# Organization of heat sink for powerful microcircuits (microwave boards)

## Introduction

I develop microwave devices, many of them are powerful enough, so I have accumulated some experience in solving heat removal problems.

In this article, we will focus mainly on small heat-generating microcircuits installed on a board with plated holes – for example, amplifiers up to 10-20 W or powerful diodes, which also require heat dissipation. For more powerful amplifiers, amplifier units and microwave modules, thermal calculation is required, as well as the use of radiators and possibly even special copper plates and heat pipes.

*Note: Habré has **publication** about heat dissipation in printed circuit boards. My article contains other information.*

## Heat-generating components

The most obvious component is the RF amplifier. Amplifiers are available in different cases – more powerful in ceramic, less in plastic. Also, amplifiers come in the form of a bare crystal. The preferred type of heat sink for the amplifier is a pedestal made of copper or aluminum, on which the amplifier is installed (with thermal grease), amplifiers of hundreds of watts are additionally pressed with screws.

But it is not only active elements that suffer from excess heat. Diodes as part of microwave switch and phase shifter circuits, due to imperfection, can be presented in the form of an equivalent circuit, which will necessarily include a resistor. It is on this resistor that heat is generated, the more, the greater the power of the input signal.

Many RF circuits involve the use of microwave EOL resistors (loads, terminators). For example, balanced amplifier circuits:

In the proposed scheme, two quadrature bridges are used, the isolated ports of which must be loaded with a matched impedance. Resistors R7 0805 smd 47 ohms and R9 2512 smd 51 ohms.

I do not use ordinary resistors in my designs, only special 50-ohm RF resistors. For example such:

In the recommendations at the end of the PBX catalog, you can find a lot of interesting things, including the following pictures:

## Calculations and formulas

So, it is necessary to estimate the amount of heat that can be removed from the fuel element installed on a single-layer board with plated holes. *The calculations below are valid for components with a metallized base.*

There are three ways to dissipate heat through the board to the base:

thermal conductivity of the board material

copper plating of holes

air inside the hole (or special compound, or solder)

First, you need to determine the thermal resistance, which can be calculated using the formula (simplified):

where K is the value of thermal conductivity, A is the area, L is the thickness of the board. Kcu – thermal conductivity of copper 394 W / (m * K) – from the table, Acu – area of copper equal to the cross-sectional area of metallization in the hole (rim area between D1 and D2), Ka – thermal conductivity of air, which is close to zero (or solder), Aa is the area of a circle with a diameter of D2, N is the number of holes, Kro is the thermal conductivity of the board material, I have it written Ro, since I used Rogers. This value is indicated in the datasheet, for example, the thermal conductivity of Rogers 4350 is 0.69 W / (m * K). Aro – the area of the board (respectively, equal to the area under the microcircuit minus the area of all via).

The formula does not take into account the spreading of heat in the horizontal direction of the board. Next, you can calculate the maximum amount of heat (power) by the formula:

where Tmax is the maximum temperature of the component (see datasheet), T is the ambient temperature, R (pcb) is calculated in the previous step, R is the thermal resistance of the microcircuit case (see the datasheet).

## Example

Chip topology – QFN 4×4

D1 = 0.3 mm – via diameter

d = 0.038 mm – deposited copper thickness

L = 0.762 mm – board thickness from Ro4350 laminate

K1 = 0.394 W / (mm * deg) – thermal conductivity of copper

K2 = 0.000026 W / (mm * deg) – thermal conductivity of air

K3 = 0.00069 W / (mm * deg) – thermal conductivity Ro4350

Т1 = 125 degrees – the maximum temperature of the microcircuit (from the datasheet)

T2 = 25 degrees – ambient temperature

Rk = 10 deg / W – thermal resistance of the microcircuit case (from the datasheet)

N = 16 – number of via

S = 20 mm^{2} – area under the component (approximate)

D2 = D1-d * 2 = 0.224 mm – inner diameter of the hole

A1 = p / 4 (D1^{2}-D2^{2}) = 0.031 mm^{2} – copper area

A2 = n / 4 * D2^{2}= 0.039 mm^{2} – air area

A3 = 18.8 mm^{2} – the area of the substrate material

Rpcb = L / ((K1 * A1 + K2 * A2) N + K3 * A3) = 3.6 deg / W – thermal resistance

Qmax = (T1-T2) / (Rk + Rpcb) = 7.35 W

If there were no plated holes, then the maximum power would be only 1.53 W

In the above calculations, only one of the three types of heat transfer was used – thermal conductivity. There are also convection and radiation phenomena.

The amount of radiated heat depends on the blackness of the microcircuit case. The black plastic case has a coefficient of about 0.95, the ceramic case is about 0.9.

It is also obvious that heat is transferred to the pads and then to the supply strips, and from them it also “leaves” by convection, radiation into the air, and also by heat transfer through the board.

The additional heat removal by the described methods is negligible and is equal to about 0.25 W for a QFN package, provided that there are ideally infinite supply strips.

## Conclusion

This article should not be considered strictly scientific, but rather a small help to the development engineer. Although the formulas can be found on the internet, I hope this article is helpful; formulas are simplified and analyzed by me.

*Thank you for the attention! Read my **past articles** and subscribe to my **Instagram**!*