In the last article he has been briefly mentioned the use of the diode for rectification of alternating current; Now we take the topic to examine in detail the operation of the half-wave rectifier circuit.
Let us refer again to the circuit considered in the previous lesson, whose schema is again shown in Fig. 1.
Recall that during the half cycle in which the positive pole of the generator connected to the anode of the diode (fig. 1-a), in the circuit passes a current through the diode and the resistor in series with it in the conventional sense indicated by the arrows in Fig. 1-a.
Because the diode and the resistor are connected in series, the alternating voltage V supplied by the generator is divided between these two elements: across the diode establishes the anode voltage Va required to do so by the cross current, while across the resistor you have the tension that has been indicated by Vu (useful voltage) as for now it is assumed that the resistor constitutes the load circuit, ie both the element in which the current is used.
To cross the load by this current usually requires a voltage Vu of several hundred volts, while to make cross the diode by the same current must be a voltage Va of just a few tens of volts, as it is seen in the preceding article; the alternating voltage V is therefore applied largely across the resistor.
Since the voltage Vu is slightly lower than the voltage V, it can be assumed that the latter voltage is applied entirely across the resistor, neglecting the voltage Va, much less, that in reality it has across the diode; henceforth it will be considered, therefore, that the entire voltage V is applied across the resistor, during the half-cycle in which the circuit is traversed by the current.
instead, during the half-period in which the negative pole of the generator connected to the anode of the diode (fig. 1-b), this electrode can no longer attract the electrons emitted from the cathode and therefore has no current flow because the circuit is interrupted between the two electrodes of the tube.
There being no current in the circuit, across the resistor does not occur any voltage drop and consequently between the two electrodes of the diode there is the same voltage V supplied by the generator, as it is shown in fig. 1-b.
To better understand the operation of the half-wave rectifier, should see how they vary in time the above test voltages and current.
Since the voltage V supplied by the generator is alternating with sinusoidal, we can certainly represent it as has been done, for two complete periods, in Figure. 2-a, assuming that the voltage has the maximum value of 300 V shown in the same figure.
This current can be considered as an alternating current of the negative half-waves missing, because the diode does not allow its circulation in correspondence with the negative half-waves of the voltage V.
Assuming that the load resistor has a resistance of 5 kΩ and recalling that at its ends is considered the entire applied voltage V of the maximum value of 300 V, it appears that the current should have a maximum value of 300 : 5 = 60 mA, as it is shown in fig. 2-b.
Even the rectified voltage Vu, present across the load, It may thus be regarded as an alternating voltage negative half-waves of the missing: the rectifier is precisely said A half-wave, because at each period of the alternating voltage it allows to obtain a single half-wave, the positive, eliminating the negative.
Keep in mind, at this point, that, practically, a half-wave rectifier is a little’ different from that of fig. 1.
First, the AC voltage is normally obtained from the electricity distribution network; because usually the value of the mains voltage is different from that required, it feeds the circuit via a transformer, from whose secondary is obtained the desired alternating voltage and to be sure there will also be a circuit of the same separation from the network.
It should be noted, Furthermore, that very often all the elements constituting the circuit are fixed to a single metal support, generally in iron or aluminum, which it is said frame.
Since the chassis is made of metal and hence can conduct electrical current, It is used in replacement of one of the circuit conductors which serve to connect together the various elements. Keep the discovery and accessible mains voltage would not be safe in any case.
Taking this into account, the rectifier circuit can be represented as shown in the diagrams of fig. 3.
All the currents that circulate in the various circuits of an apparatus returning to the transformer through the frame, it may therefore be regarded as a solid conductor, threadlike anziché; It says that the frame currently constitutes the mass of the apparatus. The grounding of the frame is performed for safety reasons, in cases where it is believed that the same frame can be touched by some person. It is thereby avoided that whoever touches the frame, having your feet in contact with the ground, may lie with his body between two points at different potential, and then receive an electric shock.
2 – FULL WAVE RECTIFIER
To understand the operation of a full-wave rectifier, it is appropriate first to consider the transformer with which the circuit itself is fed, it has a secondary fitted with a central outlet.
