JP2554593B2 - Power supply circuit - Google Patents

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a power supply circuit, and more particularly to a power supply circuit suitable for use in a characteristic test of loads such as semiconductors.

[0002]

2. Description of the Related Art FIG. 6 shows a conventional voltage source / current source circuit. In this circuit, the variable voltage generator 10 outputs a voltage according to control of a microprocessor, for example, to an operational amplifier 12
Supply to the non-inverting input terminal of. The inverting input terminal of the operational amplifier 12 is connected to the movable contact of the double throw switch 14. The output signal of the operational amplifier 12 is connected to the output terminal 18 via the detection resistor 16. The output terminal 18 is connected to the fixed contact a of the double throw switch 14 via the buffer amplifier 20. The voltage across the detection resistor 16 is the differential amplifier 2
2 is supplied to two input terminals, and the output voltage of the differential amplifier 22 is connected to the fixed contact b of the double throw switch 14. The output terminal 18 is connected to the other end of a load 24 such as a semiconductor whose one end is grounded.

The movable contact of the double throw switch 14 is a fixed contact a.
When connected to, the voltage source / current source circuit is in the voltage source mode. In the voltage source mode, the load 24 is supplied with the output voltage Vo equal to the input voltage Vi. In this mode, the input voltage Vi can be changed and the current flowing through the detection resistor 16 can be measured to measure the voltage / current characteristic of the load. When the movable contact of the double throw switch 14 is connected to the fixed contact b, the voltage source / current source circuit is in the current source mode. In the current source mode, the voltage across the detection resistor 16 becomes equal to the input voltage Vi, and the output current Io equal to Vi / R obtained by dividing the input voltage Vi by the resistance value R of the detection resistor 16 is supplied to the load 24. . In this mode, the output current Io is changed by changing the input voltage Vi and the voltage across the load can be measured to measure the current / voltage characteristic of the load.

[0004]

As described above, the voltage source /
In the load voltage / current measurement using the current source circuit, the output voltage Vo is determined by the input voltage Vi, and a current corresponding to the output voltage Vo flows through the load 24. In the voltage / current measurement, when the input voltage Vi is increased, the current Io flowing through the load 24 is increased.
May suddenly increase and destroy the load 24. One method of preventing such load breakdown is to connect a current limiting circuit (not shown) capable of limiting the voltage supply to the operational amplifier 12 to the output terminal of the variable voltage generator 10. It is detected that the output voltage of the differential amplifier 24 exceeds a predetermined level as the current Io increases, and in response to the detection signal, the current limiting circuit cuts off the voltage supply. However, since a normal current limiting circuit has a response delay time because it performs a non-linear operation, if the load current increases rapidly, the load will be destroyed before the voltage supply is cut off. Another way to prevent the load from being destroyed is to insert a protective resistor 26 between the output terminal 18 and the load 24 as shown on the right side of the arrow in FIG. However, the protection resistor 26
Is effective for preventing the load from being destroyed by suppressing the current Io flowing through the load, but lowers the voltage supplied to the load by the voltage generated across the load. Therefore, when the protective resistor 26 is used, it is necessary to correct the measurement value, and the range of selection of the resistance value is limited.

Therefore, an object of the present invention is to provide a power supply circuit which can immediately prevent the load from being destroyed when an excessive current flows in the load without requiring the above-mentioned current limiting circuit and protection resistor.

[0006]

According to the power supply circuit of the present invention, the input voltage is supplied from the input terminal to the voltage amplifier 40 through the input resistor 38, and the voltage between the output terminal and the output terminal of the voltage amplifier is first. One resistor 42 is connected, and this first resistor 42
The voltage between both ends of the subtraction circuit is detected by the subtraction circuit 50, and the output voltage of the subtraction circuit is multiplied by the first multiplier M by the first multiplication circuit 60.
The voltage of the output terminal is multiplied by the second multiplier N by the second multiplier circuit 64, and the outputs of the first and second multiplier circuits are input to the input terminal of the voltage amplifier 40 via the second and third resistors 62 and 66. Each is characterized by returning.

