World's Smartest Test Engineer—Quiz Archive
April 2005
March 2005
February 2005
January 2005
December 2004
November 2004
October 2004
September 2004
April 2005
Guard-banding
You have a lot of resistors whose values are evenly distributed over the range from 9900 to 10,100 ohms. You define a "good" resistor as having a value between 9950 and 10,050 ohms. For separating "good" from "bad" resistors, you have available an ohmmeter that exhibits ±1% (of reading) accuracy.
Question
What percentage of "good" resistors must you reject to prevent test escapes?
Answer to the April 2005 Quiz
100%. This instrument is not accurate enough to guarantee any good parts.Answer to the April 2005 Quiz
Winner
From all of the respondents who correctly answered the April question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
D. Way, San Diego, CA
Circuit Design
The circuit below is the output stage ofa digital logic circuit.
Question
Why do you need a decoupling capacitor on the above circuit. Wherewould you place it? What will it do for you?
Answer to the March 2005 Quiz
Vcc to ground, to provide high instantaneous currents during switching (or decouplethe power supply).
Explanation: When Vin is high,Q1 is on, which creates a low resistance path from R1 tothe base of Q4. Thus, Q4 is on, which pulls Voutlow. Q2 and Q3 make a Darlington pair. With Q1on, Q3 is off. When Vin is low, Q1 becomes ahigh-resistance path, cutting off the current to the base of Q4,which turns it off. R1 provides a base current for Q2,which also turn on Q3. Thus, Vout is high.
But, during the transition times, both Q3and Q4 will be partially high. Thus, a relatively large amount ofcurrent with flow through Q3 and Q4. Because the transitiontime can be less than a nanosecond, inductance in the leads can cause a largevoltage drop in the circuit. (V = L*di/dt). That limits the power supply'sability to provide the instantaneous current. The sharp rise in current canproduce a radiated emission from the circuit.
A coupling capacitor, connected from Vccto ground, can provide the additional current to keep the path of the largecurrent small. The capacitor must reside close enough to the circuit so theinductance in the circuit doesn't negate its ability to supply the current.
Winner
From all of the respondents who correctly answered the March question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
James Mcdonald, Sterling, VA
S-Parameter Measurements
Atwo-port vector network analyzer and S-parameter test set exhibit a 50-Ocharacteristic impedance. Connect a 3-in. length of lossless 50-O coaxialcable between the test set's two ports.
Question
At 1 GHz, what values for S-parameterss11, s12, s21, and s22 will the analyzer indicate?
Answer to the February 2005 Quiz
s11=s22=0;s21=s12=1 at 90 degrees.
Winner
From all of the respondents who correctly answered the February question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
Jeffrey A. Gruber,Kenosha, WI
January 2005
Data-Communications Test
As reported recently in Test & Measurement World, a panelist at a trade show alluded to a particular instrument type as the ultimate arbiter of high-speed serial data-link performance.
Question
To what type of instrument was thispanelist referring?
Answer to the January 2005 Quiz
Bit-error-rate tester, or BERT
Winner
From all of the respondents who correctly answered the January question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
Andrew Flint, Redmond, WA
December 2004
Electronic Design Automation/
Automatic Test Equipment
EDA companies and ATE makers have been cooperating to facilitate the exchange of design and test data to support test-program generation and failure analysis for digital ICs. Efforts at defining digital test-vector formats have been codified in an IEEE standard.
Question
What is the number of this standard, what acronym is commonly used to describe it, and what does the acronym stand for?
Answer to the December 2004 Quiz
IEEE 1450 STIL (Standard Test Interface Language).
For more information, see grouper.ieee.org/groups/1450.
Winner
From all of the respondents who correctly answered the December question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
Donald W. Kane, P.E., Brentwood, NY
Instrumentation Amplifiers
The following circuit is a two op-ampdifferential amplifier that uses a single supply voltage. The circuit uses fivegain resistors (R1 through R4 and RG) to control its voltage gain.
