Renat Terlecki, "DC Powered Power Amplifier RA360ST"
In recent times, radio engineering has gained a completely new field of activity in the form of high-power loudspeaker installations. Apart from the sound films of high-power loudspeakers, we also often meet in a whole range of public institutions such as railway stations, guest houses, hotels, sports grounds, concert halls, and even churches and parliaments.
A few years of practice has crystallized the concept of a modern power amplifier and today we require such a device:
- Complete electrification.
- Uniformity of amplification and reproduction of a wide frequency range.
- Versatile use, and therefore for both adapter, microphone, photocell, radios, etc.
- Possibly high efficiency.
- Transparent design.
- Simple operation and minimal maintenance, and finally...
- Carefully matched equipment.
All this works, however, when we have a source of alternating current, usually in the form of a lighting network. Alternating current, as we know, can be transformed into any high or low voltage; so we can construct economical heating circuits without losses on reducing resistors, and with any high anode voltage we use more efficient systems, eg "C-Class" or directly coupled.
In many places, however, we have power plants, supplying for lighting or industrial DC, and in this case building a high power amplifier, we face special difficulties. In fact, for the reasons given above, high power amplifiers are usually calculated on the pr. variable, so you want to power them from the pr network. fixed, we must use converters, the cost of which in small installations may exceed the value of the amplifier. Sometimes it will pay off to give up the economics of the system and to settle for a high power amp with a relatively low anode voltage of 150 ÷ 200 Volts.
The below described amplifier d. M. Is designed for power supply from the pr. st. voltage 250 ÷ 150 Volts. It is understandable that by reducing the value of the main reduction resistance "R" we can connect the amplifier to the network with lower voltage, but then the power of the output stage would drop too much.
Before we get to the right amplifier, let's start with the study of the main of the system - the heat supply circuit. This circuit, as we can see from the simplified diagram in Figure 1, consists of an adjustable resistance R5, at the ends of which, according to Ohm's law, flowing current (1.4 Amp.) Generates the potential difference "Eg", used here as a negative grid pre-induction of output tubes. Then, the filament current branches off after 0.65 Amp. on two parallel cathodes of output tubes. The rest of the current according to Kirchhoff's law flows through regulated by-pass resistance R4, which regulates the glow voltage of both tubes. Further, the accumulated glow current flows through the indirectly heated entrance tube (ca 1 Amp.), Which is regulated by the bypass resistance R3. Next, we have an iron-hydrogen tube "1331" automatically adjusting the incandescence intensity to 1,3A - and finally the main reduction resistance R.
It should be noted that the sum of resistance of this circuit between points A and B must be a constant value for a given network voltage. Not too big fluctuations in the network voltage and resistance changes R5 and R will automatically align the resistance tube, but for example. Larger changes in R5 resistance must be compensated by resistor R. As for the essential part of the amplifier, as we can see from the complete diagram in Figure 2, it is very simple .
At the input we have a switch that allows a quick transfer of the amplifier from one pair of input sockets to the other. From the switch, the audible frequency current runs through the potentiometer Pot, which regulates the volume, from where it goes to the primary winding of the transformer Tr1, whose secondary winding is the grid circuit of the input tube of the amplifier. The potentiometer can be successfully lowered if the applied source of audible currents (adapter-microphone) has its own volume regulation. The resistance R2 inserted into the cathode circuit of the first tube gives it the necessary negative biasing. The input tube's anode through the primary winding Tr2 is connected to the Dł coil, which together with the C2 capacitor is used to smooth the pulses of the mains current. The secondary winding Tr2 has three ends. A negative net voltage of the output tubes is supplied inside this winding, while the remaining terminals, on which we receive variable voltages of currents amplified with a phase shift of 1800, are connected respectively to the grid of these tubes. From the above we can see that the output tubes work here in a symmetrical push-pull system, i.e. 'class A'.
The anode currents of these tubes, or actually the variable components of these currents, add up in the output transformer "Tr. W.", The primary winding of which is also divided into two sections, connected to the anodes of both tubes in opposite directions. Since the phases of these currents are also opposite, so the magnetic fluxes produced by them in the core have the directions consistent, and thus add. On the secondary side, the output transformer has 2 windings: for a low and high resistance loudspeaker with branches, which again allows the transformer to be adapted to the loudspeaker. The system of output tubes described above is known for its exceptional uniformity of amplification, and most importantly, it does not require high anode voltages. In the above considerations, we omitted R1-C1 and R6-C3, which, as shown in the diagram, are used to filter individual grid voltages; it should be noted that the anode voltage of the end tubes is not filtered here at all, because its pulsations cancel each other on the output transformer, and even omitting the resistance R6 and direct connection of the center of the secondary winding Tr2 to the minus of the network simply does not feel.