If you connect the center tap to ground, You can refer to this the potentials of the two ends of the secondary and then consider the tensions that exist between these extremes and the same mass.
As seen in Fig. 4 the full-wave rectifier circuit may thus be regarded formed by the union of two half-wave rectifier circuits. The two rectifier circuits together so they do not disturb each other because they never work simultaneously.
It may be noted that the two diodes have its cathode connected to the same point of the circuit: from this it is understood that, instead of using two completely separate diodes, you can use a single tube which encloses the same bulb two anodes and a single cathode. These diodes are said A double anode, or even biplacca precisely because they are fitted with two anodes or plaques.
In fig. 5-to you you can see that the two anodes are arranged around the cathode cylinder one above the other and are each connected to a right leg; in the wiring diagrams the double diode is represented by the graph shown in Fig sign. 5-b.
Usually a diagram of a rectifier circuit provided with a double diode drawing as seen in Fig. 6; regarding links, this scheme does not differ from those shown in fig. 4.
Indeed, in both cases the anodes are connected to the secondary extremes, while the cathode is connected to the resistor connected to ground.
3. – LEVELING FILTERS
The circuits that allow to make the intensity of the current in the constant load resistor, thus transforming the current into direct current button, They are called leveling filters; it is more accurate to say that these filters provides continuous tension, applied to a resistor, It will circulate in it the direct current.
First of all, it is worth considering what happens when connecting a capacitor in parallel to the resistor placed in series with a rectifier diode half-wave, as seen in the diagrams of fig. 7.
When the diode is traversed by current (fig. 7-a), a part of this current charges capacitor, in such a way that its armature connected to the cathode of the diode becomes positive relative to the other plate connected to ground; the remaining part of the current flowing through the diode instead normally runs through the resistor. When the diode is not crossed by the current (fig. 7-b), the capacitor discharges through the resistor and constituting the discharge current that circulates in the direction indicated by the arrows in fig. 7-b.
In this way the resistor path is always in the same direction by a current that comes from the diode now now from the condenser. To see what this current trend has, It must be considered as it varies the voltage across the resistor and therefore also across the capacitor connected in parallel to it.
For this purpose, let us refer to Figure. 8, in which they are shown diagrams of the AC voltage V supplied by the secondary of the transformer (fig. 8-a) and the voltage Vu (fig. 8-b) already indicated in fig. 2-c, in case between the cathode and the ground there is a single resistor without the capacitor in parallel: positive half waves constituents voltage Vu, now drawn with a dashed line, allow to see how their performance is changed by the capacitor in parallel to the resistor.
Dinstant t1 onwards the alternating voltage (fig. 8-a) decreases and then, if there were the condenser, Vu should also decrease the voltage across the resistor with the trend indicated by the half-wave dotted in Fig. 8-b.
For the heads of the resistor is connected, But, the capacitor, who, with decreasing voltage, It must be discharged giving rise to the discharge current through the resistor itself.
Because of the resistance of the resistor to the discharge current, It takes a certain time to discharge: Consequently, the voltage Vu decreases more slowly, with the trend shown in fig. 8-b from the area in green between the instants t1 and t2.
Between these two instants the voltage Vu is always positive and greater value than that of the AC voltage V, that in the meantime it reaches zero and then the negative maximum and then start another positive half-wave.
Since the AC voltage V is applied between the anode and mass, while the voltage Vu is obtained between the cathode and mass, the fact that in the time between the instants t1 and t2 the diode is cut off and can not be crossed by the current.
From the instant t2 onwards, But, the alternating voltage, that is increasing because it accomplishes the second positive half-wave, exceeds the value of the voltage Vu bringing the anode at a higher potential than the cathode, and then the diode is traversed by the current that charges the capacitor.
The voltage across this element therefore increases together with the alternating voltage until it reaches again the maximum value at the instant t3, from which he repeats the same pattern described starting from the instant t1.
It is seen as well as the trend of the voltage Vu is changed across the resistor due to the capacitor, it is easy to deduce that the current circulating in the same resistor should have the same performance, as it is indicated by bold line in Fig. 8-c.
