JPS5922112A - Current source circuit - Google Patents

Abstract

PURPOSE:To realize high precision, stability, and small power consumption by obtaining a current whose value is determined by only the values of a resistance and a positive source voltage through single conversion from a current source circuit which has the positive power source and a negative power source. CONSTITUTION:The bases and collectors of diode-connected NPN and transistors (TR) Q6 and Q5 are connected in common respectively to constitute a current mirror circuit, so the collector currents of the TRs Q5 and Q6 are equal and currents flowing through NPNTRs Q1 and Q2 are also equal. Since the base of the TRQ1 is connected, V1=V2=0, where V2 is the base potential of the TRQ1 and V1 is the potential at the base (collector) point A of the TRQ2. Therefore, a standard current Iref is determined only by the values of the voltage VCC of the positive power source 1 and the resistance R. A required output current Iout is generated by the single conversion of the current mirror circuit of TRs Q7 and Q8, and its value depens on the emitter area ratio of the TRs Q7 and Q8, the voltage of the positive power source VCC, and resistance R.

Description

[Detailed Description of the Invention] [Technical Field of the Invention] The present invention relates to a current source circuit having both positive and negative power supplies. An example of a conventional current source circuit in which the current value is determined by a power source is shown in FIG.

In this current source circuit, one reference current is created between the positive power supply l and the ground point O, and a diode-connected PNP
Another PNPI with transistor Qa and base connected
- A current is passed between the positive power supply 1 and the negative power supply 2 by the current mirror circuit constituted by the transistor QA, and then the necessary current is generated by the diode-connected NPN transistor Qn and the NPN transistor Qc whose bases are connected together. The circuit configuration is to take out Iout.

In this conventional circuit, a PNP transistor must be used, and this PNP transistor is used to determine the value of the reference current.
) Characteristics of transistors were often a problem.

The biggest problem is that because the β of the PNP transistor is low, the influence of the base current is large, making it impossible to obtain the designed current value.

Furthermore, when determining the value of the reference current, the base-emitter voltage VBE of the PNP transistor must be taken into consideration in addition to the values of the electric current @ l and the resistor R, which reduces the degree of freedom in designing the reference current. There is a drawback.

Furthermore, the reference current is obtained as described above, other currents are caused to flow through the current mirror circuit, and further the desired current Io is obtained.
Since conversion is performed twice to obtain ut, there is a disadvantage that not only conversion errors are added and become large, but also power consumption increases.

[Purpose of the invention]

The present invention has been made to solve the above-mentioned drawbacks.
A resistor and a positive power supply ~ A current source whose value is determined only by the ccO value can be created with a single conversion, and is stable with lower power consumption than conventional methods. The purpose is to provide a circuit.

[Summary of the invention]

The present invention provides a current source that can determine a desired current Iout regardless of the negative power source -VBB by providing a reference point with zero potential and connecting an arbitrary element to this reference point and 101 of the positive power source Vcc. It is a food circuit.

[Embodiments of the invention]

Hereinafter, embodiments of the present invention will be described in detail with reference to the drawings. FIG. 2 shows the basic configuration of the circuit of the present invention. The collectors of an NPN transistor Q1 whose base is grounded and a diode-connected NPN transistor Q2 are connected to each other. Diode connected PIN
The emitter of the P transistor Q is connected to the emitter of the C transistor, and similarly the emitter of the P N I) transistor Q4 and the emitter of the P transistor Ql are connected. Furthermore, the bases of Ql and Q4 are connected to each other. The collector Q of the NPN transistor Q (connected to the collector of i3, diode-connected,
The collector of transistor Qa is connected to the collector of transistor Q. Furthermore, the bases of Ql and Q6 are connected together, and the bases thereof are connected together. The collector of the diode-connected NPN transistor Q8 is l! IJ iselfQ
, and the emitter of Q8, and the emitter is connected to the negative power supply 2. NPN l-transistor Q
The emitter of 7 is also connected to the negative power supply 2, and the bases of Q7 and (λ8 are connected to each other. Among the transistors connected as above, the collectors of NPN l-transistor Q and Q2 are Resistor R [connected to the more positive power supply ■.

The above is the basic configuration of the present invention. Here, NPN transistor Q, and Qt-Qs and Q, , PNP) transistor Q,
and Q4 are transistors of the same shape designed with bareness in mind.

Next, the operation of the current source circuit of the present invention will be explained while ignoring the base current. The diode-connected NPN transistor Q6 and the NPN transistor Q form a current mirror with a common base and common emitter, so the collector currents of Q and q are equal. Therefore, PNP transistors Q3 and q
The currents flowing through the NPN transistors Q1 and Q,t- are also equal.

