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LOG114 - Rref in Figure 2 in the datasheet

Hiroshi Katsunaga
Mastermind 6480 points
Other Parts Discussed in Thread: LOG114, REF3240, REF3040

Hello,

My customer has a question about LOG114.

[Q]

Why is Rref 1.62Mohm in Figure 2 in the datasheet ?

Is this correct ?

<My understanding>

Rref = (Vref - Vcmin) / Iref

If the system is dual supply configuration, Vref=2.5V, Vcmin=0V, Iref=1uA (The use case of Figure 1), then Rref = (2.5 - 0) / 1u = 2.5Mohm

If the system is single supply configuration, Vref=4.096V, Vcmin=2.5V, Iref=1uA (The use case of Figure 2), then Rref = (4.096 - 2.5) / 1u = 1.596Mohm

If the system is single supply configuration, Vref=4.096V, Vcmin=2.5V, Rref=1.62Mohm (The use case of Figure 2), then Iref = (4.096 - 2.5) / 1.62M = 0.985uA

Is my understanding correct ?

Best Regards,

Hiroshi Katsunaga

  • 0
    TI__Guru* 93105 points

    Katsunaga-san,

    The VCMIN applied to the LOG114 A1 amplifier in Fig. 2 is the internal 2.5 V reference level. The external reference voltage is 4.096 V which is provided by the REF3040, or REF3240. The current I1 is (VREFEXT - VCMIN) / Rref,or as you calculated  I1 = (4.096 - 2.5) / 1.62M = 0.985uA. Indeed, if you need I1 to be more precise then, Rref = (4.096 V - 2.5 V) / 1uA = 1.596 Megohm (~1.6 Megohm).

    I suspect the datasheet author decided to use a 1.62 Megohm resistor because it is a standard 1 % value close to 1.6 Megohm. They could have selected a 1.58 Megohm, 1 % resistor just as well. There is a 1.6 Megohm 0.1 % standard value as well if one is willing to pay more for the Rref resistor.

    Regards, Thomas

    PA - Linear Applications Engineering

  • 0 in reply to Thomas Kuehl
    Mastermind 6480 points
    Hi Thomas-san,

    Thank you for your response.
    I understood your all comments.

    Thank you for your support !

    Best Regards,
    Hiroshi Katsunaga