Q. on Diode Equation

Recently I was looking into the use of a diode as a temperature sensor. I
decided to start with the "Diode Equation":

Id=Is*(exp(qVd/NKT)-1)

This equation predicts that an increase in temperature T will result in an
increase in diode voltage Vd, assuming constant current. But I know from
experience that the opposite is true. Diode voltage drops as temperature
increases.

What gives? Is saturation current (Is) a function of temperature, strong
enough to reverse the upward dVd/dT predicted by the diode equation? Or do I
completely misunderstand the basics here?

On Wed, 25 May 2005 19:31:22 -0700, "Blake"
<(E-Mail Removed)> wrote:

>Recently I was looking into the use of a diode as a temperature sensor. I
>decided to start with the "Diode Equation":
>
>Id=Is*(exp(qVd/NKT)-1)
>
>This equation predicts that an increase in temperature T will result in an
>increase in diode voltage Vd, assuming constant current. But I know from
>experience that the opposite is true. Diode voltage drops as temperature
>increases.
>
>What gives? Is saturation current (Is) a function of temperature, strong
>enough to reverse the upward dVd/dT predicted by the diode equation? Or do I
>completely misunderstand the basics here?

Yes, that's why. If you are interested, I'll post that equation, as
well.

Jon

Blake wrote:
> Recently I was looking into the use of a diode as a temperature
> sensor. I decided to start with the "Diode Equation":
>
> Id=Is*(exp(qVd/NKT)-1)
>
> This equation predicts that an increase in temperature T will result
> in an increase in diode voltage Vd, assuming constant current. But I
> know from experience that the opposite is true. Diode voltage drops
> as temperature increases.
>
> What gives? Is saturation current (Is) a function of temperature,
> strong enough to reverse the upward dVd/dT predicted by the diode
> equation? Or do I completely misunderstand the basics here?

Yes you misunderstand, look at the equation. The temperature T is appearing
in the denominator making the *current* increase with constant Vd, or in
other words you need less voltage for the same current. The relationship
dVd/dT (partial d) is roughly -2mV/K.
--
ciao Ban
Bordighera, Italy

On Wed, 25 May 2005 19:31:22 -0700, "Blake"
<(E-Mail Removed)> wrote:

>Recently I was looking into the use of a diode as a temperature sensor.

If anybody is interested (to get maybe some further ideas) of analog
converter for diode to thermistor read-outs:
how to make a gadget that can do that for Cpu onDie diode for a PC
MoBo that is natively not supporting that feature:
my project is on my site under electronics ...
may be of use for some other purpose for someone ... :-)
--
Regards , SPAJKY ®
mail addr. @ my site @ http://www.spajky.vze.com
3rd Ann.: - "Tualatin OC-ed / BX-Slot1 / inaudible setup!"

On Thu, 26 May 2005 03:18:28 GMT, "Ban" <(E-Mail Removed)> wrote:

>Blake wrote:
>> Recently I was looking into the use of a diode as a temperature
>> sensor. I decided to start with the "Diode Equation":
>>
>> Id=Is*(exp(qVd/NKT)-1)
>>
>> This equation predicts that an increase in temperature T will result
>> in an increase in diode voltage Vd, assuming constant current. But I
>> know from experience that the opposite is true. Diode voltage drops
>> as temperature increases.
>>
>> What gives? Is saturation current (Is) a function of temperature,
>> strong enough to reverse the upward dVd/dT predicted by the diode
>> equation? Or do I completely misunderstand the basics here?
>
>Yes you misunderstand, look at the equation. The temperature T is appearing
>in the denominator making the *current* increase with constant Vd, or in
>other words you need less voltage for the same current. The relationship
>dVd/dT (partial d) is roughly -2mV/K.

I think you may be wrong in your rationalizing, here. An increase in
T would cause a _decrease_ in Id, not an increase. Thus, it would
require an increase in Vd in order to keep Id unchanged.

In this case, one needs to solve for Vd before taking the derivative
with respect to T. Thus:

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

The derivative (assuming 'Is' is constant relative to T) is then
trivially:

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

and thus positive-going with respect to T.

I believe the OP was not mixed up about the question. See my other
reply.

Jon

On Thu, 26 May 2005 03:17:38 GMT, Jonathan Kirwan
<(E-Mail Removed)> wrote:

>Yes, that's why. If you are interested, I'll post that equation, as
>well.

Ignoring the emission coefficient (taken as 1, for now), the equation
is:

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

which becomes:

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

The derivative is 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:

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

which complicates things.