Next, the base potential v(2) of the NPN transistor Q1 and the NPN transistor Q! The relationship between the potential v(1) at the base (collector) point A of is expressed by the following equation.

'V(2)-YBE(Ql)-VBK(Q3)=
V(1)-VBE(Q2)-VBE(Q4)-
(For example, Vaic(qt) represents the voltage drop between the base and emitter of the transistor Q.

is a function of temperature only, Is is the reverse saturation current and is determined by the transistor geometry and manufacturing process, and IC is the current flowing through the collector of the transistor.

KNPN l-transistors Q1 and Q, as mentioned before.

The currents flowing through transistors Q and Q4 are equal, and the currents flowing through transistors Qs and Q4. The shape of the NPN transistor Q1 and the dovetail are the same, and the operating temperatures are the same.VBE(Ql) =VBE(Q2), vBx(qa
)=VB++: (q4) Tearu. Yotte (1) formula YoI
J V (1) = V (2) theal, V (2) if N
P N is the base potential of transistor Q, and since the base of Ql is grounded as shown in Figure 2, V(1) =
V(2)=O, and the point A becomes the ground potential, that is, OV.

Therefore, the current Iref flowing through the resistor is. In this way, in the example of FIG. 2, the standard current Iref can be determined only by the voltage Vcc of the positive power supply 1 and the value of the resistor R.

By utilizing the fact that point A becomes Ov, the reference current can be designed freely without any special restrictions. NPN)
The current flowing through the transistor Q3 is the current I flowing through the transistor Q according to the law of conservation of current.

and the current I flowing through the transistor Q, I
, == I, -1-1, = Iref. The required output current Iout f, the emitter surface 4X of the transistor Q (Q?) and the emitter area A (Q) of the transistor Q8 are expressed as follows.

The current required in this way can be created by converting the standard 1st current 1ref once using a current mirror composed of transistors Q and Q8, and its value is Q, and Q.
It is determined only by the emitter area ratio of , the voltage of the positive t source Vcc, and the value of the resistance. By arranging the emitter area ratio in consideration of pairability, the error from the set value can be considerably reduced.

〔Effect of the invention〕

This circuit also uses one PNP I-transistor, but this is different from the conventional usage. Conventionally, a PNP transistor was used as a current mirror.

Since β was low, the influence of the current pairing caused by the pace current was large. In this circuit, the PNP transistor is used for the purpose of equalizing the base potential of transistor Q and the base potential of transistor Q2. In other words, what is essentially required of the PNP transistor in the circuit of the present invention is VBE (Q3) and VBK (Q4).
are equal. As stated in c above, VBE = VTJn-, so
The influence of the Sumy flow is reduced by the function of ln, and VBK becomes approximately equal. Therefore, the conventional P N l)
The error from the threatened design value due to the use of different transistors is significantly improved.

In Fig. 2, the flowing current 1d (Ire (and Iou
t, the circuit has the advantage of lower power consumption than the conventional circuit shown in FIG.

As described above, the circuit of the present invention is a highly accurate, stable, and power-saving current source circuit.

[Other embodiments of the invention]

The current source circuit of this invention basically has the configuration shown in FIG. 2, but when it is actually used, a starter circuit is added as shown in FIG. 3. For example, the starter circuit
As shown in the figure, the resistor Rs and the N P N transistor Qs
+1 diode-connected N P N transistor Q
s2. It is composed of Qs3.

There are two stable states for the current source in the m2 diagram.

It is. Therefore, let a current flow in advance 1ref=
It is necessary to draw out the voltage from the stable state of 0, so that it becomes 1rcf-Vcc/R. This is done by the starter circuit.

Due to the scooter circuit, a current starts flowing through the current source, and Ire
When transitioning from f=o to Iref = Vcc/R, transistors Q and Q are diode-connected, so the emitter potential Ve (Qs+) of transistor Qs+
is Ve (Qs+) = VBz(Q,) + va
E (Qs) (2). Also, the base potential vb (QSI) of the transistor Qs+ is Vb (Q81) = VBE (Q!!2) + V
Bg (Qsa) (3). Equation (2), (
3) From the formula, Ve(Qsl)2Vb(Qsl) and f, K
Therefore, the transistor QSI is turned off, and the starter circuit is disconnected from the current source circuit. Therefore, the starter circuit essentially has no influence on the current source circuit7. do not have.

Now, there are some applications of non-invention as shown below.

(1) It is desired to increase the ratio between the standard current Lref and the output current 1out.As shown in FIG.

A resistor I is connected between the emitter of the transistor Q and the negative power supply.
Connect your device. 1out at this time is as follows.

Here, the emitter area of transistor Q7 and transistor Q is 1-1. Equal to the current ratio, considering the pair property,
It needs to be shaped. Also, equal resistances Rs, l(
The accuracy of the current mirror can also be improved by inserting 4, R, , r-.

