# Existe um efeito de flutuação entre calor, resistência e corrente?

15

Dizem-nos que o calor aumenta a resistência de um resistor (ou diminui sua condutância) e a corrente diminui quando a resistência é aumentada.

Assim, com menos corrente, menos calor seria dissipado, o que diminui a resistência e faz com que mais corrente flua, e novamente, mais corrente, mais calor ... Parece um ciclo interminável.

Essa flutuação ocorre em circuitos reais? Para em algum momento?

(Estou me referindo a circuitos DC, pois isso provavelmente seria muito mais complicado em circuitos CA)

Por que os engenheiros projetam circuitos oscilantes inteligentes quando podem simplesmente acionar um resistor? / sarcasmo
Dmitry Grigoryev

4
@DmitryGrigoryev: porque tal como sons de oscilador seria muito sensível ao calor ambiente (assumindo que iria funcionar)
MSalters

o que você está descrevendo é o caso em que o resistor é acionado por uma fonte de corrente constante -> P = R * I². Isso pode acontecer e é chamado de fuga térmica. Isso também significa que a fonte atual precisa fornecer cada vez mais energia (na realidade, você tem um limite, ou talvez o resistor flua ou fume.) No entanto, na maioria dos casos, você terá uma fonte de tensão. Nesse caso, P = U ^ 2 / R, isso significa que quanto maior o R, menor a energia que a fonte tem para fornecer. Isso é estabilizá-lo, se o coeficiente de temperatura for positivo
#

1
Eu sempre me perguntei sobre duas lâmpadas incandescentes idênticas em série acionadas por uma fonte de tensão. Aquele com uma resistência um pouco mais alta poderia roubar o outro de poder, e eles teriam brilho desigual. Mas um impulso momentâneo para a lâmpada fraca ou uma fome momentânea de corrente para a brilhante reverteria tal flip-flop.
Richard1941

Respostas:

10

Eu acredito que é possível construir um modelo físico simples com as idéias que você forneceu.

Em um circuito CC simples, sob uma tensão constante V e resistência ôhmica R, é possível usar a equação de potência:

$P=Vi=\frac{{V}^{2}}{R}$

Se supusermos que o sistema é constituído por um fio com comprimento constante L e área de seção transversal A, a resistência R pode ser:

$R=\rho \frac{L}{A},\phantom{\rule{mediummathspace}{0ex}}where\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\rho =resistivity$

Para pequenas oscilações T de temperatura, a resistividade pode ser aproximada para:

$\rho ={\rho }_{0}\left(1+\alpha \left(T-{T}_{0}\right)\right)={\rho }_{0}\left(1+\alpha \mathrm{\Delta }T\right)$

E como existe apenas aquecimento de material sólido, a energia recebida pelo fio é: Finalmente, todo esse conjunto se torna:

$P=\frac{dQ}{dt}=\frac{d}{dt}\left(mcT\right)=mc\stackrel{˙}{T}=mc\mathrm{\Delta }\stackrel{˙}{T},\phantom{\rule{mediummathspace}{0ex}}where\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\mathrm{\Delta }\stackrel{˙}{T}=\frac{d\mathrm{\Delta }T}{dt}=\frac{dT}{dt}$
Não sei como resolver isso analiticamente, mas há uma aproximação válida, pois estou trabalhando com pequenas flutuações de temperatura:
$mc\mathrm{\Delta }\stackrel{˙}{T}=\frac{{V}^{2}A}{{\rho }_{0}L}\frac{1}{1+\alpha \mathrm{\Delta }T}⇒\frac{mc{\rho }_{0}L}{{V}^{2}A}\mathrm{\Delta }\stackrel{˙}{T}=\frac{1}{1+\alpha \mathrm{\Delta }T}$
Agora, podemos resolvê-lo:
$\frac{1}{1+\alpha \mathrm{\Delta }T}\approx 1-\alpha \mathrm{\Delta }T$
$\frac{mc{\rho }_{0}L}{{V}^{2}A}\mathrm{\Delta }\stackrel{˙}{T}+\alpha \mathrm{\Delta }T-1=0$

$\mathrm{\Delta }T=C{e}^{-t/\tau }+\frac{1}{\alpha }\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{mediummathspace}{0ex}}where\phantom{\rule{mediummathspace}{0ex}}\tau =\frac{mcL{\rho }_{0}}{\alpha A{V}^{2}}\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}and\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}C=cte$

In this model, we see a transient solution followed by a constant one. But remember this is valid just for small temperature fluctuations.