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

On Wed, 25 May 2005 19:31:22 -0700, "Blake"
<(E-Mail Removed)> wrote:

>Recently I was looking into the use of a diode as a temperature sensor. I
>decided to start with the "Diode Equation":
>
>Id=Is*(exp(qVd/NKT)-1)
>
>This equation predicts that an increase in temperature T will result in an
>increase in diode voltage Vd, assuming constant current. But I know from
>experience that the opposite is true. Diode voltage drops as temperature
>increases.
>
<snip>
Check it out. Use an standard DMM on the 2 kOhm range to put
a constant current through the diode, while reading the voltage
across it. Now try warming and cooling the diode and you will
see that the voltage is directly proportional to temperature.
In fact, it is *so* proportional that this is one of the most linear
temperature sensors you will find, from near absolute zero until
the junction melts.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Home of DaqGen, the FREEWARE signal generator

"Bob Masta" <(E-Mail Removed)> wrote in message news:(E-Mail Removed)...
> On Wed, 25 May 2005 19:31:22 -0700, "Blake"
> <(E-Mail Removed)> wrote:
>
>>Recently I was looking into the use of a diode as a temperature sensor. I
>>decided to start with the "Diode Equation":
>>
>>Id=Is*(exp(qVd/NKT)-1)
>>
>>This equation predicts that an increase in temperature T will result in an
>>increase in diode voltage Vd, assuming constant current. But I know from
>>experience that the opposite is true. Diode voltage drops as temperature
>>increases.
>>
> <snip>
> Check it out. Use an standard DMM on the 2 kOhm range to put
> a constant current through the diode, while reading the voltage
> across it. Now try warming and cooling the diode and you will
> see that the voltage is directly proportional to temperature.

I suggest you take your own advice and report back.
Then read Mr. Kirwan's contribution to see why you
are seriously mistaken with your assertion.

> In fact, it is *so* proportional that this is one of the most linear
> temperature sensors you will find, from near absolute zero until
> the junction melts.

I expect the leads will melt first. And long
before that happens, the approximations
underlying the diode equation will fail to
be useful predictors of the junction voltage.

--
--Larry Brasfield
email: (E-Mail Removed)
Above views may belong only to me.

Jonathan Kirwan wrote:
> On Thu, 26 May 2005 03:18:28 GMT, "Ban" <(E-Mail Removed)> wrote:
>
>> Blake wrote:
>>> Recently I was looking into the use of a diode as a temperature
>>> sensor. I decided to start with the "Diode Equation":
>>>
>>> Id=Is*(exp(qVd/NKT)-1)
>>>
>>> This equation predicts that an increase in temperature T will result
>>> in an increase in diode voltage Vd, assuming constant current. But I
>>> know from experience that the opposite is true. Diode voltage drops
>>> as temperature increases.
>>>
>>> What gives? Is saturation current (Is) a function of temperature,
>>> strong enough to reverse the upward dVd/dT predicted by the diode
>>> equation? Or do I completely misunderstand the basics here?
>>
>> Yes you misunderstand, look at the equation. The temperature T is
>> appearing in the denominator making the *current* increase with
>> constant Vd, or in other words you need less voltage for the same
>> current. The relationship dVd/dT (partial d) is roughly -2mV/K.
>
> I think you may be wrong in your rationalizing, here. An increase in
> T would cause a _decrease_ in Id, not an increase. Thus, it would
> require an increase in Vd in order to keep Id unchanged.
>
> In this case, one needs to solve for Vd before taking the derivative
> with respect to T. Thus:
>
> Vd(T) = (kT/q) * ln( 1+Ic/Is )
>
> The derivative (assuming 'Is' is constant relative to T) is then
> trivially:
>
> d Vd(T) = (k/q) * ln( 1+Ic/Is ) dT
>
> and thus positive-going with respect to T.
>
> I believe the OP was not mixed up about the question. See my other
> reply.
>
> Jon

I throw in the formula for Is(T) = Isk * e(-Eg/mkT)
Eg = 1.11eV bandgap of silicon; m=q=1.1; Isk= 1E6A; k=1.38E-23Ws/K
Boltzmann

for 300K we get Is = 11.2pA; for 310K Is= 39.3pA
We put that into your formula and we get for 1mA current @300K 520.7mV and
@310K 501.0mV
-19.7mV/10K. The influence of Is is much bigger than the small change in
Vt= from 25.9mV to 26,7mV.
THX for the correction, Jon

--
ciao Ban
Bordighera, Italy

On Thu, 26 May 2005 13:39:00 GMT, "Ban" <(E-Mail Removed)> wrote:

>THX for the correction, Jon

No problem. I'm just a hobbyist in electronics, so I rarely get to
say all that much about it. But I'm okay in math, so I can work a
model once in a while, here and there. It's nice when I can say
something but it doesn't mean I know a darned thing about the reality,
since my experience is so very limited.

Jon