LIST OF PARTS
Pot = 50000ohm; Transformer Tr1=interstage 1:3 - 1:6, high class;Tr2 = Polton, Pusch-Pull, 1:4; Tr. w. = output Polton Pusch-Pull, type WDM3; Dł = choke Polton D3530; B = fuse 3,5V-2A. Capacitors: C1 = C3 = 2µF; C2 = 4µF 1000V; C4 = C5 = 2=4µF. Resistors: R1 = 0,1meg; R2 = 1000 ohm wired; R3 = 20 ohm variable; R4 = 30 ohm variable; R6 = 0,01 meg; R5 = 20 ohm according to description; R = 150 ohm. W1 = power swiitch...
In the model amplifier with excellent results, I used the following tubess: "AR4101" in the input stage and in the output stage 2 x "P460" (Tungsram) as the voltage compensator of the network, or actually the heater current regulator tube "1331" (Philips).
We start with the chassis (figures 3, 4 and 5). From a piece of 400x300x10mm plywood cut the assembly board with dimensions 400x170 with the projection in the left rear corner 130x70 and put them on two frames made of thick iron sheet. The remaining part of the plywood is laid on the one hand with asbestos and placed perpendicular to the assembly board. In this way we get a complicated chassis, because in the front part we have "two floors", while in the back there is a place "on the first floor" for a resistance lamp and, well cooled and thermally insulated from the proper amplifier, space for resistance R and R5. These resistances fix the spaces in the manner given in the figures using threaded rods and metal strips.
The resistances R and R5 mentioned above will have to be done by ourselves. From purchased for 40gr. roof tiles cut out 7 bars with dimensions 340x50mm and drilled in them two holes for rods for fixing on rods, and two for the ends of the resistance wire. Then, on one such bar, wind up a 0.5mm nickel-based wire over the entire length of 10mm - it will be the R5 resistance. On the remaining six, we wind up 14 meters of the same wire. After joining them in series, you will get resistance. When winding, the wire should be strongly tensioned so that it does not slip during heating due to heating. Then, we make two buckles from the copper plate with which we will regulate the resistances of R and R5.
On the bakelite face plate, mount the switch, potentiometer (if it is used) from the left, milliammeter in the middle, then fuse and switches W1 and W2.
We put all transformers and tube bases on the assembly board. The stand for the resistance tube is mounted on a special projection of this board in the left corner. The rest of the part will fit underneath.
We will assemble the assembly with 1.5mm2 wire in rubber insulation, whereas the cables covered with shielded wire in figure 2. Armors of cables and solid capacitor boxes are connected to each other and possibly grounded. Grounding the transformer cores with the earthed plus network gave a negative result.
After mounting the amplifier and checking the connections according to the diagram in Fig.2, connect it to the network. However, it is not recommended to use a cord with a plug, because it will be easy to mistake the poles, although there is a risk of damage to the apparatus, but it is better to recognize the poles of the network with a polar voltmeter and lead it directly to the W1 switch. Because it is a bipolar switch, so after switching off the current the amplifier is completely disconnected from the mains and there is no fear of electric shock, which again is particularly important in the first adjustment.
Before switching on the electricity, we insert an ammeter of such a scale in addition to the negative wire of the mains, which would enable us to read the intensity of 1.4 Amp; then set resistance feathers R3 and R4 on half of the windings, set the buckle at R5 resistance so that about 3/4 of its resistance wire is connected, finally set the buckle on the resistance R so that the resistance is on. Now set the tubes and making sure that W2 is off; turn on the network by turning the W1 switch. The ammeter tip will move to a certain position indicating about 1 Amp. Now we look at the P460 tubes or by a cathode which one of them does not shine bright red heat. If this happens, immediately turn off W1 and find the break in the glow circuit of the second tube, because the cathode of these tubes do not shine during operation, but if one starts to shine then it would be proof that the whole glow current (and in this case about 1 Amp.) Only flows through this one shining tube. After removing any possible damage, we can proceed to the proper adjustment, so we turn on the current (W1).
Suppose we read the Amp value from 1.1 Amp. Turn off the network and move the buckle at the resistance R, so as to slightly reduce its value and again turn on W1 and read the value of current flowing from the ammeter. In this way, we proceed until 1.3 Amp is obtained. This will be the right amount of current flowing in the glow circuit (Fig.1).
