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More About Attenuators
January 1954 Radio-Electronics Article

January 1954 Radio-Electronics

January 1954 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.

I didn't know that another name for a nomogram (or nomograph) is an "abac," but thanks to Mr. Crowhust in this 1954 Radio-Electronics magazine article I now do. At first I thought maybe it was a shortened version of abacus, but I couldn't find anything suggesting that. Anyway, this article presents a couple abacs you do not often (if ever) see - the attenuation and reflection levels for resistive attenuators when the impedance(s) are incorrect for the designed values. Keep in mind, particularly when using off-the-shelf components like attenuators, filters, power couplers and splitters, PIN switches, etc., that if your source and load impedances are not exactly the same as the component input and output impedance (usually 50 Ω or 75 Ω), the response will not be as advertised. Achieving the proper performance will require inserting an impedance matching section between the circuit and the component.

Maintaining proper attenuation and impedance reflection when connecting two different impedances.

Ratio between the impedances and attenuation resulting from mismatch - RF Cafe

Chart I - Reference line in the middle gives the ratio between the impedances and attenuation resulting from mismatch.

By N. H. Crowhurst

In the preceding article (page 36, December) on the design of constant impedance attenuators, the author stated that attenuators perform correctly only when properly terminated. If terminated with impedances other than those on which the design is based, both attenuation and impedance reflection through the attenuator become incorrect. Two of the three charts here deal with the result of using an attenuator of the constant-impedance type with incorrect impedances, while the third shows how to design matching pads for connecting together two different impedances so that both terminating impedances are matched.

Chart I is a simple alignment abac (alignment chart or monogram) with two impedance scales on the outside and a reference line in the middle, giving on the left, the ratio between the impedances on the outside scales, and on the right, the mismatch loss resulting from connecting these impedances together.

If a constant-impedance attenuator is connected to one of the working impedances, and is correctly designed for that impedance, the presence of the attenuator will not affect the mismatch loss, because the impedance reflected to the opposite terminals will be the same as the correct impedance terminating it. This applies to only those attenuators, either where the constant impedance characteristic applies for both directions, or, in the case of L pads, where the attenuator is correctly terminated in the direction for which it is designed to have constant impedance. It would be pointless to extend the chart for other cases using L pads, because there seems to be no reason why the correct matching type L pad could not be used.

Mismatch reflection through the attenuator - RF Cafe

Chart II - For determining mismatch reflection through the attenuator when attenuator is improperly terminated.

Minimum attenuation necessary for correct matching - RF Cafe

Chart III - To determine minimum attenuation necessary for correct matching, and resistor values required for L-pads.

The mismatch loss, over and above the attenuation of the attenuator, will be quite independent of either the type of attenuator or the amount of attenuation for which it is designed.

Chart II deals with the mismatch reflection through the attenuator. The diagram on page 119 shows that the attenuator is designed for a working impedance Z0 and a specific attenuation in db. It is, however, improperly terminated by the impedance Z2. As a result, the chart shows what impedance will appear at the opposite end of the attenuator, represented by the symbol Z2'. The type of attenuator is unimportant. What affects the mismatch reflection is the design figure of attenuation for the particular unit.

It is also unimportant whether the designed attenuation is achieved in one stage or in a number of separate stages. For example, the mismatch reflection would be the same through 40 db of constant-impedance attenuation, whether this was made up in a single pad, or of a number of separate pads of, say, 20 plus 10 plus 5 plus 2 plus 2 plus 1.

Example 1

A constant-impedance attenuator designed for 5,000 ohms is terminated by a resistance load of 5 ohms resulting in a mismatch loss.

To find the mismatch loss: Chart I, using Z2 5,000 ohms and Z1 5 ohms, gives 24 db. Using Chart II to find the mismatch reflected impedance, Z2/Z0, as given from Chart I, is 0.001, which will be indistinguishable from zero at the right-hand side of Chart II. This means, that, to all intents and purposes, a 5-ohm resistance connected to a 5,000-ohm attenuator is a short circuit. Suppose the attenuation is designed for 20 db, then by using the straight-edge, or ruler, as shown in the key on Chart II, Z2'/Z0 is given as 0.98, which means the input impedance is 0.98 x 5,000 ohms, or 4,900 ohms.

If the attenuation was 10 db, Z2'/Z0 becomes 0.82, making the impedance at the input end of the attenuator 4,100 ohms.

Example 2

An attenuator is designed for 250 ohms and is actually terminated by 600 ohms. Using Chart I, the ratio Z1/Z2 is 2.4, and the mismatch loss attenuation is slightly less than 1 db.

Turning to the mismatch impedance transfer, using 2.4, Z2/Z0, which is in the left hand part of the arc on Chart II, and supposing the attenuator is designed for an attenuation of 6 db, alignment gives a reading of 1.24 for Z2'/Z0. So, the reflected impedance is 1.24 x 250, or 310 ohms.

If the attenuation is 20 db, the mismatch drops to a ratio of 1.01, representing 252.5 ohms.

To avoid any trouble due to mismatch, which in r.f. can result in reflections and their consequent troubles as well as just losses (as in television attenuators), it is advisable to design a matching pad so that both impedances are correctly matched. This can be done quite easily with Chart III.

The chart gives the minimum attenuation necessary to produce correct matching, and the resistor values for an L-pad to produce such matching. This pad can of course be combined with further attenuators of the constant-resistance type, designed at either of the working impedances, using one of the charts in the earlier article.

To use Chart III, the two impedances to be matched, denoted by Z1 and Z2, are aligned; and the db attenuation can be found on the left-hand side of the center reference line. The figure from this scale is then transferred to the scale on the right-hand side of the same reference line, and aligned with Z1 and Z2. This locates the values of R1 and R2 as shown in the key on the chart.

Example 3

Take the case of 5,000 ohms being matched down to 5 ohms. To do this correctly, using Z1 5,000 ohms and Z2 5 ohms, the attenuation is 36 db. On the right-hand center scale, 36 db is so close to the infinity point as to be indistinguishable from it. In aligning 5,000 ohms with this point to find the value of R1 the straight-edge falls beyond the bottom of the right-hand line; similarly for the value of R2. This can be overcome by dividing by 10 on the Z scales; then converting the R value back again by multiplying by 10. Aligning Z1 at 500 with the infinity mark gives R1 as 500, which means its actual value will be 5,000 ohms; similarly R2 is 5 ohms.

Now consider the case where the impedances are closer-the second example given above - working between 250 and 600 ohms. Using Z1 as 600 and Z2 as 250, the attenuation is about 8.8 db. Transferring this to the right-hand reference scale on the center reference line, R1 is 440 ohms and R2 is 325 ohms.

Values as calculated from the chart are the exact values required for perfect matching. In practice, the closest preferred value resistor in the 5% or 10% range will probably give a good enough match for all practical purposes. It will almost certainly result in obtaining much closer matching than using an attenuator without giving proper attention to matching. In most cases 10% resistors will do.

 

 

Posted January 28, 2022

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