What's Your EQ?
June 1964 Radio-Electronics

June 1964 Radio-Electronics

June 1964 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.

One of the nice things about these "What's Your EQ?" challenges that appeared in Radio-Electronics magazine is that they tax your ability to recall basic electronics circuit theory. The first one in the June 1964 issue requires you (spoiler here) to apply Thévenin's theorem in order to arrive at the solution. You also need to know about maximum power transfer which (another spoiler) requires the load impedance to be the complex conjugate of the source impedance. "No Volts" will make you very appreciative of today's high input impedance voltmeters; the problem statement itself made my head hurt. Being aware of such issues often meant the difference between success and failure when assessing television and radio circuits. "Music-Intercom Trouble" almost certainly was inspired by a 1960 Electronics World episode of "Mac's Radio Service Shop," entitled "Technician or Consulting Engineer?." Have at it.

What's Your EQ?

Conducted by E. D. Clark

Three puzzlers for the student, theoretician and practical man. Simple? Double-check your answers before you soy 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

 

Maximum Power

Load RO is matched for maximum power output from the generator or battery. However, it is desired to obtain maximum power in a new load to be connected between A and B. What must be the value of the new load, and what is the power it expends?

- H. D. Varadarajan

 

No Volts

A full-wave power supply has two silicon rectifiers in series on each side. Each diode has a PIV rating of 400 volts. A 1,000-ohms-per-volt, rectifier type ac voltmeter on its 300-volt range, with the hot test lead plugged into the OUTPUT jack to block the dc component, is connected in turn across each diode to determine whether the ac voltages across the two diodes are equal. They seem to be: the meter reads zero across each diode. Why?

- Basil Barbee

 

Music-Intercom Trouble

In a combination background music and intercom system, the speaker is connected so that when switch S1 is depressed it will stop the music and connect the speaker to the input of the intercom amplifier to initiate a call. However, the circuit as shown here is not usable in the CALL position. What is the trouble symptom and why does it happen? What can be done to correct it?

Capacitance effects of wiring and switches are unobjectionable. For simplicity, switching for talkback from the intercom is not shown.

- Wayne Lemons

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.

RF Cafe Quizzes

Vintage Electronics Magazine Quizzes

Vintage Electronics Magazine Quizzes

 

Answers to What's Your EQ?

 

Maximum Power

The first operation is to reduce the circuit to its equivalent Thévenin generator. To do this, we first find the Thévenin voltage at point A.

EA = E x 10/(10 + 10) = 50 volts = ETH

Next, looking back into the circuit from points A and B, with E reduced to zero, we see RO and RInt in parallel.

(10 x 10)/(10 + 10) = 5 ohms = RTH

Since the "internal resistance" is now 5 ohms, the new load must be 5 ohms for maximum power transfer. Using any of the power formulas, (I2R, for instance) the power in the new load is now found to be:

P = (50/10)2 x 5 = 125 watts

 

No Volts

Silicon diodes have a very high back resistance, much higher than the 300,000-ohm resistance of the meter. The diode of a series pair that is not shunted by the meter has to withstand virtually all of the drop, since its back resistance is so much higher than that of the meter. The shunted diode then has a negligible drop across it. While the PIV rating of the unshunted diode is temporarily exceeded during the test, it is undamaged due to its conservative rating.

 

Music-Intercom Trouble

Background music will overload the intercom amplifier when S is in CALL position. This occurs because of a built-in ground loop and the all-too-prevalent idea that a common or ground lead is "cold" just because it happens to be in the return circuit.

To make it easier to see how this ground loop occurs, we have taken the circuit and substituted a battery for the background music amplifier and a voltmeter for the intercom amplifier. Resistors are substituted for the speaker loads. With the arbitrary values shown, we can see that at point A there is a 1-volt drop across the parallel resistances of the ground wires. With the switch in the CALL position, this voltage drop will be indicated on the meter.  

In the music-intercom system this voltage is fed directly into the intercom amplifier and amplified several hundred times.

The obvious and simple solution is to use a DPDT switch for S so that both sides of the speaker will be disconnected when a call is made.

 

 

Posted July 24, 2023