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TLV3501: TLV3501 SHDN Output

Prodigy 140 points

Replies: 6

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Part Number: TLV3501

Hello,

I'm simulating the TLV3501 in TINA and when I apply the V+ (5V) to SHDN pin the input feeds through to the output.  Is this what happens or is this a simulation issue?  The datasheet is confusing because at one place it says if the SHDN pin is within 0.9V of V+ the part is disabled, which I would interpret as the output would be off (zero volts).  At another part is says that the comparator is disabled, which might mean that the input feeds through.  Here is my circuit VF4 is a 1V peak sine wave on a 2.5VDC offset being compared with a 2.5VDC offset .  I also flipped the input signals, which is how it is done in the datasheet, it still feeds through but I get a 2.5VDC on the output instead of my offset sine wave  I was wanting to use the SHDN pin to turn off the output (0V) for my application, which is why I'm asking the question.  It seems like this would only work if the vin+ was 0V.

  • Hello BTC,

    Can you re-attach your simulation? Click the paperclip icon to attach a file. Can you define what you mean by  'feed through"?

    When shutdown, the output goes high-impedance (neither sinks or sources current).

    The TLV3501 will be ENabled when the enable input voltage is < Vcc-1.7V. And will be DISabled when the input is > Vcc-0.9V. Between Vcc-1.7 and Vcc-0.9 is "gray" zone and should be avoided.

    So if you have a 5V supply, the shutdown input should be between 0 to 3.3V to "enable", and between 4.1V to 5V to disable. Between 3.3 and 4.1V should be avoided as it is "undetermined". See note 4 in the datasheet.


    Regards,

    Paul Grohe

    TI Comparators (CMPS) Applications Group

  • In reply to Paul Grohe:

    b

  • In reply to btc:

    You'll notice that VF4 is equal to VF5 if you tie pin 8 of the TLV3501 to the 5V powering the TLV3501.  I was calling this feedthrough since the output and input are equal.   The question is if this is normal.  I was hoping that disabling the device would make the output zero volts.  I sent the simulation to you with it grounded (enabled) so you can see normal operation. Just connect pin 8 to the 5V to see what I'm calling feedthrough.

  • In reply to btc:

    Hello Btc,

    Yes. This is normal and expected. When the output is disabled, it "lets go" (opens) and neither sinks or sources current.

    So the output will "float" the the input potential via R16 and R17. This is how I would expect the real device to behave.

    If you want the output to go to near zero, you can add a pull-down resistor to the output (output to GND) to pull it low during shutdown. But there will still be an influence of R16/R17 creating a voltage divider.

    Is it possible to change the reference voltage to force a high or low and not rely on the shutdown function?

    Regards,

    Paul Grohe

    TI Comparators (CMPS) Applications Group

  • In reply to Paul Grohe:

    Thanks for clarifying.  That might be possible.  I'll probably just add a logic buffer to the output that can be enabled/disabled. I just wanted to verify that was normal and not some simulation artifact.

  • In reply to btc:

    Thanks for following up and glad Paul was able to assist.  I will close the thread.

    Chuck

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