In studying estimation methods for control systems I once had a professor say "use all available information". This may seem trivial but it has served as very good advice over many years. In a recent system I have been designing a sensor array to detect light from a display for purposes of calibrating pixel timing versus RBG colors. This method is sensitive to ambient light corrupting the SNR of the signal. However it was observed that the measurement scheme is taking readings in the KHz time span but most ambient light is DC.
As an aside even a florescent bulb is primarily DC ambient luminance. The 60Hz AC mains is applied across a phosphor tube exciting the phosphor to emit light. But due to the persistence the luminance is primarily DC.
The scheme used for correction was to detect when the measurements are not being made. This was readily done by monitoring the control signal to an ADC. When the measurements have been idle for a period defined by a one-shot delay an analog multiplexer circuit integrates the output signal and applies a DC bias to the sensor transimpedance amplifier input. When the output is at zero DC the integrator stops moving and the output stays at zero volts. While a measurement is being made this circuit is opened and the integrator value is held during the measurement time. The result is a sample & hold feedback circuit that dynamically corrects for DC ambient light and prevents saturation of the input stages.
Sunday, July 17, 2011
Thursday, November 4, 2010
CPC Optic
Lately I have been doing some simulation of a neat little optic device called a CPC. Short for Compound Parabolic Concentrator. This is a class of optic device known as a non-imaging optic. As the name implies it is used to concentrate a beam into a small area. It does so much more efficiently than imaging optics. Here is a picture of a CPC:
My Matlab program treats the CPC as a solid of revolution made of a material with index n2 and surrounding air index n1. The program performs ray tracing to find which rays reach the detector at the CPC exit aperture.
The figure below shows a Matlab result of several ray traces. Note the rays enter the entrance aperture and may bounce around several times, or not at all, before hitting the sensor at the exit aperture on the left.
CPC's can be made of molded plastic (Polystyrene for example) and therefore can be made very inexpensive in large volumes.
Because the incident angle can exceed the critical TIR angle, asin(n1/n2), the outer surface is metalized to preserve 100% TIR. Otherwise drop outs would occur at the sensor. This is shown in the angle scan of the next figure. In this figure a point light source is placed at 10mm in front of the entrance aperture on the CPC centerline and the angle is rotated through the full sweep to characterize the CPC.
Another item of note is that if the surface is not metalized then touching the surface will create evanescent coupling that will degrade the optical coupling
My Matlab program treats the CPC as a solid of revolution made of a material with index n2 and surrounding air index n1. The program performs ray tracing to find which rays reach the detector at the CPC exit aperture.
The figure below shows a Matlab result of several ray traces. Note the rays enter the entrance aperture and may bounce around several times, or not at all, before hitting the sensor at the exit aperture on the left.
CPC's can be made of molded plastic (Polystyrene for example) and therefore can be made very inexpensive in large volumes.
Because the incident angle can exceed the critical TIR angle, asin(n1/n2), the outer surface is metalized to preserve 100% TIR. Otherwise drop outs would occur at the sensor. This is shown in the angle scan of the next figure. In this figure a point light source is placed at 10mm in front of the entrance aperture on the CPC centerline and the angle is rotated through the full sweep to characterize the CPC.
Another item of note is that if the surface is not metalized then touching the surface will create evanescent coupling that will degrade the optical coupling
This has been my fun with CPC optics. I hope you found this interesting and maybe even useful.
Monday, December 22, 2008
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=I0en·(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= VrevØ·n·I0en·(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.
The reverse leakage current of a Schottky diode is very sensitive to temperature. In fact it is exponential as follows Ir=I0en·(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= VrevØ·n·I0en·(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?
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.
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.
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