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RF Workbench  (shareware)

The electrical circuit entitled "Resistors Galore," which was part of the collection of posers in the June 1966 issue of Radio-Electronics magazine's "What's Your EQ" (EQ = Electronics Quotient, a la IQ = Intelligence Quotient) feature, resulted in an interesting response from a reader. Mr. Milton Badt submitted the bit shown to the left ("Ladder Lingo") in the following January edition of the magazine. Interestingly, while he pointed out the significance to the relation to phi (φ), defined as (1+√5)/2, he did not also note that the fraction is commonly referred to as the Golden Ratio, and its result, 1.618034... is called the Golden Number. A rectangle with side lengths who's proportions are according to a/b = φ is called a Golden Rectangle. There is also a resistor | capacitor voltage divider, and a mystery power source challenge.

Conducted by E. D. Clark

Three puzzlers for the student, theoretician and practical man. Simple? Double-check your answers before you say you've solved them. If you have an interesting or unusual puzzle (with an answer) send it to us. We will pay \$10 for each one accepted. We're especially interested in service stinkers or engineering stumpers on actual electronic equipment. We get so many letters we can't answer individual ones, but we'll print the more interesting solutions - ones the original authors never thought of.

Write EQ Editor, Radio-Electronics, 154 West 14th Street, New York, N. Y. 10011.

How Many Volts?

In the circuit shown, it is assumed that sufficient time has elapsed to completely charge the three capacitors.

Find the voltage across the 3-μf capacitor.

-Robert L. David

Resistors Galore

The diagram shows an infinitely long triangular network of 1-ohm resistors. What is the resistance (R) looking into the network?

Since the network cannot be broken down in pi or T junctions, existing textbook formulas for long attenuator networks do not apply here. However, with a little ingenuity and basic algebra, the problem can be solved.

-William Uhlenhoff.

Phantom Power Source?

The photodiode circuit shown here was being tested in the lab. A strobe light was used as the light source, generating a pulse of 100 microseconds' duration every 10 milliseconds. Everything seemed normal until the 90-volt battery was disconnected. With the battery removed, the observed waveform did not disappear but, instead, nearly double in amplitude. This performance continued for several minutes before a significant amplitude decay was noted.

What is the mysterious source of power? (Hint: output disappears instantly on removing the battery if the scope is decoupled).

-G. Robert Wisner

Quizzes from vintage electronics magazines such as Popular Electronics, Electronics-World, QST, and Radio News were published over the years - some really simple and others not so simple. Robert P. Balin created most of the quizzes for Popular Electronics. This is a listing of all I have posted thus far.

How Many Volts?

Neglecting leakage currents in the capacitors, we see that the resistive voltage divider produces 91 volts on the top plate of the 2-μf capacitor. (Fig. 1)

We'll use the Thévenin's Theorem approach to the solution of the problem. Calling the 3-μf capacitor the load, temporarily disconnect it and determine the voltage between points A and B. The 9 volts across the 2-μf in series with the 1-μf produces 6 volts across the 1-μf and 3 volts across the 2-μf. Therefore, the voltage between points A and B is 91V + 3V = 94V. (Fig. 1)

Looking into the circuit at A and B, we see 1-μf in parallel with 2-μf, with power supplies considered a short circuit. Fig. 2 shows the Thévenin's equivalent.

By inspection, 3-μf in series with 3-ftf across 94 volts equals 94/2 = 47V.

Resistors Galore

Since the network is infinitely long, the first two resistors can be temporarily removed without changing the resistance value of the remaining network (still R ohms).

The first two resistors are now reconnected as shown in the equivalent-circuit diagram. Solving for the series-parallel resistive circuit, we get:

R = (R1R2)/(R1+R2) = 1(1+R)/(1+(1+R))

R2 + R - 1 = 0

Solving for R:

R = (-1 + √5)/2 = 0.618 ohm.

Phantom Power Source?

The scope's AC coupling capacitor (C1-0.1 μF) supplies the power. It is initially charged by the 90-volt battery. With the battery removed, the photo-diode, with its high internal impedance, has only the 10 megohms of the scope's probe and input as a load. The low duty cycle of the light pulse results in very little average current being drawn from C1.

The time constant of the input circuit (R1, R2 and C1) is extended from its calculated value of 1 second to 100 seconds - thanks to the 1% duty cycle of the strobe light. When the scope is switched to dc, the not-so-mysterious power source (C1) is shorted and circuit operation stops immediately.

Posted April 3, 2023

 About RF Cafe Copyright: 1996 - 2024 Webmaster:     BSEE - KB3UON RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The World Wide Web (Internet) was largely an unknown entity at the time and bandwidth was a scarce commodity. Dial-up modems blazed along at 14.4 kbps while tying up your telephone line, and a nice lady's voice announced "You've Got Mail" when a new message arrived... Copyright  1996 - 2026 All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged. All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged. My Hobby Website: AirplanesAndRockets.com My Daughter's Website: EquineKingdom