Avoid noises in mixed signal design

Today most of all embedded systems consist of two part circuitry – digital and analog. Digital part is usually controller, its timing circuit and other input output devices. Frequently there is an analog part on same board like ADC, OP amplifiers, sensors and other analog circuitry. Such designs are called mixed-signal designs. Where digital and analog part meats – the grounding problems starts. Fact is that each conductor has its own impedance, so any current flowing result in voltage drops. Ground wires and planes isn’t exceptions. Digital and analog grounds can generate significant electromagnetic radiation that adds noises to signals we need. So the over all system quality drops because o poor design.   In a good design analog ground palane and digital ground plane should be separated. With multilayer PCB this can be done very easy. Another issue is that digital signal traces shouldn’t cross analog ground and analog signal wires shouldn’t cross digital ground plane area. Well of course and try to avoid aligning digital and analog wires as they can catch each other radiated noise. How to deal with these problems. Well first thing is to recognize problematic areas correctly. Then you will be able to implement…

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Eagle CAD PCB footprints for audio processors TDA7313 and TDA7315

TDA7313 and TDA7315 are digital controlled audio processors that can be used in applications where digital control of audio is needed. TDA7313 and TDA7315 are are pretty same just TDA7313 is more complex by hawing 3 stereo input channels and two outputs (front and rare). Both chips have volume control (step=1.25dB), tone (bass and treble) balance, independent faders for each output processors. Also there are loudness functionality. TDA7313 and TDA7315 have low distortion, low noise, that makes them ideal for quality audio applications for car audio and Hi-FI systems.   Both chips can be controlled by MCU via I2C(TWI) protocol. This makes easy to interface microcontrollers with additional features like LCD, buttons. Or simply control directly from PC using RS232 or other interface. Both chips can be found in DIP or SO packages. Here are Eagle libraries of both digital audio processors that I have made. tda7315.zip(2kB) tda7313.zip(2kB) I have intent to make control board with TDA7315 with LCD and button control using Atmega8 microcontroller.  

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Calculating of electrical heating elements

Electrical heating elements are often used for teapots, irons, electric ovens, soldering irons etc. When projecting designs with electric heating elements you need to do some calculations that may seem difficult at first glance. But when looking more deeply this becomes simple task. We know that electric heating is a result of current flow in wire with some resistance. Resulting heat is work done by electric current. Work A(J) can be calculated by formula: A=U·I·t Where U – Voltage(V), I – current(A), t – time(s). Then amount of heat produced in wire where electric current flows is calculated as follows: Q=I2·R·t Lets calculate simple problem. How much time we need to boil 2 liters of water? Lets take voltage U=220V; Heating element requires current I=4A; Efficiency coefficient is 80%; Starting temperature of water is 20°C; Water specific heat – 4200(kJ/kgK). First we have to calculate the amount of heat required to boil a water: Qn=C·m·(tboil-t0)=4200·2·(100-20)=672000J; With efficiency of 80% we get that wee neet to produce Q=Qn/η=672000/0.8=840000J. So we get that Q=A=U·I·t where time needed to boil: t=Q/(U·I)=840000/(220·4)=954s=15min and 54s. When we know work produced by current during time period – we can find the power: P=A/t = UI =…

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Calculating wire diameter for maximum current

When selecting wire diameter we usually lok for cross reference tables where you can find recommended wire diameter for maximum current drive. But sometimes may be more useful to calculate by formula than look in to the tables. This way you can have more accurate results. There is nothing new just simple physics. Wire resistance (Ω)is calculated as follows: R = ρ·l/S or R = (1.27·ρ·l )/d2; Where ρ – resistivity of material (Ω·m), found in tables, l – wire length (m), S – cross selection area(m2), d – wire diameter (m). According to this we can calculate wire length, when other parameters are known: l = R·S/ρ or l = (0.785·R·d2)/ρ Cross selection area can by found as follows: S=0.785·d2. Wire resistance depends on temperature. Lets say R2 is resistance for temperature t2 and R1 is for t1(usually t1 = 18°C), then: R2=R1·(1+α·(t2-t1)) Where α – temperature coefficient (K-1). Lets say that maximum current for cross selection area is marked as Δ (A/mm2), then maximal current can be found I=0.785·Δ·d2. Needed diameter can be found by formula(√ – square root symbol): d=√((1.27·I)/Δ) So if allowed load is Δ=2A/mm2, then d=0.8√I. If diameter is les than 0.2mm, then melting current…

