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MAX3232E: Current consumption

user5163865
Genius 3980 points
Part Number: MAX3232E

HI,

Could you provide information on the current consumption of the transceiver IC under the following conditions?

Devices              : MAX3232EIPWR and MAX3232IPWR (probably the same current consumption)
ON Channel       : 2TX / 2RX
Capacitive load  : 100pF
Resistive load    : Connect two TX-RXs of the same device.
                             I think that only the built-in pull-down 5k is available in this case. Correct?
Data rate            : 36800bps / 250kbps

Looking at another thread, I found out that TI engineers can provide similar information.

I would be grateful if you could give me an answer.

Best regards,
Hiroshi

  • 0
    TI__Guru 53555 points

    Hi Hiroshi,

    The capacitive load of 100pF is very small, so the extra current required for signal switching will not be very large in this case. At 500kHz, this will likely be around 3mA extra Icc pessimistically. Factoring in the DC component from the resistive load (3k-ohm pessimistically) and the unloaded current needed for the IC, the total current for both TX channels switching under these conditions should be around 10mA.

    Let me know if you have any more questions.

    Regards,
    Eric Schott

  • 0 in reply to Eric Schott1
    Genius 3980 points

    Hi,

    Thank you for your reply.
    Please tell me one more.

    To calculate the power consumption, for example, when Vdd = 3.3V
    I understand that P = 3.3V * 10mA = 33mW, is that correct?

    The customer thinks that it is 66mW because "the driver voltage is doubled by the charge pump".

    The customer wants TI confirmation.
    We apologize for the inconvenience.

    Best regards,
    Hiroshi

  • 0 in reply to user5163865
    Genius 3980 points

    Sorry for posting in quick succession.

    This calculation of power consumption is related to the change of thermal resistance value due to the revision of the data sheet in the past.

    Tj = Tt + (ΨJT x Power)

    from

    Tt (temp at pkg top): 25 ℃
    ΨJT: 3.3 ℃ / W (TSSOP)
    Power: 10mA * 3.3V = 33mW

    Then

    Tj = 25 ＋ 3.3 * 33/1000 ＝ 25.1089 ℃

    I understand that there is almost no temperature rise.
    Is it correct?

    Please let me know if there are any mistakes.

    Best regards,
    Hiroshi

  • 0 in reply to user5163865
    TI__Guru 53555 points

    Hi Hiroshi,

    The current measured here as Icc is what is drawn from the Vcc pin, so this voltage should be used to calculate power consumption. However, keep in mind that not all of this power is dissipated in the same place. The DC current through the 3k-ohm load for example will be dissipated in the receiving device as well as the driver (from driving resistance). To simplify these calculations, we  can assume each receiver is also dissipating the expect DC power while both drivers are active as well. We'll also ignore the losses from the charge pump circuit for simplicity. 

    With these thermal calculations, it appears you are trying to estimate the junction temperature Tj by measuring the case temperature Tc and the power dissipation of the device. If so, this equation is correct. Because the thermal resistance between the junction and case is so small, there will be minimal difference between these two temperatures. 

    For more info on these thermal specifications and how they may be used, see this application report on the topic. 

    Regards,
    Eric Schott

  • 0 in reply to Eric Schott1
    Genius 3980 points

    Hi Eric-san,

    Thanks a lot.
    I understood.

    Best regards,
    Hiroshi