RF Cafe Homepage
RF Cascade Workbook 2018 by RF Cafe



Please Support RF Cafe by purchasing my  ridiculously low-priced products, all of which I created.

RF Cascade Workbook for Excel

RF & Electronics Symbols for Visio

RF & Electronics Symbols for Office

RF & Electronics Stencils for Visio

RF Workbench

T-Shirts, Mugs, Cups, Ball Caps, Mouse Pads

These Are Available for Free

Espresso Engineering Workbook™

Smith Chart™ for Excel

Amplifier Solutions Corporation (ASC) - RF Cafe

What's Your EQ?
October 1966 Radio-Electronics

October 1966 Radio-Electronics

October 1966 Radio-Electronics Cover - RF Cafe[Table of Contents]

Wax nostalgic about and learn from the history of early electronics. See articles from Radio-Electronics, published 1930-1988. All copyrights hereby acknowledged.

These two puzzlers for the student, theoretician and practical man, appeared in the October 1966 issue of Radio-Electronics magazine. A wide variety of difficulty in problems exist. "Double-check your answers before you say you've solved them," says editor Clark. Readers submitted most of the "What's Your EQ?" problems. The magazine paid $10 ($92 in 2025 money*) for each one accepted. "We're especially interested in service stinkers or engineering stumpers on actual electronic equipment." See the huge list below of others I have posted over the years. *Another way of looking at it is that $10 in 2025 is worth about $1.03 in 1966.

What's Your EQ?

"What's Your EQ?, October 1966 Radio-Electronics - RF CafeConducted by E. D. Clark

Two 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.

Answers to this month's puzzles are on page 105.

Diode Motor Control? - RF CafeDiode Motor Control?

This black box is powered by a 117-volt ac line and has an ac-motor load. If the switch is opened or the diode D1 reversed, the motor does not run. The box contains four components; no tubes, transistors, relays or resistors are

used. Can you determine these components?

- Thomas Mayfield

 

Too-Bright Lamp

Too-Bright Lamp - RF CafeA 150-mA, 6-volt pilot lamp is to be powered from a 120-volt, 60-Hz power line through a dropping resistor. In an attempt to reduce power dissipation in the resistor, a diode is added to allow conduction only on half-cycles. This resistor is 380 ohms which, without the diode, would allow 300 mA to flow on each half cycle.

Addition of the diode should cut the average current in half, to 150 mA. However, the lamp burns too brightly and fails. What's wrong?

 - Donald E. Phillips


Quizzes from vintage electronics magazines such as Popular Electronics, Electronics-World, QST, Radio-Electronics, 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.

RF Cafe Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

Solutions

These are the answers. Puzzles are on page 60.

Diode Motor Control? Solution - RF CafeDiode Motor Control?

The box contains four diodes connected in a bridge-rectifier configuration. These, in addition to D1, allow ac to flow to the motor load. During the positive alternation, current flows through D4, D1 and D3. During the negative alternation, current flows through D5, D1 and D2. If D1 is reversed or switch opened, no current flows to the motor.

Too-Bright Lamp

The diode cuts the average current in half but the lighting effect of the lamp depends upon the rms or effective value. Without the diode, doubling the current to 300 mA increases the lamp dissipation four times. The diode cuts this in half but the lamp is still averaging twice its rated power. The rms current is proportional to the square root of the power so the required rms current is: I = 0.15 x √12 = 0.2121 amp.

Therefore, the dropping resistor needs to be about 537.5 ohms. (120 - 6)/0.2121 = 537.5  

Amplifier Solutions Corporation (ASC) - RF Cafe