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TL7660: Output Current

Part Number: TL7660

Hi All,

I have aquestion about TL7660.

1) Can the TL7660 pass an output current (Iout) of 100mA?


2) Does RSW fluctuate depending on the specification conditions?
    Please tell me Min and Max of RSW.


3) I have a question about the graph below.
    Am I correct in understanding that when the IL current increases, the output voltage increases proportionally?


4) I have a question about the graph below.
    Since IL=25mA when Icc=50mA, I think the efficiency η=50%, but the graph shows 90%.   
     Is the efficiency stated in the graph the input/output power efficiency? or efficiency by another calculation?


Best Regards,
Ishiwata

  • Hello Shuji,

    Thanks for reaching out to us via e2e.

    1) Can the TL7660 pass an output current (Iout) of 100mA?
    In theory, yes. But please keep in mind that the output of the TL7660 behaves like a power source with a much higher output impedance as for example a lab supply.
    Therefore, when you draw 100mA, the output voltage will drop to around 0.6V.

    2) Does RSW fluctuate depending on the specification conditions?
    Please tell me Min and Max of RSW.

    RSW is just used to show the principle.
    ROUT is more important, It depends on input voltage, frequency and temperature (see diagrams in the datasheet.
    As a simplification you can use the table below and add the influence of the ESR of the capacitor:
    RO ≈ ROUT from the table below + 5 * ESR_C

    3) I have a question about the graph below.
    Am I correct in understanding that when the IL current increases, the output voltage increases proportionally?
    The diagram shows negative voltages.
    The magnitude of the voltage actually decreases with higher load current. (From -5V decreasing to only 3.25V)

    4) I have a question about the graph below.
    Since IL=25mA when Icc=50mA, I think the efficiency η=50%, but the graph shows 90%.
    Is the efficiency stated in the graph the input/output power efficiency? or efficiency by another calculation?
    This graph is meant for the voltage doubler use case.
    The inupt power equals the output power: Uin * Iin = Uout * Iout.
    Output voltage = 2 * input voltage. But therefore, output current = 1/2 * input current

    Best regards
    Harry

  • Hello Harry,

    Thank you for answering and support.

    I have an additional question about 2).

    In the case of the calculation formula used here, Ro = 66 + 5 (ESRC), but in the graph below, it may be less than Ro = 66Ω.
    Please tell me the reason why it becomes 66Ω or less.

    Best Regards,
    Ishiwata

  • Hi Ishiwata,

    If I understand the datasheet correctly, the graph shows the Rout behavior for different Vcc and Io values based on the temperature.
    The calculation with Ro = 46 + 20 + 5 (ESRc) is just an example for the given parameters for a first order approximation.

    Please let me know if you have additional questions.

    Best regards,
    Niklas