The Reason for Matching Impedances
- Category: Radio USA
- Published: Tuesday, 28 August 2018 19:19
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The Reason for Matching Impedances
By Samuel C. Miller (Radio for May, 1924)
Here is a simple explanation of why the input impedance of a transformer or phone should equal the output impedance of the vacuum tube. In view of the more.intelligènt design of a receiving set thus made possible, it is to be hoped that transformer manufacturers will include this information in their specifications.
The constants given on the data sheets accompanying a new tube are the amplification factor, the plate impedance and the mutual conductance. To the average radio fan the seemingly more important one is the amplification factor because it is believed that this constant alone determines the value of the tube as an amplifier. The impression seems to be that if the amplification factor is high then the actual amplification of a transformer-coupled tube will be high in proportion.
But this is not correct because the plate impedance plays just as important a part in the operation of a tube as does the amplification factor and both factors must be taken into account when considering the value of the tube as an amplifier. It is not difficult to manufacture a tube that will have an amplification constant of 50 to 100 instead of 6 to 8 as in the standard tubes; but in the high amplification factor tube the plate impedance will also be very high, somewhere around 200,000 to 500,000 ohms. This high plate impedance makes the tube practically useless when connected in circuit with an audio-frequency transformer. Let us connect a generator G, designed to be coupled to the standard tube which has a plate impedance of 20,000 ohms.
The maximum output that a tube is capable of supplying is obtained only when it is connected to a circuit of equal impedance. The reason for this can be more readily seen when the tube is considered as a generator feeding the transformer. Let us connect a generator G, Fig. 1, to various values of external resistance, R1.
Fig. 1. Generator with Variable Load
The generator has an internal resistance R of 20 ohms and generates a constant voltage, E, of 100 volts. The current that G will supply to R, depends on the total resistance in the circuit, which is the sum of R and R1. When an external resistance of 1 ohm is connected across the generator, the total resistance R+R1, is 20+1=21 ohms. With 100 volts generated, the current I flowing in the circuit is 100/21=4.76 amperes. The power in watts supplied to R1, is found by the formula W=I2xR or 4.762x1=22.6 watts. The total power generated by the generator generator is 4.762x21=474.6 of which 4.762x20=452 watts is lost in the resistance R of the generator.
Substituting for the resistance of 1 ohm in the above example different values for R1 a table is obtained which gives for every value of R1, the total power generated, the power supplied to R1 and the power consumed within the generator. It is to be noticed that, when the external resistance R1 is 20 ohms or equal to the resistance R of the generator, the power supplied to R1, is greatest. While the efficiency is not greatest at this value we are more concerned with output than efficiency in a tube circuit. This example dearly indicates the importance of matching resistances, or impedances, to obtain the maximum output.
A tube when connected to a circuit may be used either as a device to impress on its output load a maximum voltage, or as a device which supplies the circuit with the maximum amount of power it is capable of delivering. In the first case, it is used as a voltage amplifier and in the second case as a power amplifier. In Fig. 2, where tube I is used as a voltage amplifier, it is desired that it impress a maxiumum voltage on the transformer which steps up and impresses a higher voltage e1, on the grid of tube 2. In Fig. 3, the tube is connected to a pair of phones, the object being to supply the maximum amount of power for conversion into sound waves of maximum intensity.
Fig. 2. Voltage Amplifier Circuit
With a standard power transformer such as is used for house lighting, bell ringing, etc., all that is required is a step up or step down in voltage at the highest possible efficiency for 60 cycles. But when dealing with transformers for radio-telephony, where the frequency varies from 60 cycles to 5000 cycles for high quality transmission, the problem becomes more complex because it is desired to amplify all this band of frequencies with equal intensity.
Fig. 3. Power Amplifier Circuit
If any of these frequencies are amplified by the transformer with a difference in intensity in relation to any other frequency, distortion will occur unless the same relative difference in frequency intensity exists when the sounds are put into the microphone at the broadcasting station. This means that if two instruments such as the violin and bass viol are playing a duet, the transformer may amplify the low bass viol notes a great deal less than the high violin notes and thus produce an effect different from that actually being played.
