About Circuit Cellar Staff

Circuit Cellar's editorial team comprises professional engineers, technical editors, and digital media specialists. You can reach the Editorial Department at editorial@circuitcellar.com, @circuitcellar, and facebook.com/circuitcellar

Electrical Engineering Crossword (Issue 294)

The answers to Circuit Cellar’s January 2015 electronics engineering crossword puzzle are now available.294-crossword-(key)


  1. SHIELD—Conductive cover that isolates equipment from electromagnetic interference
  3. NINE—A nonet is a grouping of what?
  4. LIMITER—A device that prevents signal peaks from exceeding max levels
  5. HERTZ—One cycle per second.
  6. DIP—Dual inline package
  8. MOCKUP—Non-working model of a system
  9. PETABYTE—1 quadrillion bytes
  10. PATCH—Reroute a signal to a different circuit
  11. RF—Electromagnetic signals with frequency > 70 kHz


  1. SMOOTHINGCIRCUIT—A filter for removing spurious noise from a power supply
  3. POINTSOURCE—Small energy source that has no effect on its directivity
  4. SHUNT—Component that bypasses another component
  5. GERMANIUM—Semiconductor material used in solid-state diodes and photocells
  6. DAMP—Suppress resonance
  7. DYNE—10 micronewtons
  8. CBAR—0.01 bar
  9. ERG—10–7 joules

Electrical Engineering Crossword (Issue 293)

The answers to Circuit Cellar’s December electronics engineering crossword puzzle are now available.293-crossword-(key)


  1. AMPACITY—Max electrical current
  2. HUMREDUCTION—Use a bridge rectifier driven from an 8- to 12-V transformer winding, a capacitive filter, and a three-terminal IC voltage regulator to achieve this [two words]
  3. EMI—Radiated spurious EM energy
  4. BINARY—Offs and Ons
  5. PASCAL—1 newton/cm2
  6. QUARTZ—Timing crystal
  7. ETCHING—The production of a printed circuit through the removal of unwanted areas of copper foil from a circuit board
  8. CROSSTALK—Caused when one circuit’s signal creates an unwanted effect on another
  9. BUCK—Switched-mode power supply converter
  10. BETATRON—Designed to accelerate electron
  11. BIDIRECTIONAL—Radiating toward or receiving from the front and back only
  12. CRYOTRON—Operates via superconductivity


  1. STOKESSHIFT—Can reduce photon energy [two words]
  2. INPUT—Signal in
  3. ATTO—0.000000000000000001
  4. NOISECANCELLATION—Eliminates out-of-phase information [two words]
  5. MAL—10 Decimals equals?
  6. RAMP—Linearly rising signal
  7. FORCE—Newton
  8. KEYED—OOK is on/off what?
  9. BUS—Common path for several signals

Electrical Engineering Crossword (Issue 292)

The answers to Circuit Cellar’s November electronics engineering crossword puzzle are now available.292-crossword-(key)


  1. BITS—A nibble is 4 of these
  3. MICRO—Metric Prefix for 0.000001
  4. PATCH—To re-route a signal to a different circuit
  5. TOKEN—Used for authentication
  6. RELAY—A switch that is actuated by another electrical signal
  7. JITTER—The deviation of some aspect of a digital signal’s pulses
  8. RHEOSTAT—A variable resistor
  9. LUX—Lx
  10. VOLTA—Italian physicist who invented the first batteries
  12. CONDUIT—Wire piping
  13. PIGTAIL—Short wire connecting components
  14. RECTIFIER—A diode, used for converting AC into DC


  1. SCHOTTKY—High-speed diode that has very little junction capacitance
  3. INRUSH—A sudden input current surge
  4. CRESTFACTOR—The ratio of the peak value to the RMS value (two words)
  5. MICROFARAD—1,000,000 pF
  6. ANALOG—Constant signal processing



Issue 294: EQ Answers

Problem 1—Let’s get back to basics and talk about the operation of a capacitor. Suppose you have two large, flat plates that are close to each other (with respect to their diameter). If you charge them up to a given voltage, and then physically move the plates away from each other, what happens to the voltage? What happens to the strength of the electric field between them?

Answer 1—The capacitance of the plates drops with increasing distance, so the voltage between them rises, because the charge doesn’t change and the voltage is equal to the charge divided by the capacitance. At first, while the plate spacing is still small relative to their diameter, The capacitance is proportional to the inverse of the spacing, so the voltage rises linearly with the spacing. However, as the spacing becomes larger, the capacitance drops more slowly and the voltage rises at a lower rate as well.

While the plate spacing is small, the electric field is almost entirely directly between the two plates, with only minor “fringing” effects at the edges. Since the voltage rise is proportional to the distance in this regime, the electric field (e.g., in volts per meter) remains essentially constant. However, once the plate spacing becomes comparable to the diameter of the plates, and fringing effects begin to dominate, the field begins to spread out and weaken. Ultimately, at very large distances, at which the plates themselves can be considered points, the voltage is essentially constant, and the field strength directly between them becomes proportional to the inverse of the distance.

Problem 2—If you double the spacing between the plates of a charged capapcitor, the capacitance is cut in half, and the voltage is doubled. However, the energy stored in the capacitor is defined to be E = 0.5 C V2. This means that at the wider spacing, the capacitor has twice the energy that it had to start with. Where did the extra energy come from?

