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TPSI3052-Q1 TIDA-050063: Can using this circuit to give a constant current to a resistor?

Part Number: TIDA-050063
Other Parts Discussed in Thread: TPSI3052-Q1

Hi Team,

Is it suitable that i put a heating resistor between V_Cap and -HV for battery pre-heating?

Will the switching frequency too low to keep a constant current?

The battery voltage around 120V and output current for heater around 7A

if it works, how can i choose the inductor?

Thank you

  • Hello Jianmin,

    Thanks for joining E2E and reaching out to our team!

    Is it suitable that i put a heating resistor between V_Cap and -HV for battery pre-heating?

    I'm unfamiliar with battery pre-heating, what are the requirements here?

    Are you asking if you can add a heating resistor in parallel to the DC-link capacitor? At a glance, this looks like it would increase the pre-charge time since we send a fixed amount of average current through the inductor and part of that would be getting burned in the resistor. 

    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer

  • Hi Tilden,

    Thanks for reply.

    part of that would be getting burned in the resistor

    what is it meaning?   Does it mean that  "part of the energy from average current will be converted into the heat in heating resistor"?

    Are you asking if you can add a heating resistor in parallel to the DC-link capacitor?

    Yes, I want to put a heating resistor in parallel to the DC-link capacitor.
    My ideal is using this circuit to output constant current in order to maintain a constant power on the heating resistor rather than using it for pre-charging.

    And the input of +HV  & -HV will be connected to output of a battery pack ( with Vmax. around 120V)

     

    furthermore, I see that TIDA-050063 is not available for sale, do you offer assembled board for testing?

    Once again, thank you for your response.

  • Hello Jianmin,

    Thanks for your reply here, need some more time to look into this. So I can run some calculations/simulations - what size is your DC-link capacitor?

    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer

  • Hi Tilden,

    We are planning to place 180uF or 560uF in parallel to heating resistor.

    i have used PspiceTI run sim with circuit like TIDA-050063 but with 180uF// 10 ohms heating resistor.

    And the current pass the resistor was around 3A



    If i want to create 400W heating power and output current below 7A
    How to choose the resistors and inductor?

    Thank you.

  • Hello Jianmin,

    Thanks for the update here. We need 6.32-A to create 400-W of heat through a 10-Ω resistor. To charge a 560-uF capacitor to 120-V in 100-ms (meant to ask what your charging time requirement is earlier), we need 0.672-A of average current. So, if we set peak current here to 7-A, we could use a 100-uH inductor (Isat = 8-A). 

    The TIDA-050063 project is available for PSpice here. (more information can be found in this thread). You can adjust the component values and check if it fits your requirements in case anything changes. Let me know what values you are considering and I can take another look.

    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer

  • Hi Tilden,

    Thanks for reply. We are planning to make a PCBA to testing it.

    If I keep all the values the same except change capacitor from 560-uF to 180uF, will the current through 10-Ω resistor change?

  • Hello Jianmin,

    Thanks for your reply. Typically the pre-charge time requirement and DC-link capacitance will determine the average charging current. What is the pre-charge time requirement here?

    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer

  • Hi Tilden,

    We have not gotten a actually pre-charge time requirement yet, but I think <1000ms would fulfill our clients requirement.

    Thank you.

  • Hello Jianmin,

    I'm also seeing the issue when simulating this circuit in Pspice, still trying to understand what is happening. But it looks like the DC-link capacitor (56uH) is not charging up to the V_battery voltage (120-V) once we add the heating resistor (10-Ω) in parallel and the pre-charge becomes stuck. Since the DC-link capacitor voltage cannot charge up further but is not yet up to V_battery voltage (120-V), it becomes stuck at the end.. I need some more time to think about this, will try to respond by Friday (17 Nov).





    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer

  • Hello Jianmin,

    Thanks for your patience here, I understand some of the issue now. The problem was that the average current sent through the inductor was less than the current drawn through the 10-ohm heating resistor. So earlier, we were sending an average of 4-A through the inductor (blue). Once V_cap = 40-V (pink), the current through the 10-ohm heating resistor (red) becomes equal to the current coming in (blue), so the capacitor cannot charge up further and becomes stuck at around 40-V.

    We can fix this by making sure I_AVG ≥ I_charge + I_heat. 

    For example, to determine I_charge

    Let's say we want to charge 180-uF to 120-V in 100 ms.

