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57

Before I dive into relays, let me give you a primer on switches in general. There are lots of different kinds of manually operated switches, all being useful for different purposes. Here are a few, with their names and some nomenclature: simulate this circuit – Schematic created using CircuitLab The number of "poles" refers to the number of ...


8

Image from here. The pin marked "30" corresponds with the middle contact in the diagram in your question. The pin marked "87a" is normally closed when the relay is unpowered. The pin marked "87" is normally open and becomes closed when the relay is energized. If you want to know more about relays, follow the link where the image ...


4

Image source: Instrumentation tools. What isn't quite clear from the image in your question is that the armature pivots at the top right. The animation shows the pivoting action and almost exactly matches the correct illustration you supplied. The part that's missing from the illustration is that the relay's ferrous cylindrical core extends down to the ...


3

Relays can be used to switch between two circuits if needed. In your case, when the relay is powered, the connection between 2 and 3 shifts to the connection between 3 and 4. Essentially these are different circuits. Circuit#1 is powered when 2 and 3 are in contact and circuit#2 is powered when 3 and 4 are in contact. If you are interested, you can also ...


3

It is safely rated for 20A 240VAC, 1HP 16A 120VAC But you can extend lifespan using an RC snubber with a small plastic cap and power resistor. This a frequent question and often accepted answers overlook details. OMRON the best company for RELAYs in the world has even gone so far as to remove all their relay snubber design info off the website in ...


2

simulate this circuit – Schematic created using CircuitLab Figure 1. Modified circuit. If your fan can be controlled that way then you only need two resistors. RLY3 slow. RLY2 medium. RLY1 fast. simulate this circuit Figure 2. Low-side switching of relays. The diodes aren't needed as there are diodes in the ULN2003. simulate this circuit Figure 3. ...


2

Absolutely no problem with the brief loss of power. All 12VDC compressor fridges I have seen are made for use in vehicles with their less than smooth system voltage. Drops down to 5V (when starting the engine) are common and expected. Jumps up to 16V and even higher transients are common as well. Losy and sparky cigarette socket plugs are everywhere, too. ...


2

Your circuit should work as drawn, but with a standard, non-latching relay. Just make sure its solenoid/coil is rated for 12V, and its contacts for at least 10A. A latching relay would remain stuck in its ON position until another pulse of current is sent to its solenoid, so it wouldn't serve your stated purpose.


2

You could set you buck converter to 12V but puth a capacitor between the feedback pin and ground so that it starts passing full power. you'll probably need a resistor in series with the capacitor to prevent voltage oscillations. simulate this circuit – Schematic created using CircuitLab but it's probably simpler to to use a series resistor with a ...


2

You did it right, the relay side circuitry. Tie pin9 of ULN2003 to the +5V of the relay. Arrange the 5V and GND connection, in the way the relay side current does not pass the logic side. Problem happened, hypothetically, when relays are driven, low voltage is coming through the ULN internal re-circulation/clamping diode then driving the inactive relay. That ...


1

The pushbutton only needs to drive current to relay coil to actuate it. That might be in the order of 0.1A. The relay contacts are the ones that need to handle the 80A load. So of course you can use the same 12V supply to drive a 12V relay coil.


1

The way they've drawn it doesn't really make sense. To make sense, imagine pin 3 being moved to the right, so it's in contact with pin 4 when it's at rest. Then when the magnet is actuated, it pulls down in the grey arm. The end of the pin from the grey arm extends through pin 2, and is connected to pin 3, so when the grey arm is pulled down, it pulls pin 3 ...


1

This is a general question, so the answer, as always, is that it depends. When disconnected, the coil will indeed try to the same current through it and will create spikes as the field in it collapses; that is the reason for D2. But in case of a "bounce" you will get some transients on your supply line and possibly your ground line. But whether ...


1

Below is a simple simulation of using ideal diode rectifier that has 30mV voltage drop. When the aux bank is drained, it will take all the current. Connecting together aux. bank with a start battery could boil your start battery - only the voltage difference and internal resistance would limit the current. simulate this circuit – Schematic created ...


1

https://www.jimellisvwparts.com/products/Volkswagen-VW/WIRE-SET/5131548/000979144E.html Although I couldn't source the manufacturer of the connector, this matches your part and description.


1

You are correct, Audi uses TE/Tyco connectors but not always nor exclusively. You may check Bosch. The year and model will have some bearing on the connector as well. What I have is probably not good news but Audi uses custom connectors made by several terminal manufacturers as do most of the automotive manufacturers. Sometimes several use the same connector....


1

Adding a pulldown resistor to the base should solve your problem. Something like 10k would do. You don’t want the value too low as it forms a voltage divider with R8. Also consider if you really need the opto coupler. Many assume that these have magic properties that will ‘protect’ the microcontroller. If the gnd/0V is shared between your 12V supply and ...


1

Just look at a standard stop-start circuit: - Now you don't appear to need the stop contact so just imagine that shorted out then, re-arrange in your mind that your relay contact is more clearly in parallel with the start switch like this: - And you have the basic start-circuit used in industry.


1

This will work. The ‘off’ normally-closed switch must carry the entire load current, in the circuit shown, of course. The relay is a normal non-latching type and must have the same coil voltage as the load. You also have the option of just interrupting the coil current to the relay, which should cause it to drop out (provided the contacts have not welded). ...


1

Energy requirement would be minimum with a magnetic latch relay. Only a single pulse would be required for 'On' / 'Off'. The 'NO' contact, of the latch relay K1, would drive the motor.


1

I would suggest the following. Do not direct connect AC200V+ to logic control circuits. Especially connect PC814A input direct to L and N is dangerous. I would suggest to use any wall wart or smart phone charger to get around 5VDC from AC mains. Actually if you don't use AC mains as control signal, you can forget the optocoupler. In other words, just ...


1

An inductor has NO DC inrush current. Current in inductor current cannot change instantaneously, so if there is no current in the coil at turnon current will start at zero and increase. Diode current is L_coil l at turnoff. 350 mA in this case. Any 1N400x diode rated higher in voltage than Vsupply will easily suffice.


1

Your optimal solution is to keep running off battery with a float charger at its normal resting voltage. Then periodically it will never fail on power outage unless cutout for a long spell. if you do not have this type then get one sized to your requirements. smart chargers may or may not have the hysteresis to float CV after decay from charge CV.


1

The easiest way to control a relay with any microcontroller is via a MOSFET. You have two choices, one is to try and drive the MOSFET directly from the Jetson output pin or use a chip that helps drive the MOSFET. The output pins of the Jetson are either 3.3V or 1.8V, so your best bet is to go with the 3.3V outputs. Each pin can produce 1 mA which is plenty ...


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