Radio Broadcast (May 1929-Apr 1930)

RADIO BROADCAST No. 22 Radio Broadcast's Home-Study Sheets RADIO-FREQUENCY AMPLIFIERS May 1929 A RADIO-FREQUENCY amplifier consists of a tube and some means for connecting that tube to an output circuit. In this "HomesStudy Sheet" we| shall learn how to design a [single stage of radio-frequency amplification. The circuit diagram of such a stage is shown in Fig. 1. Evidently there are several variable factors. We can use a low or high-mu tube; the number of turns on the secondary, the size of the tuning condenser, the number of primary turns, and the coupling to the secondary are all variable and at the control of the designer. Tuning Capacity The size of the tuning condenser and the secondary inductance are related definitely to the frequency band to be covered. The present broadcasting band covers a range of from 200 to 550 meters, or about three to one. If the wavelength varies as the square root of the tuning condenser capacity, what range in capacity must be available to tune from 200 to 550 meters? As shown in " Home Study Sheet" No. 19 the wavelength is proportional to the square root of the product of E and C. That is,J>00 meters = 1.884 j/LCi and 200 meters =1.884 I/LC2 where the two values of C (in mmfd.) are the capacities required to tune the fixed inductance (in [Ah) to 200 and 600 meters. Dividing one equation by the other we get 3 or 9 = S-' which L.2 states that the capacity for 600 meters must be 9 times the capacity for 200 meters. This is the first design problem. We must be able to use a condenser that has a capacity ratio of at least 9 to 1. If the maximum capacity is 0.00035 mfd., the minimum must not be greater than 0.000039. Because of (a) the input capacity of the tube, (b) the capacity of leads to the coil and to the tube, and (c) the distributed capacity of the coil to which the condenser is attached, all of which are in parallel with the condenser, the minimum capacity of the condenser must be less than this amount by at least 30 mmfd., or it must be as small as 9 or 10 mmfd. If it is larger, or if the distributed capacity of the coil is high, the lower wavelengths cannot be reached. A large-diameter coil with large wire closely wound will have a large distributed capacity. This is one reason why commercial manufacturers use small coils wound with fine wire carefully spaced. A short single-layer solenoid has a distributed capacity in mmfd. nearly equal to 7 per cent, of its circumference in centimeters. (Breit, Physical Review, Aug. 1921.) Secondary Inductance The greatest percentage of the amplification factor of a tube will be obtained when the load into which it works is high. The effective resistance of a coil-condenser combination at resonance (see "Home-Study Sheet" No. 12) is equal to „ L2(rf2 L Reff = ■ = r Cr Where L is the inductance of the coil w is 6.28 times the frequency r is the series high-frequency resistance of the coil. This shows that the effective resistance, which is the load presented to the tube, increases as the square of the inductance, and so if we can keep the resistance of the coil fairly low, a large inductance is better than a small one. The selectivity, S, of such a tuned circuit is proportional to Lw/r, and the width of the resonance curve, at a point where the response is 0.707 of the value at exact resonance, is related to this ratio, i.e., S = — = j j (see Fig. 2) which is another r 12 — 11 reason for using as large a coil as possible. Problem 1. Assume a fixed capacity of 50 mmfd. across which the condenser is to be placed. This includes the minimum capacity of the condenser. Calculate the inductance of the coils to be used to cover the broadcasting band if the maximum capacity of the condenser is 250, 350, and 500 mmfd. 100 90 80 70 60 h S50 %o 30 20 10 ft ffc «C — \ \ «-f2 r f r \ 40 30 20 10 0 10 20 30 40 K.C.0FF RESONANCE Fig. 2 If they have the same resistance (assume 20 ohms) at 1000 kc, calculate the effective resistance, or the load presented to the tube in each case. Calculate the width of the resonance curve and the selectivity factor, S. Coupling to Previous Tube How saall we couple the coil-condenser combination to the previous tube? One way is by means of a transformer. At first let us use an auto-transformer as in Fig. 3. Also, let us for the moment consider a tube with an amplification factor of 8 and a plate resistance of 6000 ohms (a 112-t.ype tube), and connect it across the entire coil. What is the amplification and effect on selectivity? The voltage amplification of such a tube and coupling device cannot be greater than the mu of the tube, and will be more nearly equal to it the larger the effective resistance of the tuned circuit becomes compared to 6000 ohms. As a matter of fact the voltage amplification will be equal to ft X Reff; in this case it will be equal to G = Reff+ Rp 8 X Reff Reff + 6000 so that, if the effective resistance of the tuned