[0007]

When M = 0 and N ≠ 0, the power supply circuit operates in the voltage source mode. When M ≠ 0 and N = 0, the operation is in the current source mode. When M ≠ 0 and N = 1, it operates equivalently as a voltage source having an internal resistance value that is M times the resistance value of the first resistor, and due to the internal resistance value, it becomes excessive when a load is short-circuited due to an increase in output voltage. It can be suppressed so that a large current does not flow.

[0008]

1 is a circuit showing a power supply circuit of the present invention. The input terminal 30 includes, for example, a digital-analog converter (not shown) controlled by a microprocessor.
The input voltage Vi is supplied from a DAC). The input terminal 30 is connected to an inverting input terminal of an operational amplifier 34 via a resistor 32, and the inverting input terminal is connected to a resistor 36.
Is connected to the output terminal of the operational amplifier 34 via. The non-inverting input terminal of the operational amplifier 34 is connected to a reference potential source which is, for example, a ground potential source. The resistors 32 and 36 have the same resistance value r, and together with the operational amplifier 34, form an inverting amplifier having a gain of -1. Therefore, the output voltage of this inverting amplifier is -Vi. However, the resistance values of the resistors 32 and 36 do not have to be equal, and may have any relationship as needed.

The output terminal of the operational amplifier 34 is connected to the inverting input terminal of the operational amplifier 40 via the resistor 38. The non-inverting input terminal of the operational amplifier 40 is connected to the reference potential source. The output terminal of the operational amplifier 40 is connected to an output terminal 44 via a resistor 42 having a known resistance value RS. The output terminal 44 has a resistance value RL and is connected to the other end of the load resistor 46, one end of which is connected to the reference potential source. At the output terminal 44, an output voltage Vo equal to the product of the resistance value RL of the load resistor 46 and the output current Io is generated. Vo = RLIo (1)

The resistor 42 is a current detecting resistor for detecting a current Io flowing through it from a voltage generated across the resistor 42, and has a relatively high resistance value so that a small current can be detected. Is desirable. The output terminal of the operational amplifier 40, that is, the input side terminal of the resistor 42 is connected to the non-inverting input terminal of the operational amplifier 50 via the resistor 48, and the non-inverting input terminal is further connected to the reference potential via the resistor 52. Connected to the source. The output terminal 44, that is, the output side end of the resistor 42 is connected to the inverting input end of the operational amplifier 50 via the buffer amplifier 54 and the resistor 56, and this inverting input end is further connected to the resistor 58.
Is connected to the output terminal of the operational amplifier 50 via. The resistors 48, 52, 56 and 58 have the same resistance value r, and together with the operational amplifier 48 form a subtraction circuit with an amplification factor of 1, the output voltage Vs thereof is equal to the voltage RSI0 across the resistor 42. Vs = RSI0 (2)

The output terminal of the operational amplifier 50 is commonly connected to the inverting input terminal of the operational amplifier 40 through the multiplier 60 and the resistor 62, and the output terminal of the buffer amplifier 54 is commonly connected through the multiplier 64 and the resistor 66. It The multipliers 60 and 64 have multipliers M and N, respectively, and the output voltages of the multipliers 60 and 64 are MVS = MRSIo and NVo = NRLIo, respectively. Multiplier 6
0 and 64 preferably use a multiplying DAC to perform high precision multiplication. Multiplying D
When AC is used, the range of the multipliers M and N is 0 ≦ M ≦
1 and 0 ≦ N ≦ 1. Resistors 38, 62 and 66
For the sake of simplicity of explanation, it is assumed that they have the same resistance value r.