To analyze the operation of a tube coupled to an audio-frequency transformer and its circuit, it is necessary to present an equivalent circuit. This is shown in Fig. 4 and represents the circuit of Fig. 2.
Fig. 4. Equivalent to Voltage Amplifier Circuit
E is the voltage generated by tube 1 and R1, is the internal plate impedance of the tube. L is the mutual inductance of the transformer and is equal to the primary inductance due to the coupling being pratically unity. R3 is the primary ohmic resistance of the primary and R4 the ohmic resistance of the secondary. L2 and L3 are the leakage inductance of the primary and secondary. R is the load corresponding to the input of tube 2 and is practically infinite resistance. C is the distributed capacity of the secondary of the transformer and, when converted back across R1, becomes higher in value, depending on the ratio of turns.
When dealing with a wide band of frequencies which are all to be amplified with equal intensity, as in the circuit described above, there are two main points to be taken into consideration. First, L (primary inductance) should be high so that its impedance will be higher than the internal plate impedance R1 of the tube in order to prevent the lower frequencies from being eliminated. This impedance should be about three times R1. And, second, the distributed capacity C should be kept low to prevent the higher frequencies from being bypassed.
Fig. 5. Curve of Transformer Performance
When the primary impedance is higher than R1, the greater part of the voltage E generated by the tube is impressed across L. This relation should be obtained even on the lower frequencies. If R1 is greater than the primary impedance, then most of the generated voltage E is across R1 and the voltage across L is lower than it is at the higher frequencies. This will give a smaller increase in amplification at the lower
frequencies and in some cases, where the primary impedance becomes practically negligible to R1, the lower frequencies are eliminated entirely. In a good transformer the primary impedance is made high by using iron of high permeability and copper wire of a fine gauge such as No. 40.
On the other hand, the distributed capacity C, which limits the higher frequencies by by-passing them, can only be reduced by proper design in the winding of the transformer. A curve showing the performance of a transformer is shown in Fig. 5, where across R (input of table 2). It is to be noticed that at zero frequency the voltage across R is zero. As the frequency is increased the voltage increases until a point is reached where the line becomes merely horizontal. It is at this saturation point that the primary impedance becomes larger than the internal plate impedance of the tube, with the result that the maximum voltage is across R and continues so at all the higher frequencies. This steady state remains until the effect of the distributed capacity becames apparent at the higher frequencies.
The result is that the voltage begins to drop with increase of frequency. The peculiuar hump marked X is due to the resonance circuit formed by the leakage inductance L2 and L3 and distributed capacity C. It is interesting to note that the broadness of this resonant hump depends on the ohmic resistance of the primary and secondary windings. The higher the resistance of these windings the less noticeable will be this resonance effect as shown by the dotted line in the cutve.
When a tube is connected to a loud speaker, the circuit is similar to the example of the generator coupled to a resistance of equal value for maximum output. To get the maximum vibration of the diaphragm, a maximum amount of power must be delivered, which means that the impedance of the loud speaker should be equal to the plate impedance of the tube. But as the loud speaker impedance varies with frequency, the power output of the tube would also vary, with a maximum at that point where the plate impedance equals the loud speaker impedance. However, this is not actually the case when the loud speaker is operating, because both the diaphragm resonance and the shape and size of horn are also considerations in its performance. In Fig. 6 the curve A is that of the power output of the tube into the loud speaker, curve B is the resonance effect of the diaphragm and curve C is the effect of the resonant characteristics of the horn, w hich mainly compensate for the lower frequencies lostt by the low impedance of the loudspeaker. The resultant curve is as in Fig. 7 where the amplitude is nearly of the same value for all frequencies except for the very low and very high frequencies, where the amplitude begins to drop off to a great extent.
Fig. 7. Resultant of Curves in Fig. 6
The reason for distortion in the great number of the loud speakers is due to the horn not being properly designed. The result is usually a tinny effect because of the loss of the lower frequencies which should have been compensated for and accentuated by a properly designed horn.