Answer 2—There is an attractive force between the plates of a capacitor created by the electric field. Physically moving the plates apart requires doing work against this force, and this work becomes the additional potential energy that is stored in the capacitor.

Question 3—What happens when a dielectric is placed in an electric field? Why does the capacitance of pair of plates increase when the space betwenn them is filled with a dielectric?

Answer 3—Dielectric materials are made of atoms, and the atoms contain both positive and negative charges. Although neither the positive nor the negative charges are free to move about in the material (which is what makes it an insulator), they can be shifted to varying degress with respect to each other. An electric field causes this shift, and the shift in turn creates an opposing field that partially cancels the original field. Part of the field’s energy is absorbed by the dielectric.

In a capacitor, the energy absorbed by the dielectric reduces the field between the plates, and therefore reduces the voltage that is created by a given amount of charge. Since capacitance is defined to be the charge divided by the voltage, this means that the capacitance is higher with the dielectric than without it.

Problem 4—What is the piezoelectric effect?

Answer 4—With certain dielectrics, most notably quartz and certain ceramics, the displacement of charge also causes a significant mechanical strain (physical movement) of the crystal lattice. This effect works two ways — a physical strain also causes a shift in electric charges, creating an electric field. This effect can be exploited in a number of ways, including transducers for vibration and sound (microphones and speakers), as well as devices that have a strong mechanical resonance (e.g., crystals) that can be used to create oscillators and filters.

Contributed by David Tweed

Issue 292: EQ Answers

Problem 1—Let’s talk about noise! There are different types of noise that might be present in a system, and it’s important to understand how to deal with them.

For example, analog sensors and other types of active devices will often have AWGN, or Additive White Gaussian Noise, at their outputs. Any sort of analog-to-digital converter will add quantization noise to the data. What is the key difference between these two types of noise?

Answer 1—The key difference between AWGN and quantization noise is the PDF, or Probability Density Function, which is a description of how the values (voltage or current levels in analog systems, or data values in digital systems) are distributed.

The values from AWGN have a bell-shaped distribution, known variously as a Gaussian or Normal distribution. The formula for this distribution is:292-EQ-equation

µ represents the mean value, which we take to be zero in discussions about noise. σ is known as the “standard deviation” of the distribution, and is a way to characterize the “width” of the distribution.

It looks like this:


Source: Wikipedia (en.wikipedia.org/wiki/File:Standard_deviation_diagram.svg)

While the curve is nonzero everywhere (from –∞ to +∞) it is important to note that the values will be within ±1 σ of the mean 68% of the time, within ±2 σ of the mean 95% of the time, and within ±3 σ of the mean 99.7% of the time. In other words, although the peak-to-peak value of this kind of noise is theoretically infinite, you can treat it as being less than 4σ 95% of the time.

On the other hand, the values from quantization noise have a uniform distribution — the values are equally probable, but only over a fixed span that’s equal to the quantization step size of the converter. The peak-to-peak range of this noise is equal to the converter’s step size (resolution).

However, it’s important to note that both sources of noise are “white”, which is a shorthand way of saying that their effects are uniformly distributed across the frequency spectrum.

Problem 2—Signal-to-noise ratios are most usefully described as power ratios. How does one characterize the power levels for both AWGN and quantization noise?

Answer 2—The power of a noise signal is proportional to the square of its RMS value.

The RMS value of AWGN is numerically equal to its standard deviation.

The RMS value of quantization noise is simply the peak-to-peak value (the step size of the converter) divided by √12, or VRMS = 0.2887 VPP. This is easily derived if you characterize the quantization noise signal as a small sawtooth wave that gets added to the analog signal.

Question 3—When you have multiple sources of noise in a system, how can you characterize their combined effect on the overall system performance?

Answer 3—When combining noise sources, you can’t simply add their RMS voltage or current values together. From one sample to the next, one noise source might partially cancel the effects of the other noise source(s).

Instead, you add the individual noise power levels to come up with an overall noise power level. Since power is proportional to voltage (or current) squared, this means that you need to square the individual RMS measurements, add them together, and then take the square root of the result in order to get an equivalent overall RMS value.

VRMS(total) = √(VRMS(n1)2 + VRMS(n2)2 + …)

Problem 4—Broadband analog sensors and other active devices often specify their noise levels in units of “microvolts per root-Hertz” (µV/√Hz) or “nanoamps per root-Hertz” (nA/√Hz). Where does this strange unit come from, and how do you use it?

Answer 4—As described in the previous answer, uncorrelated noise sources are added based on their power. With AWGN, the noise in one “segment” of the frequency spectrum is not correlated with another segment of the spectrum, so if you have a particular voltage level of noise in, say, a 1-Hz band of frequencies, you’ll have √2 times as much noise in a 2-Hz band of frequencies. In general, the RMS noise level for any particular bandwidth is going to be proportional to the square root of that bandwidth, which is why the devices are characterized that way.

So, if you have an opamp that’s characterized as having a noise level of 2 µV/√Hz, and you want to use this in an audio application with a bandwidth of 20 kHz, the overall noise at the output of the opamp will be 2 µV × √20000, or about 283 µVRMS. If your signal is a sinewave with a peak-to-peak value of 1V (353 mVRMS), you’ll have a signal-to-noise ratio of about 124 dB.

Contributed by David Tweed