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Need N.C. SPST switch recommendation

Other Parts Discussed in Thread: SN74AUP1G80, TS12A4515, TS5A2053, SN74LVC1G3157

Hi,

I need a recommendation for a low-cost, low-power, low-speed N.C. SPST switch.

I have a circuit that I want to switch off the battery power to when an external 5.0V is applied from a USB cable:

The switch (shown in the red box) needs to connect the Li-Ion battery (3.2-4.1V) to power the small circuit (with only 2 active components a schottky diode and a low-power D latch SN74AUP1G80.  The switch would be SPST normally closed and would open when the USB power is supplied (~5.0V but could be 4.5-5.5 per USB spec).

Thanks in advance!

Kind regards,

Kevin Kreger

  • Hello Kevin,

    I would recommend the TS12A4515 for your application as long as the current through the switch does not reach the 20 mA maximum current through any port as specified by the datasheet.

    Regards,

    Andrew

  • Thank you Andrew!

    Kind regards,

    Kevin

  • Hi Andrew

     The LVC1G80 has the Ioff feature that sets the I/O's to high impedance when VCC is 0V. It would not work if it is floated. I would suggest a SPDT instead and switch Vcc to 0V.  This would help keep you current low.   http://www.ti.com/product/sn74lvc1g3157 should work

  • Hello all,

    Thank you Chris that was a good catch!

    Kevin in light of Chris' insight I would recommend you look into the TS5A2053 as a SPDT solution instead of a SPST, with battery tied to NC and 0V tied to NO.

    Regards,

    Andrew 

  • Hi Chris,

    Thanks so much for your design advice! I have incorporated the switch and just have a few questions and comments:

    • The switch should close the power connection from the battery to the circuit below if the USB 5.0V is floating Is this a problem?
    • Should I add a bypass cap to the Vcc pin of the switch
    • Should I use the battery to power the switch? At first it seems the power to the switch can be from either the battery (which ranges up to 4.1V)  or USB 5.0V. I have chosen the power to be from the battery with the following reasoning: If the battery is so exhausted it cannot drive the switch then we do not care if the M.O.switch in our circuit works. USB power would have to be applied which charges the battery. Only with a sufficiently charged battery is the M.O. switch (which turns the rest of our circuit board ON) of any use. To reiterate, the purpose of the switch you recommend is to make sure the circuit power is turned OFF when USB 5.0V is applied and turned ON when USB 5.0V is absent (floating).

    Here is how I believe the circuit should be implemented based upon your advice:

    Chris, I'm not sure I understand all of your comment to Andrew. However, based on the reasoning above I do not have to worry about the case of Vcc = 0.

    Thanks again to both of you for helping out with this design!!!!

    Kind regards,

    Kevin Kreger

    PS: Which forum is appropriate to get a recommendation for ~0.7V surface mount diode (D2)

  • Hi Guys,

    I guess I totally f*d that up!   because Andrew advised that I tie 0V to NO and battery to NC. Here is something closer to your advice:

    In addition, here are my newly configured questions

    • The switch should close the power connection from the battery to the circuit if the USB 5.0V is floating. Is this a problem?
    • Should I add a bypass cap to the Vcc pin of the switch
    • Should I use the battery to power the switch?
    • I'm not sure if I should use the SN74LVC1G3157 or the TSA5A2053 (but the circuit above shows a SN74LVC1G3157)
    • Chris are you advising I tie the power of the SN74LVC1G3157 to USB 5.0?

    THANKS!

    Kind regards,

    Kevin

    • The switch should close the power connection from the battery to the circuit if the USB 5.0V is floating. Is this a problem?  when these switches have no power they are open to all pins
    • Should I add a bypass cap to the Vcc pin of the switch.  yes
    • Should I use the battery to power the switch? If you will loose the USB 5V then yes. If the battery gets too low to power the switch then the switch will open floating Vcc. Since the battery is already drained you dont care.
    • I'm not sure if I should use the SN74LVC1G3157 or the TS5A2053 (but the circuit above shows a SN74LVC1G3157) either would work
    • Chris are you advising I tie the power of the SN74LVC1G3157 to USB 5.0?  It is a tough decision. If you use the usb and you loose the 5v then you will float Vcc. This could cause batt current drain from the AUP. If you use the battery to power the switch then it will use part of the battery budget.. one thought is to use a weak pulldown on the on the AUP vcc pin and use USB power. that way when you loose USB power the Vcc pin is pulled to gnd and hopefully keeps the Ioff circuit working. When the usb power on, the pulldown across its 5v and not draining your batt. 