The operation of the above power supply circuit will be described below.
One end of each of the resistors 38, 62 and 66 has an operational amplifier 4
0 is commonly connected to the inverting input terminal, and the other end has a voltage of −
Since Vi, MRSIo and NRLIo are supplied, the voltage Vc at the common connection point is expressed by the following equation. Vc = 1 / 3.multidot. (MVS + NVo-Vi) = 1 / 3.multidot. (MRSIo + NRLIo-Vi) (3) No current flows through the inverting input terminal of the operational amplifier 40, that is, V
The condition for c = 0 is expressed by the following equation. Vi = MRSIo + NRLIo (4) When the output current Io is obtained from the equation (4), Io = Vi / (MRS + NRL) (5) When the output voltage Vo is obtained from the equations (1) and (5), Vo = ViRL / (MRS + NRL) (6)

If M = 0, N ≠ 0, then Io and Vo
From the equation (5), Io = Vi / NRL (7) From the equation (6), Vo = Vi / N (8), and the output voltage Vo is related to the resistance value of the load resistor 46. Instead, the input voltage becomes 1 / N times Vi, and the power supply circuit operates in the voltage source mode. Here, if N = 1, the output voltage Vo becomes equal to the input voltage Vi. Moreover, since the output voltage can be changed by changing N, for example, a stepped output voltage according to the value of N can be generated by periodically decreasing or increasing N with a constant input voltage.

On the contrary, when M ≠ 0 and N = 0, Io and Vo are Io = Vi / MRS from the equation (5) ... (9) From the equation (6), Vo = ViRL / MRS, The current Io is determined by the constants of the input voltage Vi, the resistor RS and the multiplier M, regardless of the resistance value of the load resistor 46, and the power supply circuit operates in the current source mode. Where M =
If it is 1, the current Io is determined only by the input voltage Vi and the resistor RS. In addition, the resistance value of the resistor 42 can be changed equivalently according to the value of M, and when a multiplier that obtains a multiplier greater than 1 is used instead of the multiplying DAC, the value of M By increasing, the resistance value of the resistor 42 can be equivalently increased.

The main feature of the present invention resides in the case where M ≠ 0 and N = 1. Since N = 1, it is equivalent to removing the multiplier 64 and directly connecting the output terminal of the operational amplifier 54 to the resistor 66. If M = 1, the multiplier 60
Is eliminated and the output terminal of the operational amplifier 50 is directly connected to the resistor 62. When Io and Vo in this case are obtained, Io = Vi / (MRS + RL) ... (10) from the equation (5) and Vo = ViRL / (MRS + RL) ... (11) from the equation (6). As shown in FIG. 2, these equations are equivalent to a voltage source 70 having a voltage value Vi having an internal resistance 72 having a resistance value MRS.
Is equivalent to connecting the load resistor 46 to the resistance value MRS. This can be considered as voltage source mode operation. The internal resistor 72 functions as a protection resistor, and the protection resistor 72 can limit an excessive increase in the load current due to an increase in the input voltage Vi. Moreover, the power supply circuit of FIG.
Unlike the conventional circuit, the output terminal 44 has a load resistor 46
Since the output voltage Vo is directly connected to the load resistor 46, the output voltage Vo is equal to the supply voltage to the load resistor 46, and the output voltage Vo from the output terminal of the buffer amplifier 54 to the load resistor 46 does not require correction as in the conventional case. Can be obtained. Further, since the current limiting operation of the power supply circuit of FIG. 1 is a linear operation, the delay time is the response speed of the power supply circuit itself. Therefore, the delay time is higher than that of the conventional current limiting circuit, and there is a possibility that the element is destroyed. Low.

Io and Vo when M ≠ 0 and N ≠ 0 are expressed by equations (4) and (5), respectively. Formula (5) representing Vo
When is transformed, it becomes as follows. Vo = (Vi / N) .RL / {(MRS / N) + RL} (12) This equation is equivalent to the resistance value MRS / N as shown in FIG.
This is equivalent to connecting the load resistor 46 to a voltage source having a voltage value Vi / N having an internal resistance of. In this case, when N decreases, the output voltage of the voltage source 74 increases, and in proportion to this, the resistance value of the resistor 76, which is the output resistance value of the power supply circuit, increases. That is, when the output voltage doubles, the output resistance value also doubles. As a result, the short-circuit current when the load resistor 46 is short-circuited is kept constant even if the output voltage is increased, and a destructive short-circuit accident can be avoided. For example,
When such a power supply circuit is used as a collector voltage source when measuring transistor characteristics, if the collector voltage doubles from the state shown in FIG. 5A to the one shown in FIG. 5B, the load resistance also doubles and a short circuit occurs. The current is constant.

