JPH0844441A - Constant voltage circuit - Google Patents

Abstract

PURPOSE:To sufficiently suppress the base current when an output transistor(TR) is in a saturated area and to reduce the input current. CONSTITUTION:This constant voltage circuit has a TR Q1a connected to an output TR Q1 so that their bases and emitters are connected in common and having an emitter area ratio of N1:1, a current mirror circuit Q4, Q4a connecting the TR Q4 to the collector of the TR Q1a and havinq an emitter area ratio of N1:1, a TR Q3 for generating a collector current corresponding to 1/N2 of the base current of the TR Q1 in accordance with an output from an error amplifier A1, a TR Q3a connected to the TR Q3 so that their bases and emitters are connected in common, connecting its collector to the TR Q4a and having an emitter area ratio of N3:1 with the TR Q3, and a diode D1 connected between the collector of the TR Q3a and the input terminal of the amplifier A1 and capable of supplying a drive current controlling current when the TR Q1 is saturated.

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a constant voltage circuit, and more particularly to a constant voltage circuit composed of an integrated circuit.

[0002]

2. Description of the Related Art FIG. 6 shows a conventional constant voltage circuit 31 as an example.
A circuit diagram is shown. The constant voltage circuit 31 has an input voltage V_inSupply
The specified constant voltage V_out= V_OCGenerate a load circuit
It is a circuit to supply to. Reference voltage V_ZConstant voltage to generate
Diode D_ZThe anode is grounded and the cathode is
Constant current I from constant current source 21_ZIs supplied. Reference voltage V
_ZIs the error amplifier A₁It is supplied to the inverting input terminal of.

The voltage V _{S obtained} by dividing the output voltage V _out by the resistors R ₁ and R ₂ is supplied to the non-inverting input terminal of the error amplifier A ₁ . The ratio of the resistors R ₁ and R ₂ is set so that the voltage V _S = V _Z when the output voltage V _out has the specified value V _OC .

The error amplifier A ₁ is composed of, for example, a differential amplifier, and the output voltage is supplied to the base of the PNP transistor Q ₁₂ . The output current of the error amplifier A ₁ flows into the output terminal as the base current of the transistor Q ₁₂ . The emitter of the transistor Q ₁₂ is connected to the input terminal T ₁ to which the input voltage V _in is supplied.

Diode connected NPN transistor Q
₁₃ , the emitter is grounded, the collector is a transistor Q
Connected to ₁₂ collectors. The NPN type transistors Q ₁₄ and Q ₁₅ are mirror-connected to the transistor Q ₁₃ ,
The base and the emitter are commonly connected to the transistor Q ₁₃ .

In the PNP type output transistor Q ₁₁ , the emitter is connected to the input terminal T ₁ supplied with the input voltage V _in , and the collector is connected to the output terminal T ₂ . PNP
The type transistor Q _11a is mirror-connected to the transistor Q ₁₁ , and the emitter and the base are commonly connected to the transistor Q ₁₁ . Further, between the base and the emitter of the transistor Q ₁₁ is the resistance R ₃ is connected.

The base of the transistor Q ₁₁ is connected to the collector of the transistor Q ₁₄ , and
The collector of _11a is connected to the collector of the transistor Q _15.

The anode of the diode D ₁₁ is connected to the collector of the transistor Q ₁₂ , and the cathode is connected to the collector of the transistor Q ₁₅ .

Transistors Q _11a and Q ₁₅ , diode D
₁₁ is a control circuit of the base current I _B11 of the transistor Q _11, as described later, the input voltage V _in is small, the transistor Q ₁₁ when the transistor Q ₁₁ is in the saturation region
It is provided to suppress the base current I _B11 of.

Mirror-connected transistor Q₁₁And Transi
Star Q_11aIs the emitter area ratio m ₁₁: N₁₁Set to
is there. In addition, a mirror-connected transistor Q₁₄And Transi
Star Q_FifteenIs the emitter area ratio m₁₂: N₁₂Set to
It

Further, the transistor Q ₁₃ and the transistor Q
₁₄ , the emitter area ratio is set to, for example, 1: m ₁₂ .

FIG. 7A is an explanatory diagram showing the relationship between the input voltage V _in and the hfe ₁₁ of the transistor Q ₁₁ . Input voltage V _in
Is small and equal to or less than the specified value V _OC of the output voltage, the transistor Q ₁₁ is in the saturation region. In this case, the hfe _{11 of the} transistor Q ₁₁ becomes smaller than the value hfe _11S in the normal operation.

In FIG. 7A, when the input voltage V _in is V _in0 or more, the transistor Q ₁₁ goes out of the saturation region and hf
e ₁₁ substantially matches the value hfe _11S during normal operation. When the input voltage V _in is V _in0 or less, the transistor Q ₁₁
Enters the saturated region, and hfe ₁₁ is lowered.