We observe that in the time between the instants t1 and t2 the current through the resistor is supplied by the capacitor which is discharged thereby losing a certain amount of electricity; this same amount of electricity is supplied again to the capacitor by the current flowing through the diode in the time between the instants t2 and t3.
Since this time, during which the capacitor is charged, It is much shorter than that between the instants t1 and t2, during which the capacitor discharges, the charging current must be much more intense than that of discharge, since the amount of electricity is always the same.
In fig. 8-c is also shown as it varies the current through the diode to charge the capacitor: due to the characteristic pattern of this current, it is said that in the diode will have current peaks.
In the previous article it is mentioned that the characteristic of a diode comprises a first portion to entire line and a second portion in dotted line: This second portion serves to precisely know the behavior in correspondence of the tube to the said current peaks.
It is also said that the tube may operate under such conditions without having to exceed the maximum power that can be dissipated, and this is confirmed by the fact that, as seen in Fig. 8-c, between a peak current and the other one you have a rather long period of time during which no current flows through the diode, the anode of which can therefore dissipate the heat.
However, it must make sure that during the current peaks is not so intense as to produce excessive dissipation anodic.
It should be noted, Furthermore, that the cathode of the diode must not emit current exceeding a certain value, because beyond this value can be damaged.
Since the peak current is all the more intense the greater the capacitance of the capacitor that must be loaded, builders of the diodes generally indicate the maximum capacitance of the capacitor which can be connected in parallel to the load without that the current peaks prove dangerous for the integrity of the cathode.
Even in the case of a full-wave rectifier, connecting a capacitor in parallel to the resistor is obtained a voltage which, as seen in Fig. 9, It has a trend similar to that considered for the half-wave rectifier.
In the full-wave rectifier positive half-waves are no longer separated by an interval of one half-cycle, but follow one another without interruption, as it is indicated with dashed line in fig. 9.
Consequently, while the voltage obtained by half-wave rectifier varies recovering the same values after a time equal to the period of the alternating voltage, as seen in Fig. 8-b, the voltage obtained from the full-wave rectifier resumes the same values after a time which is half of the previous period and therefore its, shown in fig. 9, is half that of the alternating voltage.
Because of this, also the time during which the capacitor discharges is halved and therefore the decrease in the voltage Vu and thus its variation is smaller in the case of a full-wave rectifier. This can be seen by comparing the diagrams of fig. 8-b and fig. 9, in which such variation, denoted by Vr, It is given by the difference between the maximum and minimum values of the voltage Vu.
If this voltage Vu is nourished a radio receiver, It would get a reproduction accompanied by an annoying buzz, for the reason that we will see later.
Since the buzz is due to the fact that the voltage Vu varies between its maximum value and its minimum value, It is called ripple voltage Vr the difference between these values.
The ripple voltage, It is due to the successive charges and discharges the capacitor, It must depend on both the capacitance of the capacitor itself, both the current that this element provides the load-store rendered during its download.
E’ clearly, the greater the current absorbed by the load, the greater the discharge of the capacitor and thus the decrease in the voltage at its ends, which constitutes the ripple voltage Vr. , the less the greater the capacitance of the capacitor.
To eliminate the buzz that accompanies the reproduction is not necessary that the Vr voltage is exactly equal to zero, because small variations in the voltage Vu not substantially disturb the reproduction.
They are generally tolerable variations 1 V or even 2 V for each 100 Vu of the voltage V; in short we can say that the voltage Vr can be l’ 1 % or even the 2 % the voltage Vu.
From the foregoing it is understood that in order to reduce the ripple voltage to these values would be necessary to use a high-capacitance capacitor, thus incurring the danger of having excessive current peaks; It is therefore resort to the use of a filter cell disposed between the condenser and the rectifier circuit resistor.
This filter cell is constituted by an inductor fitted with a ferromagnetic core and a capacitor, connected as shown in fig. 10-a; the inductor is said more precisely impedance because its winding has a certain ohmic resistance, It is formed by several coils.
