+ Post New Thread
Results 1 to 7 of 7

2nd July 2019, 23:58 #1
 Join Date
 Aug 2016
 Posts
 498
 Helped
 4 / 4
 Points
 1,877
 Level
 10
obtaining thermal rise in diode
This diode specifies Average Rectified Output Current @ TT = +75°C but referencing the If  Vf graph Figure 2 its Vf for 5A is ~1.15v power dissipated is 5.75W and Junction to lead thermal resistance is give as 10 C/W which would cause the junction to rise 57.5C not the stated 75C. https://www.diodes.com/assets/Datasheets/ds16007.pdf
Manufacturer specified 75C thermal rise for 5A is not accounted for even if its the sum of power dissipation and typical 25C ambient temperature which would be 82.5C. Isn't power dissipation for diodes the product of its forward voltage and average conducted current?Last edited by Zak28; 3rd July 2019 at 00:08.

Advertisement

3rd July 2019, 01:54 #2
Awards:
 Join Date
 Apr 2014
 Posts
 15,457
 Helped
 3513 / 3513
 Points
 76,224
 Level
 67
Re: obtaining thermal rise in diode
Hi,
Junction to lead....temperature ruse is 57°C.
Maybe the lead temperature risees by 18°C
Making a total junction temperature rise of 75°C.
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
1 members found this post helpful.

Advertisement

3rd July 2019, 01:58 #3
 Join Date
 Aug 2016
 Posts
 498
 Helped
 4 / 4
 Points
 1,877
 Level
 10

Advertisement

3rd July 2019, 01:58 #4
 Join Date
 Mar 2008
 Location
 USA
 Posts
 6,356
 Helped
 1851 / 1851
 Points
 39,312
 Level
 48
Re: obtaining thermal rise in diode
We call this "sandbagging the spec". Conservative only costs
the customer money but tends to keep them safe; aggressive
makes field returns and ruins AOQ statistics.
There is also that diode I*V(I) is nonlinear and working with
averages embeds some assumptions about linearity and
superposition. Which may or may not be true / accurate.
Average of products being different than product of averages
in many cases.
1 members found this post helpful.

3rd July 2019, 02:34 #5
 Join Date
 Nov 2012
 Posts
 3,056
 Helped
 741 / 741
 Points
 16,666
 Level
 31
Re: obtaining thermal rise in diode
Vf for 5A is ~1.15v power dissipated is 5.75W and Junction to lead thermal resistance is give as 10 C/W which would cause the junction to rise 57.5C...
The resistance has two ends; one end will be 57.5C if the other end is 0; if the exposed end is at 20C, the junction will be 77C. That is how I see it.
In addition, the heat transfer process is like diffusion and is highly nonlinear. The thermal resistance is a function of temperature.
Fig 1 in the datasheet you have attached says that at 5A current, max terminal temp acceptable is 75C. The junction temp will be 75+57=132C! (good for frying an egg)
Max power dissipation can always be obtained as V.I; the exact power dissipation can always be obtained as the integral of V(t).I(t).dt over a complete cycle.
1 members found this post helpful.

Advertisement

3rd July 2019, 02:40 #6
 Join Date
 Jul 2010
 Location
 Sweden
 Posts
 950
 Helped
 369 / 369
 Points
 7,193
 Level
 20
Re: obtaining thermal rise in diode
@TT = +75C is the terminal temperature. It is in the "maximum" section, so the diode is rated for max 5A when the terminal temperature is +75C.
Vf @5A according to table 2 is about 1.05V @25C. Power dissipation = 1.05V * 5A = 5.25W
Temp rise = 10 * 5.25 = 52.5C
Junction temp = +75C + 52.5C = 127.5C but it will be less since the forward voltage decreases with increasing temp.
Approximate with 2mV/C so Vf @5A @+125C is about 0.85V.
Power dissipation = 0.85V * 5A = 4.25W
Temp rise = 10 * 4.25 = 42.5C
Junction temp = +75C + 42.5C = 117.5C but it will be a little higher since the temp is less than 125C, so Vf is a little higher than 0.85V.
Expect the junction temperature to be about +120C when the terminal temperature is +75C and the average current is 5A.
1 members found this post helpful.

3rd July 2019, 04:57 #7
 Join Date
 Nov 2012
 Posts
 3,056
 Helped
 741 / 741
 Points
 16,666
 Level
 31
Re: obtaining thermal rise in diode
but referencing the If  Vf graph Figure 2 its Vf for 5A is ~1.15v power dissipated is 5.75W...
What I mean is that all the heat produced at the junction must be dissipated via the terminal ends.
About 5W of heat must be conducted away from the pads at 75C by radiation, conduction, etc etc...
Because the junction temp is not directly accessible to the end user, you need to use the terminal temp as the base for all your calculations.
If you are planning to use the diode at 5A for extended periods, you surely need some healthy heat sinking. Radiation will perhaps help about 1W/cm2 at 75C (guess value)
But if you increase the pad size, the terminal temp will perhaps decrease but the dissipation will remain the same. You have to get rid of the heat somewhere else somehow.
1 members found this post helpful.
+ Post New Thread
Please login