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June 1932 Radio-Craft [Table of Contents]People old and young enjoy waxing nostalgic about and learning some of the history of early electronics. Radio-Craft was published from 1929 through 1953. All copyrights are hereby acknowledged. See all articles from Radio-Craft. |
It seems most of the articles we see on the subject of attenuator pads are based on signal reduction in terms of decibels for units of power. Although it is a simple matter to convert power decibels to voltage decibels, it would be more convenient if you are working with voltage to have formulas and tables of values based on voltage ratios. This article does just that.
As a reminder, the decibel representation of a ratio is always 10 * log_{10} (x). If you have a voltage ratio of V_{1}/V_{2 }= 0.5, then 10 * log_{10} (0.5) = -3.01 dB. If you have a power ratio of P_{1}/P_{2} = 0.5, then 10 * log_{10} (0.5) = -3.01 dB. Does that mean that -3.01 dB of voltage attenuation is the same as 3.01 dB of power attenuation? Confusingly, no.
Let's say for the sake of simplicity that you have 1 V across a 1 Ω resistor, yielding a power dissipation of P = V^{2}/R = 1^{2}/1 = 1 W. Now, reduce the voltage to 0.5 V and the power dissipation is 0.5^{2}/1 = 0.25 W. The ratio of the voltage change is 1:2 but the ratio of power change is1:4. In decibels of voltage that represents 10 * log_{10} (0.5) = -3.01 dB, but in decibels of power it represents 10 * log_{10} (0.25) = -6.02 dB. A multiplication factor of 2 therefore exists between voltage ratio and power ratio when applied to the same system.
Now, let's work it in the opposite direction. For the sake of simplicity again, you have 1 W being dissipated by a 1 Ω resistor, yielding a voltage across it of V = sqrt(P*R) = sqrt(1*1) = 1 V. Now, reduce the power to 0.5 W and the voltage across the resistor is sqrt(0.5*1) = 0.7071 V. The ratio of the power change is 1:2 but the ratio of voltage change is1:0.7071. In decibels of power that represents 10 * log_{10} (0.5) = -3.010 dB, but in decibels of voltage it represents 10 * log_{10} (0.7071) = -1.505 dB. Once again as you would expect, a multiplication factor of 2 exists between voltage ratio and power ratio when applied to the same system.
Part I appeared in the May 1932 issue of Radio-Craft.
(Part II)
In this second of a series of articles, the author will discuss additional design considerations of attenuators, illustrating them with numerical examples.
By Hy Levy, B.S.
Last month, some of the more pertinent problems of transmission lines (with special reference to "pads" or "attenuators") were outlined. In this issue, we will continue with a discussion of actual design considerations.
The two extreme conditions that a transmission line could be subjected to are an open-circuit and a short-circuit condition; in which cases, the greatest reflection of energy will take place, and abnormal voltages and currents will be set up from small impulses originally sent into the line. This is the same as saying that infinite impedance terminates the line on open circuit, and zero impedance terminates the line on short circuit. These two conditions are very rarely encountered in practice, unless the line should accidently become open or shorted at the far end.
Assume a transmission line in which the line conductors have negligible attenuation (zero losses) and in which the generator or source impedance is of known value. Then, if a terminating impedance is inserted at the far end of the line, and starting from the open-circuit condition of infinite impedance, this terminating impedance is gradually made smaller and smaller, the reflected energy will also become smaller, and as the terminating impedance approaches the value of the source impedance, the reflected energy rapidly decreases. At the point where the value of the terminating impedance exactly equals the source impedance, the reflected energy will be zero (no reflection losses), and maximum energy will be obtained at the load.
Likewise, starting with the short-circuit condition of zero terminating impedance, and increasing the load impedance, the reflected energy becomes smaller, so that when the point is reached where both the terminating and the source impedances are exactly equal, the reflected energy will be zero (no reflection losses), and maximum energy will again be fed to the load.
When the terminating or load impedance exactly equals the source impedance, the line has the same characteristics as a line of infinite length. If a line of infinite length could be constructed, any energy sent into this line would never be reflected back to the source. By inserting at the far end of a transmission line a load or terminating impedance equal to the source or generator impedance, the characteristics of a line of infinite length are immediately obtained (assuming zero losses in the circuit) and consequently no reflection losses are set up in such a circuit.