First, the input voltage V_inIs the specified value V of the output voltage
_OCIn the normal operation, which is sufficiently large,
Transistor Q₁₂Collector current I₁₂Is as it is
Langista Q₁₃Collector current I₁₃Becomes Transis
Q₁₄Has a transistor Q ₁₃Collector current I₁₃M
₁₂Double the collector current I₁₄Flows. This collector
Flow I₁₄But transistor Q₁₁Base current of_B11Tona
It Note that m₁₁>> n₁₁Since it is set to
Star Q_11aBase current of_B11aIs I_B11Enough compared to
It becomes smaller and can be ignored. Transistor Q₁₁Has
Base current I_B11Hfe₁₁Double collector current I_C11But
Flowing. Collector current I_C11Is the output terminal T₁Connected to
Load circuit and resistor R₁, R₂Flowing in the series circuit.

The base current of the transistor Q ₁₂ is controlled by the operation of the error amplifier A ₁ so that the output voltage V _out maintains the specified voltage V _OC irrespective of load fluctuations and the like.

[0016] The collector current I ₁₅ of the transistor Q ₁₅ is,
The value is I ₁₅ = I ₁₄ · (n ₁₂ / m ₁₂ ). This current I
₁₅ becomes the collector current I _11a of the transistor Q _11a .

During normal operation, the value of hfe _11S = I _C11 / I _B11 of the transistor Q ₁₁ becomes a large value, for example, a value of about 100 to 200. As described below,
The emitter area ratio m ₁₁ / n ₁₁ is set to a value smaller than hfe _11S , for example, 100, and the emitter area ratio m ₁₂
/ N ₁₂ is set to a value larger than 1, for example, 2.

[0018] transistor Q _11a, the emitter area ratio m _11: current determined by _{n 11 I 11a = I C11 ·} (n 11 /
m ₁₁₎ becomes the value of the current tends to flow, from the transistor Q _15, than _{I C11 · (n 11 / m} 11), much smaller current _{I 15 = I 14 · (n} 12 / m 12) = (I _C11 / hf
Only e _11S ). (n ₁₂ / m ₁₂ ) is supplied. Therefore, the transistor Q _11a is in a saturated state during steady operation. At this time, the collector voltage of the transistor Q ₁₅ has a value close to the input voltage V _in, the diode D ₁₁ are turned off.

Next, the input voltage V _in is the specified output voltage V _OC
The operation when the transistor Q ₁₁ is in the saturation region will be described below.

The input voltage V _{in falls} below the voltage V _{in0 at} which the transistor Q ₁₁ enters the saturation region (see FIG. 7A), h
When fe ₁₁ becomes small to some extent, the transistor Q
Collector-emitter voltage of _11a becomes large, the transistor Q _11a enters the active region. At this time, as for the collector current I _11a of the transistor Q _11a, the following expression (1) is established.

[0021] _{I 11a = I C11 · (n} 11 / m 11) = I C11 / N 11 (1) In addition, the collector current I ₁₅ of the collector current I ₁₄ and the transistor Q ₁₅ of the transistor Q ₁₄ is represented by the following (2 ) The formula holds.

I_Fifteen= I₁₄・ (N₁₂/ M₁₂) = I₁₄/ N₁₂ (2) Diode D₁₁Is off, diode D₁₁Current I_D11
= 0, the current I _11a= I_FifteenTherefore, equation (1),
From (2), the following equation (3) is established.

I _C11 / N ₁₁ = I ₁₄ / N ₁₂ (3) Here, since m ₁₁ >> n ₁₁ is set, the transistor Q
The base current I _{B1 1a} of _11a is sufficiently smaller than the base current I _B11 of the transistor Q ₁₁ . Therefore, I ₁₄ =
I _B11 + I _{B11a ≈I B11} . Therefore, equation (3) becomes the following equation (4).

I _C11 / N ₁₁ = I _B11 / N ₁₂ (4) Therefore, the hfe ₁₁ of the transistor Q ₁₁ at this time is
It can be expressed as in equation (5) below.

The hfe ₁₁ of _{hfe 11 = I C11 / I B11} = N 11 / N 12 = hfe 11L (5) transistor Q ₁₁ is reduced, when it becomes a hfe _11L, as follows, the diode D ₁₁ switches from off to on. The voltage of the anode of the diode D ₁₁ is equal to the forward voltage V _F (about 0.6) of the transistor Q _13.
V). On the other hand, when hfe ₁₁ decreases and becomes about hfe _11L or less, the transistor Q ₁₅ is saturated and the collector voltage of the transistor Q ₁₅ becomes almost the ground potential.
Therefore, the diode D ₁₁ is turned on.

As described above, when the diode D ₁₁ is turned on, the current I _D11 flows from the collector of the transistor Q _{12 to} the collector of the transistor Q ₁₅ , and the transistor Q ₁₃
Collector current I ₁₃ of the transistor Q ₁₄ decreases, and accordingly, the collector current I ₁₄ of the transistor Q ₁₄ , that is, the base current I _B11 of the transistor Q ₁₁ decreases.

