I have the question about ADG5208F.

Q.How does it work, if it has follow status?

 Vcc =24V

 Input = 5V

 Signal = switch ON

 D = GND

In this case, it connected for the OPAMP .

The customer want to not get the damage, when the OPAMP's voltage is GND level by supply voltage is shutdown.

best regards

  • 0
    •  Analog Employees 
    on Jul 22, 2015 9:11 AM

    Hi Hayashi san,

    Just to confirm, your supply voltages to the switch are Vdd=24V and Vss=0V and you are inputting 5V through the switch and connecting the output to GND.


    So you have 5V going through the on switch which has approximately 300ohm resistance and then shorting to ground.  If the input voltage to the switch goes above Vdd then the switch will turn off.  Also is the supply voltage to the switch is shutdown then the switch will turn off. 





  • Hi Eric-san

    Thank you for the advice.

    So I have the question for your comment.

    Would you teach me the follow?

    Where can I find the 300 ohm resistance is set?

    Is it okay because the resistance is set, if the output voltage becomes ground level suddenly?



  • 0
    •  Analog Employees 
    on Jul 23, 2015 5:19 PM

    Hi Hayashi san,

    The 300ohms I am quoting is a very approximate figure for the Ron of the switch when Vdd is 36V,see Table 4 in the datasheet.  Actually its 310ohms typical at +25C.

    For lower Vdd single supply levels Ron will increase, see Table 3 for the 12V case.

    But I think I am not understanding your use case correctly. Can you confirm the Vdd and Vss of your setup. 



  • Hi Eric-san

    Thank you for your reply.

    So、My question is to became a just simple.

    Therefore、my question is to become a simple.

    Can it sink without sourcing output to circuit behind, if output device become short?

    Best regards,


  • For the conditions you mentioned in your original posting you can model the D pin of the switch as a resistor of about 300 Ohm connected to 5V. If your D pin is shorted to ground then roughly 16 mA will flow through the switch into that ground.

    If the "ground" is really a body diode of an OPAMP input with a grounded supply then that diode would conduct slightly less current due to the diode drop. Your OPAMP might or might not like this...

    If this explanation does not help, because I misunderstood the question then maybe you could attach a screen shot of the schematics of your circuit to help us understand what you are after...