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LM27761: output voltage

Taito Takemura
Intellectual 1610 points
Part Number: LM27761
Other Parts Discussed in Thread: LM4766,

Hi Support Team,

Could you please let me confirm how to set the output voltage of the LM4766? I'd like to make the output voltage as close as possible to -5V with an input voltage of 5V.

According to Equation (5) In the datasheet, It seems possible by setting R1 to 155kOhm and R2 to 50kOhm. (Vout = -1.22 x (155+50)/50 = -5.0 [V] ) ()

However, according to section 8.2.2.2 of the data sheet, "the output characteristic of this circuit can be approximated by an ideal voltage source in series with a resistance. The voltage source equals –(VIN)."  I think it says that it can be simplified as a series of voltage source -(Vin) and Rout. This indicates that Vout does not become -5V when Vin = 5V due to the voltage drop of "i x Rout". (I'd like to know also Rsw because it is essential for calculating Rout but it isn't written in the data sheet.)

In this way, it seems that the descriptions in Equation (5) and 8.2.2.2 are inconsistent. Could you please explain it? In addition, what is the minimum value of Vout that LM4766 can output when Vin = 5V?

Thank you for your help.

Best Regards,

Taito Takemura

  • 0
    TI__Genius 13160 points

    Hi Taito,

    For LM27761, there will always be a voltage drop across the charge pump switches and the FET of the integrated LDO. You need to give some headroom in your design for the charge pump output resistance drop as well as the LDO dropout voltage. These voltage drops also depend on the load current. You can find the charge pump output resistance in figure 19 and the LDO dropout voltage in figure 12 of the datasheet.

    Equation 5 is applicable when you have accounted for this headroom in your design. For example you want to convert +5V to -4.5V and the combined charge pump and LDO voltage drops is 100mV (assumption), then you have enough headroom between the absolute values of input and output voltage to take care of this drop. In this case you can set the -4.5V output using equation 5. In case you wanted an output of -5V, you can still use equation 5, to set the output voltage. But there wouldn't be any headroom and the output will only reach -4.9V.

    Best regards,

    Varun

  • 0 in reply to Varun John
    TI__Intellectual 1610 points

    Hi Varun,

    Thank you for your reply.

    I understood that headroom is required when we design the LM27661 using Equation 5.

    I have an additional question in the absence of headroom.

    - Could you please tell me about the lower limit of Vout? I would be grateful if you could tell me how to calculate it. For example, if 5 V is input, what voltage can Vout be set at the minimum?

    I consider I can calculate it if I know the expression of a voltage drop across the charge pump switches and the FET of the integrated LDO. ( I think the Rout expression (1) represents  a voltage drop across the charge pump switches , but I don't know the numbers such as Rsw)

    Best Regards,

    Taito Takemura

  • 0 in reply to Taito Takemura
    TI__Genius 13160 points

    Hi Taito,

    What is the max load current in the application?

    The ROUT in figure 19 already includes the RSW. You just need to find the ROUT corresponding to the max load current from this graph. From figure 12 you can find out the LDO dropout corresponding to the max load current. The total voltage drop would then be ROUT * IOUT_max + V_DROPOUT. The minimum VOUT then would be = - (VIN - ROUT*IOUT_max - V_DROPOUT)

    Once you know the minimum VOUT, you can use equation 5 to set R1 and R2.

    Best regards,

    Varun

  • 0 in reply to Varun John
    TI__Intellectual 1610 points

    Hi Varun,

    My customer would like to know how far the absolute values of Vin and Vout are in the worst case.
    So it is the value of Rout (max) * Iout (max) + V_Dropout (max). Iout (max) is 250mA from the performance of LM27661 itself.
    Could you please tell me about the other values, Rout (max) and V_Dropout (max)?

    They would prefera formula to calculate without going through TI (to save steps) than a result value.

    Best Regards,

    Taito Takemura

  • 0 in reply to Taito Takemura
    TI__Genius 13160 points

    Hi Taito,

    Rout and V_Dropout have a dependence on IOUT. We don't have a formula to show this dependence. 

    If you consider IOUT_max is 250mA, then from figure 12, V_Dropout_max = 64mV. From figure 19, Rout_max = 2 Ohm.

    So VOUT = - (VIN - ROUT*IOUT_max - V_DROPOUT) = - 4.436V

    Best regards,

    Varun

  • 0 in reply to Varun John
    TI__Intellectual 1610 points

    Hi Varun,

    Thank you for your reply.

    I think the Rout obtained from Fig. 19 and V_dropout obtained from Fig. 12 are typical values. Is there a way to calculate the max value for Rout or V_dropout?  For example, llike the calculation the max value will be considered as 1.1 times the typical value.

    Best Regards,

    Taito Takemura

  • +1 in reply to Taito Takemura
    TI__Genius 13160 points

    Hi Taito,

    Unfortunately there are no max values defined for these parameters. I checked internal characterization data and it was done at 3.6V VIN and lower load currents. So there is no way to define max values.

    You could ask the customer to add an extra safety margin like 150mV to take care of process shifts.

    Best regards,

    Varun