In this way, when the transistor Q ₁₁ enters the saturation region and hfe ₁₁ drops to hfe _{11 L} = N ₁₁ / N ₁₂ , a current I _D11 flows through the diode D ₁₁ ,
The base current I _B11 of the transistor Q ₁₁ is suppressed.

FIG. 7B shows the relationship between the input voltage V _in and the current I _D11 of the diode D ₁₁ . Note that the input voltage V _in has the same scale as that in FIG. Figure 7
As shown in (B), when the input voltage V _in is V _in0 or less, the transistor Q ₁₁ enters the saturation region, and hfe ₁₁ is N ₁₁ / N ₁₂
The current I _D11 has started to flow in the diode D ₁₁ when the voltage drops to, and the peak value is reached at the voltage of V _in1 .

The value of N ₁₁ / N _{12 is} the value of h of transistor Q ₁₁ .
Set according to the fe characteristic. For example, h during normal operation
When fe ₁₁ is about 100 to 200, for example, N ₁₁ / N ₁₂
= 50, and when hfe ₁₁ drops to 50, the control circuit for the base current I _B11 starts to operate.

FIG. 8 shows the input voltage V in the circuit of FIG.
It is an explanatory diagram of the relationship _in the input current I _in. FIG. 8 shows the input voltage V when the load current of the load circuit is 0 (no load).
The relationship between _in and the input current I _in is shown.

Transistor Q₁₁Is in the saturation region,
Transistor Q₁₁Collector-emitter voltage V_CE11But
Extremely small, hfe₁₁Becomes smaller, so
Q ₁₁Most of the emitter current of the
I will As a result, the transistor Q₁₄Drive capacity of
Collector current I₁₄Is transistor Q₁₁Flow to the base of
Would.

Therefore, the base current I_B11Control circuit
If not, V_in1Base current I at_B11The peak of
The value becomes extremely large, and the constant voltage circuit 3
Input current I of 1_inAlso shows that the peak value is
It will be an extremely large value. On the other hand, the base current I
_B11By providing the control circuit of
_B11The peak value of can be suppressed and is shown by the solid line in Fig. 8.
Input current I_inCan also be suppressed.

For example, when the control circuit for the base current I _B11 is not provided, the peak value of the base current I _B11 is several mA, whereas the control circuit for the base current I _B11 is provided. The peak value of I _B11 can be suppressed to several hundred μA.

FIG. 9A shows a relational diagram between the collector current I _c11 and hfe ₁₁ of the transistor Q ₁₁ in the saturation region. FIG. 9B shows the transistor Q ₁₁ in the saturation region.
A relational diagram of the collector current I _{c11 of the above} and the current I _D11 of the diode D ₁₁ is shown. When the collector current I _c11 increases, h
When fe ₁₁ approaches the set value of hfe _11L = N ₁ / N ₂ and the collector current I _c11 decreases, hfe ₁₁
The value of ₁₁ is smaller than the theoretical value N ₁ / N ₂ .

As a concrete example of the circuit of FIG. 6, when the hfe ₁₁ of the transistor Q ₁₁ in the normal operation is about 100 to 200, m ₁₁ = 100, n ₁₁ = 1 (N ₁₁ = 100),
An example in which m ₁₂ = 8 and n ₁₂ = 4 (N ₁₂ = 2) are set will be described. In this case, N ₁ / N ₂ = 50 is set.

V _in is a specified output voltage V _OC (for example, 5 V)
Collector current I _C11
The following measured values are obtained under the condition of = 5 mA.

I _B11 = I ₁₄ = 42 μA I ₁₅ = I _11a = 22 μA Transistor Q ₁₁ hfe ₁₁ = 5 mA / 42 μA = 11
9 At this time, the diode D ₁₁ is off. Further, the transistor Q _11a is in a saturated state, and the value of I _C11 / I _11a is larger than N ₁₁ .

V _in is less than or equal to the specified output voltage V _out ,
When transistor Q ₁₁ is in the saturation region, the following measured value is obtained under the condition of collector current I _C11 = 5 mA.

The hfe of _{I B11 = I 14 = 200μA I} 15 = 100μA I 11a = 60μA I D11 = 40μA transistor _{Q 11 11 = 5mA / 200μA =} 2
5 Thus, when the transistor Q ₁₁ is in the saturation region,
The base current I _B11 is suppressed by the transistor Q ₁₅ drawing the current I _D11 from the transistor Q ₁₂ via the diode D ₁₁ . However, compared with the current I ₁₅ when the diode D ₁₁ is turned from OFF to ON, the base current I
The value of I _D11 for controlling _B11 increases. As a result, the suppression of the base current I _B11 becomes insufficient, and hfe
The measured value of ₁₁ is smaller than the theoretical value N ₁ / N ₂ = 50.