This entire discussion on reflection losses can be summed up by stating that the cause of these losses is improper impedance matching, and it is for this reason that the subject of proper impedance matching is so strongly stressed and observed in voice transmission circuits. In order to avoid reflection losses, care must be taken to see that, at any point in a transmission line where an impedance is to be inserted, this terminating or load impedance is made equal to the impedance working into it. Reflection losses, as previously stated, give rise to abnormal voltages and currents; harmonics are set up, and the wave form of the impressed original signals are distorted, with the result that straight-line frequency-response characteristics are not obtained from such a circuit.
If the load impedance Z1 (see Fig. 3) exactly equals the source impedance Zs, then the source impedance Zs is said to be working into its "image impedance," for in such a case, there is no reflection of energy from the load impedance back into the line.
If an attenuator is now inserted into the transmission line between the source and the load, it must be so designed that the impedance of the attenuator must in no way upset the impedance match between the source and the load. If then, when the attenuator is working in the transmission line, the impedance "looking" into the source (from the line) exactly equals the impedance "looking" into the load (from the line), then the pad is said to be working between its "image impedances," and no energy will be reflected back at any point along the line, as the impedances looking in either direction along the line are exactly equal. In such a case, no reflection losses will occur at any junction points throughout the line, and practically ideal transmission characteristics are obtained, except for such losses as arise from the effects due to the length of line over which the energy is fed. (The losses due to length of line will not be considered in these papers. It is assumed that the voltage at the source is the voltage appearing across the 1-2 terminals, (See Fig. 7) of the pad, and the voltage at the 3-4 terminals is the voltage across the load).
Design Considerations
In order to maintain the same impedances in either direction, the pad must be designed in such a manner, that when looking into the 1-2 terminals, the combined (resultant) impedance of the attenuator and the load must exactly equal the source impedance. Also, when looking into the 3-4 terminals, the combined (resultant) impedance of the attenuator and the source must exactly equal the load impedance. These last two statements described the "characteristic impedance" of the circuit when looking into both the source and load ends of the line.
In order therefore, to fulfill the circuit requirements for image impedance, and characteristic impedance, the two must be equal, and upon completion of the design of the pads, this will readily be seen.
Having discussed the various problems that must be considered from an impedance matching standpoint, our next problem is to analyze the manner in which pads function in the transmission circuit.
It is the purpose of the writer to lead up to the actual design of the pads by the method of "approach." For, having threshed out the various problems associated with the circuit, the design of the pads is then reduced to the simple process of substitution into the working formulae for the pad under consideration.
Continuing now with an analysis on pads, it was stated in the first part of this paper, (May, 1932 issue) that pads are used to control the level of the energy entering the load. We are principally interested then, in determining the relative amount of current and voltage appearing at the load. In Fig. 8 is shown an "H"-type network working between a source impedance Zo and a load impedance Zo. Now, referring to Fig. 9, which is the equivalent circuit of Fig. 8, it can be seen that if we wish to keep the current I2 and the voltage V2 at the load at certain fixed values, with a constant voltage V1 impressed across the input terminals of the pad, the values of the resistances used in the pad would have certain predetermined values in order to give the correct loss between the input and output terminals. The values of these resistances are determined by the impedances between which the pad is working and the ratio of the current or voltage on the input side, to the current or voltage on the output side of the pad.
Let us return now to our own problem under consideration, that of designing a pad to give a 20·decibel loss. We have determined that if the voltage is reduced to 1/10 of its original value, or in the order of 10 to 1, the voltage ratio is 10, and expressed logarithmically this loss is 20 decibels.
Suppose we wish to reduce the voltage only a small amount, or in the ratio of 1.12 to 1. We would then have only a 1-decibel loss in the circuit, which is comparatively small. Now suppose we wish to set up a large loss, in the order of 100 to 1. We then would have a 40-decibel loss. Taking these three cases of 1-, 20-, and 40·decibel losses, and impressing various values of voltage V1 at the input terminals, let us see how the pads that we would have to design would vary, in order to keep a constant voltage V2 at the output terminals of the pad. In Table 3 are given the values of resistances as used in the series and shunt arms of "H"-type pads, when working between 200-, 500-, and 600 ohm impedances.
