# Calculate thermal resistance of heatsink required

1. Jan 9, 2016

### pbonesteak

1. The problem statement, all variables and given/known data
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.

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

3. 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

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

Regards

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2. Jan 9, 2016

### pbonesteak

Just noticed the max 'Iout' is actually 7.5A not 1A

3. Jan 9, 2016

### Staff: Mentor

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.

4. Jan 9, 2016

### pbonesteak

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

5. Apr 2, 2016

### StripesUK

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?

6. Apr 16, 2016

### gazp1988

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.