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TCAN1043-Q1: Questions about R_serise and R_bias values

Part Number: TCAN1043-Q1

Hi team

My customer asks for confirmation of the resistance value connected to the WAKE PIN of TCAN1043.

The using formula calculated is shown below.

VSUP = 26.5V.

R_serise = Vsup/Io(wake) = 26.5V/3mA =8.4kohm.

R_serise tries to apply 10 kohm.(More than 8.4 kohm)

R_bias < ((VSUP – Vih)/Iih) – R_series = ((26.5 – (26.5-1.9V))/Iih) – Rseries = 1.9V/25uA – 10KΩ = 76KΩ

R_bias tries to apply 10 kohm.(Less than 76 kohm)

Question : Is it OK to apply both R_serise and R_bias at 10KOHM?

  • Charles,

    Are you doing this to create a resistor divider on the WAKE pin for threshold crossing as shown in the datasheet?

    If so, then putting 10k at both will create a 50% voltage divider, leaving the high voltage on the WAKE pin at 13.25V and never allowing it to cross the high threshold. The calculation for the series resistance is correct, but the combination of R_bias and R_series should limit the current into the pin at 3mA (absolute maximum), not just the one resistor, and I would recommend limiting it to something much lower than 3mA.

    Because the high level input voltage has to be a minimum of VSUP - 1.9, to cross the high logic threshold, the voltage seen by the WAKE pin will need to be at least 24.6V in this case, and the low level input will need to be at most 23.1V. 

    If my understanding of your WAKE pin configuration was incorrect, can you please share a schematic?

    Regards,

    Eric Hackett

    Texas Instruments | INT-TRX| Applications Engineer

  • In reply to Eric Hackett:

    Hi Eric.

    I can not understand being a voltage distribution.

    Customer will configure same as above TI.(Low side switch input)

    When the switch is open, the internal current source (25uA) will flow to R_Series.

    Since the switch is open, I think the voltage of the Wake Pin is ((R_bias+R_series) * 25uA)+Vsup.

    So, V_wake pin_switch open = ((10k+10k)*25uA)+26.5V = 27V

    When the switch is closed, internal current will flow through R_S and switch to GND.

    So, V_wake pin_switch closed = 10k*25uA = 250mV (If Switch impedance is 0ohm)

    For the above reason, I think that R_bias and R_series both use 10k.

    If you think I'm wrong, please tell me.

    ----------------------------------------------------------------------------------------------------------------------------------

    And I have additional questions.

    In the datasheet, we use the following formula to calculate Rb for a high side switch configuration.

    RBIAS < ((VSUP - VIH) / IIH) - RSERIES

    The customer wants to design the circuit in a Low Side Switch configuration.

    When configuring with a low side switch, is it right to change the formula as shown below?

    RBIAS < ((VSUP - VIL) / IIL) - RSERIES

  • In reply to Charles Jeon:

    Charles,

    You are correct, I got confused with the circuit, there isn't a voltage distribution happening, that's my mistake. Your original calculation of 10k for both RBIAS and RSERIES will work. 

    For the low side configuration (with the switch connected to GND), the sum of RBIAS and RSERIES can't be high enough to cause a 3.5V difference between VSUP and the WAKE pin with the current source driving 25uA through both resistors. The formula will work, but the 10k ohm RSERIES and RBIAS are sufficient for this case as well. When the switch is activated, as you said, the voltage at WAKE drops to ~250mV.

    Again, my apologies for the confusion, and please let me know if you have any other questions or concerns.

    Regards,

    Eric Hackett

    Texas Instruments | INT-TRX| Applications Engineer

  • In reply to Eric Hackett:

    Thank you