Remeber:

Apllying a current to a passive sensor (or even applying a current/voltage in general) will result in heat-dispersion, so if the temperature of the sensor is a relavant factor for determening the measure of the target, self-heating can be a problem.

To reduce this problem we have seen a method, that we will call “Train of Pulses”, which consists in changing the source from a DC Voltage to a repeating square wave function or “train of pulses”.
This special function has 2 distinct periods:

$T$ : full period or normal defined period.

$T_{ON}$ : period in which the source feeds current into the system.

$δ$ called the “duty cycle” is the ratio between this two periods.

==In this case we need to read the output only during the $T_{ON}$ periods, otherwise we’ll measuring nothing==.
So we need to have an electrical system fast enough to adapt to this pulses, which is usually possible.

To evaluate how much this approach reduces the effect of self-heating we need to:

Construct the thermal lumped parameter system:

NOTE: The heat flux $\dot{Q}_{S}$ (the flow quantity of the thermal lumped parameter system), since we consider only the electronic circuit (we are evaluationg only the self-heating effect) is exaclty the power of the electronic circuit.

To find exaclty how much the self-heating value is we need to find using the laplace tranformation $T (s)$ and isolate the part containg $\dot{Q}_{S}$ , the result will be: $self-heating ≜ Δ T_{S} (s) = \dot{Q}_{S} \frac{R _{XS}}{1 + R _{XS} C _{S}}$ (This is not so imporant)
==While it’sIMPORTANTE to remeber that the PT100 as any other sensor suffers form self heating, in actual number the measured value changes of about $≃ 0.5 \frac{°C}{mW}$ ==.

Since the self-heating depends directly to $P (t)$ , if we reduce the power or medium power, we will reduce the self-heating effect, so we can calculate:

So the final result is that, if we use a train of pulses instead of a DC input, we can reduce the self-heating effect by a factor of $δ$ (the duty cycle):

Often you find this kind of diagram to relate in logaritmic form the voltage and current:

Each diagonal line represent a different resistance value, or a value for power following the formulas: $lo g V = lo g R + lo g I$ and: $lo g V = lo g P - lo g I$

This diagonal lines are ideal, but due to self-heating the relationship between $I$ and $V$ is not linear, as show under.

If I draw lines which describes different constant powers, you will have this behavior here, due to self-heating:

> - You see that resistance remains with the same value until something happens due to the fact that the power is too high (self-heating), and the resistence value tend to diminsh.

Usually these diagrams are drawn wiht a $T_{x}$ constant, usally taken at ambient temperature ( $25° C$ ), the medium being air.

Memory Card

Index

Train of Pulses
Thermal Circuit
Low Pass Function
Fundamental Frequency on Train of Pulses
DC Errors on Train of Pulses
~ Real World Example • How Data Sheet Represent Self-Heating

Train of Pulses

The idea of having variables power for the resistance can be useful also to reduce the effect of self-eating.
In fact, you can use for instance a current source which generates a train of pulses.
==So we can draw our current which I input to the resistance, with a defined “full period” $T$ from one pulse to the other, and another much shorter period $T_{ON}$ where the generator is on (the period of a pulse of current)==.
==So, the duty cycle $δ$ is defined as the $T_{ON}$ divided by the full period $T$ ==.
And obviously you have to perform the reading during the period $T_{ON}$ .
So you should have an AD converter, which converts only in periods where the power is on.
And this means that you need something which synchronizes the excitation and the reading of the resistance.

Definition of Duty Cycle: ==The duty cycle is a term commonly used in electronics and signal processing to describe the ratio of the time a signal is ON (active) compared to the total period of the signal.
It is often expressed as a percentage==.

Thermal Circuit

Here you have an equivalent thermal system (or circuit), where you have your sensor, with temperature $T_{s}$ , and the thermal capacitance $C_{s}$ of the sensor,

==We have that the power $P (t)$ which is the heat flux ( $\dot{Q}_{s}$ ) that we give to our resistance is equal to $R I^{2} (t)$ .
$R$ will vary with temperature but we consider it constant or almost constant.
So the temperature of the sensor ase we have already seen it is $T_{s} (s)$ (in Laplace domain)
==The first component of the $T_{s} (s)$ equation is the error we have for self heating==.
==We can go to the freqency domain and write the error as $Δ T_{s} (ω)$ ==.
== $τ$ recall is due to the product of the thermal resistance and the sensor’s thermal capacitance==.
As an example we consider $τ$ as a PT-100 surface film resistance, that is something which is approximately $0.5 \frac{° C}{mW}$ , which is typical value that you can find.

So we have that the self heating effect (for a simple body+sensor example) is defined as: $Δ T_{S} (ω) = \frac{P ( ω ) R _{XS}}{1 + jω s}$

Low Pass Function

The error (self-heating effect $Δ T_{S}$ ) divided by the power you give to your sensor, in the frequency domain, it’s a low pass function.

Fundamental Frequency on Train of Pulses

We have defined the power as $P (t) = \dot{Q}_{S} = R I^{2}$ Where $I (t)$ is defined as a train of pulses:

==So you will have a fundamental frequency $f_{0}$ ==

Let’s represent our $f_{0}$ , and obviously also the higher harmonics ( $2 f_{0}$ , $3 f_{0}$ , …) all above the cutoff frequency of the thermal filter:

Then the only thing which you will have, as a steady state response, is the average value, (the delta at $ω = 0$ ), which represents the DC component of the power.

The dotted line represents the DC component of the generator:

We can evaluate it:

==So you have reduced the average power, with respect to the DC, by a factor which is given by the duty cycle $δ$ ==.

DC Errors on Train of Pulses

Let’s compere now the self-heating error on two different exitation, one pulsed (train of pulses we have seen now) and the other given by a DC power source:

Both error are considered as “DC errors”, what changes is the exitacion
So by having a pulsed exitacion, instead of a DC exitation we will have reduced the error, by a factor of $δ$ the duty cycle.

I need to read the resistance at the points where the generator is on, where for a brief moment the currnet is costant, that's why we call it still a DC error:

We need to have an electrical system fast enough to adapt to this pulses, which is usually possible.

~ Real World Example • How Data Sheet Represent Self-Heating

Often you find this kind of diagram to relate in logaritmic form the voltage and current:

Each diagonal line represent a different resistance value, or a value for power following the formulas: $lo g V = lo g R + lo g I$ and: $lo g V = lo g P - lo g I$
This diagonal lines are ideal, but due to self-heating the relationship between $I$ and $V$ is not linear, as show under.

If I draw lines which describes different constant powers, you will have this behavior here, due to self-heating:

You see that resistance remains with the same value until something happens due to the fact that the power is too high (self-heating), and the resistence value tend to diminsh.
Usually these diagrams are drawn wiht a $T_{x}$ constant, usally taken at ambient temperature ( $25° C$ ), the medium being air.

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SaM - Reduce the Effect of Self-Heating