JPS6158046B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to an amplifier protection circuit. If excessive voltage and current are applied to the output transistor in an amplifier, the transistor will be destroyed. Figure 2A shows the relationship between the collector-emitter voltage and the collector current that causes this output transistor to break down.
It is commonly called the ASO line. Conventionally, a transistor protection circuit as shown in FIG. 1 has been used to prevent the transistor from operating outside the area indicated by the ASO line A (shaded area). In the figure, input terminals 1 and 2 are NPN and
connected to the bases of PNP transistors 3 and 4 and NPN and PNP transistors 15 and 16;
connected to. The collectors of transistors 3 and 4 are connected to power supplies +B and -B, respectively, and the emitters are connected to load end 7 via resistors 5 and 6, respectively. The bases of transistors 15 and 16 are connected to the emitters of transistors 3 and 4 through resistors 9 and 10, respectively, and to the anodes and cathodes of diodes 13 and 14 through resistors 11 and 12, respectively. The cathodes and anodes of diodes 13 and 14 are grounded. The emitters of transistors 15 and 16 are connected to load terminal 7, and load 8 is connected between load terminal 7 and ground.

In the conventional array as described above, the diodes 13 and 14 are connected when the emitter voltages of the transistors 3 and 4 are higher than the forward voltage of the diodes, that is,
The collector-emitter voltage of transistors 3 and 4 (hereinafter referred to as the operating voltage) V _CE is the power supply +B and -B.
It turns on when the voltage is below _VB . In this case, since the impedance of the diode is small, it is ignored, and a first bridge circuit is formed by the resistors 5, 9, 11 and the load 8, and a second bridge circuit is formed by the resistors 6, 10, 12 and the load 8. If the balance of the first and second bridges is disrupted due to a short circuit in the load 8, a detection voltage is generated between the base emitters of the transistors 15 and 16, respectively, and the transistors 15 and 16 are turned on, and the input terminals 1, 16 are turned on. The input signal to be applied to the transistors 3 and 4 is cut off by the transistor 2, and the transistors 3 and 4 are protected. Now let the resistance values of resistors 5, 9 and 11 and the resistance values of resistors 6, 10 and 12 be R _E , R ₁ and R ₂ respectively, and the base-emitter voltage of transistors 15 and 16 be V _BE .
The collector current I _C of transistors 3 and 4 for turning on transistors 5 and 16 is I _C ≧R ₁・V _B +R ₂・V _BE / _RE (R ₁ +R ₂ ) −R ₁ /R _E (R ₁ +R ₂ )・V _CE (1). Furthermore, when the emitter voltages of the transistors 3 and 4 are lower than the forward voltage of the diodes, the operating voltage V _CE of the transistors 3 and 4 is equal to the voltage V of the power supply +B and -B.
Turns off when _B or higher. In this case, transistors 15 and 16 are turned on when the voltage drop across resistors 5 and 6 due to the collector current I _C flowing through transistors 3 and 4 exceeds the base-emitter voltage V _BE of transistors 15 and 16. Therefore, the collector voltage I _C in this case becomes I ₁ expressed by the following equation.

I ₁ =V _BE /R _E (2) _The protection line B shown in FIG _. This diagram illustrates the relationship between Here, when the load is inductive, the load curve L becomes an ellipse as shown in the figure, so if the conventional protection circuit as described above is used, the protection line B will meet the above load curve at the bending point. Intersects with L. However, this part
Although it is located inside the ASO line and does not originally require protection, the protection circuit operates as described above, resulting in disadvantages such as distortion in the output signal.

The present invention has been made to eliminate these drawbacks, and by bringing the protection line closer to the ASO line, the range in which a connection with a protection circuit can operate normally is expanded.

FIG. 3 shows an embodiment of the present invention. In the figure, the same parts as in FIG. 1 are given the same reference numerals and their explanations are omitted, but a voltage dividing circuit connected in series with resistors 27, 28 and 29 is provided between power supplies +B and -B. Here, the connection points 30 and 31 between the respective resistors are used as divided voltage output terminals. The bases of the transistors 15 and 16 are connected to the divided voltage output terminal 31 through a series circuit of a resistor 23 and a diode 25 and a series circuit of a resistor 24 and a diode 26, respectively, and the bases of the transistors 15 and 16 are connected to a resistor 19, respectively. It is connected to the divided voltage output terminal 30 through a series circuit of a diode 21 and a series circuit of a resistor 20 and a diode 22, and further connects the bases of transistors 15 and 16 to power supplies -B and +B through resistors 17 and 18, respectively. do.

In the above configuration, the divided voltage output terminals 30 and 3
Let the voltages of 1 be V ₂ and −V ₂ .

Now, in order for both diodes 21 and 25 to turn off, the voltage -V ₂ at the voltage-divided output terminal 31 must be higher than the emitter voltage (V _B -V _CE ) of the transistor 3. Therefore, V _CE >V _B +V ₂ (2), and if V _B +V ₂ =V _T , the range is as shown in Fig. 4C. In this range, one side of the bridge becomes a resistor 17 connected to the negative power supply -B instead of the grounded resistor 11.If this resistance value is _R3 , then in the above equation (1), _R2 Let be R ₃ and V _B be
Since it is sufficient to set 2V _B , I _C ≧2R ₁ V _B +R ₃ V _BE /R _E (R ₁ +R ₃ )−
R ₁ /R _E (R ₁ +R ₃ )V _CE (3), and the protection line C becomes gentle as shown.