circuit is 50,000 ohms, the voltage amplification, G, will be about 7.3. What happens to the selectivity of the tuned circuit? We have placed directly across this circuit, which has an effective resistance of 50,000 ohms, a 6000-ohm resistance. The resistance of these parallel resistances is now equal to (50,000 X 6000) -r (50,000 + 6000) ohms or 5860 ohms. Now what resistance added to the series resistance of the tuned circuit would reduce its effective resistance to this low value? Let us put down two equations, assuming the series resistance equal to 20 ohms L2a>2 ~20~ 50,000 • (1) L2o>2 = 5860 (2) 20 + R dividing (1) by (2) we get 20 5860 20 + H. ~ 50,000 Fig. 1 from which R (added resistance) = 150 ohms so the total series resistance is now 170 ohms and the selectivity is where it was equal to — for the coil-condenser alone. From this we learn that the selectivity has been reduced 170 -r 20 or a factor of 8.5. Many investigators have experimented with this problem and the solution is to use a two-winding transformer with a primary of such an inductance and so coupled to the secondary that the effective resistance of the tuned circuit is stepped down so that it "looks," to the tube, like a resistance equal to the plate resistance. What is the result? We can see that, if the tuned circuit is shunted, not by 6000 ohms but by its equivalent, 50,000 ohms, the selectivity has been decreased only by a factor of two, and we get. some voltage step-up because of the transformer. Whereas it was possible to get an amplification of only 7.3 without the transformer, it is now possible to get more than the mu of the tube. Where the selectivity was cut down by a factor of 8.5 it is now reduced by a factor of only 2. What should the turns ratio be? From transformer design theory the proper turns ratio is N = "V ~ff . * Rp 50,000 6000 voltage gain, so far as the tube is concerned is, as before, mu times the ratio between the load resistance (as stepped down by the transformer) and the sum of the load and plate resistances. This voltage gain is stepped up by the turns ratio of the transformer and becomes = 2.9 for the given conditions. The V X Reft/N2 8 X 6000 Reff/N2 -f Rp 6000 + 6000 = 11.6 u.La> 4 X 2.9 2 j/rXv Rp Problem 2. Consider a coil of 180 microhenries inductance and a high-frequency resistance of 15 ohms at 1500 kc, 10 ohms at 1000 kc, and 7 ohms at 505 kc, and a 112-type tube. Calculate the proper turns ratio for maximum voltage gain at 1500 kc, 1000 kc, and 550 kc. Assume that the proper ratio for maximum amplification at 550 kc. is employed. Calculate the voltage gain at this frequency, and using the same turns ratio, calculate the voltage gain at other frequencies in the broadcast band. Then, remembering that the resistance across the secondary is the tube plate resistance times the turns ratio squared, calculate the increase in series resistance of the tuned circuit at this frequency and the selectivity factor, Lfi>/r. Calculate the width of the frequency band at the 0.707 point at 550, 1000, and 1500 kc. Plot all of this data against frequency in kc. Do you see why most tuned r.f. receivers do not have equal amplification at all frequencies? Why the selectivity is poor at high frequencies, the gain low at low frequencies? Problem 3. Repeat using at 201a tube, a 240 tube, and a 222 tube. The constants are for the 201a, [A = 8, Rp = 12000 ohms; for the 240, ft = 30, Rp = 150,000 ohms, and for the 222 screen-grid tube, ■J. = 200, Rp = 200000 ohms. Problem 4. A 600-turn honeycomb coil has an inductance of 20 millihenries. A variable condenser is placed across it, and the circuit is coupled to a 40 kc. circuit. The voltage across the coil is measured as the tuning condenser is varied. Then a pair of steel wire cutters is placed near the coil and another resonance curve is plotted. The results are given below. In such a circuit the high-frequency resistance can be obtained when the capacity is the variable factor, just as when the frequency is variable. It is only necessary to plot the resonance curve showing response against capacity and to ascertain the two values of capacity which make the response 0.707 of its value at resonance. Then, if Cr and C2 are these two capacities and Cr is the capacity at resonance, 2xfL _ 2 Cr L&> _ fr R ~~ C2 — Cx ~ r ~~ f2 — fi Plot the results given in the table, and find out the resistance of the coil in the two cases; calculate the width of frequency band passed at the 0.707 response point; calculate the value of the circuit's effective resistance, L2o>2/r; calculate the proper turns ratio if this coil is to be tapped and used between a 227-type tube and the grid-filament input of a following tube; calculate the voltage gain at 40 kc. Would such an amplifier be a highquality unit? f=40 kc. C mmfd. E E when pliers are near coil 780 0.40 760 0.68 0.45 750 1.43 0.82 740 2.30 1.26 735 1.53 1.22 720 0.60 0.60 Fig. 3 • may, 1929 page 30 •