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LM4880: Can I connect the shutdown pin directly to a ucontroler or you be better to put a resistor in parallel with the shutdown pin?

Part Number: LM4880

Looking at the LM4880 datasheet, it says that it is possible to connect the shutdown pin directly to a GPIO pin of an ucontroller respecting the voltage of the GPIO. My question is if it is not better to put a referencial resistor in parallel to this pin. What would be a good pratice in this case?

Thanks in advance.

Best regards.

  • Hi Rafael,

    It is best if you place a resistor to ensure a known state of the shutdown pin, eg. you can place a pull-up resistor along with uC GPIO such that LM4880 is shutdown by default, and it's only enabled when uC pulls the pin down to GND.

    Best regards,
    -Ivan Salazar
    Applications Engineer

  • Hi Ivan

    How are you? Thanks for your fast reply. Let see if I get it right. The ideia is to connect the GPIO (HsMute) in serie with a resistor, connected to the shutdown pin and the GND, is that correct? See the figure below.

    Thanks in advance.

    Best regards.

  • Hi Rafael,

    That is a way to do it, in that case HsMute would need to pull-up the pin to shutdown LM4880.
    What I suggested is something like this:

    Best regards,
    -Ivan Salazar
    Applications Engineer

  • Hello Ivan

    Thanks again for the feedback. So let me see if I understood correctly. The HsMute is a GPIO configure to ouput so I control it digitaly. If HsMute is LOW, the current between VDD and Shutdown will be very low thus the Shutdownpin will have almost 5 V on it. But if the HsMute is HIGH the shutdownpin will be 3.3 V right?

    Best regards.

  • Hi Rafael,

    You're correct, although I don't know what is your expected VDD for LM4880 and for the uC. Let me add more details below.
    Usually you can configure GPIO outputs from uC as open-drain, then you'd have two cases:

    • GPIO LO means uC will pull down the node, then Shutdown will see GND potential (Low state, enabling the LM4880)
    • GPIO HI means uC will release the node, then Shutdown will see VDD potential from pull-up resistor (High state, disabling the LM4880)

    If uC is push-pull instead of open-drain, it would work similarly:

    • GPIO LO means uC will pull down the node, then Shutdown will see GND potential (Low state, enabling the LM4880)
    • GPIO HI means uC will force voltage to the node (be careful to force only voltage that is supported by LM4880, this is recommended only if both devices have the same VDD reference), then Shutdown will see VDD potential from pull-up resistor (High state, disabling the LM4880)

    Best regards,
    -Ivan Salazar
    Applications Engineer

  • Hello Ivan

    Thanks for your reply and attention. Being a little more specific, I'm using a Raspberry Pico uC, with 3.3V level on the GPIOs and a VDD of 5V. Considering the push-pull config. When the GPIO LO means the uC will pull down the node and the Shutdown will see the GND as you said and I agree with you. But when the GPIO HI means the Shutdown will see 3.3V and will have something about 1.7V drop on the resistor, but it this configuration (push-pull) I don't see the need of the VDD and resistor, I could have only the GPIO and shutdown node. Do you agree?

    As I saw in the datasheet, the use of a pull-up will be in handy when I want to control the shutdown pin whit a button, then it makes sense.

    Best regards.

  • Hi Rafael,

    You're correct on your analysis, the only benefit from using external resistor with a push-pull controller would be to have a known state in the situation that amplifier is powered but the controller is not. If that is not a concern for your application then you can just go without the resistor.

    Best regards,
    -Ivan Salazar
    Applications Engineer

  • HI Ivan

    Thanks again. Indeed the source of Vdd that is 5V is the same that will feed the uC, so I don't think this concern my application.

    Thanks very much for all your answers.

    Best regards.