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I want to make circuit which can working like a Latching Relay which Hold Current using SPDT Relays.

So i can use this circuit to Pules ON/OFF switch.

Using limit switch when pressed a pulse goes and the circuit ON and if re-pressed the same Limit switch the circuit OFF.

Other option Using Two Limit Switch First for ON and Second For OFF.

Please help me to make this type of circuit.

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  • \$\begingroup\$ Is this for an assignment or homework? You can still get help here BUT you must say. If not, what is the application? \$\endgroup\$
    – Russell McMahon ♦
    Commented Jan 9, 2015 at 4:52
  • \$\begingroup\$ It is my assignment, so please help me...I want to make some big circuit it is a part of that circuit. \$\endgroup\$
    – ANKIT JAIN
    Commented Jan 9, 2015 at 14:34

3 Answers 3

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Both of the two solutions below use one button for turning on, and a second button for turning off. The difference between the two, is thr first uses a NC button for turning off, which may not be available. The second solution uses NO buttons for both actions.

In both cases, when the first button is pressed, the relay activates, and the contact is closed holding the relay in the closed position until the circuit is interrupted.

enter image description here

Doing both the on and off actions with one button is considerably more complicated.

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  • \$\begingroup\$ Link to simulation of circuit 2: bit.ly/2bEtcsh \$\endgroup\$
    – Jonathan S. Fisher
    Commented Aug 30, 2016 at 0:01
  • \$\begingroup\$ Just what I needed, thanks. I have a NC momentary switch to release the latch. \$\endgroup\$
    – SDsolar
    Commented Apr 17, 2017 at 9:06
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enter image description here

The off limit switch is wired as a normally closed, the on as a normally open.

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  • \$\begingroup\$ This is the ladder logic for PLC, But we do not use PLC in our circuit. We need hard circuit. \$\endgroup\$
    – ANKIT JAIN
    Commented Jan 9, 2015 at 4:43
  • \$\begingroup\$ It is an equivalent to a real circuit, one side of the ladder is high voltage, the other is ground. \$\endgroup\$
    – Lior Bilia
    Commented Jan 9, 2015 at 6:25
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    \$\begingroup\$ This is the same circuit as solution 1 from @tcrosley . \$\endgroup\$
    – EM Fields
    Commented Jan 9, 2015 at 10:49
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Use a D type flip flop with Q bar wired to the D input. Each time the clock pin is activated by a button press it flips the output Q pin. This, via a transistor can drive a relay. You also need to have some contact denounce circuitry to prevent relay chatter. This is a one switch solution.

enter image description here
(source: electronics-tutorials.ws)

Above is the basic circuit and input frequency (CLK) is your debounced switch opening and closing. So, you press the switch once to bring the logic level to 1 and the output on Q latches and stays latched even when you deactivate the input switch on CLK. Next time you press your input switch the output toggles state.

Here's a different way of doing it with two inverters: -

enter image description here

The circuit also has debounce circuitry (220k and 47nF) and an output transistor drive to an LED. The LED and series resistor can be replaced by a low power relay coil fed from 12 volts.

If you want to implement the D type circuit, here's a contact debounce circuit that can be used with it. This uses a changeover switch as the input device: -

enter image description here

You can even get away from a relay and use a MOSFET in this type of circuit: -

enter image description here

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    \$\begingroup\$ Where are the relays? (See title of question). \$\endgroup\$
    – tcrosley
    Commented Jan 9, 2015 at 11:05
  • \$\begingroup\$ @tcrosley without wishing to insult the op (who clearly knows about relays hence the term SPDT in the title), I offered examples of how to drive said relay. One of the circuits I explained where to add the relay and on another I've shown a circuit that performs like a relay by using a MOSFET as a "contact" suitable for DC circuits. \$\endgroup\$
    – Andy aka
    Commented Jan 9, 2015 at 11:37
  • \$\begingroup\$ For circuit with two inverters, how can we determine initial state on BJT base (at t = 0, capa is discharged and pushbutton opened)? \$\endgroup\$
    – rem
    Commented Jul 10, 2017 at 15:08
  • \$\begingroup\$ @rem the inverter with the lightest output load normally wins but, if in doubt, add a 33pF capacitor (or greater) to one of the outputs and connect the cap to ground. That output will be the slower and lose the initial battle to reach Vcc. \$\endgroup\$
    – Andy aka
    Commented Jul 10, 2017 at 16:32

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