Question about diode temperature and forward voltage

[[email protected]]

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature. This doesn't seem to agree with the diode equation, which
suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

 

[Rich Grise]

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature. This doesn't seem to agree with the diode equation, which
suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

Where did you get this equation, and what do all of those other variables
represent?

Thanks,
Rich

 

[[email protected]]

Diode equation:

If = Is*exp( (Vf * q) / (k*T))

k = Boltzman constant
T = Temperature in Kelvin
q = charge of an electron, 1.6*10^-19 Coulombs
If = forward current
Is = saturation current
Vf = forward voltage

 

[[email protected]]

Thanks but do you know how to derive an equation so I can get dVf / dT
= -2mv / C ?

 

[Jamie]

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature. This doesn't seem to agree with the diode equation, which
suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

what your missing is the way it's being used
in the circuit.
diodes start conducting better as they get
warmer.
if you were to have a diode in a series
circuit and measure voltage that way, then you
would get the increase as indicated how ever
what i think your looking at is the diode being
used as a shunt load..
as the diode warms up it conducts better and there
for dropping the voltage on the load.
normally 2 diodes in series are used into a pot that
drive's the base of a transistor.
the pot, is to calibrate the bias point.

 

[Phil Allison]

** GG.

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature.

** Notice how only one significant digit is given.

Means actual tempco can be between -1.5 and - 2.5 mV per C

A diode needs to be calibrated to use as a thermometer.

This doesn't seem to agree with the diode equation,

** The "diode equation " has NOTHING to do with the matter.

That is a PHYSICS equation for a hypothetical IDEAL diode with NO forward
voltage drop ( hence no forward drop tempco) and no internal resistance or
other imperfections all real diodes have.




........ Phil

 

[G. Schindler]

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature. This doesn't seem to agree with the diode equation, which
suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

I think what is being missed is that in this application the diode is
usually fed with a constant current source. When this is done
everything except Vf and T become constant.

The bias current, in this case If, is selected to be much lower than Is
to avoid self heating.

When If/Is <1 then the resulting value of ln(If / Is) is negative.

I'd have to do the math to figure out the ratio for the 2mV/degree C but
I suspect that you can find that anyhow.

 

[Bob]

I think what is being missed is that in this application the diode is
usually fed with a constant current source. When this is done everything
except Vf and T become constant.

The bias current, in this case If, is selected to be much lower than Is to
avoid self heating.

When If/Is <1 then the resulting value of ln(If / Is) is negative.

I'd have to do the math to figure out the ratio for the 2mV/degree C but I
suspect that you can find that anyhow.

How do you get If/Is < 1? *Is* is REALLY small, isn't it (pico amps)?

Bob

 

[Jasen Betts]

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature. This doesn't seem to agree with the diode equation, which
suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

k is negative.

 

[[email protected]]

the diode equation is

i = i0((exp)(v/nVt)-1)

i=diode current
v=applied voltage
n=1(for germanium diode):2(for silicon diode)
i0=reverse leakage current
Vt=kt/q

k=boltzmann const
t=temp of diode junction
q=charge of electron

this eqn gives the relation between current and temp

 

[[email protected]]

the diode equation is

i = i0((exp)(v/nVt)-1)

i=diode current
v=applied voltage
n=1(for germanium diode):2(for silicon diode)
i0=reverse leakage current
Vt=kt/q

k=boltzmann const
t=temp of diode junction
q=charge of electron

this eqn gives the relation between current and temp

 

[Phil Allison]

the diode equation is

i = i0((exp)(v/nVt)-1)

i=diode current
v=applied voltage
n=1(for germanium diode):2(for silicon diode)
i0=reverse leakage current
Vt=kt/q

k=boltzmann const
t=temp of diode junction
q=charge of electron

this eqn gives the relation between current and temp

** But as any businessman knows, never **** around with your current "
temp ".





...... Phil

 

[[email protected]]

PHYSICS is the study of PHYSICAL PHENOMENA, such as the operating
characteristics of a p-n junction. That's where the diode equation
comes from. To say that the diode equation has nothing to do with the
physical operation of the device is incorrect.

 

[John Fields]

PHYSICS is the study of PHYSICAL PHENOMENA, such as the operating
characteristics of a p-n junction. That's where the diode equation
comes from. To say that the diode equation has nothing to do with the
physical operation of the device is incorrect.

---
Please read:

http://groups.google.com/support/bin/answer.py?answer=12348&topic=250

In particular:

"Summarize what you're following up.

When you click "Reply" under "show options" to follow up an existing
article, Google Groups includes the full article in quotes, with the
cursor at the top of the article. Tempting though it is to just
start
typing your message, please STOP and do two things first.
Look at the quoted text and remove parts that are irrelevant.
Then, go to the BOTTOM of the article and start typing there.
Doing this makes it much easier for your readers to get through your
post. They'll have a reminder of the relevant text before your
comment, but won't have to re-read the entire article.
And if your reply appears on a site before the original article
does,
they'll get the gist of what you're talking about."


Thank you.

 

[Ban]

I'm trying to understand how a diode can be used to detect changes in
temperature. According to all the literature I've read, a diode's
forward voltage, Vf, falls by 2 mV for every 1 degree C increase in
temperature. This doesn't seem to agree with the diode equation, which
suggests that Vf should increase with temperature.

