LDO Thermal Design Concept - Thermal Resistance, Copper Area

Typically when LDOs are used, most users will take the maximum current on the datasheet while designing the current of the system without considering the amount of voltage conversion (V

Why is this so? Heat is mainly the reason, as the wider the difference in the Drop Voltage (V

Before investigating the effects of temperature, the concept of thermal resistance will have be the grasped. Its unit, θ, or°C/W, represent the lower the resistance, the better the cooling. A few types of thermal resistance will be generated during the packaging of an IC.

θja = (Tj-Ta)/P is the thermal resistance between the surface of the die to the environment

θjc = (Tj-Tc)/P is the thermal resistance between the surface of the die to the surface of the IC

θca = (Tc-Ta)/P is the thermal resistance between the surface of the IC to the environment

In the image, P is the power of the IC in Watts (W)

Tj: Temperature of the die’s surface

Ta: Temperature of the environment

Tc: Temperature of the IC’s surface

Calculating the power dissipation (PD ) of a typical LDO (image 3) involves multiplying the voltage differences (V

PD = (V

From the formula above, two additional formulae can be derived, which can be used to determine the output current of the LDO:

PD = (V

PD = (TJ−TA)/θJA…………………………………..Formula 2

Taking the example of the GStek LDO GS7130, we will be investigating the effects of thermal resistance θJA with different copper area (image 4), and in turn, the output current with different thermal resistance.

First we will look at the thermal resistance θJAfor the smallest copper area (25 mm

The maximum power dissipation can be found with formula (1)

PD(max) = (TJ(max)−TA)/θJA

The maximum power dissipation at TA = 25°C

can be calculated by following formula:

PD(max) = (125°C − 25°C)/(75°C/W)=1.33W

The maximum power dissipation at TA = 85°C

can be calculated by following formula:

PD(max) = (125°C − 85°C)/(75°C/W)=0.53W

TJ (max) represents the highest tolerable temperature for the die of the IC, which is 125°C (image 5)

Graph for the calculated PD (Image 6)

Using formula (2) PD~~VDD x IQ ~~(IQ is negligible in image 7), the maximum output current I

Assuming V

If the maximum 6 A output as reflected on the datasheet in used, PD will be (5-3.3) X 6A = 10.2W, exceeding the PD

Therefore, the output current of the LDO will be affected by the temperature, which is also the maximum PD

1.33W/I

I

0.53W/I

I

As the result, for converting 5 V to 3.3 V, the maximum current can only be 0.78 A at 25°C and 0.3 A at the high temperature of 85°C. When the surface temperature goes beyond 125°C, unexpected results may occur.

Next we will look at the thermal resistance θJA for the biggest copper area (70 mm

Using formula (1), the maximum PD can be found.

The maximum power dissipation at TA = 25°C

can be calculated by following formula:

PD

The maximum power dissipation at TA = 85°C

can be calculated by following formula:

PD

I

2.04W/I

I

0.81W/I

I

As tabulated in table 1, the increase in the IC’s heat dissipation will influence the output current when converting 5 V to 3.3 V.

Using the above formula, the results from converting 5 V to 4.76 V (smallest dropout voltage) is shown in table 2.

From table 1 and 2, we see that conversion of voltages will influence the output power, which will in turn affect the output current.

Typically when LDOs are used, most users will take the maximum current on the datasheet while designing the current of the system without considering the amount of voltage conversion (V

_{in}– V_{out}) involved, an action that will lead to problems as the maximum current is conditional. The LDO will only work within the dropout voltage when a small conversion difference (V_{in}– V_{out}) exists. If a user wishes the LDO to output 6A of current (as shown in image 1), the difference in the input and output (V_{in}– V_{out}) must fall within the range of 240 ~ 400 mV. In other words, if the input is 5 V (V_{in}= 5V), 6 A of current can only be achieve with an output of 4.76 ~ 4.6 V (V_{out }= 4.76 V ~ 4.6 V).Why is this so? Heat is mainly the reason, as the wider the difference in the Drop Voltage (V

_{in}– V_{out}), the greater the power consumption. If heat is out of the picture, regardless of the voltage to be stepped-down, the outputted current can be achieved as the maximum current reflected on the datasheet. But in reality every LDO will carry its highest operating temperature, which will act as a restrictive element to the output power and current.Before investigating the effects of temperature, the concept of thermal resistance will have be the grasped. Its unit, θ, or°C/W, represent the lower the resistance, the better the cooling. A few types of thermal resistance will be generated during the packaging of an IC.

