Schottky Diode Thermal Runaway

The problem looked at here is a real-world case of thermal runaway in a Schottky diode. I will use some ideas from closed loop feedback control to analyze the problem. Schottky diodes are used extensively in switching regulators that power nearly every piece of high technology electronics. Putting this in perspective consider the case of a switching supply that converts 5Vdc to -5Vdc in a inverting Cuk topology. Given typical numbers for the output as -5Vdc @ 30ma you might think a Schottky diode rated at 2Amps forward current and 20V reverse breakdown would be perfectly safe. However, Schottky diodes can suffer failure due to thermal runaway caused by excessive reverse current under certain conditions.

The reverse leakage current of a Schottky diode is very sensitive to temperature. In fact it is exponential as follows Ir=I₀e^n·(Tj-To). The power dissipated due to this current is P = Ir·Vrev and the steady-state temperature rise is given to be dT = P·Theta. You can see this is a case of positive feedback. As the temperature rises so does the current. This in turn increases the power dissipation which increases the temperature. This can be analyzed using small signal gain theory applied to the following block diagram.

The small signal loop gain is given by:

g= V_revØ·n·I₀e^n·(Tj-To)

The temperature at which runaway will commence is when the loop gain is greater than unity. Using the small signal gain equation the critical junction temperature is

Tj= T0 + (1/n)·ln(1/(Ø·n·Vrev·Ir0))

As an example consider the MBRM140 diode operating at 60C ambient environment. From the data sheet the maximum leakage Ir(10V,25C)= 0.1mA and Ir(10V,85C)= 10mA . Fitting this data to the exponential results in μ equal to 0.07675. In the inverting -5V application the reverse voltage will be nominally Vrev = 11V. The thermal resistance on the PC board is estimated to be 25C/W. From this data the junction temperature at onset of thermal runaway is Tj= 105.2C. This is a surprisingly low junction temperature that can lead to thermal run away. The reverse voltage is well below specification and the forward current was not even considered in this example. It is a simple exercise to set up an example in an Excel spreadsheet and view the iterations as they evolve.

As noted this analysis included only the reverse leakage current at 100% duty cycle. If there is significant power dissipated in the forward direction that too should be included in the calculation using the appropriate duty cycle. The objective of this note was to show how closed loop control theory can be applied to seemingly non-control problems.

Self-heating Feedback

The basic feedback equation encountered is Gcl = Gf/(1 + Gol) where Gcl is the closed loop gain of a system that has open loop gain Gol and forward loop gain Gf. Consider the case of a resistive heater. The source of power could be either a constant current source or a constant voltage. In this problem we are interested in the steady state temperature of the load.

Constant Voltage Source

For the first case lets look at constant voltage. The power dissipated in the load is P = V²/R. Here R is the load resistance and V is the applied voltage. However, R is a function of temperature, T, given by R = Ro*(1 + a*(T-To) ) where the resistance is linear with temperature and has a value Ro at temperature To. The steady-state temperature is also a function of the temperature given by the dissipation characteristics of the thermal system T = Tamb + theta*P where theta is the thermal resistance (C/W) and Tamb is the ambient temperature. Intuitively you can see that when the voltage is applied the power increases the temperature which in turn increases the resistance (assuming a positive temperature coefficient, a). The increase in resistance then reduces the power. This is inherent negative feedback at play that prevents a runaway thermal condition.
Combining these three equations we can solve for the temperature as

(dT)² + (1/a)*(dT) - V²*Theta/(a*R0) = 0

Where dT = (T-Tamb). This is a non-linear system. The block diagram is shown below as a feedback network. I also added a thermal time constant to the thermal resistance block. The loop gain is found using small signal gain theory to be -(a*Ro*Theta*I²(dT)) which is negative for a>0.









Constant Current Source

The second case we will look at is a constant current supply. In this case the power dissipation is given by P = I² * R. We can see intuitively that power dissipation raises the temperature which raises the resistance and therefore increases the power dissipation. This is a case of positive feedback and hence has the potential for thermal runaway. Solving the three equations for power, resistance and temperature we have the result for temperature rise

(T-Tamb) = (I²*R0*Theta) / (1 - (a*Theta*I²*R0))

This is a linear system and has the form of a closed loop equation with open loop gain a*Ro*Theta*I². In the block diagram below I added a thermal time constant to the thermal resistance block.


Note that for positive temperature coeffficents (a>0) this system has positive feedback. If the loop gain gets too close to unity there will be a thermal runaway condition. Some boundary conditions are: 1. If the temperature coefficient is negative (a<0) dt=I²*R0*Theta, 2. Perfect Heat sink (Theta=0) then dT= 0 and 3. Perfect Short R=0 then dT= 0. The loop gain can be used as a measure of the safety factor in a system design.

This basic analysis of the feedback mechanism in a self heating system has many applications. Who would of thought that a handful of variables would say so much about a system?

Introduction

This is my introduction for this blog. I enjoy engineering design very much; especially describing a problem solution analytically which can then be viewed in the 'real world' as a working mechanism. The link between abstract mathematics and physical reality is fascinating. There is something beautiful in the simplicity of equations. On these pages I plan to periodically post solutions to engineering problems that strike me as interesting. I hope you find it interesting and useful as well. For more information about me visit OmnificSolutions.com


The first topic I plan to write about is how feedback shows up in the equations for thermal heating using a either a constant voltage or current source. An admittedly simple but interesting problem.