The equation for the voltage gain is:
Question
Calculatethe value of RG under the following conditions:
R1 = R4 = 33 k
R2 = R3 = 3.3 k
VREF = 6 V
Vin = 0.1 V
Vout = 8.1 V
Resistors Ra and Rb are current-limitingresistors and don’t figure into the gain calculations.
(To learnthe details of this circuit, see “The Two Op Amp Instrumentation AmplifierTopology,” Practical Analog Design Techniques, Analog Devices, 1995, p.1-10. www.analog.com.)
Answer to the November 2004 Quiz
RG = 6.6 k
Winner
From all of the respondents who correctlyanswered the November question, the following person was randomly chosen toreceive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
Allen Bennink, Corinth, MS
RF/Microwave Measurements
In the figure below, vector OL represents a normalized complex load impedance:
zL = 2.5 – j1
What does the green circle (centered about the origin) represent? Please provide the parameter name and its numerical value.
Answer to October 2004 Quiz
The green circle represents the locus of points for which the standing wave ratio equals 3, or SWR = 3.
The diagram is a Smith chart. The point L resides at the intersection of two circles: one is the locus of points representing a resistance of 2.5 and the other is the locus of points representing a reactance of –1.
Circles centered about the origin represent contestant standing wave ratios. Note that the green circle containing point L shares a tangent with the circle representing resistances of 3. That provides a quick and easy way of reading SWR from the Smith chart.
For more, see "How does a Smith chart work?" Test & Measurement World, July 1, 2001, in our Article Archives .
Winner
From all of the respondents who correctly answered the October question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
Darrin Daigle, Lake Charles, LA
September 2004
DC Resistance Measurements
For the circuit shown below, you wish to measure RDUT in the presence of the parallel network of RP1 and RP2 connected in series.
You know that RP1 is 2000 ohms and that RP2 is 100 ohms. You set the adjustable current source iTEST to 1 mA and find that the DMM indicates 1 V.
Assume that the voltage, current, and resistance values cited above are exact. Further, assume that the op amp exhibits ideal performance (for example, zero input bias current, infinite gain-bandwidth product, and zero output impedance) except for one specification: It suffers from a ±1-mV input offset voltage.
Question
Based on the measurement values cited:
What is the nominal value of RDUT, and what, if any, is the measurement uncertainty contributed by the op amp's input-offset voltage specification?
Express your answer in this format: X ohms plus or minus Y percent.
Use the response form below to submit your answer; we cannot accept answers submitted through any other method.
Hint: An article in the Test & Measurement World archives describes the operation of this test circuit. If you know a common term for the test-point connection shown at the right-hand side of the diagram, you can quickly search the archives to locate the article.
Answer to September 2004 Quiz
The circuit-common connection at the right-hand side of the diagram is called an “ohms guard.” An ideal amplifier will keep each side of RP2 at 0 V. In the ideal case, no current flows through RP2, and all of iTEST flows through RDUT. RDUT is therefore 1 V divided by 1 mA, or 1000 ohms.
Amplifier input offset voltage. results in a worst-case plus or minus 1 mV across RP2, resulting in a current of 1 mV divided by 100 ohms, or 10 microamps, flowing through RP2. If the current flows left to right, the current through RDUT must equal iTEST + 10 microamps, or 1.01 mA. An RDUT producing a 1-V drop in response to 1.01 mA would equal 990 ohms.. If the current flows right to left, the current through RDUT must equal iTEST - 10 microamps, or 0.99 mA. An RDUT producing a 1-V drop in response to 0.99 mA would equal 1010 ohms.
The correct answer is 1000 ohms plus or minus 1 percent.
Winner
From all of the respondents who correctly answered the September question, the following person was randomly chosen to receive an MP3 player, courtesy of contest sponsor, Agilent Technologies:
Steve Petermann, Mendota Heights, MN
