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Related

Switch ONLY when powered off

atmn
atmn over 7 years ago

Hi!

 

I wonder if there are any curcuits that allow me to only switch when device is powered off.

 

I select between 2 chips with their 5V pin and I want to prevent to be able to "hot swap". If I would toggle the switch when power on it will stay on the state it was when the system was powered on.

 

Is there any solution, any chip I could use?

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  • dougw
    dougw over 7 years ago in reply to atmn +6
    This is a quick example that doesn't need any chips... Q1 and Q3 are P channel FETs like BSS84, Q2 and Q4 are N channel FETs like BSS138. With the switch in the capacitor position, Q2 will turn on Q1 and…
  • dougw
    dougw over 7 years ago +2
    You could use an DPST switch to select which chip gets power on power-up. The DPST switch state gets latched on power-up by a D-flip-flop. The D-flip-flop clock can be a simple capacitor to the power rail…
  • Robert Peter Oakes
    Robert Peter Oakes over 7 years ago +1
    OK, so a slightly more complicated approach would be to use a very low cost microcontroller (ATTiny or similar), have two power transistors, one for each unit and a button. When you press the button, the…
  • RParkerE
    RParkerE over 7 years ago

    I do not know if there is any switch that will do what you need, but you should look into Hot Swap Controllers.

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  • atmn
    atmn over 7 years ago in reply to RParkerE

    Thanks, will look into that.

     

    To clearify. The only way to switch to the other chip so to speak is to power off the device, and then toggle a switch and last power on again.

     

    I want it to just "check" upon power on.

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  • niteowl12
    niteowl12 over 7 years ago

    I would say the best way for implementing this is through a circuit loop for sensing connection and a transistor to to turn power on/off. Then have a section of code that looks for the sense loops during start up.

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  • dougw
    dougw over 7 years ago

    You could use an DPST switch to select which chip gets power on power-up. The DPST switch state gets latched on power-up by a D-flip-flop. The D-flip-flop clock can be a simple capacitor to the power rail, with a pull down resistor on the clock pin (creating a pulse on power-up). The flip-flop output turns on power to one chip or the other depending on its state (using transistors).

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  • jw0752
    jw0752 over 7 years ago

    Hi atmn,

    Perhaps you could use an SPDT switch as a programming switch to a microprocessor. The uP reads the switch position once at power up and pulls the corresponding output High to power your IC "A" or IC "B". Since the program switch is only read once in the setup of the uP program nothing would happen if you switched it while the unit is powered up. You would have to power down reset the program switch and power up to move to the other IC.

    John

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  • atmn
    atmn over 7 years ago in reply to dougw

    This sounds good.

    Are there any schematics for a curcuit like this? This must have been done before I would assume.

     

    The easier solution the better. image

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  • Robert Peter Oakes
    Robert Peter Oakes over 7 years ago

    OK, so a slightly more complicated approach would be to use a very low cost microcontroller (ATTiny or similar), have two power transistors, one for each unit and a button. When you press the button, the micro will power off the first device, wait a period of time, then power on the other. this approach can be applied to more than two devies in this case.

     

    Yes you can do it with simple ttl chips and passives but it will take just as long and less fun. A clunky switch will give no time for one to power off before the other starts powering up. You could also include a GPIO signal to the target device to tell it to go to sleep or some other pre-defined condition prior to the power being removed.

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  • dougw
    dougw over 7 years ago in reply to atmn

    This is a quick example that doesn't need any chips...

    image

    Q1 and Q3 are P channel FETs like BSS84, Q2 and Q4 are N channel FETs like BSS138.

    With the switch in the capacitor position, Q2 will turn on Q1 and turn off Q4 upon power up. Q3 stays off because R4 pulls the Q3 gate up. The circuit will then latch in this state as R3 will hold Q1 on. Switching the switch will not change the state.

    With the switch in the grounded position at power on, Q2 and Q1 never turn on since Q2 gate is always grounded. This Allows R2 to pull up the Q4 gate, turning on Q4 which turns on Q3. Again the switch has no effect after initial power up.

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  • atmn
    atmn over 7 years ago in reply to dougw

    Looks great!

    I really wanted to go analogue with my solution so this looks doable. image

     

    Gonna order some components and try it out!, again thanks!

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