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Calibration and monitoring sensors

Every sensor that measure real world things have real tolerances. There is no absolut accurate sensors as there are many factors that influence tolerances and reliability. If you ever tried to measure simple resistor resistance you noticed that 1kΩ resistor actually is between 990Ω to 1,1kΩ. This means that tolerance of resistor is 1%. if resistor tolerance would be 5% then this range would be much bigger. Also these values vary due to temperature. Lets say wee want to measure temperature with 1°C accuracy. When you buy temperature sensor you cant get this accuracy as its tolerance vary. Especially cheap sensors can’t be trusted. For instance you may get 98°C for boiling water and so on. In this case ve need to adjust sensor to give 100°C on boiling water. Of course this is simple example. As sensor is a part of whole measuring system where are board, other components. We need to adjust sensor to work as part of system. First thing you have to do isto exclude environment factors affecting sensor like board temperature or temperature of sensor itself. This is called compensation. Compensation doesn’t calibrate sensor as it only exclude environment factors, but doesn’t adjust sensor itself.…

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Electric circuits analogy to water pipes

Sometimes calculating parameters and behaviour of some objects may be much easier when using analogy to objects with well developed theory and calculation methodology. In earlier article we analysed power dissipation of electronic devices using Ohms law where Voltage=temperature, Current=Dissipation and Resistance=Thermal resistance. This time lets look how can electronic devices transformed in water pipes and vice versa. Lets take Voltage source. A simple battery is like a water pump which provides a Pressure (a voltege analogy): Then second electronic element is a resistor. Resistor can be imagined as water pipe with smaller aperture. The higher resistance is – the smaller aperture of pipe: Now we have the elements needed for constructing a simple circuit. We have Pressure = Voltage, Resistance=Flow resistance of pipe. Current = Water flow. The electric current and water flow can be calculated using the same Ohms Law formula: I=V/R In reality there are many limitations of such approach as operating temperatures, power dissipation and power limits. But with some corrections to model calculations can be quite accurate. Ok lets see how semiconductors can be converted to water pipe system elements. Diode is imagined as valve allowing current flow to one direction. In this case diode…

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Thermal analysis of semiconductor devices-when do you need heat-sinks

Every electronics constructor and engineer knows that electronic devices like IC’s voltage regulators and CPU heats up. Temperature is critical parameter for reliable system operation. System reliability is very closely related to device temperature high or low. While temperature increases the reliability drops exponentially. If you look in device datasheets you will find a recommended Operating temperature range. For instance Voltage regulator 7805 can normally operate at 0 – 70°C temperature range. When maximum Junction Temperature can reach up to 150°C. So how is operating temperature range calculated. It is not as difficult as it may look like. We know, that temperature is analogous to voltage, thermal resistance changed to electric resistance, power dissipation is as current, then: Temperature=Power*Thermal_Resistance; Temperature means the rise in °C, Power in Watts and Thermal_Resistance (°C/Watts). The calculation can be done using simple Ohms law. Lets analyse real example by using simple 7805 voltage regulator. We know that output produces 5V. Lets assume that input is 9V and output provides 1A current. Then regulators power dissipation is: P=V·I=(9V-5V)·1A=4 Watts So datasheet says that thermal resistance of junction to case (TO-220) is 4°C/W and case(without heat-sink) to ambient is 50°C/W. Then total thermal resistance is Rt=4+50=54°C/W.…

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