Further, the current I ₁₅ for controlling the base current I _B11 has a large value of 100 μA.

When the collector current I _C11 becomes small, the ability to control the base current I _B11 is insufficient, and the base current I _B11 is about 200 μA even when there is no load.
The current I ₁₅ flows about 100 μA.

[0042]

As described above, although the conventional circuit of FIG. 6 can suppress the base current I _B11 when the output transistor Q ₁₁ is in the saturation region,
The suppression is still insufficient, and there is a problem that the current of the control circuit of the base current I _B11 is large.

The present invention has been made in view of the above points, and it is possible to sufficiently suppress the base current of the output transistor when the output transistor is in the saturation region, and to reduce the current required for controlling the base current. An object is to provide a constant voltage circuit that can reduce the input current.

[0044]

A constant voltage according to the invention of claim 1
The circuit is an output transistor that supplies the output voltage to the load.
And an output signal according to the difference between the output voltage and the specified voltage.
Depending on the error amplifier to be generated and the output signal of the error amplifier,
The base side drive current of the output transistor
In a constant voltage circuit that has a drive current supply circuit to supply
The output transistor and the base and emitter are
Connected to the output transistor and the emitter of the output transistor
Area ratio is N₁A first transistor set to: 1;
The input side transistor is connected to the collector of the first transistor.
The input side transistor is connected.
The emitter area ratio of the output transistor to the output transistor is N _Four: 1
The set first current mirror circuit and the drive
Output signal of the error amplifier provided in the current supply circuit
According to 1 / of the drive current of the output transistor
N₂Second transistor that produces double collector current
And the second transistor and the base and emitter are
And a collector connected to the first current mirror.
-Connected to the collector of the output transistor of the circuit
And the emitter area ratio with the second transistor is
N₃The third transistor set to: 1 and the first transistor
The output side transistor of the current mirror circuit and the third
Connection point of the transistor and the input terminal of the error amplifier
Connected between the output transistor and the output transistor
Drive to the input terminal of the error amplifier
A structure having a switch element for supplying a current control current;
I do.

According to a second aspect of the present invention, in the drive current supply circuit, the emitter area ratio of the output side transistor and the input side transistor for supplying the base side drive current of the output transistor is set to N ₂ : 1. Of the current mirror circuit and the collector of the input side transistor of the second current mirror circuit are connected to each other, and the collector current of the output transistor is 1 / N ₂ times the drive current of the output transistor according to the output signal of the error amplifier. And a second transistor that generates a collector current.

According to a third aspect of the invention, in the constant voltage circuit according to the second aspect, a resistor is provided between the emitter of the input side transistor of the second current mirror circuit and the power supply terminal.

[0047]

According to the first aspect of the present invention, when the output transistor is in the saturation region and the current amplification factor is lower than that in the normal operation, the value becomes (N ₁ · N ₄ ) / (N ₂ · N ₃ ). ,
The drive current control current flows into the input terminal of the error amplifier via the switch element. As a result, the base side drive current of the output transistor is suppressed.

The values of N ₁ · N ₄ and N ₂ · N ₃ can be set large, and the values of the currents flowing through the first current mirror circuit and the third transistor can be reduced. The current required to control the drive current can be reduced. Moreover, the output signal of the error amplifier can be greatly changed and the drive current of the output transistor can be sufficiently suppressed by passing a drive current control current of a very small value through the switch element.

As described above, since the current required for controlling the drive current can be reduced and the drive current can be sufficiently suppressed, the input current of the circuit when the output transistor is in the saturation region as compared with the conventional circuit. It is possible to reduce.

According to the second aspect of the present invention, it is possible to realize, with a simple circuit configuration, a constant voltage circuit capable of suppressing the drive current with a small input current when the output transistor is in the saturation region.

According to the third aspect of the invention, the change in the current amplification factor of the output transistor when the output transistor is in the saturation region can be reduced, and the circuit design can be made easier.

[0052]

1 is a circuit diagram of a constant voltage circuit 11 according to a first embodiment of the present invention.
The circuit diagram of is shown. In FIG. 1, the same components as those in FIG. 6 are designated by the same reference numerals, and description thereof will be omitted as appropriate.

The constant voltage circuit 11 is a circuit which is supplied with an input voltage V _in , generates a prescribed constant voltage V _out = V _OC , and supplies the voltage to a load circuit. The anode of the constant voltage diode D _Z that generates a reference voltage V _Z is grounded, the cathode, constant-current I _Z is supplied from the constant current source 21. Reference voltage V _Z
Is supplied to the inverting input terminal of the error amplifier A ₁ .

The voltage V _{S obtained} by dividing the output voltage V _out by the resistors R ₁ and R ₂ is supplied to the non-inverting input terminal of the error amplifier A ₁ . The ratio of the resistors R ₁ and R ₂ is set so that the voltage V _S = V _Z when the output voltage V _out has the specified value V _OC .