Now, using the 20-decibel loss as a reference level and referring to this table, it is seen that to obtain a 20·decibel loss the series arm, Z1 is 82 ohms, and the shunt arm Z2 is 40.4- ohms, when working between two 200-ohm impedances. With these constants, the output voltage would be 1/10 the value of the input voltage. Or, if the input voltage is 1.5 volts, the output voltage would be 0.15-volts, and the voltage would have been dropped 1.35 volts.
Now assume that the input voltage is 0.168 volts and we still wish to maintain the output voltage V2 at 0.15-volts. The voltage ratio, is 0.168/0.15 or 1.12; or, a 1-decibel loss is obtained in the circuit. From Table 3, for a 1-decibel loss, the series arm Z1 is only 5.65 ohms, whereas, the shunt arm Z2 goes up to 1760 ohms.
From the simple laws of current flow, this can be understood; for to obtain only a 1-decibel loss, we do not wish to introduce a large amount of series resistance, as this would impede the current flow, which is what we wish to avoid. The voltage drop in this case is only 0.168 minus 0.15 or 0.018-volts. Therefore, the series arm Z1 must be very small, and is but 5.65 ohms. Also, the shunt arm Z2 must be very large, so that very little current will be bypassed or shunted from the load, and as can be seen, its value is large, being 1760 ohms.
Now comparing the 20-decibel loss to the 1-decibel loss, in order to obtain the same output voltage with different input voltages, the series arm Z1 for a 1-decibel loss is only 5.65 ohms, and for a 20-decibel loss is 82 ohms. The shunt arm Z2 for a 1-decibel loss is 1760 ohms, and for a 20-decibel loss is only 40.4 ohms. Therefore, to obtain a fairly large loss, (such as 20 decibels), the series arm is quite large and the shunt arm comparatively small (compared to the load impedance.) In this case, the current will be impeded by the large series arm, thereby dropping the voltage considerably, and at the same time shunting it from the load by the comparatively small shunt arm, thus creating the fairly large loss desired.
For a small loss such as a 1 decibel, the series arm is very small and the shunt arm very large as compared to the load impedance. In this case, very little voltage drop is encountered in the series arm, and since the shunt arm is very large, very little current will be shunted from the load. Therefore, practically all of the original current will enter the load, thereby obtaining the small loss desired.
Now if the input voltage is 15 volts and we want to maintain the output voltage V2 15 at 0.15-volts, the voltage ratio is 15/0.15 or 100, and the decibel loss is 40. This means we need to set up a very large loss, for the voltage is to be dropped from 15 volts to .15-volts, a decrease of 14.85 volts. To do this, we should need a very large series resistance and a very small shunt resistance, so that practically all of the current will be prevented from entering the load by (1) the large series arm, and (2) the very small shunt arm. Referring to Table 3, it can be seen that to cause a 40-decibel loss, the series arm Z1 is 99 ohms and the shunt arm Z2 is but 2 ohms, which exactly follows the reasoning as given above.
Comparing this 40 decibel loss to the 20 decibel loss, the series arms for the 20- and 40-decibel losses are 82 ohms and 99 ohms respectively, and the shunt arms 40.4 and 2 ohms respectively. The values for the series arms are such that a large series voltage drop is encountered in them, being greater in the 40 decibel case, as would be expected. The shunt arms are quite different.
In the 20-decibel loss, the shunt arm is practically 1/5 of the value of the load impedance, so that only a comparatively large bypass effect of current from the load is obtained; whereas, 1 in the 40-decibel loss, the shunt arm is only 1/50 of the value of the load impedance, thereby shunting practically all of the current from the load. For a very large loss then, the series arms are very large and the shunt arms very small in comparison to the load impedance. By inspecting Table 3, it will be seen that the various values of the series and shunt arms necessary to give certain losses follow the simple laws of current flow; in that large-decibel losses require large series arms and small shunt arms in comparison to the value of the load impedance, and that small decibel losses require small series arms and large shunt arms in comparison to the load impedance.
Posted September 15, 2015