Measure temperature and power losses?

[mading2018]

I know that it is possible to measure the power dissipation in LTspice according to this source:
"To measure the wattage dissipation of a component , place the cursor on the components and press and hold down the ALT key, a thermometer cursor should appear. Read the Watts on the bottom screen screen status text"

But do you think it is okay just to take average value of the losses? Cause it is really a large variation I think, like the figure shows the losses over the capaictor?

And about the temperature? It is not possible to measure the temperature right?
I think I read somewhere that ronsimpson said it can't be done. Are there any other way maybe?

 

[Pommie]

I'm guessing you're using a software simulation. Care to give us a little more information?

Mike

 

[mading2018]

It is in LTspice sorry, I should have mention it.

--edit--
I attached the file where I want to measure the power losses and the temperature (if that this possible)

 

[ronsimpson]

I think I read somewhere that ronsimpson said it can't be done.

Don't believe what that old guy said. He lies too much.

You know how to get
instantaneous watts. Now average, which might be hard in that LTSPICE will not hand you a number.
Temperature is thermal resistance and watts. Look in the data sheet and see thermal resistance or it might say "100C/1watt". (die to air)
Watch out for die to infinite heat sink. (unless you have one) lol
If you do have a heat sink then there are more numbers to think about.

Bottom line, it is possible. Not easy.

 

[Ratchit]

I know that it is possible to measure the power dissipation in LTspice according to this source:
"To measure the wattage dissipation of a component , place the cursor on the components and press and hold down the ALT key, a thermometer cursor should appear. Read the Watts on the bottom screen screen status text"

If LTspice uses a thermometer to show wattage, that is confusing. Temperature is the amount of heat energy in a object per mass. If a small mass object has the same amount of heat energy as an object with a large mass, the temperature of the small mass object will be higher. Wattage is the rate energy is acquired or dissipated. Without knowing the wattage at equilibrium (equal input and output energy per time) and the mass of the object, you cannot know the temperature. By the way, "watts" is not capitalized.

But do you think it is okay just to take average value of the losses? Cause it is really a large variation I think, like the figure shows the losses over the capaictor?

What losses? Quantity of hear or its rate of dissipation?

And about the temperature? It is not possible to measure the temperature right?

Ever hear of a thermometer?

I think I read somewhere that ronsimpson said it can't be done. Are there any other way maybe?

Temperature measurement is done all the time.

Ratch

 

[ronsimpson]

Temperature is the amount of heat energy in a object per mass.

No. temperature rise is the energy/mass/time. (temp increase)

In electronics we are looking for the temp of a part not how fast the temp comes up. Heat goes into the air. We need to know how much the heat leaves the part. and We need to know how much energy goes into the part. Mass is not part of that. If you are talking about a transistor the little piece of silicon is so small it's mass approaches zero.

 

[Ratchit]

No. temperature rise is the energy/mass/time. (temp increase)

I stand by my definition of temperature. A constant temperature means that thermal equilibrium has been achieved; that is, an equal amount of heat energy per unit time is being dissipated as is being received. If more heat energy is being received than dissipated, the temperature will rise until a new equilibrium is reached, and that temperature will be higher.

In electronics we are looking for the temp of a part not how fast the temp comes up.

It depends on what you are concerned about. If you want to know what the different values of a component will be at a different temperature, then you will be interested in temperature. If you want to know if a component will burn up, then you want to know its rate of heat dissipation (1 watt, 2 watt, etc.).

Heat goes into the air.

It goes into and through the space occupied by the air and beyond.

We need to know how much the heat leaves the part. and We need to know how much energy goes into the part.

Indeed, that is so, as I have said previously.

Mass is not part of that. If you are talking about a transistor the little piece of silicon is so small it's mass approaches zero.

Mass and heat energy are interrelated with respect to temperature. The small piece of silicon is not zero and has to follow the same rules of physics as any larger masses.

Ratch

 

[mading2018]

You know how to get
instantaneous watts. Now average, which might be hard in that LTSPICE will not hand you a number

Yes, instantaneous watts. So it would be alright to use the average, if I look for the same time period every time for each component of 10 ms? In this case, I have 0.074 W?