The error amplifier A ₁ is composed of, for example, a differential amplifier, and the output voltage is supplied to the base of the PNP transistor Q ₃ (second transistor). Error amplifier A ₁
Output current flows into the output terminal as the base current of the transistors Q ₃ and Q _3a . The emitter of the transistor Q ₃ is connected to the input terminal T ₁₁ to which the input voltage V _in is supplied. Transistor Q _3a (third transistor)
Is a mirror connection with the transistor Q ₃ , and its base and emitter are connected in common.

Diode connected NPN transistor Q
_{In 2a} , the emitter is grounded and the collector is a transistor Q.
Connected to ₃ collectors. The NPN transistor Q ₂ is mirror-connected to the transistor Q _2a , and its base and emitter are commonly connected. Transistor Q
_2a, constitutes a second current mirror circuit by Q _2.

In the PNP type output transistor Q ₁ , the emitter is connected to the input terminal T ₁₁ to which the input voltage V _in is supplied, and the collector is connected to the output terminal T ₁₂ . PNP
The type transistor Q _1a (first transistor) is mirror-connected to the transistor Q ₁ , and the emitter and base are commonly connected to the transistor Q ₁ . Further, between the base and the emitter of the transistor Q ₁ is, the resistance R ₃ is connected.

The base of the transistor Q ₁ is connected to the collector of the transistor Q _2a . The diode-connected NPN transistor Q ₄ has an emitter grounded and a collector connected to the collector of the transistor Q _1a . The NPN transistor Q _4a is mirror-connected to the transistor Q ₄ , and its base and emitter are commonly connected. The transistors Q ₄ and Q _{4a form a} first current mirror circuit.

The collector of the transistor Q _4a is connected to the collector of the transistor Q _3a .

The anode of the diode D ₁ (switch element) is connected to the collector of the transistor Q _3a , and the cathode is connected to the non-inverting input terminal of the error amplifier A ₁ .

Transistors Q ₃ , Q _2a and Q ₂ form a drive current supply circuit for supplying base current I _B1 of transistor Q ₁ .

The transistors Q _1a , Q ₄ , Q _4a , Q _3a and the diode D ₁ form a control circuit for the base current I _B1 of the transistor Q ₁ . The control circuit of the base current I _B1, as described later, the input voltage V _in is small, suppress the base current I _B1 of the transistor to Q ₁ when the transistor Q ₁ is in the saturation region.

The emitter area ratio of the mirror-connected transistor Q ₁ and transistor Q _1a is set to m ₁ : n ₁ = N ₁ : 1. The mirror-connected transistor Q ₂ and transistor Q _2a have an emitter area ratio of m ₂ : n ₂ = N
It is set to ₂ : 1. The transistor Q ₃ and the transistor Q _3a of the mirror connection, the emitter area ratio m _3:
The setting is n ₃ = N ₃ : 1. The transistor Q ₄ and the transistor Q _4a of the mirror connection, the emitter area ratio _{m 4: n 4 = N 4} : is set to 1.

FIG. 2A is an explanatory diagram showing the relationship between the input voltage V _in and the hfe ₁ of the transistor Q ₁ . Input voltage V _in
Is smaller than the specified value V _OC of the output voltage, the transistor Q ₁ is in the saturation region. In this case, the hfe _{1 of the} transistor Q ₁ becomes smaller than the value hfe _1S in the normal operation.

In FIG. 2A, when the input voltage V _in is V _in0 or more, the transistor Q ₁ leaves the saturation region and hf
e ₁ substantially matches the value hfe _1S during normal operation.
When the input voltage V _in is V _in0 or less, the transistor Q ₁ enters the saturation region and hfe ₁ is lowered.

First, the input voltage V_inIs the specified value V of the output voltage
_OCIn the normal operation, which is sufficiently large,
Transistor Q₃Collector current I₃Is as it is
Langista Q_2aCollector current I_2aBecomes Transis
Q₂Has a transistor Q _2aCollector current I_2aM
₂/ N₂Double the collector current I₂Flows. This
Current I₂But transistor Q₁Base current of
_B1(Drive current). Note that m₁>> n₁Set to
Therefore, the transistor Q_1aBase current of
_{B1 a}Is I_B1Can be ignored because it is much smaller than
It Transistor Q₁Has a base current I_B1Hfe₁
Double collector current I_C1Flows. Collector current I_C1Is
Output terminal T₁₂And the load circuit connected to the resistor R₁, R₂
Flowing in the series circuit.

The base current of the transistor Q ₃ is controlled by the action of the error amplifier A ₁ so that the output voltage V _out maintains the specified voltage V _OC irrespective of load fluctuations and the like.

The transistor Q ₄ includes a transistor Q _1a.
Collector current I _1a flows as it is as collector current I ₄ .

