LM340 temp coefficient question

I have a question about temperature coefficients.
I'm using an LM340 (5 volt regulator) and don't understand how much heat it
will disapate and how much it will change over temp.
The datasheet specs 0.8mV/c but doesn't show a junction to case spec.
My questions:
Is 0.8mV/c positive or negative from what... 25 degrees C?
So if change my temperature 10 degrees C, my output voltage will be 8mv higher (or lower) and from what external temp?
Also, how do I calcualte an accurate junction to case number?
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Peter wrote:

You perhaps missed the figures on the data sheet:
http://www.ecst.csuchico.edu/ieee/pdfs/LM78XX.pdf
Hint: The junction-to-case thermal resistance (θJC) for TO-3 and TO-220 package is 4°C/W.

The average temperature coefficient of output voltage is just that, an average slope. It is only at 0.8mV/c at one temperature (from the graph mentioned below, at about 60C).
Output voltage falls with increasingly positive temperature and with increasingly negative temperature, either side of its design temperature of 25C. The graph of "Output Voltage (Normalized to 1V at TJ = 25°C)" shows the effect.

See the graph mentioned above. It depends on the value of the initial temperature as well as the change in temperature.

No need, read the data sheet (more carefully, perhaps).
--
Sue

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.