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LM26001 Vbias

Other Parts Discussed in Thread: LM26001

I would like to confirm Vbias of LM26001.

In the case Vout = 12V, do we need connect the voltage 3V < and 10V>?

I think we don' need it because VDD is 5.95V typ. in the case Vout = 12V (Vin > 12V).

Could you please advise the connection of Vbias?

Best regards,

Atsushi Yamauchi

  • Hi Atsushi,

    For Vout = 12V, you can connect Vbias to ground to use the internal VDD, or if available, you can connect Vbias to an external source between 3V and 10V. Since Vout is above 10V, you cannot connect Vbias to Vout.

    The benefit of connecting Vbias to external source is to reduce power loss on the internal VDD regulator. Power consumed on VDD is (VIN * Ibias). Power consumed on external voltage source is (Vext * Ibias).

    If power consumption difference is not critical in your application, it is fine to connect Vbias to ground.

    regards,

    Yang

  • Dear Yang-san,

    Thank you for your reply.

    When we use Vbias with arranged Vout =12V, which is better the below conditions?

    1. making 5V by the divider of resisters

    2. making 5V by zener diode of VZD = around 5V

    I think the divider is better because zener diode has a capacitance and it impacts output voltage, gain-phase

    and so on.

    Best regards,

    Atsushi Yamauchi

  • If you do not have a regulated 5V rail, I'd suggest just tie Vbias to ground and use the internal bias voltage.

    If you have resistor divider pair from Vout to ground, there's always leakage current through the divider. The middle point voltage will move when the current through Vbias changes. There's also I2R loss on the dividers as well. Similar with the zenor diode approach, there are probably more losses added than saved. The easiest way is to use the internal bias voltage in my opinion.

    Regards,

    Yang

  • Dear Yang-san,

    Thank you for your reply.

    I have a question.

    How much is IBIAS of VBIAS = GND?

    Is it Iq_PWM_VDD?

    Best regards,

    Atsushi Yamauchi

  • Hello,

    Ibias in PWM mode is the same no matter Vbias is connected to external rail or to ground. Ibias = IBIAS_PWM. How much current the control circuitry consumes is not going to change. The difference is which supply is supplying Ibias.

    When Vbias is connected to ground, Ibias (IBIAS_PWM) is supplied by VIN. Together with leakage current (Iq_PWM_VB) through VIN, the quiescent current on VIN becomes Iq_PWM_VDD.

    When Vbias is connected to external voltage rail, Ibias (IBIAS_PWM) is supplied by the external rail. Only leakage current (Iq_PWM_VB) goes through VIN.

    regards,

    Yang

  • Dear Yang-san,

    I don't understand when Vbias is connected to ground.

    I don't know Iq_PWM_VB for VBIAS = GND.

    Is Iq_PWM_VB for VBIAS = GND the same of it for VBIAS = 5V?

    I want to calculate power consumption in the case of VBIAS = 5V and 0V.

    In the case VBIAS = 0V: Pd = IBIAS_PWM_VB (for 5V, because I only can find it in the datasheet) *VIN +

                                                    Iq_PWM_VDD * VIN

    In the case VBIAS = 5V: Pd =  IBIAS_PWM_VB (for 5V)*VIN

    Are they correct?

    Best regards,

    Atsushi Yamauchi

     

  • Hello,

    You are correct in the case of Vbias = Gnd, power loss = Iq_PWM_VDD * VIN.

    When Vbias = 5V (external rail), power loss = Iq_PWM_VB * VIN + Ibias_PWM * 5V

    This is when you have an external 5V rail. If you are dividing 12V Vout to generate 5V, power loss on the resistors includes leakage current on R1 and R2, and Ibias^2*R loss on the top divider, say R1.

    additional power loss = 12*12 / (R1+R2) + (Ibias_PWM^2)*R1.

    If you have a zener diode, it also has these two parts of losses, not the same formula.

    Yang