[0069] collector current I _3a of the transistor Q _3a is,
I _3a = I ₃ · (n ₃ / m ₃ ) = _{IB 1} · (n ₂ / m ₂ ) ·
(N ₃ / m ₃ ) = (I _C1 / hfe _1S ) · (n ₂ / m ₂ ).
・ (N ₃ / m ₃ ) = (I _C1 / hfe _1S ) / (N ₂・
N ₃ ).

The value of hfe _1S = I _C1 / I _B1 of the transistor Q ₁ at the time of steady operation is large, for example, 10
The value is about 0 to 200. Further, as will be described later, the emitter area ratios N ₁ , N ₂ , N ₃ and N ₄ are hfe _1S ·
Set so that (N ₂ · N ₃ )> N ₁ · N ₄ .

For the transistor Q _4a , I _4a = I ₄ · (n
₄ / m ₄ ) = I _C1 · (n ₁ / m ₁ ) · (n ₄ / m ₄ ) =
A collector current I _4a having a value of I _C1 / (N ₁ · N ₄ ) is about to flow. However, as mentioned above, hfe _1S · (N
₂ · N ₃ )> N ₁ · N ₄ , so transistor Q _4a
Flows only a current I _3a = (I _C1 / hfe _1S ) / (N ₂ · N ₃ ), which is considerably smaller than I _C1 / (N ₁ · N ₄ ). Therefore, the transistor Q _4a is in a saturated state during steady operation. At this time, the transistor Q
The collector voltage of _3a is close to the ground potential,
The diode D ₁ is off.

Next, the input voltage V _in is the specified output voltage V _OC.
The operation when the transistor Q ₁ is in the saturation region will be described below.

The input voltage V _in becomes less than the voltage V _{in0 at} which the transistor Q ₁ enters the saturation region (see FIG. 2A), h
When fe ₁ becomes small to some extent, the transistor Q
Collector-emitter voltage of _4a increases, transistor Q _4a enters the active region. At this time, the following relationships are established.

Regarding the collector current I _1a of the transistor Q _1a, the following expression (1) is established.

[0075] _{I 1a = I C1 · (n} 1 / m 1) = I C1 / N 1 (6) Further, the collector current I _4a of the collector current I ₄ and transistors Q _4a of the transistor Q ₄ are the following (7 ) The formula holds.

[0076] _{I 4a = I 4 · (n} 4 / m 4) = I 4 / N 4 = I 1a / N 4 (7) The collector current I _2a of the collector current I ₂ and the transistor Q _2a of the transistor Q ₂ , The following equation (8) is established.

[0077] _{I 2a = I 2 · (n} 2 / m 2) = I 2 / N 2 = I B1 / N 2 (8) The collector current I _3a of the collector current I ₃ and the transistor Q _3a of the transistor Q ₃ , The following equation (9) is established.

I _3a = I ₃ · (n ₃ / m ₃ ) = I ₃ / N ₃ = I _2a / N ₃ (9) The diode D ₁ is off and the current I _D1 of the diode D ₁ =
If 0, the current I _{3a a} = I _4a . At this time, the expression
From (6) ~ (9), hfe 1 of the transistor Q ₁ is represented by the following
It can be expressed by equation (10).

[0079] _{I C1 / I B1 = (N} 1 · N 4) / (N 2 · N 3) = hfe 1L (10) hfe 1 of the transistor Q ₁ is reduced, when it becomes a hfe _1L, less Thus, the diode D ₁ is switched from off to on.

When the diode D ₁ is off, the cathode voltage of the diode D ₁ is approximately V _Z. When hfe ₁ decreases and becomes about hfe _1L or less, the base current I _B1 increases and the collector current I _{3a of the} transistor Q _3a increases.
Will try to increase. As a result, the collector voltage of the transistor Q _3a rises to the voltage of V _Z + V _F or more, and the diode D ₁ is turned on.

In this way, when the diode D ₁ is turned on, the current I _D1 (drive current control current) flows from the collector of the transistor Q _3a to the connection point of the resistors R ₁ and R ₂ , and the error amplifier A ₁ is turned off. The voltage at the inverting input terminal rises. This reduces the current flowing into the output of the error amplifier A ₁ , that is, the base current of the transistor Q ₃ .
Therefore, the collector current I _2a of the transistor Q _2a is reduced, Accordingly, the collector current I ₂ of transistor Q _2, i.e., the base current I _B1 of the transistor Q ₁ is reduced.

In this way, the transistor Q₁Is saturated
Enter the area and hfe₁Hfe_1L= (N₁・ N_Four) /
(N₂・ N₃), The diode D₁Electric
Flow I _D1Flows, transistor Q₁Base current of_B1But
Suppressed.

FIG. 2B shows the relationship between the input voltage V _in and the current I _D1 of the diode D ₁ . Note that the input voltage V _in has the same scale as that in FIG. Figure 2 (B)
As shown in, when the input voltage V _in is less than V _in0 , the transistor Q ₁ enters the saturation region, and when hfe ₁ drops to hfe _1L , the current I _D1 starts to flow in the diode D ₁ , and the voltage of V _in1 , The peak value has been reached.

