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4

@Andyaka Thanks for the tip! This seems to get me so close to what I need, with the exception of requiring two switches to control it. The functionality I need is a single momentary switch latching the relay on, then latching it off again on the next press. I imagine I'm missing something fairly obvious in making this happen, so will keep playing, ...


0

To day I use two power sources and a relay between RPI and the remote control(green card) since I don't want to accidentally fry my RPI. That's a good policy. Most cell-phone 'chargers' can't actually deliver their rated current reliably - if at all. Using a dedicated supply for the Pi makes it much less likely to suffer from brownouts or surges caused by ...


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probably all that you need to do is convert the 24V output from the pump hub down to the 4 to 9V that your rain pump wants. for that a DC to DC converter module like LM2596 or XL5015 seems ideal, if you're not using the wave pump you could use its cable and plug for the connection from from the pump hub to the dc-dc converter.


1

Just stick to electromechanical relays for simplicity. Use your little relay to drive the coil of an even bigger relay. Don't parallel both relays' main contacts so they share the load since you can't guarantee they switch at the same time. simulate this circuit – Schematic created using CircuitLab If a relay has two pins it's probably a reed ...


0

Use the wireless receiver to drive a relay with contacts rated at 20A or more, then drive the actuators via those contacts. You don't say what happens when the open signal stops, Do the gates close under some spring action or are they driven shut by reversing the actuators? In either case the contacts on the 20A relay need mimic those on the 6A part. 14A is ...


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You need a relay that's rated for 48VDC or more, and 32A or more, with form A (normally open) or form C (transfer -- this is the sort that @transistor talked about). Note that relay currents get complicated -- AC current ratings are usually higher, and for DC, "making" or "holding" current is higher than "breaking" current. If you're going to break a 32A ...


4

The relay will control a 48v 32a circuit. Your relay will require a coil voltage to suit your control circuit voltage. You haven't stated this as other than "signal". The contact(s) will require a rating of ≥ 48 V DC and ≥ 32 A DC. (note capital 'V' for volt and 'A' for ampere.) It should be normally open unless a signal is present. Almost all ...


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From my experience with good power supply designs, they always have enough storage energy for a 1 cycle dropout at rated load, so if using less than rated load, you may get more. The “one cycle” rule of thumb depends on your grid frequency and voltage and if it is a universal supply. The other no-name supplies that do not fall in this “good design” category ...


3

Holdup is dependant on stored energy. It should not be necessary to hold-up the heaters over such a short period - only the electronics. 5A x 220 V x 10 ms = E = V.I. t = 220 x 5 x 0.010 = 11 joules. Sounds doable - lets see ... In the following I calculate the capacitance required to contain the whole hold-up energy. However, in real-world applications ...


3

you can halve the relay current with a capacitor and a resistor. the capacitor feeds the relay on startup, the resistor reduces the current on hold.


1

Unless you actually measure the current out of the cell charger under load, don't count on the printed current rating. I have seen these types of chargers/ext power supplies go bad and unable to supply required load current but voltage readings are still good. Measure your load current draw when powered from your laptop. In the USB 1.0 and 2.0 specs, a ...


1

but i changed the source to a mobile charger Aha! Many mobile chargers (power banks) switch off when the load only draws a small current. This is likely the reason why the power is switched off. The Arduino + relay simply doesn't draw enough current to keep the power bank switched on. The power bank "thinks" there's nothing connected so it switches off. ...


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W.r.t. diagram in question, the following connections must be made (in addition to existing connections) for the NO contactor to close when there is no fault detected by VPR and for the NO contactor to open when there is fault detected by VPR. L1 to COM NO to A1 N to A2


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You could use ORing diodes1 instead of a DPDT relay. Then the transient will be smooth. When (in the schematic below) D1 has a drop of 0.7V and D2 of 0.3V, the 12V adapter will take over when the battery voltage drops below about 11.6V. You can change the voltage where the adapter takes over by replacing D2 by a standard diode or by adding Schottky diode in ...


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perhaps something like this: simulate this circuit – Schematic created using CircuitLab improved version: simulate this circuit The relay is just an ordinary automotive headlamp/horn relay the 30A fuse prevents damage to the cable to the machine from starting a fire and the 1A fuses do the same for the control circuit (allowing the use of smaller ...


2

Connect the pushbutton to drive a relay coil. Then connect the relay's NC and common primary contacts where you would normally connect the NO push button terminals. Note that the relay coil will always consume power when not pushed though...The idle current will be whatever the relay's coil current is which could be some tens of milliamps. simulate this ...


1

Since uC's are all made from CMOS FET switches that need bias from supply to make low resistance "0" & "1" they cannot conduct current to the coil to a shared ground. However when power is applied with an external coil pullup to a port, this causes the pin to exceed the supply voltage and induces the dreaded SCR effect of latchup faults of all CMOS.


3

If the Arduino is disconnected from power then the voltage at the Arduino outputs is undefined...it could be anything. The output pins, even if they happen to be at 0V, will not be able to sink enough current to activate the relay. This is in fact a dangerous situation for the Arduino. Applying a voltage to the pins of an unpowered microcontroller may ...


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Use mechanical relay. Input AC and output AC or DC


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The datasheet doesn't explain it but does say Level sensor probe volts 5 VAC @ 1.5 mA. The AC part is to be expected as DC would cause plating of the probes due to electrolysis. 5 V @ 1.5 mA suggests a source impedance of 5 / 1.5m = 3k3. It then goes on to say Adjustable sensitivity: 0 - 50 kV which doesn't make any sense to me so I'd be interested ...


