, for example ) circuit, keeping the Power supply ON. Any further button press will be detected my MCU as interrupts If ti is along press, MCU can detect that too. In case ...

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Let me address a couple things as a pilot... Why? Well, for example, if the pressure switch in the gear up part of the circuit should fail, I can still get the gear to retract by activating the override push button. This is a big no-no as a pilot. The last thing you want to do is to bypass something that is failing. If your gear doesn't retract, leave it ...

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I guess the 1us RC lowpass filtering tries to keep off false keyhits caused by phones and other radio transmitters. The debouncing is performed with software. Do not add slowness to the circuit which detects if a key is pressed. Players hate delays. There's enough it already (MIDI transmission,communication between the MIDI interface and the computer ...

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simulate this circuit – Schematic created using CircuitLab Figure 1. (a) Typical 2-way switching as used on stairways. (b) OR switching. If you want either room to be able to switch on or off then use (a). If you want the heater to be on if either or both rooms require it then use (b). If you want to switch the input to the digital system then: ...

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This is a bit of a creepy circuit. It appears to work on the relay hysteresis. So the relay coil, in series with a diode D2 and the 100$\Omega$ resistor, receives enough current to hold the relay in, but not to actuate it. Full voltage is applied to the relay when TP1 is bridged, so it pulls in. When TP2 is bridged, the 100$\Omega$ resistor receives ...

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I think you are mixing up the ideas of the transient behavior of the circuit, and the steady state behavior in the frequency domain (by using Laplace methods). Transient response would have to do with closing the switch, and seeing the current/ voltage in the inductor as time goes on from t=0. The steady state behavior in the frequency domain would have ...

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I traced the output section of your RADAR module from the product photo. It appears to look like this:- simulate this circuit – Schematic created using CircuitLab The BISS0001 is designed for use in PIR motion detectors, and performs a similar function in the RADAR detector. When a signal is detected pin 12 goes high, turning on Q3 through R11. This ...

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Digikey has a good search engine with filters electromechanical>switches>Rocker> illuminated > Ctrl+ select after sort by illumination colour, Choose in stock only Results e.g.

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Infinite switches are not thermostatic controls. They regulate the percent of time the controlled heating element is on. All infinite switches use a nichrome heater wire or ribbon and a bimetal strip to open and close the electrical connection to the heating element being controlled. The shaft of the control has an eccentric that varies pressure on the ...

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