The value of (N ₁ · N ₄ ) / (N ₂ · N ₃ ) is
It is set according to the hfe characteristic of the transistor Q ₁ . For example, when hfe _{1 in the} normal operation is about 100 to 200, for example, (N ₁ · N ₄ ) / (N ₂ · N ₃ ) = 40 is set, and when hfe ₁ drops to 40, the base The control circuit for the current I _B1 is set to start operating.

FIG. 3 shows the input voltage V in the circuit of FIG.
_in a diagram for explaining the relationship between the input current I _in. FIG. 3 shows the input voltage V when the load current of the load circuit is 0 (no load).
The relationship between _in and the input current I _in is shown.

[0086] In Figure 3, shows the input current I _in of the present embodiment by a solid line shows the input current I _in of the conventional circuit of FIG. 6 in broken lines.

In this embodiment, N₁・ N_FourValue of N₂
・ N₃You can set a large value for
Current I of control circuit_Four＋ I_3aThe value of can be reduced
It And the diode D₁Has a very small current I_D1
Error amplifier A ₁Change the output voltage of
And transistor Q₃Collector current I₃Control
Base current I_B1Can be sufficiently suppressed.

Therefore, in this embodiment, the base current I _B1 can be sufficiently suppressed and the base current I _B1 of the control circuit can be made smaller than that of the conventional circuit. Even in the region where the collector current I _C1 is small, the base current I
_B1 can be suppressed and hfe ₁ can be made larger than that of the conventional circuit. Therefore, the input current I _in when the output transistor Q ₁ is in the saturation region can be reduced as compared with the conventional circuit.

[0089] For example, in the case of the conventional circuit of FIG. 6, whereas the peak value of the input current I _in at no load is hundreds .mu.A, in this embodiment, the number of peak values of the input current I _in ten μA
Can be suppressed to

As a concrete example of the circuit of FIG. 1, when hfe ₁ of the transistor Q ₁ in the normal operation is about 100 to 200, m ₁ = 200, n ₁ = 1 (N ₁ = 200),
m ₄ = 3, n ₄ = 1 (N ₄ = 3), m ₂ = 16, n ₂ =
An example in which 1 (N ₂ = 16) and m ₃ = 1 and n ₃ = 1 (N ₃ = 1) are set will be described. In this case, hfe _1L =
(N ₁ · N ₄ ) / (N ₂ · N ₃ ) = 37.5 is set.

V _in is a specified output voltage V _OC (for example, 5 V)
Collector current I _C1 =
The following measured values are obtained under the condition of 5 mA.

I _B1 = I ₂ = 25 μA I _2a = 1.6 μA I ₃ = I _3a = I _4a = 1.8 μA I ₄ = 19 μA I _D1 = 0 μA Transistor Q ₁ hfe ₁ = 5 mA / 25 μA = 20
0 At this time, the diode D ₁₁ is off. Further, the transistor Q _4a is in a saturated state, and the value of I ₄ / I _4a is N.
_It is larger than ₄ .

V _in is less than or equal to the specified output voltage V _out ,
When the transistor Q ₁ is in the saturation region, the following measured value is obtained under the condition of collector current I _C1 = 5 mA.

[0094] _{I B1 = I 2 = 150μA I} 2a = 9μA I 3 = 10μA I 3a = 10.1μA I 4 = 30μA I 4a = 10μA I D1 = 0.1μA hfe of the transistor _{Q 1 1 = 5mA / 150μA =} 3
3.3 Thus, when the transistor Q ₁ is in the saturation region,
The base current I _B1 is suppressed by the current I _D1 flowing from the transistor Q _3a to the diode D ₁ .

In this embodiment, as in this example, the slave shown in FIG.
Compared to the conventional circuit, the base current I_B1Control circuit current I_Four
＋ I_3aThe value of can be reduced. And Daio
De D ₁Has a very small current I_D1Just pour
Current I_B1Can be sufficiently suppressed.

As a result, the base current I _B1 can be suppressed sufficiently, and the measured value of hfe ₁ is the theoretical value (N ₁ ·
The value can be made quite close to N ₄ ) / (N ₂ · N ₃ ) = 37.5.

[0097] Also, even when the collector current I _C1 is reduced, there ability to control the base current I _B1 is sufficiently, in the case of no load, it is possible to suppress the base current I _B1 of about 10 .mu.A, the base current I _B1 The value of the current I ₄ + I _3a of the control circuit can be reduced to about several μA.

FIG. 4 is a circuit diagram of the constant voltage circuit 11 _{A according to} the second embodiment of the present invention. 4, the same components as those in FIG. 1 are designated by the same reference numerals, and the description thereof will be appropriately omitted.

