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Current protection of amplifier.

Aspect 1. Little of theory. Prologue.

The usual protection circuits of amplifiers work on exceeding of output voltage. The threshold of the wear and tear of protection can be advanced either on the peak load or on the nominal, but with the long exposure (2… 5 seconds). In the essence, the protection on the stress is ineffective with the work of amplifier on the high-resistance load and does not work at all with the low-resistance load. The greatest effect of the work of protection circuit on stress is achieved only at the nominal load.

Example 1.
Amplifier with the maximum power of 100W for the load of 4Ohm. Protection circuit relies on wear and tear with the output voltage of 20V (P=U^2/R). But if we to the amplifier connect the load of 8Ohm, protection circuit will operate only with 50Watts despite the fact that, as a rule, the output power of amplifier for the load doubly higher resistance it composes not 50% of the power to the rating, but approximately 60-65% (i.e., 100W/4Ohm, but 60W/8 Ohm) but if we load amplifier for the load of 2 Ohm, we will obtain the overload of amplifier to the wear and tear of protection. 100 W to 2 Ohm is achieved by the output voltage of 14 W. If this amplifier is capable of working on the two-ohm load, then with the wear and tear of protection power there will be 20^2/2=200W, but this is fraught with the failure of amplifier, if it is not calculated for such working conditions.

The second type of protection – current – most practical in the work ensures the fitness for work of amplifier with any load. This protection limits output current at the level of the maximum permissible current with the maximum power and operates both with the nominal load and with that exceeded.

Example 2.
Amplifier with the maximum power of 100W for the load of 4Ohm. Maximum output current there will be 5A (P=I^2*R). With the work to 4Ohm the protection will operate at the power of 100W, with 8Ohm of current overload simply it will not be, with 2Ohm the protection will operate at the power of 50W, i.e., it is accurate at the moment of the overload of amplifier. But if amplifier is not calculated for the two-ohm load, then neither the total nor exceeded power from the amplifier to remove will come out. Failure of amplifier is improbable.

The elements of current protection most frequently are connected into the emitter or discharge circuits of output transistors. With the appearance of current overload the protection locks output transistors, limiting output current. This is sufficiently effective and is reliable, if amplifier is executed on the discrete elements, t to the elements of protection are immune to heating from the heat-releasing components, and their operating conditions practically do not change. But if amplifier integral (microcircuit), then there are no these elements either or they are subjected to general heating together with the crystal of microcircuit, and the stabilization of the modes of operation of these elements is hindered. It is completely possible that these elements are disposed to the wear and tear with the maximum permissible values of the output current, when microcircuit works on the limit. By these, I think, it is possible to explain the frequent failures of integral amplifiers, and the, as a result, extended opinion about their unreliability.

As “experimental rabbit” it was selected WITH TDA7294. With its operation in the normal modes it is sufficiently steady in the work, suffers overloads and overheating. The wear and tear of heat shielding was observed neither in me, and I think nor in one user. There are debatable questions about a sufficient supply voltage – someone exploits it from + -30V, someone from + -40V. I will not tie my opinion; in the experiments was selected the stress + -16,5V as sufficient for its work and is far from maximum. As it follows from [datashita], minimum load on 7294 is permitted not lower than 4Ohm. With these parameters and voltage of supply + -16,5V from the microcircuit it is possible to obtain not less than 25W of the undistorted signal and 35W of signal with the cut tops of sinusoid, what also was confirmed in the course of experiments (12,5[volt] to 4Ohm – 40W) unfortunately, in the work was used the not entirely disposed generator; therefore waveform not entirely sinusoidal.

Output signal without trimming of the apexes of sinus.+/-16,5V nourishment, 1[kHz] signal, 4Ohm the equivalent of load, 5V/ [del] the development Cut of [verkhi].+/-16,5V nourishment, 1[kHz] signal, 4Ohm the equivalent of load, 5V/[del] the development

But which did prove to be?
The test of microcircuit for the load of 2Ohm showed that the tops of sinusoid are cut with the somewhat smaller signal levels. This speaks, that the microcircuit holds the almost previous signal level, but for the dual load! In the conversion to the power this there will approximately be the doubled power. But the here short circuit of the output of microcircuit led to the unexpected result – in 80-watts power unit it operated protection and it was opened! It leaves, that with the short circuit entire power of the source of nourishment is scattered nowhere otherwise as in microcircuit itself! But if power unit to 200W? 300? It is easy whether by it to derive this power to the heat-transferring flange and to transmit to radiator?
Let us be turned to datasheet.
Thermal resistance passage- housing composes 1,5  Sec/W. By passage is implied the crystal of microcircuit, which diffuses heat-. By housing is implied not entirely the housing, but the heat-transferring flange of microcircuit, on which is fixed the crystal. What does mean this number? But means it that the fact that with each scattered watt of power the temperature of crystal rises to 1,5 Sec. In my case, 80W from the power unit approached the burning-out of crystal. This neither much nor is small as 120 degrees (1,5*80) the difference between the flange with the warm radiator and the crystal, which for the fractions of a second here was heated to 160 degrees. Thought immediately appears – microcircuit could not limit current surge, which led to the heat emission inside the microcircuit. Why? Let us leave this on the conscience of producer. What must be the temperature of flange, in order to ensure the normal operation of microcircuit? The temperature of crystal is limited 150[S]. the 13th graph of [datashita] shows that with the nourishment of +30V and power of 50-60W the scattered power composes approximately 50W. Then we should ensure this cooling that the flange would not be heated higher than 150-50*1,5=75[gradusov]. Certainly, such conditions are possible only on the flat signal, everything will be somewhat more coldly with the music. In the example, described above, it was clearly shown that the fitness for work of current protection was required in any amplifier for its reliable work.

