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TPS2H160-Q1: Current Limit working on TPS2H160-Q1

Part Number: TPS2H160-Q1

Hi Team,

One of my customer requested for additional information on how the current limit is working for TPS2H160-Q1 and on below points. 

1)    During tests observed voltage of 0.78V when there is over current as well as when there is set limit current flowing and also in Open Load condition when no current is flowing through the load? How is this possible? 

2)    In Datasheet 8.3.6.1, it says that the MCU should turn off the switch in overload condition (highlighted in yellow). Can you clarify if the OUTx pins will go low by itself or will the MCU be required to make the output low? Right now, the statements are contradicting each other. 

3)    But in Table2 Fault table it is mentioned as in overload condition, the OUTx pins will go to low by itself.

Thanks in advance for the help and support.

Regards, Shinu Mathew.

  • Hi Shinu,

    For your question about the response of the device during overload condition:

    The switch will turn off automatically during an overload condition. It does this by the switch reaching thermal shutdown. On the other hand, if you have an MCU parsing the current sense value and an overload condition is seen, the MCU could respond by turning off the switch before thermal shutdown is reached. Basically, the switch will protect itself but an MCU checking the current flowing through the switch can add a second layer of protection to the device and downstream loads.

    For your question about the observed voltage, I have a few follow up queries:

    1. What version of the device is your customer using? I am assuming Version B.

    2. What supply and sense resistor is being used?

    3. Can you share a little more information about the connections around the CS pin?

    4. Is 0.78V the expected fault voltage?

    Assuming Version B, Open load and current limit fault triggering will result in the voltage on the CS being pulled high. In the linear reporting range (current below limit) the voltage is below the fault voltage. Please refer to Sections 8.3.5.2 and 8.3.5.3 in the datasheet to see the muxing of the current sense pin.

  • Hi Shreyas,

    Thanks for the reply and inputs. Please find the below answers,

    1. What version of the device is your customer using? I am assuming Version B.                                                                                                                       Customer is using version A
    2. What supply and sense resistor is being used?                                                                                                                                                                       Supply = 12V and current limit resistor, Rcl = 806 ohms (implies 2.5A limit per channel)
    3. Can you share a little more information about the connections around the CS pin?                                                                                                                      No CS Pin in B Version
    4. Is 0.78V the expected fault voltage?                                                                                                                                                                                              No, as per datasheet, as the Current increases beyond current limit, the voltage across Rcl should increase and then only will it cross 0.8V from which the internal op-amp shall determine it is a fault (this is customer understanding)

    Regards, Shinu Mathew.

  • Hi Shinu,

    Apologies for my previous post, I thought your customer is using version B devices. 

    The voltage on the CL pin is a reference voltage set with the help of the current limit resistor for the internal resistor divider network. Irrespective of your loading condition, you will see this voltage on the CL pin.  You cannot use the CL pin to sense load current flowing through the device. This is performed with the CS pin(not available on version A). Please refer to Figure 25 on the datasheet for a pictorial representation.

    Typically this reference will be set at 0.8V but we do expect slight variation between parts. In your customer's situation, the reference is dropped by 20mV (2.5% decrease) which causes the Current Limit to be 2.42A instead of 2.5A with 800Ohms. By decreasing resistance of the current limit resistor to around 780 Ohms, the limit can be upped back to 2.5A. The device is behaving as expected.

    I hope this answers all your questions.

  • Hi Shreyas,

    Could you explain with a 780 ohms Resistor (rated current limit of 2.5A), if our load draws 5A amps of current, how will the driver know its an over current. Could you provide a more clear explanation of the detection mechanism?

    Regards, Shinu Mathew. 

  • Hi Shinu,

    With your customer's device, the V_CL(th) is 0.78V instead of 0.8V due to slight variation. Using the same formula for R_CL(Equation 4) provided in the datasheet, this should set your current limit at 2.5A. 

    The current limit methodology can be inferred from Figure 25 in the datasheet. A reference voltage(0.78V in your case) is set on the CL pin which changes the amount of current flowing through the internal divider network. This provides a voltage value which is compared with the voltage gained through the mirror network and the comparison is used to limit the current.

    As soon as any current over 2.5A is sensed, the current limit is triggered. 

  • Hello Shreyas,

    Thanks for the inputs.

    The datasheet of TPS2H160BQPWPRQ1 mentions of the CS pin being internally pulled up to a range between 4.5V and 6.5V during a fault condition. 

    What is the best suggested method to clamp the voltage to 3.3V or lesser?

    This little strange and would like to understand why the voltage levels are incompatible as these signals will be connected to MCU and  most of the MCU are running at 3.3V. Why in the design it is connected in a way that will generate voltages that are non standard and can lead to additional implementation in order to bring them down to MCU levels.

     Ideally customer would like to avoid anything extra just to do level translation.

    Please help with your comments and inputs.

    Regards, Shinu Mathew.

  • Hi Shinu,

    To clamp the voltage a simple 3.3V zener diode can be used on this pin. A low power zener is sufficient as the amount of current flowing through this pin is small. We suggest this to users who are running a 3.3V input vs a 5V input. 

    The pull up to high voltage exists so that there is sufficient separation between normal operation and fault reporting. This device was also built with the intention of working with 5V inputs and so the internal pull up pulls the voltage to that value.

  • Hi Shreyas,

    Using a Zener is not a neat solution as it takes time to start conducting and before that MCU pin will see a higher voltage. So the concern is we see voltage of 6V and higher, so intent of working at 5V doesn't do work out well.

     Please help with your inputs.

    Regards, Shinu Mathew. 

  • Hi Shinu,

    There is no way to disable the internal pullup.

    Using a Zener clamp is a solution we have seen and recommended to the field on multiple occasion with no issues. It is an easy addition to the circuit in place. I don't think a TVS diode would work here since the high voltage is not a transient.

    Another solution is to use a resistor divider network on the CS pin with the node in between used as the voltage input to the MCU. But this again complicates the circuitry and changes the R_CS.

    My recommendation would still be to try using a zener clamp.