θja = (Tj-Ta)/P is the thermal resistance between the surface of the die to the environment

θjc = (Tj-Tc)/P is the thermal resistance between the surface of the die to the surface of the IC

θca = (Tc-Ta)/P is the thermal resistance between the surface of the IC to the environment

In the image, P is the power of the IC in Watts (W)

Tj: Temperature of the die’s surface

Ta: Temperature of the environment

Tc: Temperature of the IC’s surface

Calculating the power dissipation (PD ) of a typical LDO (image 3) involves multiplying the voltage differences (V

_{in}-V_{out}) with the current (I_{out}) and adding the power of the static circuit. The formula is as shown:PD = (V

_{in}−V_{out}) x I_{out}+ VDD x IQFrom the formula above, two additional formulae can be derived, which can be used to determine the output current of the LDO:

PD = (V

_{in}−V_{out}) x I_{out}+ VDD x IQ…………Formula 1PD = (TJ−TA)/θJA…………………………………..Formula 2

Taking the example of the GStek LDO GS7130, we will be investigating the effects of thermal resistance θJA with different copper area (image 4), and in turn, the output current with different thermal resistance.

First we will look at the thermal resistance θJAfor the smallest copper area (25 mm

^{2})in the PSOB-8 package, which is 75°C/W (image 4)The maximum power dissipation can be found with formula (1)

PD(max) = (TJ(max)−TA)/θJA

The maximum power dissipation at TA = 25°C

can be calculated by following formula:

PD(max) = (125°C − 25°C)/(75°C/W)=1.33W

The maximum power dissipation at TA = 85°C

can be calculated by following formula:

PD(max) = (125°C − 85°C)/(75°C/W)=0.53W

TJ (max) represents the highest tolerable temperature for the die of the IC, which is 125°C (image 5)

Graph for the calculated PD (Image 6)

Using formula (2) PD

_{(max)}= (V_{in}−V_{out}) x I_{out}+_{out (max) }can be derived with the known PD_{(max) }Assuming V

_{in}is 5 V, converted V_{out}is 3.3 V (at 25°C)If the maximum 6 A output as reflected on the datasheet in used, PD will be (5-3.3) X 6A = 10.2W, exceeding the PD

_{(max)}of 1.33 V and the surface temperature of 125°C, a situation that will lead to shutdowns as well as other problems.Therefore, the output current of the LDO will be affected by the temperature, which is also the maximum PD

_{(max)}.1.33W/I

_{out(max)}=(5V-3.3V) (at 25°C)I

_{out(max)}=1.33W/1.7V = 0.78A0.53W/I

_{out(max)}=(5V-3.3V) (at 85°C)I

_{out(max)}=0.3AAs the result, for converting 5 V to 3.3 V, the maximum current can only be 0.78 A at 25°C and 0.3 A at the high temperature of 85°C. When the surface temperature goes beyond 125°C, unexpected results may occur.

Next we will look at the thermal resistance θJA for the biggest copper area (70 mm

^{2}) in the PSOP-8 package, which is49°C/W.Using formula (1), the maximum PD can be found.

The maximum power dissipation at TA = 25°C

can be calculated by following formula:

PD

_{(MAX)}= (125°C − 25°C)/(49°C/W)=2.04WThe maximum power dissipation at TA = 85°C

can be calculated by following formula:

PD

_{(MAX)}= (125°C − 85°C)/(49°C/W)=0.81WI

_{out(max) }can be found with formula (2)2.04W/I

_{out(max)}=(5V-3.3V) (at 25°C)I

_{out(max)}= 2.04W/1.7V = 1.2A0.81W/I

_{out(max)}=(5V-3.3V) (at 85°C)I

_{out(max)}=0.47AAs tabulated in table 1, the increase in the IC’s heat dissipation will influence the output current when converting 5 V to 3.3 V.

Using the above formula, the results from converting 5 V to 4.76 V (smallest dropout voltage) is shown in table 2.

From table 1 and 2, we see that conversion of voltages will influence the output power, which will in turn affect the output current.

**Conclusion**- The smaller the LDO’s thermal resistance θ, the better the heat dissipation and the larger the output current. The copper area can be used to lower thermal resistance and amplify the output current
- LDOs are extremely heat-sensitive, and it will influence the output current (A difference of 3 A is observed at the ambient temperature of 25°C and 85°C when converting from 5V to 4.76 V)
- Obtaining the desired current as well as operating temperature with the above formulae will prevent the IC from overheating (and the lack of output power)
- LDOs are heat-generating components, do not place them with other heat generating and heat sensitive sources to prevent cross influence.

## 0 Comments