19

This could be analyzed in the same way as a control circuit with feedback. From a practical sense, the heating will be much slower than the other effects, so that will dominate the loop equations. As such, it will exponentially approach equilibrium, unless there's other elements of the system that limit its response (ridiculously enormous inductors, state machines introducing delays, etc).

15

This is something like a PTC thermistor. which will reach an equilibrium temperature.

To get oscillation you'd have to have a phase shift or delay of some kind. You could probably make an oscillator with a mass transport delay having a a heater heat water flowing in a tube which warms a thermistor downstream and increases the heat to the upstream heater.

8

Does this fluctuation ever occur in real circuits?

I don't think this is exactly what you were asking for, but just in case, turn signal flashers depend on this behavior.

From the 1933 Patent:

A thermostatic switch closes and opens the secondary circuit. When current flows a metal strip in the switch heats up, expands and eventually opens the circuit. When it cools down it shrinks and closes again.

Some modern ones (especially when low-current LED bulbs are used) are digital/solid state, but lots of cars still use the same exact principle.

1
Turn signals rely on the temperature change making and breaking a contact, not merely altering the resistance of a homogenous conductor.
Peter Green

True, though the flasher does depend on the current draw from the bulbs for the flash rate.
Nick

I suspect "bimetallic" might be more accurate than "metallic", but don't know for sure
Scott Seidman

3

That depends on heat capacity of the element. Lower the heat capacity, more like a resistive feedbacked opamp circuit where the temperature will converge. Heat capacity acts like reactive elements and will cause oscillations. Element's heat conductivity (heat transfer speed to outside) will determine if it is going to be damped or diverged.

3

For the record, I loved Pedro Henrique Vaz Valois' answer and upvoted it.

Said simply: Yes there are transients.

You can think of this in the same way you would an RLC step-function circuit. Apply blow dryer, throw the switch, see transients on oscilloscope, watch flat line appear as all the energy balances out to a steady state. Turn the switch into an oscillating voltage and watch the resistance swing back and forth for as long as the oscillating voltage exists.

And it's a very real problem

One of many reasons why big honking cooling systems are attached to CPUs and other high-density/high-frequency chips is that we don't (we desperately don't) want to deal with heating effects. Resistor manufacturers go to great lengths to minimize the resistance variability in their products.

It's worth your time to read "Non-Linearity of Resistance/Temperature Characteristic: Its Influence on Performance of Precision Resistors" published earlier this year from Dr. Felix Zandman and Joseph Szwarc of Vishay Foil Resistors.

2

We are told that heat increases a resistor's resistance (or decreases its conductance) and current decreases when resistance is increased.

Depends on what the resistor is made of. Most of them have a positive temperature coefficient but it is quite possible to make one with a negative temperature coefficient.

Does this fluctuation ever occur in real circuits?

In general no, normally they just gradually tend towards a steady state temperature.

1

No. The temperature approaches an equilibrium, but does not overshoot it such that it must then change directions and come back.

Consider a resistor that's initially at room temperature with no current.

Then, it's connected to a constant voltage. Immediately the current increases to some value determined by Ohm's law:

$\begin{array}{}\text{(1)}& I=\frac{E}{R}\end{array}$

The resistor converts electrical energy into thermal energy through Joule heating:

$\begin{array}{}\text{(2)}& {P}_{J}=\frac{{E}^{2}}{R}\end{array}$

It also loses heat to its environment at a rate proportional to its temperature. The size, geometry, airflow and so on can be combined and characterized as a thermal resistance ${R}_{\theta }$$R_\theta$ in units kelvin per watt. If $\mathrm{\Delta }T$$\Delta T$ is the temperature of the resistor above the ambient temperature, the rate of thermal energy lost to the environment is given by:

$\begin{array}{}\text{(3)}& {P}_{C}=\frac{\mathrm{\Delta }T}{{R}_{\theta }}\end{array}$

As the resistor becomes warmer, it loses thermal energy to the environment faster due to an increasing $\mathrm{\Delta }T$$\Delta T$. When that rate of loss (equation 3) equals the rate of energy gain by joule heating (equation 2), the resistor has reached temperature equilibrium.