Vf = k*T/q * ln( If / Is)

What am I missing?

Good question,
You are right with your idea, exept that you didn't consider the temperature
dependence of Is, which doubles every 10 degrees and thus together they
roughly decrease by 2mV/K

 

[[email protected]]

PHYSICS is the study of PHYSICAL PHENOMENA, such as the operating
characteristics of a p-n junction. That's where the diode equation
comes from. To say that the diode equation has nothing to do with the
physical operation of the device is incorrect.

 

[[email protected]]

Good question,
You are right with your idea, exept that you didn't consider the temperature
dependence of Is, which doubles every 10 degrees and thus together they
roughly decrease by 2mV/K

Ban,

Thank you for your insightful reply. Your approach seems correct. If I
keep If constant and increase Is with temperature, dVf/dT is indeed
negative for the rate of change that you mentioned.

I had considered that before, but I thought the change in Is would be
too small to make a difference. How are you solving for Is? I used:

Is = q*A*(Dp/Lp*pn), for a p+ n diode

A = cross-sectional area
Dp = Drift current coefficient
Lp = recombination length
pn = minority carrier concentration on the n side

Can you please tell me how you arrived at Is doubling every 10 degrees?

Thank you

 

[Rich Grise]

How do you get If/Is < 1? *Is* is REALLY small, isn't it (pico amps)?

Bob

I guess that would depend on what the definition of 'Is' is. ;-)

Sorry, couldn't help myself.

 

[Jonathan Kirwan]

Good question,
You are right with your idea, exept that you didn't consider the temperature
dependence of Is, which doubles every 10 degrees and thus together they
roughly decrease by 2mV/K

Yes, though none of the other respondents appear to know anything
about it.

Here is the EM model equation for how Is varies with temperature,
ignoring the emission coefficient (taken as 1, for now), the equation
is, for the OP:

Id(T) = Is(T) * ( e^( q*Vd / (k*T) ) - 1 )

which becomes:

Vd(T) = (k*T/q) * ln( 1 + Ic/Is(T) )

The derivative is then trivially:

d Vd(T) = (k/q) * ln( 1 + Ic/Is(T) ) dT

which is a positive trend, very nearly +2mV/K for modest Ic... but
__positive__.

However, that isn't the whole picture. Is also varies with temp:

Is(T) = Is(Tnom) * (T/Tnom)^3 * e^( -(q*Eg/k) * (1/T-1/Tnom) )

The new derivative is a bit large.

Assume:
X = T^3 * Isat * e^(q*Eg/(k*Tnom))
Y = Tnom^3 * Ic * e^(q*Eg/(k*T))

Then the derivative, I think, is:

X+Y
k*Tnom*T*( (X+Y) * ln( -------- ) - 3*Y ) - q*Eg*( X*T+Y*T+Y*Tnom )
Isat*T^3
---------------------------------------------------------------------
q * Tnom * T * (X+Y)

Tnom is the nominal temperature (Kelvin, of course) at which the
device data is taken and Eg is the effective energy gap in electron
volts for the semiconductor material. Of course, 'k' is Boltzmann's
constant and T is the temperature of interest.

Eg defaults to 1.11 eV in spice, I think. For an Ic=10uA and a stock
Isat of about 1E-15, the figure comes out to about -2.07mV/K in the
vicinity of 20 Celsius ambient.

Jon

 

[[email protected]]

Yes, though none of the other respondents appear to know anything
about it.

Here is the EM model equation for how Is varies with temperature,
ignoring the emission coefficient (taken as 1, for now), the equation
is, for the OP:

Id(T) = Is(T) * ( e^( q*Vd / (k*T) ) - 1 )

which becomes:

Vd(T) = (k*T/q) * ln( 1 + Ic/Is(T) )

The derivative is then trivially:

d Vd(T) = (k/q) * ln( 1 + Ic/Is(T) ) dT

which is a positive trend, very nearly +2mV/K for modest Ic... but
__positive__.

However, that isn't the whole picture. Is also varies with temp:

Is(T) = Is(Tnom) * (T/Tnom)^3 * e^( -(q*Eg/k) * (1/T-1/Tnom) )

The new derivative is a bit large.

Assume:
X = T^3 * Isat * e^(q*Eg/(k*Tnom))
Y = Tnom^3 * Ic * e^(q*Eg/(k*T))

Then the derivative, I think, is:

X+Y
k*Tnom*T*( (X+Y) * ln( -------- ) - 3*Y ) - q*Eg*( X*T+Y*T+Y*Tnom )
Isat*T^3
---------------------------------------------------------------------
q * Tnom * T * (X+Y)

Tnom is the nominal temperature (Kelvin, of course) at which the
device data is taken and Eg is the effective energy gap in electron
volts for the semiconductor material. Of course, 'k' is Boltzmann's
constant and T is the temperature of interest.

Eg defaults to 1.11 eV in spice, I think. For an Ic=10uA and a stock
Isat of about 1E-15, the figure comes out to about -2.07mV/K in the
vicinity of 20 Celsius ambient.

Jon

Jon,

Great. I've never seen Isat in that form. I'll do some work in matlab
and see if it works out. I've checked the spice results and it is about
what you said. Thank you for taking the time to delve into this.

Anthony