FIG. 4 shows another embodiment of the power supply circuit of the present invention. This circuit differs from the circuit of FIG. 1 only in the input section.
As in FIG. 1, the input voltage from the DAC is supplied to the input terminal, but the input terminal 30 is directly connected to the non-inverting input terminal of the operational amplifier 40. The inverting input of operational amplifier 40 is connected to the common connection point of resistors 60 and 66. The voltage Vc at the common connection point is expressed by the following equation. Vc = (NRLIo + MIoRS) / 2 (13) No current flows to the inverting input terminal of the operational amplifier 40, that is, V
When i = Vc, Vi = (NRLIo + MIoRS) / 2 (14) When Io is calculated from the equation (14), Io = 2Vi / (NRL + MRS) (15) Equations (1) and (15) When Vo is calculated from the above, Vo = 2Vi
If RL / (NRL + MRS) M = 0, N ≠ 0, then Io
And Vo, Io = 2Vi / NRL (16) Vo = 2Vi / N (17), and the output voltage Vo is the same as the input voltage Vi regardless of the resistance value of the load resistor 46. 2 / N times, and the power supply circuit operates in the voltage source mode. On the contrary, when M ≠ 0 and N = 0, when Io and Vo are obtained, Io = 2Vi / MRS (18) Vo = 2ViRL / MRS (19), and the current Io is the load. Regardless of the resistance value of the resistor 46, the power supply circuit operates in the current source mode, which is determined by the double voltage of the input voltage Vi, the resistor RS and the constant of the multiplier M.

When M ≠ 0 and N = 1, Io = 2Vi / (MRS + RL) (20) Vo = 2ViRL / (MRS + RL) (21) These equations are equivalent to the case where the load resistor 46 having the resistance value MRS is connected to the voltage source having the voltage value 2Vi equivalently having the internal resistance having the resistance value MRS. Therefore,
The operation of the circuit of FIG. 4 is substantially the same as that of the circuit of FIG.

[0019]

The power supply circuit of the present invention operates as a voltage source having an internal resistance value equivalent to M times the resistance value of the first resistor even if the protection resistor is not connected in series to the load, and The resistance value can prevent an excessive current from flowing when the load is short-circuited due to an increase in the output voltage. Since the output terminal is directly connected to the load, the output voltage Vo is equal to the supply voltage to the load, and the supply voltage from the output terminal to the load can be obtained without the need for correction as in the conventional case.

[Brief description of drawings]

FIG. 1 is a circuit diagram showing a power supply circuit of the present invention.

FIG. 2 is an equivalent circuit diagram for explaining the operation of the circuit of FIG.

FIG. 3 is an equivalent circuit diagram for explaining the operation of the circuit of FIG.

FIG. 4 is a circuit diagram showing another embodiment of the power supply circuit of the present invention.

FIG. 5 is a graph diagram for explaining an operation when the power supply circuit of the present invention is used for measuring transistor characteristics.

FIG. 6 is a circuit diagram showing a conventional voltage source / current source circuit.

[Explanation of symbols]

 40 Voltage Amplifier 42 First Resistor 60 Multiplier 62 Second Resistor 66 Third Resistor

Claims (1)

(57) [Claims] 1. A voltage amplifier to which an input voltage is supplied from an input terminal through an input resistor, a first resistor connected between an output terminal of the voltage amplifier and an output terminal, and the first resistor. A subtraction circuit for detecting a voltage between both ends of the voltage regulator, a first multiplication circuit for multiplying an output voltage of the subtraction circuit by a first multiplier, a second multiplication circuit for multiplying a voltage of the output terminal by a second multiplier, A power supply circuit comprising: second and third resistors for respectively feeding back outputs of the first and second multiplication circuits to the input terminals of the voltage amplifier.