Electrical Engineering Crossword (Issue 291)

The answers to Circuit Cellar’s October electronics engineering crossword puzzle are now available.291crossword (key)

Across

4.     PARAMETRON—Phase-locked oscillator

8.     UNBALANCED—Single-ended

9.     STRAY—Unwanted capacitance

10.   LEYDENJAR—Early capacitor [two words]

12.   ELECTROLYTE—Conducting fluid

14.   CROSSTALK—Caused when one circuit’s signal creates an unwanted effect on another

16.   ANECHOIC—Absorbs sound or electromagnetic wave reflections

17.   BIFILAR—Used in bipolar power-supply transformers to improve output voltage symmetry

18.   CRYSTALRECTIFIER—Semiconductor diode [two words]

19.   DOPPLEREFFECT—Frequency change that occurs when emitter and receiver move in unison [two words]

Down

1.     BLEEDER—A resistor that draws the critical amount of load current

2.     GAUSSMETER—Detects magnetic anomalies

3.     HETERODYNE—Two frequencies combine to produce new ones

5.     SURFACEMOUNT—Place components directly on PCBs [two words]

6.     HASH—Garbage or gibberish

7.     GALVANOMETER—Measures small voltages

11.   RECTIFIER—Passes current in only one direction

13.   CATWHISKER—Sharp, flexible wire that connects to a semiconductor crystal’s surface [two words]

15.   ANOTRON—Cold-cathode-glow discharge diode

 

Electrical Engineering Crossword (Issue 290)

The answers to Circuit Cellar’s September electronics engineering crossword puzzle are now available.CrosswordEmptyGrid (key)

Across 

2.     THREE—Trivalent valence

3.     PHYSICS—Kilby’s Noble Prize in 2000

5.     INVERTER—Converts DC to AC

8.     BATCH—BAT file

9.     MAXIM—Founded ARRL in 1914

10.   KEYBOARD—If you are AFK, what are you away from?

11.   UPENN—University that housed the ENIAC in a 30’ × 40’ room

12.   HERTZ—1 cycle per second

14.   NIBBLE—4 bits

17.   EXPLAINER—Asimov was the great what?

Down

1.     TRACK—PCB path

3.     PATCH—Quick fix

4.     SNIFFER—Used to monitor network traffic

6.     MAXWELL—A Gauss is one of these per square centimeter

7.     IBM—”Big Blue”

8.     BOOLE—“An Investigation of the Laws of Thought” (1854)

13.   TOGGLE—Move from setting A to B

15.   BLUE—Screen of death

16.   NINE—A nonet is a group of what?

17.   EW—Exawatt

Electrical Engineering Crossword (Issue 289)

The answers to Circuit Cellar’s August electronics engineering crossword puzzle are now available.289PuzzleGrid (key)

Across

1.     FAST—Ethernet at 100 Mbps

3.     FAB—IC factory

6.     XOR—Logic gate

7.     VERILOG—HDL created in the early 1980s by Goel and Moorby

9.     MIL—0.001 inches = 25.4 what?

10.   AMPHOUR—Current flow over time [two words]

13.   SANTOS—Greek national soccer team manager with a degree in electrical engineering

14.   NOLEAD—Quad, flat, … [two words]

16.   BUCK—Step-down

17.   FEMTO—0.000000000000001

18.   GND—Ground pin

19.   NULL—Zero

Down

2.     SLICE—Wafer or substrate

3.     FILO—Antonym for FIFO

4.     QUINARY—Base-5

5.     CODERDECODER—CODEC [two words]

7.     VERSORIUM—Gilbert’s static-detection device

8.     DISSIPATION—Release heat

11.   HAPTIC—Relates to touch

12.   JOULE—1 watt second

15.   DOPING—Process of purposely adding impurities

Electrical Engineering Crossword (Issue 288)

The answers to Circuit Cellar’s July electronics engineering crossword puzzle are now available.288Crossword (key)

Across

2.     QUIESCE—Inactive but still available

4.     GLUELOGIC—Used for circuitry interfacing [two words]

7.     AMAYA—Open-source web tool developed by members of the World Wide Web Consortium (W3C)

8.     ROUNDROBIN—A continuous sequence [two words]

9.     FATCLIENT—A tower PC, for example [two words]

11.   LOGICBOMB—Explosive code [two words]

15.   HEISENBUG—A software glitch that changes its conduct when analyzed

16.   STROBOSCOPE—Makes things appear to move slowly or not at all

17.   STATAMPERE—Approximately 0.333 nanoampere

18.   KORNSHELL—Unix command-line interpreter developed by and named after a Bell Labs employee [two words]

19.   VOXEL—Defines a point in 3-D

Down

1.     BEAMFORMING—Signal processing for sensor arrays

3.     SPIBUS—Works in double-duplex mode [two words]

4.     GREP—UNIX-based command-line utility

5.     SUPERHETERODYNE—Used to convert to intermediate frequencies

6.     ENDIAN—Creates data words

10.   PHOTOVOLTAICS—Uses solar power to create energy

12.   BITTORRENT—File sharing protocol

13.   BINARYPREFIX—E.g., gibi [two words]

14.   AUSTRUMI—Linux distribution based on Slackware

 