1

simulate this circuit – Schematic created using CircuitLab Figure 1. A double-pole switch or relay solution. If you review your schematic you will find a few problems. You have two independent AC supplies where in fact you probably only have one. You have no return path from the solenoids other than through another series-connected solenoid. The ...


2

If I do understand your question correctly, I think you can just use one relay, but "the other way around" (instead of 1 output serving 2 outputs, using it as 2 inputs serving 1 output): simulate this circuit – Schematic created using CircuitLab


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Most likely the failure is in the contact and not in the coil. At those currents contacts are likely to fuse together or get sticky and build up oxidation layers. Adding more current to the coil will probably not help actuate it if the contacts are bad. Source: https://www.pickeringtest.com/en-us/kb/hardware-topics/relay-reliability/finding-relay-failures ...


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Relays have a life span which is mostly determined by how much current and voltage the contacts are switching (and the inductance/capacitance and surge current) and how often they are called upon to switch. Increasing the coil voltage is unlikely to help if the contacts are worn out. Below is a curve showing the life span of the contacts for a common old-...


12

A relay is almost certainly NOT the right tool for the job. The first step in wiretapping an interface in order reverse-engineer it is to determine the nature of the signals you want to look at. For voltage signals, you need a high-impedance buffer amplifier (like the input of an oscilloscope) that will cause minimal disruption to the existing circuit. For ...


0

As mentioned above, an H-bridge driver with four external power MOSFETs will give you an all-electronic motor reverser that needs only one control line from the Arduino (two if you want a third state - off). The larger the MOSFETs, the lower the heat. International Rectifier is big in h-bridge control chips. Here is the general idea:


2

I'd also take a look at the IPC 2221[edit: Electrical Clearances] standards before going any further with the PCB design. For a high voltage PCB, the one thing that matters the most is the material selected for the core/prepreg. Depending on the availability and budget, you can go with extra overcoats and better alternatives/versions of FR4. Also, consider a ...


1

I am not sure what the warning from "NCD" company is about, but I am pretty sure you shouldn't worry along your lines. The referenced USB board uses a regular relay with 5-V coil that takes only 71 mA to operate, so there should be no worries whatsoever about overloading of any USB port. The relay itself provides 1.5 kV isolation between its coil and power ...


1

From looking at the diagram, there is likely no isolation on the USB board (I only see 1 USB converter chip and no optocoupolers). In the event of a short on the board, it could send the high voltage back to the USB host, or to other places. This could be unsafe. I wouldn't use this board to switch anything above 30V and it would be really unsafe to switch ...


6

That above component is called Positive Temperature Coefficient relay. In this application it is uisng as starting device for compressor in your refridgerator. It is used for powering the start winding for a short moment to help starting up the fridge compressor motor. what is the disk and how does it make this thing into a relay or equivalent? The disk ...


1

simulate this circuit – Schematic created using CircuitLab Figure 1. Pi-controlled latching relay. How it works: When SW1 is first closed RLY1 and RLY2 are off and both contacts are open. When the Pi energises RLY1 (using one of those 3.3 V or 5 V relay modules rather than directly from the GPIO pin) it energises RLY2 and switches on the load. Note ...


0

Is there such a relay type, that would be normally open with no input power, and and then latch closed? I can't speak for your diagram, but I do know about relays. Relays use a coil and spring to stay on or off, when the coil is powered they switch positions. There are normally open relays and normally closed relays, the trick is to figure out how to wire ...


1

There are latching relays that are turned on by a momentary application of power to one terminal and turned off by a momentary application of power to another terminal. If the application of power to either terminal is continuous, you could design a relay circuit that would disconnect the power automatically. There may be latching relays with that built in. ...


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From the comments: I want to have the central tap in the secondary of my balanced isolation transformer connected to the neutral an not to live. That is not the correct way to do this and is potentially dangerous as well as presenting an earth leakage which will contribute to tripping of your RCD / GFCI. In Europe where 230 V phase to neutral is the ...


3

You could but it is not a surefire technique since you could trigger an GFCI device (Ground Fault Circuit Interruptor). Moreover this could be really UNSAFE. The fact is that whatever current you draw from the live wire it ends up to earth ground through the protective conductor (earth ground wiring system - green/yellow conductor). This adds up to ...


0

Safely drive a relay OR sense a switch from a single terminal? I would suggest your circuit is not "safe". While there exists a path from the input pins to VDD through an intrinsic diode in the MCP23017, you should NEVER use this as a method to clamp the input. If you are doing just one input/output, then the current may be limited by the 10k Ohm resistor,...


0

It seems OK so far. Initialization, intermittent cable connection or faults, not-considered. ALthough you have not defined your max output current. The port is rated at 25mA max and with your driver ... I assume is more than 25mA ? or 50mA? 100mA? 1A? for a 12V relay , SSR or ? They have NCh arrays low RdsOn , not sure about Pch arrays 20V Pch 100 mΩ @ -...


0

So, why in this industrial applications they use 12V to control the relays? While 5V relays are available, they are less commonly used probably because 12V is already available for most applications (especially industrial) and relays with a lower voltage have higher current requirements. A 3.3V relay would have an even higher current requirement if the ...


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