Equation 2 decreases with increasing temperature, assuming a typical positive temperature coefficient. Equation 3 increases with increasing temperature. At some point the resistor has warmed sufficiently that they are equal. There is no mechanism by which the resistor would "overshoot" this equilibrium, thus requiring that the resistor go from warming up to cooling off. Once equations 2 and 3 are equal, the temperature, resistance, and current have reached equilibrium and there's no reason for them to change further.

1

In a simple model, the current is a direct function of the resistance and the resistance is a direct function of the temperature. But the temperature is not a direct function of the current: the current governs the amount of heat that is produced, which influences the variation of the temperature over time.

In the linear regime, this corresponds to a first order equation

$\frac{dT}{dt}=-\lambda \left(T-{T}_{0}\right).$

As the coefficient is negative (an increase of the temperature causes an increase of the current, a decrease of the amount of heat and finally a decrease of the temperature), the system is stable and will converge to a steady state.

And in any case, a first order system does not have a oscillatory mode.

For such a behavior to be possible, a source of instability is needed, such as a negative thermal coefficient, as well as a second differentiator.

"And in any case, a first order system does not have a oscillatory mode.". I am afraid that is inaccurate. First order systems can oscillate if there is a delay, even if they are linear (I just googled a paper about that), or if they are nonlinear (this is from my deep memory).
Sredni Vashtar

@SredniVashtar: I specifically said "linear regime", and "first order" implicitly excludes delay (otherwise you state it). Your comment is irrelevant.
Yves Daoust

"IN ANY CASE, A first order system dose not have a oscillatory mode". This is wrong. When you state "in any case" you are voiding all previous specifications, while using indeterminate "A first order linear system..." implies all first order system, regardless of being linear or not. So my comment still stands. You are right about the retarded system to be nonlinear, though.
Sredni Vashtar

1
@SredniVashtar: you misunderstand the meaning. In any case refers to the sign of the constant. Stop this useless argument.
Yves Daoust

I am sure that "in any case" means what you mean, in your head. And now I also know you could not possibly be wrong, ever. But I'll leave my comment for anyone else.
Sredni Vashtar

0

Different materials have different conduction properties, including their thermal profiles. That is, some materials will heat up much more than others given the same current flow. This is one reason why components such as resistors have a tolerance.

The temperature fluctuations you describe don't really happen in real circuits. Instead, the resistor would heat up as current begins to flow but would reach an equilibrium point where the amount of heat generation from the current matches the amount of heat radiated into the surrounding air. Then the temperature of the resistor remains stable, the actual resistance remains stable, and the current remains stable.

Fifty years ago in college we learned about the first law of thermodynamics. The heating of the resistor depends on the power, the time and the thermal heat capacity, not at all on the material (assuming it is not getting hot enough to melt or vaporize as in a fuze).
richard1941

And what determines the heat capacity...?
Mick

Also, AiR is not needed for a resistor to radiate heat any more than ether is required for it to radiate electromagnetic healing energy frequency vibrations of the life force. Of course heat may be transferred by conduction and convection, but that is another story for another day...
richard1941

0

Actually there was a neat application for this in the olden days. The blinkers on a car were operated by a bimetallic thermal switch. When the blinker light is on the bimetallic heats up and flexes opening the circuit. Then the heat dissipates, the switch cools and closes again.

Not sure if all cars still use the bimetallic switch, but I'd guess that some now use computer control.

I don't think a bimetallic strip thermostat is what the original poster of the question had in mind.
richard1941
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