# calculate the temperature due to Power loss over a res.

#### Merdidose

Hi All, Having the following problem I am trying to solve I am an electronics guy some of e physics eludes me in my calc.

Here goes....

I am supplying 125Vac into a R which is a 20kohm resistor where I have a 12V zener in series to GND.

P_loss = (125Vac - 5V)^2 / R
P_loss = 0.72W

So lets say we wait for 1000 secs.

Total Energy (J) = 0.72kJ

Now I am taking into consideration that the SPH of dry air is 0.716 KJ/Kg.K
and Air Density is 1.3 Kg/m3

Therefore Temperate raise should be

Tr = 0.72 / 0.716 / 1.3
= 0.77 K?

Am I missing something here, or is this not he way to calculate heat loss over a resistor?
Someone?

THKS.

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#### oz93666

Total Energy (J) = 0.72kJ

Now I am taking into consideration that the SPH of dry air is 0.716 KJ/Kg.K
and Air Density is 1.3 Kg/m3

Therefore Temperate raise should be

Tr = 0.72 / 0.716 / 1.3
= 0.77 K?
.
If the electronic components are putting out 0.77W , The temp raise should be 0.77 , if all the heat goes evenly into only 1m3 of air

Am I missing something here, or is this not he way to calculate heat loss over a resistor?
Someone?
You don't need to do this to "calculate the heat loss over (from) a resistor" .... ohms law is all you need , the temperature or specific heat of the air is immaterial.

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#### Merdidose

Hi oz93666,

0.77W is 0.77 Deg. C per second? Doesnt sound right..

#### studiot

First you need to review you power calculation because

1) The diode will block every other half cycle.

2) Why have you clamped the base at 5 volts not 12?

When you have got the electronics of power generation in the resistor correct you should look up 'themal resistance' from an electronics point of view.
The calculation of temperature rise in a component is a standard procedure using published characteristics of that component.

• 1 person

#### oz93666

Hi oz93666,

0.77W is 0.77 Deg. C per second? Doesnt sound right..
No .... it isn't ..

assuming you have components putting out 0.77W ...

Now what is your question ?

#### Merdidose

@studiot

1) I am using a bridge so its full-wave cycle. RMS is 120V

2) Yes you are right I usually use 5V zener so its abit of force of habit. My bad.

Should be (125V - 12V)^2 , but these are minor errors..

You nailed it with "thermal resistance" this is the search engine term I needed!

Thank You!

#### nhatminh61

You don't need to do this to "calculate the heat loss over (from) a resistor" .... ohms law is all you need , the temperature or specific heat of the air is immaterial.
Now what is your question ?

--------------------------------------------
best ringtones

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#### studiot

You don't need to do this to "calculate the heat loss over (from) a resistor" .... ohms law is all you need , the temperature or specific heat of the air is immaterial.
Now what is your question ?
I suggest you study the subject (a good place to start is the manufacturer's pdf I linked to) before making claims that might confuse others.

Yes the OP originally attempted to calculate thermal performance from first principles - all credit to him. It would work but, as you say, it does make things more complicated. Of course such calculations are required if the resistor is on a heatsink, and this is a situation of the mains being applied to a high power low ohmage resistor.

#### oz93666

I suggest you study the subject (a good place to start is the manufacturer's pdf I linked to) before making claims that might confuse others.

Yes the OP originally attempted to calculate thermal performance from first principles - all credit to him. It would work but, as you say, it does make things more complicated. Of course such calculations are required if the resistor is on a heatsink, and this is a situation of the mains being applied to a high power low ohmage resistor.
This is ridiculous ... no need for any of it ... just operate components within their power ratings with heat sinks if required ...

OP is all too muddled .....What exactly is the question being asked ???