 

[Ratchit]

Yes, instantaneous watts. So it would be alright to use the average, if I look for the same time period every time for each component of 10 ms? In this case, I have 0.074 W?

Make sure your circuit can withstand all the instantaneous values. As an analogy, folks have drowned in rivers that averaged 1 inch, but were 10 feet deep in a couple of spots.

Ratch

 

[ronsimpson]

I have 0.074 W?

Why are you doing
this? heat or energy??
If you want to know how hot a part is going to get ....... that is one question. If you are looking at energy in/our that is another problem. (which)
How do you get negative watts? (averaging + and - will result in 0 watts)
If you want to know how hot a inductor is going to get you need to know current and DC resistance. (V*I does not cause heat in a coil)

 

[ronsimpson]

Use a resistor because that is simple.
W=V*I simple but really -V does not get you -W. W=|V*I| Absolute is | |. New keyboard and can not find the symbols I want. lol

Capacitors and inductors act different at AC verses DC. You can put 10V on a cap and it does not heat up. (OK maybe there is 1uA of leakage that will cause heating) For a capacitor, while you are changing the voltage from one level to another, current flows. That current and the "internal series resistance" causes heating. It does not matter if the current is flowing in or out. (absolute value)

Incuctors are fare more complicated. For the first order simple example; current * resistance. (absolute value!) Voltage does not matter. (OK voltage across the internal resistance does matter)

 

[Ratchit]

Use a resistor because that is simple.
W=V*I simple but really -V does not get you -W. W=|V*I| Absolute is | |. New keyboard and can not find the symbols I want. lol

Capacitors and inductors act different at AC verses DC. You can put 10V on a cap and it does not heat up. (OK maybe there is 1uA of leakage that will cause heating) For a capacitor, while you are changing the voltage from one level to another, current flows. That current and the "internal series resistance" causes heating. It does not matter if the current is flowing in or out. (absolute value)

Incuctors are fare more complicated. For the first order simple example; current * resistance. (absolute value!) Voltage does not matter. (OK voltage across the internal resistance does matter)

Below is a plot of voltage across an inductor. The blue is the voltage, the orange is the current, and the green is the power. The voltage is one volt 0 to peak, the frequency is 1 radian/sec, and the inductance is 1 henry. Notice that the current lags the voltage by 90°. The maximum power to energize and deenergize the capacitor is one-half watt, which is taken and given back to the circuit for a zero average. The correct formula for power is V*I*Cos(θ).

Ratch

 

[ronsimpson]

We are talking apples & oranges.

It is not possible to measure the temperature right?

To measure the wattage dissipation of a component

I thought you want to know how hot a part gets. Or maybe power loss. From post #12 you have a very different question.

 

[ronsimpson]

If you look at transistors, diodes, or resistors the LT SPICE "temperature" function works. It does not measure temperature but watts. W=V*I
LTSPICE does not measure watts loss with inductors or capacitors.

 

[Ratchit]

If you look at transistors, diodes, or resistors the LT SPICE "temperature" function works. It does not measure temperature but watts. W=V*I
LTSPICE does not measure watts loss with inductors or capacitors.

Why does LTSpice call a function "Temperature" if it does not, in fact, measure temperature?

Ratch

 

[ronsimpson]

LT SPICE calls the function "instantaneous power" not temperature. Only you and I call it temperature.

 

[Ratchit]

LT SPICE calls the function "instantaneous power" not temperature. Only you and I call it temperature.

View attachment 112640

Thanks for clearing up that point. Only I would call it instantaneous power, too. Temperature depends on the thermal mass of the component and its heat dissipation rate. Just knowing the current and voltage of a component is not enough to determine temperature.

Ratch

 

[mading2018]

heat or energy??

I am interesting in power dissipation (W losses). yes, i want to know how hot parts gets.

But I notice I got way to much W losses, could that be due to that components have not "withstand all the instantaneous values"?

 

[mading2018]

What losses? Quantity of hear or its rate of dissipation?

Power losses (W losses)

Ever hear of a thermometer?

Yes, but how do I measure temperature in LTspice?

 

[ronsimpson]

Yes, but how do I measure temperature in LTspice?

You can not.
To measure temperature you need "thermal resistance" for each part.
AND
You will need to make better models for coils and capacitors.

What part do you want to find the temperature of?