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I have to give a signal to a inverter. I am trying to do this using two push button.

First pushbutton can turn on and turn off the signal(a relay coil) like a toogle switch.

Second push button can only turn off the signal (previously on by first pushbutton)

Is there any way to do this only using several relays (relay logic)? If not suggest a simple method.

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    \$\begingroup\$ Can you fix your question? The title says one button. The explanation says two buttons. Your comment to a good answer says you only want one button. Which is it?! \$\endgroup\$ Commented Dec 18, 2017 at 22:44

4 Answers 4

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Two methods:

  1. Using internal logics of the inverter. Some contain comprehensive programmable inputs and sometimes even PLC (ladder diagram) like functionality.

  2. Relay latching circuit:
    Pressing On engages the relay. The relay then bypasses the On switch.
    The Off switch disconnects the relay.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ There is no any logic system of my invrter. Above mentioned system has two pushbuttons.I need to on and off using ONLY one pushbutton. \$\endgroup\$
    – user_fs10
    Commented Dec 13, 2017 at 17:52
  • \$\begingroup\$ @NSiri Then just get a maintained pushbutton. Eg: M22-DRL-W. \$\endgroup\$
    – Jeroen3
    Commented Dec 19, 2017 at 6:32
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See this app note, it does the toggle switch you just need to add the second button to it.

https://www.maximintegrated.com/en/app-notes/index.mvp/id/4444

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    \$\begingroup\$ It may be a good idea to provide a summary / picture here in case that link ever disappears or moves. \$\endgroup\$ Commented Dec 12, 2017 at 18:58
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    \$\begingroup\$ It is a good idea. The answer should make sense on its own without following links. -1 (almost). \$\endgroup\$
    – Transistor
    Commented Dec 12, 2017 at 19:36
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Well simplest method is to use some type of latch. That utilizes two buttons - one for latching the input state and second for reseting it. But of course there are two buttons... So one of the simplest method is to use NE555 timer, as shown on the schematics:

NE555 toggle switch

It works really simple. Reset (4) pin is tied to Vcc, so the IC is not resetting. Discharge (7) is not connected anywhere, so that means the capacitor (C1) will stay charged even if you turn switch off. As soon as you pull the switch, the C1 will charge and stay charged. That will also change the state of NE555s internal comparator, because now you´ll have more than 0V on trigger (2) pin, so it latches the state of capacitor (ON). When you press the button again, capacitor will discharge and now you have 0V on treshold (6) pin, which is not higher than 0V trigger value, so the latch will reset. Of course, you have transistor and resistor on output so you can draw more than 20mA from NE555. Hope i helped :)

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Two DPDT relay toggle circuit:

enter image description here

The second function can be done by a button shorting out the coil of d1, provided a resistor in series with PB1 limits the current if both buttons were pressed at the same time. The resistor needs to be small enough to still effectively short out the relay coil in the transition state.

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