    • C = 180 uF, V = 120-V
    • Q = CV
    • Q = 0.0216 C
    • I_charge = Q/t
    • I_charge = 0.216 A

    To determine I_heat

    • We want 400-W power dissipation in a heating resistor
    • I_heat = 400-W/120-V
    • I_heat = 3.33-A
    • R_heat = 120-V/3.33-A
    • set R_heat = 36-ohm

    Determining I_AVG

    • I_AVG ≥ I_charge + I_heat
    • I_AVG ≥ 3.55-A
    • set I_AVG = 3.75-A (I_AVG_max = 7-A, I_AVG_min = 0.5-A)
      • Rbot (R6) = 2.32k
      • Rtop (R5) = 200k
      • Rhys (R7) = 15.4k

    I'm not sure why the charge time here is 16 ms, need to look into this further. I also don't understand all the limiting factors in your design (for example if the heating resistor needs to be smaller, if the inductor needs to be smaller, etc), so I've attached the Pspice project for you to experiment with. 
    TIDA-050063_Jianmin.zip

    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer

  • Hi Tilden,

    Thank you for this comprehensive answer.

    If we have much longer pre-charge time, such as 3 or 5 seconds, the I_charge will be small.

    That means  I_heat will dominant in  "I_AVG ≥ I_charge + I_heat", and the DC-link capacitance will not significantly affect the I_AVG, right?"

    We aim to set R_heat between 10- and 20-ohm

    Our target is to maintain I_AVG at 5±​​1A without DC-link capacitor voltage reach V_battery voltage( means NMOS keep switching)

    In this situation, do you have any recommendations for selecting the appropriate inductor and setting the values for Rbot, Rtop, and Rhys?

    Are there any specific considerations, besides the I-sat of the inductor, that I should keep in mind when choosing the inductor?

    Thank you.

  • Hello Jianmin,

    I'm still trying to understand how the heating resistor affects the pre-charge time. It definitely increases the time compared to pre-charge with no heating resistor, but I'm not sure what the best way is to estimate this increase. Looking at the simulation when I probe the current into the inductor, capacitor, and resistor, I get the following results:

    Trying to estimate pre-charge time while account for heating resistor

    It looks like current through the heating resistor (red) starts at 0-A and settles to 3.3-A. So let's linearize this and say I_heat_avg = 1.67-A

    The inductor current we set to begin at 3.75-A and settles to 3.33-A. So let's linearize this and say I_L_avg = 3.50-A

    The average current to the inductor must be the difference between the two. So I_cap_avg = 3.5-1.67 = 1.83-A.

    t = Q/I

    t = 0.0216 C / 1.83-A

    t = 11.8 ms (this might be okay for an estimate considering the decay and rise are nonlinear. Can use this estimation for finding initial passive component values and need to simulate for better accuracy at the end)

    That means  I_heat will dominant in  "I_AVG ≥ I_charge + I_heat", and the DC-link capacitance will not significantly affect the I_AVG, right?"

    We aim to set R_heat between 10- and 20-ohm

    Our target is to maintain I_AVG at 5±​​1A without DC-link capacitor voltage reach V_battery voltage( means NMOS keep switching)

    In this situation, do you have any recommendations for selecting the appropriate inductor and setting the values for Rbot, Rtop, and Rhys?

    Are there any specific considerations, besides the I-sat of the inductor, that I should keep in mind when choosing the inductor?

    I'll do an example and hopefully that can answer most of your questions here. 

    Design requirements

    • C = 180-uF
    • 10-Ω < R_heat < 20-Ω
    • I_AVG = 5-A ± ​​1 (settles at 5-A, DC-link does not reach V_battery)

    Design selection

    Inductor

    Disclaimer: I am not an expert on inductors. Looking at these inductors, it looks like lower inductance --> can handle more current. I think at minimum Irms > I_AVG (5-A) and Isat > I_AVG_max (6-A). So the largest inductor we can use from this list is 390-uH.


    However, we are also limited by the TPSI3052-Q1 switching frequency. Smaller inductance --> needs higher switching frequency. For the FETs used in the design guide, the TPSI3052-Q1 can switch for a maximum of 67-kHz. Let's select the 330-uH inductor for some margin and calculate the maximum switching frequency here.

    F_MAX = 120 / (4*330uH*2A)

    F_MAX45-kHz (Less than 67-kHz, so the 330-uH inductor will work here)

    Resistor Hysteresis (more information about hysteresis reference design can be found here)

    I_AVG = 5-A ± ​​1 (so 4-A to 6-A). We have a 100-mΩ sense resistor, so the voltage thresholds are 0.4-V and 0.6-V. 

    • Rbot = 2.32-kΩ
    • Rtop = 25.5-kΩ
    • Rhys = 51.1-kΩ 

    Heating Resistor

    If we want to settle at 5-A while still switching, we need to select a resistor to draw 5-A before V_cap reaches 120-V. 

    • R_heat = 20-Ω (V_cap will charge up to 100-V)

    Estimating Pre-charge Time

    • I_L_avg = 5-A
    • I_heat_avg = 2.5-A
    • I_C_avg = 2.5-A

      t = Q/I
      t = 0.018 C / 2.5-A
      t = 7.2-ms

    Simulation Results

    Schematic

    Waveforms

    Please let me know if you have further questions.

    Best regards,
    Tilden Chen


    Solid State Relays | Applications Engineer