The current due to the discharge of the capacitor connected between the cathode and mass must now cross the impedance before reaching the resistor and since the impedance has the property to counteract the variations of the current, the resistor is traversed by an almost continuous current; on the other hand, the capacitor connected across the resistor behaves towards the voltage as the capacitor connected between the cathode and mass, helping to reduce the ripple voltage.
The set formed by the two capacitors and the impedance constitutes a smoothing filter that is said to capacitive input, as the first element of the filter is precisely the capacitor connected between the cathode and mass. Due to the large capacity required, the two capacitors are always of the type called electrolytic, which has already been described in the practical lessons; as seen in Fig. 15, these capacitors have the negative plate connected to ground and the positive plate connected to the points that are located in potential higher than that of mass.
Because the filter impedance presents the drawback of being expensive and cumbersome, in many cases it is used in its place a simple resistor, who, despite not having the property of impeding the variations of the current, however, it slows down with its resistance the charge of the second capacitor, making it less variable the current trend.
These filters are also called resistance filters and capacity (abbreviated RC), while the filters equipped with impedance are said inductance and capacitance filters (abbreviated LC).
They are also used to filter said inductive input because, as seen in Fig. 10-b, their first element is precisely the inductor, lacking the capacitor connected between the cathode and mass.
For the absence of this capacitor the diode is no longer crossed by current peaks, that could be hazardous, especially in cases in which you must provide a substantial current to the load.
The inductive input filter is therefore used for the supply of equipment that requires a considerable power; these filters in the receivers are not practically used, as the output power is always somewhat reduced.
4. – POWER anodic
The assembly formed by the rectifier and the smoothing filter constitutes an anode feeder, so called because it is obtained from the DC voltage to be applied to the anodes of the electronic apparatus tubes.
In fig. 11 They are reported the complete diagrams of two types of power supplies used most commonly; since the two power supplies differ for the type of diode, in that one is an indirect ignition while another is to direct ignition, They were also below the tube filaments.
In the case feeder with an indirect switching diode (fig. 11-a), the transformer has two secondary windings: the one having the ends connected to the anodes of the diode and the center tap is connected to ground of said high voltage secondary (statements AT) because it provides a voltage value of a few hundred volts; It is said instead of low voltage secondary (condensed BT) what serves for the ignition of the filament, because its voltage is a few volts, typically 6,3 V.
As can be seen in the diagram, this secondary has one end connected to one end of the filament, while the other end is connected to ground, to which is also connected the other end of the filament: in tal modo, the filament ignition circuit is closed through the mass of the receiver.
The same secondary BT also serves for the ignition of the other receiver tubes, which also they have a head to mass.
Feeder are thus obtained two voltages: a DC voltage, positive with respect to ground, a few hundred volts, which it is said anodic because it serves to feed the anode circuits of the receiver tubes, and an alternating voltage, a few volts, with which it ignites the same tubes.
When the diode is used in the direct ignition type (fig. 11-b), the rectified voltage is obtained from the same filament, because now it is this element that, in the absence of the cathode, shall issue electronic.
Because of that, the filament can no longer have a boss connected to ground as shown in Fig. 11-a, because this connection constitutes a short circuit for the current from the diode.
Indeed, would reach such a current directly to ground by means of the aforementioned connection no longer pass through the filter and the impedance of the receiver circuits which must be sent. Not encountering the resistance offered by the receiver circuits, the current would become so intense burning filament.
To avoid this inconvenience, the ignition circuit of the diode is completely isolated from mass, having both ends of the filament connected directly to the extremes of a secondary dedicated BT, as seen in Fig. 11-b. For the ignition of the other receiver tubes instead uses another sub-BT, one end of which is connected to ground.
Designing an amplifier for power supply
A series of general concepts before moving on practical example.
The main condenser
Given that in the case should replace a rectifier valve because exhausting substitute in an old receiver with a 1N4007 diode with a resistor in series from 47 Ohm leaving the old in its place. In practice it is its equivalent circuit with semiconductors without compromising the historical value of the equipment remains with all its original parts, adding only has the mere function to restore the appliance to function.
In the design of modern amplifiers use silicon diodes directly to their easy availability and the increased efficiency.