(2) Next, negative i' is 1 determined by the source −Vcc
The circuit of the current source can be considered as shown in FIG.

The circuit configuration has a configuration in which a PNP transistor and an NPN transistor are replaced symmetrically with the basic configuration shown in FIG. (C is effective when there are restrictions on the negative m source and sufficient power supply voltage (-3V or less) cannot be used.) (3) For normal use, create a sufficient output current using the basic configuration shown in Figure 2. I can do it. However, if even more precision is required, by inserting a PNP transistor Qi between the NPN transistor Q4 and the NPN transistor q as shown in FIG.
It is possible to create a highly accurate current source by reducing the influence of

When transistor Q1 is not present, transistors Q and Q4 form a circuit generally called a Wldlar current source, and the collector current IC flowing through Qa
(Q, ) and transistor q, and in a transistor with low β such as a PNP transistor, the pairability between Ic((υ) and IC(Qs) is poor. When transistor Qi is inserted, transistors Qs and Qi 1 Qi
t/iWi 18On (constitutes a circuit called a D current source, and the pairability of IC(Qi) and 1c(Q,) can be improved by having the square term of nβ of IC(Qa) and IC(Ql) .

As described above, by adding the PNP transistor Qi, a current source with even higher accuracy can be constructed.

(4) By using this current source in a differential amplifier as shown in Figure 7 and determining the circuit constants as described below, the output DC is almost unaffected by power supply fluctuations and operating temperature changes.
It is possible to realize a circuit that can maintain approximately 1u equal to the reference voltage Vref (C).

If you want to set the output DC level to Ov with Vref = o, the current Ia flowing through the resistor hole and the current Ib flowing through the resistor Rdt of the differential amplifier are calculated as follows: RIa, magnetic pressure drop at resistor Rdt &bIb are equal. Further, the ratio of the emitter area A (Qll) of the transistor Q, □ to the emitter area A (Qi2) of the transistor Q+2 is designed as shown in the following formula.

A(Ql2) Ia Therefore transistor Q1. between base and emitter′
voltage VBE ((↓1) and the pace of transistor Q12.
The emitter voltages VBnCQA2) are equal. Using this relationship, the potential V (d) at the output end is v (d) =
VCc −Rd I b −VBE (Qll)=
Vcc - 几Ia - YBE (Qi2) - V(a) = O. Even if the power supply fluctuates here, Ia.

Although the absolute values of Ib and Ic change, the ratio of each is always constant tζ. This means that when the operating degree changes, the resistance value maintains pairing, and even when the absolute value changes, the absolute value of each current changes, but the ratio remains constant. Therefore, the DCz level of the output is always Ov with respect to changes in current fluctuation operating temperature and changes in resistance value.

Here resistance R, )1. dz, E-down 2, 1 and L are arranged considering pairability. Transistors (I, I and Q+2) are also arranged with consideration given to their pairability.

As described above, the present invention has a wide range of applications.

[Brief explanation of drawings]

Figure 1 shows an example of a conventional current source that has been used in a circuit with positive and negative side power supplies. It is. 4 to 7 are circuit diagrams showing application examples of the present invention. Ql ~QHI Qa~Qit QA-QD IQS
I~Qss-h transistor, R+, R2, R,
,R,,R,,几,,RA,几s,Rd+,)tdz,
Chill... Resistance 4-Vcc, -vB B...Power supply voltage 1
, 2...Power supply Vref...Reference voltage

Claims (1)

[Claims] Current with two sources, positive and negative? In path 1a, a base-grounded rpJl NPN transistor whose collector is connected to the positive power supply via an arbitrary element, and a second ON transistor whose collector and base are crab-connected to the collector of this first NPN l transistor. l)
N) A transistor, a third PNP transistor whose base plate is connected to the emitter of the NPN transistor of C41, and whose base wall is connected to the base and collector of the third PNP transistor and whose emitter of the second NPN transistor is connected to the emitter of the NPN transistor of C41. 'The fourth part whose emitter is connected to the ivy
and a fourth PNP I-transistor whose collector is connected to the collector of the third PNP I-transistor.
A fifth ONPN transistor whose base is connected to the collector of the I- transistor and which is connected to the G transistor;
a sixth ONPN transistor having its base and collector connected to the transistor pace and having its emitter connected to the emitter of the fifth NPN transistor;
The fifth and sixth ONPN transistors have the negative power supply connected to the emitters of the transistors via a current mirror circuit, and utilize the fact that the collector potential of the first NPN transistor is Ov to connect to the positive power supply. 1. A current source circuit characterized in that a current proportional to the current determined by the current mirror element is obtained as the output of the current mirror circuit.