Now, by moving the resistive pen R3, set it so that the voltmeter attached to the lamp legs of the incandescent tube, indirectly heated, indicated 4 Volts. In the same way, by regulating the resistance R4, we set the tubes 'P460' to 3.5 Volts.
After completing this operation, we can turn on the anode voltage of the tubes with the W2 switch, while observing the "P460" tubes. If now one of these tubes shines with bluish or violet light in the inter-electrode space, one should immediately turn off W2, then W1 and send the glowing tube to the supplier with the complaint, stating in the claim card that gas has occurred in the tube. This is a serious case, with a strong gas and a too large fuse B, a milliammeter and the right half of the output transformer winding can burn out, and finally the tube itself will be damaged due to the rapid bombardment of the cathode.
However, this phenomenon should not be equated with another harmless symptom: fluorescence of glass. Sometimes in the tubes of more power during work the balloon's glass overlaps the inner layer of vibrating, gentle celadon light, vividly reminiscent of the X-ray fluorescence of the screen.
So if gas does not occur in the tubes, in turn we read the intensity of the anode current from the milliammeter "mA". With a 220V network it should be about 100mA, or 50mA per tube. If the milliammeter shows more, the resistance value R5 should be increased by moving the buckle on its winding and vice versa, if it indicates less - then it should be reduced. After this the ammeter will show us 1.4 Amps. as the sum of the burn current = 1.3A and the anode current of the output tubes = 0.1A.
The proper value of the anode current, indicated by the milliammeter, can easily be calculated from the tube's permissible power. The permissible power of the "P460" tube is 10 Watts, so the two have 20W. Suppose that the voltage measured on the anodes of these tubes (not net!) measured with a good voltmeter is 200 Volts. We will receive 20W: 200V = 0.1A = 100mA. With other network voltages, the anode voltage will naturally change and the anode current will change accordingly.
At the end, connect the voltmeter to the legs of the "1331" tube and adjust the resistance R so that the read voltage is about 10V; adjust the resistance of the tubes with W2 to 1 Volt on the R3 and R4 resistors and switch off the ammeter that is no longer needed.
Connect the adapter cables to one pair of input sockets or directly to a pair of contacts in the switch. To the second - we can attach a second adapter or some other device, e.g. a pre-amplifier from the photocell (in the cinema) or the anode circuit of the receiver's detector tube and the like The dynamic or magnetic loudspeaker or both are connected simultaneously to the right output transformer sockets. If your transformer does not have a winding for a magnetic loudspeaker, you can attach it via two block capacitors of 2 ÷ 4μF to the terminals of the primary winding of this transformer according to Fig.2 (C4-C5); then the transformer acts as low frequency choke.
Then turn on the network by turning the W1 switch and wait about 1 minute until the cathode of the indirect heating tube warms up, because the P460 tubes will warm up earlier; after this time, turn on the anode voltage with the W2 switch - the milliammeter arrow will jump to the higher set value and the amplifier will be ready for use.
Actually, the W2 switch could be successfully omitted, because it complicates the service, but if you do not want to simplify it specially, and attach great importance to the durability of the tubes, its use is commendable.
After setting the plate and releasing the adapter, set the force regulator to max. and we are observing the milliammeter arrow: if it will go back to zero with stronger sounds, it means that we have given too little mesh retardation and vice versa if we get the arrow deflection to max. - this pre-cocking is too big. We correct this error by changing the R5 resistance. With a good adjustment, the arrow will stay in place while the amplifier is in operation.
In the above system, the directly heated output tubes are lit by direct current, which means that the negative ends of the cathodes of these tubes will be much more loaded than the positive ones, so it is recommended to use them evenly at intervals of several days (eg twice a week) the direction of the glow current flow through the tubes. It will be best to take this into account when mounting the amplifier and bring the glow current to the tube bases so that the adjustment of both tubes gives this change, so that one tube passes the current, e.g. from the left foot to the right and the second from right to left. The above was included in the assembly diagram.
The above amplifier after nearly one year of work, in very difficult local conditions, did not show any damage.
Adapters for the described amplifier should be of the high-resistance type, because with the use of low-resistance types the turn ratio of the input transformer should be changed to 1: 10 ÷ 1: 20. We take the recordings with an electric recording, because they give a natural sound to the voice and the so-called. the roundness of the throne, which can not be said of records recorded in the old way. It should be emphasized that for some discs recorded in a special way, such as "Pathé" or "Edisson", where the needle performs vibrations not horizontal but vertical, adapters on the market are not suitable at all.