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TPS27S100: Recommended Resistor for reverse polarity protection

KKF
Expert 4495 points
Part Number: TPS27S100


Hi,

When using TPS27S100 RCP circuit as shown in datasheet figure 31, does the resistor need to withstand the full reverse polarity voltage (eg for a 1k Ohm resitor and 30V reverse voltage protection, I calculate the resistor needs to withstand a Pd of ~900mW)? Is that correct? The EVM uses a 0805 resistor with only 125mW and wondering if that will be sufficient?

Thanks & BR,
KF

  • 0
    TI__Mastermind 28340 points

    Hello,

    The current flowing into the ground resistor equal to (Reverse Voltage/ground resistor). With 30V, the current is 30V/1K = 30mA. The reverse voltage protection is defined for automotive system and with limited time 60 seconds. The ground resistor can be increased if the reverse polarity takes longer time. Reverse polarity mechanism can be found in the app notes

    Regards

  • 0 in reply to Mahmoud Harmouch
    TI__Expert 4495 points

    Hi Mahmoud,

    There's an follow-up question:

    What is the best solution if a reverse polarity protection is required for nominal voltages up to ~31V and even higher voltages under surge/burst conditions? Can the resistor be increased to 4.7kOhm to avoid high power consumption during reverse polarity conditions? Which problems can occur if the value of the resistor is increased? We would like to avoid the other solution with a block diode in the power supply.

    Since this is in an industrial environment the reverse voltage cannot be limited to 60s. The load that will be connected to the highside switch has a pullin current up to 5A for 100ms and reduces to 0.5A afterwards. In the datasheet the continuous reverse current is mentioned with I(R1) = 2A and I(R2) = 3A. Since the load would switch on in a reverse polarity condition, can the output diode handle a current of 5A for 100 ms and 0.5 A continuously?

    Thanks again & best regards,
    KF

  • 0 in reply to KKF
    TI__Mastermind 28340 points
    Hi KF,
    I believe in your application the 5A reverse current is when supply voltage pin VS is connected to GND. If this is the case, you should consider I(R2) 3A typical for less than 60 seconds. I think 5A should be OK for less than 100mS but need to double check. If the peak load current is due to capacitive load and the capacitor is electrolytic, a blocking diode is a must. The electrolytic capacitor gets damaged in reverse polarity.
    For the ground resistor 4.7K is OK and nothing can occur in circuit functionality. The I/O pins resistor should be higher as well to limit the reverse current when 31V reverse polarity Occurs. The I/O resistors should be around 15K.

    Regards, Mahmoud
  • 0 in reply to Mahmoud Harmouch
    Prodigy 20 points

    Hello,

    one more question regarding reverse polarity protection. We noticed that the resistor at pin 14 (IMON) has a very high dissipation power during reverse polarity condition. At the Evalboard we set R5 to 220 Ohm and enabled GND network of the chip. But since R5 is connected to GND to avoid the offset of the GND network at the monitoring signal there are 21V across R5 when a reverse voltage of 24V is applied to the circuit. So the question is what is the best way to avoid a high power dissipation in R5 during reverse polarity condition?

    Thanks and regards!

  • 0 in reply to User_321
    TI__Genius 12410 points
    Hello,
    With R5 equal to 220 are you saying that the maximum current you are tying to measure is 9A? This is more than double the maximum normal current. The equation to pick the IMON resistor is in the datasheet.
    Most of the voltage will be dropped across the IMON resistor because there are only two body diodes connecting it to VS (Ground in reverse battery case) therefore it needs to be size appropriately to have this voltage drop.
  • 0 in reply to Cameron Phillips
    TI__Mastermind 28340 points

    Hello,

    Cameron post is correct. The FET can not handle 9A load current. If low value resistor is a must, I recommend to use more than one in series.

    Regards

  • 0 in reply to Mahmoud Harmouch
    Prodigy 20 points

    Hello,

    i did not say that there is a load current of 9A. As you can read above the load current is 5A for 100ms and 0.5A continuously. And with I_Out = 5A, K_Imon = 500, V_Imon = 2V i get a resistor value of R_Imon = 200 Ohm. V_Imon = 2V was the input range of the AD converter.

    Meanwhile i changed the input range of the AD converter and thus the resistor R_Imon to 390 Ohm. In addition i put a diode in series to R_Imon for reverse polarity protection. Since the Imon output is described as a current mirror the additional diode shouldn't cause problems and it also seems to work with the evalboard. Can you please confirm that a diode in series to R_Imon does not cause any problems?

    Thanks and Regards

  • 0 in reply to User_321
    TI__Mastermind 28340 points

    Hello,

    Thanks for the description! The Imon is a current source and the diode in series with the pull down resistor will not cause an issue. The solution should be fine.

    Regards