The second embodiment differs from the circuit of the first embodiment of FIG. 1 only in that a resistor R ₄ is provided between the emitter of the transistor Q _2a and the ground.

FIG. 5 is an explanatory view of the relationship between the current I _C1 and hfe ₁ when the transistor Q ₁ is in the saturation region.
In FIG. 5, the broken line shows the case of the first embodiment, and the solid line shows the case of the second embodiment.

Collector current I _C1 of output transistor Q ₁
When increases, the current I _B1, I ₂ is increased, the resistance R _4, increases the emitter voltage of the transistor Q _2a is, the base-emitter voltage V _BE of the transistor Q _2a is gradually compressed. This substantially increases the emitter area ratio N ₂ = m ₂ / n ₂ and hfe _1L = (N ₁ ·
The value of N ₄ ) / (N ₂ · N ₃ ) is suppressed. Therefore, h
As shown by the solid line in FIG. 5, the value of fe ₁ becomes a substantially constant value when I _C1 exceeds a certain level.

Therefore, in the second embodiment, when the transistor Q ₁ is in the saturation region, the change in the current of the diode D ₁ due to the change in the load current can be reduced. This allows
The circuit design can be made easier.

[0103]

As described above, according to the invention of claim 1,
Since the current required to control the drive current of the output transistor can be made small and the drive current can be sufficiently suppressed, the input current of the circuit when the output transistor is in the saturation region can be made smaller than that of the conventional circuit. You can

According to the invention of claim 2, a constant voltage circuit capable of suppressing the drive current with a small input current when the output transistor is in the saturation region can be realized with a simple circuit configuration.

According to the third aspect of the invention, the change in the current amplification factor of the output transistor when the output transistor is in the saturation region can be reduced, and the circuit design can be made easier.

[Brief description of drawings]

FIG. 1 is a circuit diagram of a constant voltage circuit according to a first embodiment of the present invention.

FIG. 2 is an explanatory diagram of a relationship between an input voltage, an hfe of an output transistor, and a diode current in the circuit of FIG.

3 is an explanatory diagram of a relationship between an input voltage and an input current in the circuit of FIG.

FIG. 4 is a circuit diagram of a constant voltage circuit according to a second embodiment of the present invention.

FIG. 5 is an explanatory diagram of a relationship between a collector current of an output transistor and hfe.

FIG. 6 is a circuit diagram of a conventional constant voltage circuit.

7 is an explanatory diagram of the relationship between the input voltage, the hfe of the output transistor, and the diode current in the circuit of FIG.

FIG. 8 is an explanatory diagram of a relationship between an input voltage and an input current in a conventional circuit.

9 is an explanatory diagram of a relationship between a collector current of an output transistor, hfe, and a diode current in the circuit of FIG.

[Explanation of symbols]

11, 11 _A constant voltage circuit Q ₁ output transistor Q _1a first transistor Q ₄ , Q _4a current mirror circuit Q ₂ , Q _2a current mirror circuit Q ₃ second transistor Q _3a third transistor D ₁ diode A ₁ Error amplifier

Claims (3)

[Claims] 1. An output transistor for supplying an output voltage to a load.
Output signal corresponding to the difference between the output voltage and the specified voltage.
Error amplifier for generating a signal and an output signal of the error amplifier
Drive voltage on the base side of the output transistor
Voltage circuit having drive current supply circuit for supplying current
In, the output transistor is commonly connected to the base and the emitter.
And the emitter area ratio with the output transistor
Is N₁A first transistor set to: 1 and an input-side transistor connected to the collector of the first transistor.
The input side transistor is connected.
The emitter area ratio of the output transistor to the output transistor is N _Four: 1
The set first current mirror circuit and the error amplification circuit provided in the drive current supply circuit.
Of the output transistor according to the output signal of the
1 / N of current₂A second transistor that produces double the collector current
The transistor and the second transistor are commonly connected to the base and the emitter.
And the collector is the first current mirror circuit.
Connected to the collector of the output transistor of the path,
The emitter area ratio with the second transistor is N₃: 1
And a third transistor set to the output side, and an output side transistor of the first current mirror circuit.
The connection point of the third transistor and the input of the error amplifier
Output transistor connected between the output terminal and
Is in the saturation region, the input terminal of the error amplifier
With a switching element that supplies drive current control current
A constant voltage circuit characterized in that 2. The drive current supply circuit includes a second current mirror circuit in which an emitter side area ratio of an output side transistor for supplying a base side drive current of the output transistor and an input side transistor is set to N ₂ : 1. The collector of the input side transistor of the second current mirror circuit is connected to the collector of the second current mirror circuit, and a collector current 1 / N ₂ times the drive current of the output transistor is generated according to the output signal of the error amplifier. The constant voltage circuit according to claim 1, wherein the constant voltage circuit comprises a second transistor. 3. The constant voltage circuit according to claim 2, wherein a resistor is provided between the emitter of the input side transistor of the second current mirror circuit and the power supply terminal.