AC Termination

In the case of the RS422 circuit, the differential line impedance is around 100$\Omega$ and "Standard RS-422 drivers are guaranteed to source and sink a minimum 20 mA across a 100Ω load."  (The receiver sensitivity is 200mV.)  That's 40mW dissipated on the termination resistor, and it could be 100mW dissipation at the driver if powered from a 5V supply.  Since this power dissipation is constant regardless the data pattern, the AC termination should be considered.  The AC termination draws power only at the data transition.  How effective is the AC termination?  The AC termination is a resistor and a capacitor in series.  The resistor value matches the transmission line impedance, so about 100$\Omega$.   People seem to use a capacitor value from several hundreds picofarads to a few nanofarads.  How do you select the capacitor value?  If we look at the transmission ratio (the voltage at the termination vs the input voltage), it is $$2\frac{1+j\omega R C}{1+2j\omega RC}.$$If $\omega >> 1/RC$, the ratio is 1.  If we take $R=100$ and $C=1$nF, the frequency must be much greater than 1.6MHz.  This is much more than the usual baud rate we use.  

But the steady state analysis is not the right way to go.  The step response that we are interested in is a transient.  Consider a 1V step applied to the transmission line.  Initially, a 10mA current flows into the termination resistor and capacitor.  The voltage across the capacitor rises as it is charged; and the voltage starts overshoot and reflects back to source.  It can be shown that the reflected voltage is half of the voltage across the capacitor.   The overshoot rises as 2RC time constant and reaches the peak when the reflected voltage makes the round trip.   So the overshoot is depended on the length of the transmission line.  If the transmission line delay is 10ns (~6ft), there is about 10% overshoot with 100$\Omega$ and 1nF termination.

If the transmission line delay is 100ns, the overshoot is $1-\exp(-2\cdot100n/(2\cdot 100 \cdot 1n)) = 63\%$.

TI's AN-903 erroneously stated, ; then the overshoot would be 39% or larger.

But one thing that we have not considered is the source impedance.  The driver could have 10-20$\Omega$ output impedance or series resistors may be included.   When the source impedance is greater or equal to the termination resistance, there is no overshoot, instead there is just a step followed by exponential rise.  For this case, a small capacitor gives fast edges (necessary for high speed transmission).

Christmas LED Lights

I have a string of 60 LED Christmas lights with 6 distinct colors.  I measured the RMS current with a multimeter, and found to be 27mA, so it worked out to be about 3.2W or about 54mW each, which seems reasonable if each LED takes about 20mA.  The string appears to have 3 parallel segments, with 3 lights having 3 wire connections.  That means each LED string consumes 9mA, so the 54mW number seems high for each LED.  Four of LEDs (all blue color why?) are not working, but the rest of the string works, so the connection is maintained.  I was curious to take one off; that segment was out.  So it is the LED that provides the electrical connection.  I took it apart and measured the resistance an found it to be 90 Ohms both ways.  Is there a built-in parallel resistor for fault tolerance?   Each of these resistors would take about 10mW.  The LEDs are running directly off AC, so the current is measured as true RMS, but the voltage RMS is not known.  

Anti Sulfur Chip Resistors

 Many automotive grade chip film resistors have the  anti sulfur feature.  Bourns has a white paper, "Advantages of Sulfur-Resistant Film Resistors".  It describes in good detail the construction.   In short, sulfur based gasses (such as caused pollution) can react with Ag/AgPd resistor terminal that leads to crack the package.  A Palladium (Pd) protection layer is added to the Ag terminal to eliminate the formation of Silver Sulphide and hence increases the reliability.