Next, in order for both diodes 21 and 25 to turn on, the voltage V ₂ at the divided voltage output terminal 30 must be lower than the emitter voltage (V _B −V _CE ) of the transistor 3. Therefore, V _CE <V _B −V ₂ (4), and when V _B −V ₂ =V _S , the range is as shown in FIG. 4A. In this range, one side of the bridge is replaced by resistors 19 and 23 whose ends are connected to the divided voltage output terminals 30 and 31, instead of the resistor 11. here,
Assuming that the current flowing through the resistor 9 is _IX , the base voltage _VX of the transistor 15 is _VX = _VB - _{VCE - IXR1} (5). Further, current _IX is the sum of the currents flowing through resistors 17, 19 and 23. Voltage applied to resistor 17
Since V ₃ is V _X +V _B , the current I ₃ flowing through the resistor 17 and R ₃ becomes I ₃ =V _X +V _B /R ₃ . Also, the voltage V ₄ applied to the resistor 19 is V _X −V ₂
Therefore, if the resistance value of resistor 19 is R ₄ , then
The current I ₄ flowing through this resistor is I ₄ =V _X −V ₂ /R ₄ . Also, the voltage V ₅ applied to the resistor 23 is V _X +V ₂
Therefore, if the resistance value of the resistor 23 is _R5 , then
The current I ₅ flowing through this resistance is I ₅ =V _X +V ₂ /R ₅ . Therefore, the current _IX flowing through the resistor 9 is _IX = _VX + _VB / _R3 + _{VX -V2} / _R4 + _VX
+V ₂ /R ₅ (6).

Here, in order for the transistor 15 to turn on, the voltage drop across the resistor 5 should be greater than the voltage drop across the resistor 9, and the difference between them should be greater than or equal to the base-emitter voltage V _BE of the transistor 15, so I _C R _E −I _X R ₁ ≧ V _BE (7). Therefore, by substituting V _X in equation (5) into equation (6) to obtain I _X , and substituting this into equation (7), we get becomes. Here, R _P is the parallel resistance value of the resistors 17, 19, and 22, and is 1/R _P =1/R ₃ +1/R ₄ +1/R ₅ . Therefore, since R _P <R ₃ , equations (8) and
As is clear from comparing the second terms on the right side of equation (3), the protection line C has a steep slope in this range.

Next, when the operating voltage V _CE is in the middle of the ranges A and C above, V _S <V _CE <V _T (9), only the diode 25 is turned on and the resistor 19 is disconnected. Therefore, the current I _X flowing through the resistor 9 is the sum of I ₃ and I ₅ and I _X =V _X +V _B /R ₃ +V _X +V ₂ /R ₅ (10). Therefore, by substituting V _X in Equation 5 into Equation (10) to obtain I _X , and substituting this into Equation (8), we get becomes. Here, R _Q is the parallel resistance value of the resistors 17 and 23, and is 1/R _Q =1/R ₃ +1/R ₅ . Therefore, since R _P <R _Q <R ₃ , the formula (3)
As is clear from the comparison with the second terms on the right-hand side of Equations (8) and (11), the protection line C has an intermediate slope in this range.

The above description has been made regarding the protection line for the transistor 3, but since the protection line for the transistor 4 is completely the same, detailed explanation will be omitted.

As described above, according to the embodiment shown in FIG. 3, since the protection line C has two bending points, the protection line C can be brought closer to the ASO line A more accurately.

In the above-described embodiment, there are two bending points, but the bending points are not limited to two, but a large number of bending points can be provided. That is, the voltage dividing circuit is constituted by a large number of resistors, a large number of voltage dividing output terminals are provided, and a large number of series circuits consisting of resistors and diodes are connected between each voltage dividing output terminal and the bases of the transistors 15 and 16. All you have to do is connect. In this way, the protection line C can be provided with many bending points, and the protection line C can be aligned with the ASO line A very accurately.
It can be approximated by Note that resistors 17 and 1
8 can be omitted. In this case, the slope of the range C in FIG. 4 is zero, as in the case of FIG. 2.

As described above, according to the present invention, it is possible to provide a plurality of bending points for the protection line, and the protection line can be
Since it can be placed close to the ASO line, it has the excellent effect of widening the range in which the amplifier can operate normally without undergoing unnecessary protection operations.

[Brief explanation of the drawing]

FIG. 1 is a circuit diagram showing a conventional amplifier protection circuit, FIG. 2 is a diagram explaining its operation, FIG. 3 is a circuit diagram showing an embodiment of the present invention, and FIG. 4 is a diagram explaining its operation. In the figure, 3, 4, 15 and 16 are transistors, 21, 22, 25 and 26 are diodes,
8 is the load.

Claims (1)

[Claims] 1 Connect the emitter of the output transistor to the load end via the emitter resistor, connect the collector and emitter of the protection transistor between the base of the output transistor and the load end, and connect the base of the protection transistor to the first resistor. In the amplifier connected to the emitter of the output transistor through the output transistor, a plurality of reference voltage points having different voltages selected from among voltage values whose absolute value is smaller than the collector voltage of the output transistor or whose polarity is opposite to the collector voltage. , a plurality of series-connected circuits each having a resistor and a diode connected in series and corresponding to the plurality of reference voltage points, and between the base of the protection transistor and the plurality of reference voltage points,
A protection circuit for an amplifier, characterized in that the plurality of series-connected circuits described above are connected to each other.