The first capacitor after the rectifier stores most of the energy for the whole amplifier. Each half-cycle rectified charges the capacitor voltage on the peak AC with a short current pulse but big. The voltage then decreases when the load current is continuously withdrawn from the amplifier circuit, until it can not be recharged as we have seen in Figure. 8
The ripple voltage Vr is often expressed as a percentage of the maximum DC voltage. A typical figure could be the 10% for a push-pull amplifier or 5% for a single-ended amplifier, although this depends largely on the requirements of individual circuits. Therefore, if puntassimo to a diet 400 V dc with the undulation 5%, we would not want more than 400 × 0,05 = 20 Vpp voltage ripple. The main condenser can then be chosen approximately using the following formula:
C = I / (2 f Vr)
Where I is the load current DC media, f is the mains frequency (50 O 60 Hz) and Vr is the ripple voltage peak-to-peak desired. This is a formula somewhat "conservative"; in practice the ripple voltage will prove to be a bit 'lower than this. However, a greater capacity also puts a strain on the rectifier and the transformer because it requires larger current pulses to keep load. Most of traditional amp projects use from 22uF to 60uF if a rectifier valves or up to 220uF with silicon diodes is used (rarely use higher values). The single-ended amplifiers require a greater capacity because they do not cancel out the hum as do the push-pull amplifier. If you need a high working voltage, the usual trick is to put two capacitors in series so that their voltage values are added.
However, the total capacity will be halved, then two capacitors 100uF amount to 50uF. Furthermore, the resistors must be added in parallel in order to encourage the sharing of the same voltage between the capacitors. The resistors should be equal to 50 / C o name, then two capacitors 100uF would need a resistor 500000 ohm (470k would be the most obvious choice). These they serve also download capabilities when the amplifier is switched off.
Most amplifiers feeds the primary of the output transformer directly from the main condenser. However it is not sufficient to provide the DC noise-free necessary for the screen grids and the stages of the preamplifier, then a further leveling is needed.
This is accomplished with a chain of LC or RC filters (lowpass), denominated in various ways leveling filters, bypass or decoupling. These alternative names are derived from the fact that there are actually three related jobs between them to perform:
1: filtering the residual ripple voltage;
2: provide a local energy supply for sudden current demands;
3: decouples / isolates each amplifier stage from the rest.
Each phase of RC leveling is a low-pass filter with a cutoff frequency of:
f = 1 / (2 pi RC)
obviously, the only frequency that we really want to go is 0 Hz or DC, only then do the cutoff frequency as low as possible, often below 1HZ. Normally you see the smoothing capacitors 10uF to 100uF, whatever is easily available. For a given capacity, a larger resistance lowers the cutoff frequency and hence improves the sanding. However, there is also a certain fall of direct current through the resistor due to the load current flowing in it, so there is a compromise between smoothing and voltage drop. The preamplification stages are generally polarized to the cathode and very tolerant with respect to the supply voltage, therefore usually it does not matter exactly which voltage is obtained after the dropping resistor. Anything from 250 V a 400 V is OK. The resistor must be able to withstand the full supply voltage and the capacitor charging current at start. Chaining the filters together we get a better ripple reduction. The amplifier stages less sensitive such as screen grids and the phase inverter are fed from the first sections of the filter while the more sensitive stages receive the most power filter, they are also subject to the maximum voltage drop. The inlet valve is always the last in the chain.
Many amplifiers use a choke-capacitor filter (LC) to feed the screen grids of the power valve. An LC filter is a second order filter, therefore provides more steep attenuation ripple compared to an RC filter (first order), and an inductance has only a small DC resistance, so it does not fall much DC voltage. The interest in this stage is to maximize the voltage in order to maximize the power output of clean. The exact value of the inductor is not critical, but it should be remembered that an LC filter resonates to its frequency, which is given by:
f = 1 / (2 pi sqrt [LC])
It is usually best to keep the resonance frequency below 10 Hz, beyond audio range. This will require a larger capacitor:
C = 1 / (L × [2 pi f] ^ 2)
Since the filter capacitors are available for most of the range of 10uF to 100uF, they usually are seen in the range reels 20 a 2 henry .