Issue 286: EQ Answers

Question 1—A divider is a logic module that takes two binary numbers and produces their numerical quotient (and optionally, the remainder). The basic structure is a series of subtractions and multiplexers, where the multiplexer uses the result of the subtraciton to select the value that gets passed to the next step. The quotient is formed from the bits used to control the multiplexers, and the remainder is the result of the last subtraction.

If it is implemented purely combinatorially, then the critical path through all of this logic is quite long (even with carry-lookahead in the subtractors) and the clock cycle must be very slow. What could be done to shorten the clock period without losing the ability to get a result on every clock?

Answer 1—Pretty much any large chunk of combinatorial logic can be pipelined in order to reduce the clock period. This allows it to produce more results in a given amount of time, at the expense of increasing the latency for any particular result.

Divider logic is very easy to pipeline, and the number of pipeline stages you can use is fairly arbitrary. You could insert a pipeline register after each subtract-mux pair, or you might choose to do two or more subtract-mux stages per pipeline register You could even go so far as to pipeline the subtracts and the muxes separately (or even pipeline *within* each subtract) in order to get the fastest possible clock speed, but this would be rather extreme.

The more pipeline registers you use, the shorter the critical path (and the clock period) can be, but you use more resources (the registers). Also, the overall latency goes up, since you need to account for the setup and propagation times of the pipeline registers in the clock period (in addition to the subtract-mux logic delays). This gets multiplied by the number of pipeline stages in order to compute the total latency.

Question 2—On the other hand, what could be done to reduce the amount of logic required for the divider, giving up the ability to have a result on every clock?

 

Answer 2—If you don’t need the level of performance provided by a pipelined divider, you can computes the quotient serially, one bit at a time. You would just need one subtractor and one multiplexer, along with registers to hold the input values, quotient bits and the intermediate result.

You could potentially compute more than one bit per clock period using additional subtract-mux stages. This gives you the flexibility to trade off space and time as needed for a particular application.

Question 3—An engineer wanted to build an 8-MHz filter that had a very narrow bandwidth, so he used a crystal lattice filter like this:

EQ-fig1-CC287-June14

However, when he built and tested his filter, he discovered that while it worked fine around 8 MHz, the attenuation at very high frequencies (e.g., >80 MHz) was very much reduced. What caused this?

Answer 3—The equivalent circuit for a quartz crystal is something like this:EQ-fig2-CC287-June14

The components across the bottom represent the mechanical resonance of the crystal itself, while the capacitor at the top represents the capacitance of the electrodes and holder. Typical values are:

  • Cser: 10s of fF (yes, femtofarads, 10-15F)
  • L: 10s of mH
  • R: 10s of ohms
  • Cpar: 10s of pF

The crystal has a series-resonant frequency based on just Cser and L. It has a relatively low impedance (basically just R) at this frequency.

It also has a parallel-resonant (sometimes called “antiresonant”) frequency when you consider the entire loop, including Cpar. Since Cser and Cpar are essentially in series, together they have a slightly lower capacitance than Cser alone, so the parallel-resonant frequency is slightly higher. The crystal’s impedance is very high at this frequency.

But at frequencies much higher than either of the resonant frequencies, you can see that the impedance of Cparalone dominates, and this just keeps decreasing with increasing frequency. This reduces the crystal lattice filter to a simple capacitive divider, which passes high freqeuncies with little attenuation.

Question 4—Suppose you know that a nominal 10.000 MHz crystal has a series-resonant frequency of 9.996490 MHz and a parallel-resonant frequency of 10.017730 MHz. You also know that its equivalent series capacitance is 27.1 fF. How can you calculate the value of its parallel capacitance?

Answer 4—First, calculate the crystal’s equivalent inductance, based on the series-resonant frequency:EQ-equation1-CC287-June14

Next, calculate the capacitance required to resonate with that inductance at the parallel-resonant frequency:EQ-equation2-CC287-June14

Finally, calculate the value of Cpar required to give that value of capacitance when in series with Cser:EQ-equation3-CC287-June14

Note that all three equations can be combined into one, and this reduces to:EQ-equation4-CC287-June14