LTM4657 datasheet(16/28 Pages) AD

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LTM4657

Rev. A

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Temperature Monitoring

Measuring the absolute temperature of a diode is possible

due to the relationship between current, voltage and tem-

perature described by the classic diode equation:

ID =IS •e

η• VT

⎛

⎝⎜

⎞

⎠⎟

VD = η• VT •In

where ID is the diode current, VD is the diode voltage, η

is the ideal factor (typically close to 1.0) and IS (satura-

tion current) is a process dependent parameter. VT can

be broken out to:

VT =

k • T

where T is the diode junction temperature in Kelvin, q is

the electron charge and k is Boltzmann’s constant. VT is

approximately 26mV at room temperature (298K) and

scales linearly with Kelvin temperature. It is this linear

temperature relationship that makes diodes suitable tem-

perature sensors. The IS term in the previous equation is

the extrapolated current through a diode junction when

the diode has zero volts across the terminals. The IS term

varies from process to process, varies with temperature,

and by definition must always be less than ID. Combining

all of the constants into one term:

KD =

η•k

where KD = 8.62−5, and knowing ln(ID/IS) is always pos-

itive because ID is always greater than IS, leaves us with

the equation that:

VD = T KELVIN

(

)•KD •InID

where VD appears to increase with temperature. It is

common knowledge that a silicon diode biased with a

current source has an approximate –2mV/°C temperature

relationship (Figure 8), which is at odds with the equation.

In fact, the IS term increases with temperature, reduc-

ing the ln(ID/IS) absolute value yielding an approximate

–2mV/°C composite diode voltage slope.

APPLICATIONS INFORMATION

Figure 8. Diode Voltage VD vs Temperature T(°C)

TEMPERATURE (°C)

–50 –25

0.3

0.5

0.8

0.4

0.7

0.6

100

4626 F08

125

To obtain a linear voltage proportional to temperature we

cancel the IS variable in the natural logarithm term to

remove the IS dependency from the equation 1. This is

accomplished by measuring the diode voltage at two cur-

rents I1, and I2, where I1 = 10 • I2) and subtracting we get:

∆VD = T(KELVIN)•KD•IN

– T(KELVIN)•KD•IN

Combining like terms, then simplifying the natural log

terms yields:

ΔVD = T(KELVIN) • KD • lN(10)

and redefining constant

K'D=KD•IN(10) =

198µV

yields

ΔVD = K’D • T(KELVIN)

Solving for temperature:

T(KELVIN)

∆VD

K'D

(

°CELSIUS)= T(KELVIN)–273.15

where

300°K = 27°C

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