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For behavior during input open LM2904

Tomoaki Yoshida
Expert 2200 points
Other Parts Discussed in Thread: LM2904

Hi,

Please tell me, the behavior when input is open in the following circuit.

I would like to detect the input open used this circuit.

So, when input is opened, I hope that the output voltage is settle to Low or High.
Since this device is a bipolar, I'd either convergence.
Is it true?
In this case, tell me what happens in theory
When the formula, etc. I'm glad.
Thanks.
  • 0
    TI__Guru**** 153026 points
    Tomoaki-san,

    The input is a PNP so an open input would charge the capacitors via the LM2904 input bias current.
    After that time, the output would be either VOH (about VCC -1.5V) or upper input common mode range (also about VCC -1.5V).

    Place 10nF and 10k resistor close to LM2904 to keep external noise from affecting output.
  • 0 in reply to Ron Michallick
    Expert 2200 points
    Hi Ronald,

    Thank you for your reply.
    I understood that the output voltage ends up to about Vcc-1.5.

    >The input is a PNP so an open input would charge the capacitors via the LM2904 input bias current.
    Please let confirmed for sentence above.
    The input bias current is 6uA by internal Current Regulator. Is it correct?
    If so, the current is distributed to the differential, and it is that charges the external capacitor connected to the input terminal (Vin).
    The transition time of output voltage is a time charging the external capacitor. Is it also correct?

    >Place 10nF and 10k resistor close to LM2904 to keep external noise from affecting output.
    Thank you for your advice.
    I'm going to place a resistor and capacitor in close to LM2904 with your advice.

    Thanks.
  • 0 in reply to Tomoaki Yoshida
    TI__Guru**** 153026 points
    Tomoaki-san

    Input bias is typically -20nA. It is the PNP base current which is collector current divided by Beta (current gain).
    The input collectors only get a fraction of the 6uA total current each.
    So input voltage with an open input typical increase is 20nA/11nF = 1.8V/second
  • 0 in reply to Ron Michallick
    Expert 2200 points

    Hi Ron,

    Thank you for your reply.

    I was able to understand well  the output of the behavior and mechanisms.

    Thanks.