Calculate thermal resistance of heatsink required

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Discussion Overview

The discussion revolves around calculating the thermal resistance of a heat sink required for a voltage regulator circuit, specifically the LT1083. Participants are exploring the thermal management aspects of the regulator, including power dissipation and junction temperature calculations based on various operating conditions.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a formula for power dissipation (PD) and calculates maximum power dissipation based on input and output voltages.
  • Another participant corrects the output current value from 1A to 7.5A, which may impact subsequent calculations.
  • A suggestion is made to refer to the thermal considerations section of the datasheet, emphasizing the importance of defining actual operating conditions rather than maximum values.
  • One participant calculates power dissipation for a specific scenario (Vin = 18V, Vout = 12V, Iout = 5A) and derives junction temperature, indicating a need for a heat sink.
  • Another participant calculates thermal resistance values and expresses concern about the high values for heat sinks, questioning whether the maximum load current should be used in calculations.
  • A participant presents a calculation for junction temperature based on power dissipation and thermal resistances, concluding that the calculated junction temperature is within acceptable limits.
  • There is uncertainty expressed regarding the appropriateness of the thermal resistance values and whether they should be additive.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct values to use for calculations, particularly regarding maximum load current and thermal resistance values. Multiple competing views and uncertainties remain regarding the calculations and assumptions made.

Contextual Notes

Participants highlight the need for precise definitions of operating conditions and the potential impact of varying input and output parameters on thermal management calculations. There are unresolved questions about the appropriateness of using maximum values and the implications for heat sink design.

pbonesteak
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Homework Statement


Q4) Calculate the thermal resistance of the heat sink required for the regulator of Q3 above given the information below [from the datasheet] and the data in the table.

Q3 FIGURE 2 shows an adjustable voltage regulator using the LT1083*. The LT1083 develops a 1.25V reference voltage between the output and the adjust terminal. By placing a resistor R1 between these two terminals, a constant current is caused to flow through R1 and down through R2 to set the overall output voltage. If VIN = 18 V, determine the range over which the output voltage can be varied.

Homework Equations


My formula
PD=Vin-Vout*Iout
θ Jct = T(jmax) - TA / PD

The Attempt at a Solution


PDmax = 18V+10% - 15V*1A = 4.8W
Control Section θ Jct = 125-75/4.8 = 10.4 °C/W
Power section θ Jct = 150-75/4.8 = 15.6 °C/W

The data sheet given is here

https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwiGuOeymZ3KAhUC_SwKHR6TCCAQFggdMAA&url=http://www.linear.com/docs/3741&usg=AFQjCNE8-DSdhVtLg7527HpB0xFgb3dF6g

This is as far as I get. there's so much info on the LT1083 datasheet I just need steering in a direction.
Any suggesgions much appreciated

Regards
 

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Just noticed the max 'Iout' is actually 7.5A not 1A
 
There's a discussion of Thermal Considerations in the APPLICATIONS INFORMATION section of the datasheet. It includes a calculated example that may help your understanding.

You'll need to pin down your operating conditions a bit better than just assuming a maximum input voltage and maximum device current; The device would not survive under such conditions. An actual maximum load current would be good. Look for the maximum power dissipation curve. It will tell you what the maximum case temperature can be for a given power dissipation.
 
So if i use a Vin as 18v use the adjustable resistor and set an output of 12v and am drawing 5amps load on a device that is connected on the output.

Using the formula Pd=Vin-Vout*Iout
Pd = 18v-12v*5=30W

For the control calculation: JuncT = T(amb)+Pd(θHsink+θJct-cas+θcas-Hsink)
I would get 75°+30W(1°C/W+0.6°C/W+0.2°C/W) = 105*1.8
JuncT=189° > 125° I would have 64°C that would need a heat sink to dissipate

I could do the same for the Power side
 
Okay so here is what I have so far...

Control:
ΔT=Tjmax-Ta=125-75=50°C
PD=Ilm(Vin-Vout)=1(18-15)=3W
θ=\frac{ΔT}{PD}=\frac{50}{3}=16.667°C/W
This value is well above the value of 1.6°C/W stated on the datasheet so it will require a heat sink.
θhs=θ-θjc-θchs=16.667-0.6-0.2=15.867°C/W

I have done the same for the power section, but I feel I'm not on the right track due the thermal resistance values for the heat sinks being so drastically high. Especially if I'm right in thinking they then have to be added together? The question gives the maximum load current as 1A is this the correct value to use in the PD calculation? I've also seen the mention of pinning down operating conditions in a previous post, but should a heatsink not be able to handle the maximum values of the regulator?
 
this is what i have.

PD= (Vin-Vout) x (Iout) = (18 - 15) x 1 = 3w

TJ = TA + PD (θHS + θCHS + θJC)
TJ= 75 + 3 (θHS + 0.2 + 0.6)

to work out the heat sink value

PD = ΔT/(θJC + θHS) =
3 = 50/(0.6 + θHS) =
θHS = 50/(0.6 + 3) = 13.89 C/W

TJ= 75 + 3(13.89 + 0.2 + 0.6) = 119.07°C <125°C
this shows that the calculated Tj is within the range of the maximum junc temperature.

am i on the right track here.
 

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