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DRV5013-Q1: Output Low level voltage

Part Number: DRV5013-Q1

Hi,Staff

Please tell me how to calculate Vol (Output Loe Level voltage).
I'm thinking of the following calculation, is there any problem?

Vcc=5V, Output-Pull up =1kΩ, Rds=50Ω

Vol = 5V/(1kΩ + 50Ω)×50Ω = 238mV(min)
Vol = 5V/(10kΩ + 50Ω) ×50Ω = 23.8mV(min)

※It is based on the data sheet Figure 17.

Please tell me how to calculate Vol = 50mV with the following question
e2e.ti.com/.../603195


best regards
cafain

  • Hi Cafain,

    I will need to check internally with our team on this.  I expect to be able to provide a response by Thursday.

    Regards,

    Mekre

  • Hi Cafain,

    There is still one more thing we have to verify.  I expect to have an answer Monday or Tuesday at the latest.  Sorry for the delay.

    Regards,
    Mekre

  • Hi Cafain,

    Sorry for the delay.  I was able to confirm internally that your VOL calculation is correct as long as the current stays below 10 mA. 

    Regards,

    Mekre

  • Hi, Mekre san

    Thank you for checking.

    If you refer to your answer, you will get the following results.
    When Vcc=5V and Io=10mA, Vol=50mV, so Ron=5Ω.
    When Vcc=24V and Io=10mA, Vol=240mV, so Ron=24Ω
    Is it OK to understand this calculation?

    However, from the datasheet(page6), when Vcc=3.3V and Io=10mA, Ron=50Ω, so Vol=500mV!?
    Is this idea wrong?

    Should I consider Vol=50mV at all temperatures if Vcc=5V and Io=10mA or less, regardless of the formula?

    best regards
    cafain

  • Hi Cafain,

    To elaborate, Ron should not vary too much across current unless the current goes above 10 mA since that would start activating the overcurrent protection feature on the device. I am not saying to design with 10 mA in mind.

    Since Ron shouldn’t vary that much if the current is less than 10 mA, you could use the VCC/(Rpullup + Ron) x Ron formula that you mentioned in your first post.

    I’m not sure I understand the equations in your first paragraph. Could you provide more details on your method of calculation?

    Regards,

    Mekre

  • Hell Mekre san

    Thank you for your support.

    In the previous question, I referred to e2e below.
    e2e.ti.com/.../603195

    When Ron=50Ω at 5V, (Pullup=1kΩ) Vol=50mV is answered.

    It does not reach 50mV when calculated.
    Calculated as 5V/(1kΩ + 50Ω) x 50Ω = 238mV !

    It can be calculated as Vol=238mV.
    But the answer is 50mV.

    Back-calculate Ron to get Vol=50mV.
    Assuming Io=10mA
    It can be calculated as Vol=10mA x 5Ω=50mV.

    ※To calculate 50mV, the pullup resistance value and power supply voltage cannot be considered.

    Please tell me how 50mV was calculated.
    (The same applies when V = 24V Vol = 240mV)

    Regards
    cafain

  • Hi Cafain-san,

    For this calculation, I recommend using the following formula: VCC/(Rpullup + Ron) x Ron

    Based on 5 V VCC value, 1k pull up resistor, and 50 Ohm Ron, you would get 238 mV as you originally mentioned.  

    Regards,

    Mekre

  • Hello Mekre san

    thanks you.
    I understand that the Ron resistance value changes depending on the current value (temperature), etc.

    I will consider it.

    Regards

    Cafain