     

          

    THANKS!

  • Hi Chris,

    First of all many thanks for your careful analysis! I want to go over your points and respond to each, thank you in advance for your kind patience:

    • The switch should close the power connection from the battery to the circuit if the USB 5.0V is floating. Is this a problem?  when these switches have no power they are open to all pins. Okay, that's important (more on this below).
    • Should I add a bypass cap to the Vcc pin of the switch.  yes. Okay, I'll add 0.1uF bypass to both the switch and the AUP.
    • Should I use the battery to power the switch? If you will loose the USB 5V then yes. If the battery gets too low to power the switch then the switch will open floating Vcc. Since the battery is already drained you dont care. Yes, I will lose the USB 5V (it is only connected to charge the battery) and the drained battery is a don't care, but you seem to have a workaround to the loss of USB 5V below.
    • I'm not sure if I should use the SN74LVC1G3157 or the TS5A2053 (but the circuit above shows a SN74LVC1G3157) either would work. Great, I'll decide based on power, price, and availability then :-)
    • Chris are you advising I tie the power of the SN74LVC1G3157 to USB 5.0?  It is a tough decision. If you use the usb and you loose the 5v then you will float Vcc. This could cause batt current drain from the AUP. If you use the battery to power the switch then it will use part of the battery budget.. one thought is to use a weak pulldown on the on the AUP vcc pin and use USB power. that way when you loose USB power the Vcc pin is pulled to gnd and hopefully keeps the Ioff circuit working. When the usb power on, the pulldown across its 5v and not draining your batt. 

      Hey Chris. I'm a little confused (it's my bitheadedness showing as I'm a software guy) So please bear with me while I explain some things to organize my thoughts:
       
    • I have tentatively made the circuit so that USB 5.0V is controlling the switch in my circuit diagram and power to the switch Vcc is provided by the battery.
    • These are the circuit requirements: 1) the only time the AUP circuit must have power ON is when there is no USB Power, only battery power, and 2), the power to the AUP circuit must be turned OFF when there is USB power. 3) If the battery is dead and cannot drive the switch or AUP it is a "don't care" as the USB power is eventually connected to charge the battery.
    • These are the statements and logic I want to confirm: 1) The switch is open on all pins without power applied to the switch VCC pin (i.e. there is no switch closure to either pole) 2) then the switch must have power when there is no USB power (otherwise we cannot power the AUP circuit) 3) Therefore the switch power to its Vcc pin must come from the battery. Right? However, you seem to have a workaround (see below)
    • Conversely, the circuit must be powered off when there is USB power (as the current drain makes the battery management chip under perform and keeps the charging LED status lit). The only rub is there is a small amount of current needed by the switch, hence creating a tough decision, as you say: "If you use the battery to power the switch then it will use part of the battery budget"
    • However, and this is where I am grasping for understanding and hoping we have another solution (that defies my logic in the bullet above), you note that "one thought is to use a weak pulldown on the on the AUP vcc pin and use USB power. that way when you loose USB power the Vcc pin is pulled to gnd and hopefully keeps the Ioff circuit working. When the usb power on, the pulldown across its 5v and not draining your batt. "
    • So, a weak pull-down like 1MOhm to gnd on pin 5 of the AUP (i.e. pin 4 of the switch) may keep the switch in place so pins 3 and 4 are connected?
    I see you have really put a lot of thought into this, and I thank you as much as I can with the printed word for your time! I don't know if I mentioned that I am a US guy living in India looking for startup investment for our TI-based product. TI has spent a great deal of time grooming us for success to the point where we have been through formal review process with your engineers. You guys really are the best!

    Kind regards,

    Kevin