Part 2. the practice

Since to inject protection in the existing discrete amplifier is complex, and – it is generally impossible into the integral, asserts itself the certain external block, which fulfills its functions. Let us examine the examples to the possible realization of current protection. Device commutates load and driver amplifier in the limits of 100W.

The first comparator of microcircuit IC2 follows the load resistance, the second – after the output current of amplifier. With the start of amplifier the load at first is connected to the first comparator, which compares a voltage drop across it with the reference voltage of 0,1V. The threshold of wear and tear is selected 3Ohm in order to exclude false responses with the different acoustic systems. Accordingly, with the load resistance is more than 3Ohm on conclusion 3 stress greater than on conclusion 2, comparator is switched and by relay connects load to the amplifier. The second comparator follows the current, flowing from the amplifier through the load and the current-measuring resistor R1. As soon as current will exceed that permitted, comparator with a certain delay (R11C3) with the aid of the transistor Q2 switches the first comparator into the regime of the measurement of the load resistance with its turning off from the amplifier. This delay is extended in terms of the value of the current through R1 – with the current spikes up to that permitted diagram it is located on the switching threshold, and the higher the current surge through R1, the more rapidly the load will be opened. The time of the connection of load to the amplifier is conversely assigned by ratings R10C3 and is found in the range 0,5:2 [s] in the dependence on the value of the current surge, which caused the wear and tear of diagram. The light-emitting diode, connected to the joint Of [kh]2, identifies the peaks of output signal. This function can be used as a clip- detector. Is here possible the different realization of diagram – either trigger protection (turning off to the manual discharge), or protection with the signal to reduction in the loudness. In the version in the figure higher diagram it will cyclically disconnect/to connect load to the amplifier, until the level of output signal is lowered.

Let us look by somewhat different version of the realization of the diagram:

Diagram of the trigger current protection

Diagram the same, only to second comparator is added the chain of hysteresis R15D3. With the wear and tear it itself closes before breaking of chain. Temporary delay to turning off of load is absent (it is disconnected immediately), delay to the reclosing of load to the amplifier after discharge it is preserved. Diagram with the signal to reduction in the loudness I do not lead, t to loudness to decrease is possible differently – through the processor of sound or digital of [tembroblok], by simply active level or by the sequence of pulses, if volume control is fastener. Depending on this the diagram will differ.

The exterior view of the already assembled device can be seen on picts below.

The exterior view

Printed-circuit board for the cyclic current protection.

Printed-circuit board

Installation and tuning.
[Nznachenie] of the contacts of the joints:

AMP+ To the output of driver amplifier
AMP- Earth driver amplifier
RL+ Load (acoustic system)
RL- Earth the load
+15 +15V of 0,1A
0V General (earth)
-15 -15V of 0,1A
1 + the light-emitting diode
2 – the light-emitting diode

Diagram feeds by bipolar stress +10… +15V by the current of 100[mA]. In the diagram are used two 12[voltovykh] relay for the commutation of load – their contact groups are connected in parallel, and windings – consecutively. If there are by relay to 24V by a sufficient current of commutation (15-20A), it is possible to use it. Diagram does not require special tuning, but it is necessary to verify the fitness for work of device in all regimes.
With the tuning we check the presence of the following stresses (with the off by amplifier and the load): +5V on the third conclusion IC1 of +0,1V on the second or sixth conclusion IC2 we further connect resistor of approximately 3 Ohm (2,7… 3,3Ohm) between housing and junction of resistors R7R8, measure stress at this point. If it differs from 0,1V, then by the selection of resistor R7 we advance the stress of 0,1V. Now diagram will react to reduction in the load in less than disposed. Let us open resistor and diagram. Connecting to the contacts RL+, RL – joint U1 different resistors from 2 to 5Ohm and turning on power supply, we observe the clear wear and tear of relay of higher than the disposed threshold (3 Ohm).
Let us dispose the threshold of wear and tear on the current.
Example of tuning. We have an amplifier of 100W/4Ohm. We count the threshold of wear and tear on the current – 5A. Resistor R1 according to diagram – 0,22Ohm. On the current of 5A on it falls the stress of 5*0,22=1,1V. Comparator compares this stress with the supporting of 0,1V; therefore we should lower 1,1[Volt] 11 times (to 0,1V). By this is occupied divider R2R3. On the resistor R3 must be isolated our of 0,1V; therefore everything else ( 1V ) will be isolated on the resistor R2. Since on it is developed a voltage 10 times more, its resistance must be into 10[raz] more R3, i.e., 1[kOm]*10=10[kOm]. Are in the diagram indicated the ratings of resistors R2R3 for the amplifier of 35W/4Ohm. A finer tuning can be made with the aid of the generator and the ammeter with the voltmeter.

On the replacement of elements.
[OU] 4558 can be from any producer – KA4558, NJM4558 and so forth the transistors of 2PC945 are frequently called even as 2SC945, it is possible to place them. With a change in the base will approach [KT]3102, BC547 other relay BS-115C-12V of firm Bestar can be analogous from the firms Omron, Tianbo and others being suitable with the voltage of the winding of 12V by the current of 40-50[mA] on the commutation of currents to 10A.

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