After this long but necessary introduction we pass a practical example.
Example of design
To design the power supply is necessary to know the amount of DC current media that the amplifier circuit will attract. The preamp stage will normally be single-ended (Class A) so as to draw a constant average current. If the preamp tubes are ECC83 / 12AX7 then it will usually be polarized around 1 mA for triode, by name.
If the output stage is AB class (for example, almost all of the push-pull amplifiers), its current will increase as the signal level. Therefore, the outlet valves will usually be biased at approximately 70% of the maximum anode dissipation of the data card, thereby allowing a certain power margin, so that they are not heated at full power.
Suppose an amplifier 50 W three uses ECC83 / 12AX7 and a pair of EL34.
The valves of the preamplifier will consume approximately 6 x 1 mA = 6 mA (there are two triodes for valve).
The EL34 has a nominal power of 25 watts maximum, They will most likely polarized around 0,7 x 25 W = 17,5 W. However, we should work with the average maximum figures, that is, the full current when the average dissipation increases to 25W. Therefore, if the raw DC supply voltage (first capacitor) is equal to 400 V, each of them consumes about
25 W / 400 V = 62,5 over there 125 mA for the pair.
The technical details suggests a current ratio between screen and anode 6,5, so we can expect that the current amount to the screen
125 / 6,5 = 19 mA for the pair.
The total for the entire amplifier is then
6 + 125 + 19 = 150mA.
Given that we use silicon diodes, we can use much more capacity, less buzz. If puntassimo to a ripple voltage of the 5%, then it would mean 400 × 0,05 = 20 Vpp. The main condenser would then:
C = I / (2 f Vr) = 0.15 / (2 × 50 × 20) = 75uF.
However, the most common capacitors are 450 V. If we assume a change in the network 10%, our 400 V HT could rise to 440 V, which it is within the limits. But we should also consider a further increase 5% due to the losses of the transformer when it is not at maximum load, making possible an increase of up to 462V. It would therefore be a good idea to use two capacitors from 250V series to achieve a higher working voltage. It should be torque from 150uF equivalent to 75uF, although we probably could use a pair of capacitors 100uF, certainly easier to find because the above equation is conservative. Each will require equalization resistors <50 / C o 500000 ohm, eg 470 k. We probably do not need to do the same for the preamplifier capacitors because the tensions are minor and within the limit of 450V voltage We use a coil to filter the power supply to the screen grid. This will require another pair of capacitors in series because one has a very low resistance to restrict the leakage current, then we could use another couple of electrolytic 100uF. This is equivalent to 50uF, so if we want to maintain the resonance frequency below 10 Hz, the inductance must be at least:
L = 1 / (C × [ 2 pi f ] ^ 2) = 1 / (50 × 10 ^ -6 × [2 pi × 10] ^ 2) = 5 henry.
Keep in mind that must withstand without the saturations 19 mA of current screen and 6 mA of current preamplifier.
If space is limited, we may want to use smaller electrolytic preamplifier, say 22uF. For a decent hum reduction we should push the cut-off frequency to a very low frequency, for example 1Hz. This means that the resistors in the fall must be at least:
R = 1 / (2 pi f C) = 1 / (2 pi × 1 × 22 × 10 ^ -6) = 7234 ohm.
However, Since 6 mA of the preamp current must flow into the first resistor, this would cause a voltage drop, perhaps excessive. We therefore want to lower the first resistor to 4.7k, that would just fall
6mA × 4,7k = 28V
0,006 ^ 2 × 4700 = 0,17 watt.
The subsequent fall in resistors may perhaps be bigger, although 22uF and 4.7k 1.5Hz provide a cut-off frequency which is quite respectable, so there is no harm in making them all equal. The final scheme is shown, with nominal voltages (maximum).
Of course, the diagram of an amplifier is not only constituted by these elements, they are only those needed to power only, the polarizations of the individual stages have been omitted to simplify the treatment of the subject.
With the next articles (being drafted) on successive stages, anyone can try their hand in the design and implementation of its own tube amp.