JP2663449B2 - Constant current circuit - Google Patents

Description

Description: FIELD OF THE INVENTION The present invention relates to integrated electronic circuits, and more particularly to constant current circuits. [Prior Art] Conventionally, a circuit called a Widlar constant current circuit as shown in FIG. 6 has been proposed as a constant current circuit in an integrated electronic circuit. This circuit is composed of NPN transistors Q33 and Q
In the current source circuit of 34, the collector current of the transistor Q33 is I ₁ , and the collector current of the transistor Q34 is I
Assuming that ₂ , the base-emitter voltage V _BE is expressed by the following equation. _{V BE (Q33) = V T} · ln (I 1 / I S1) V BE (Q34) = V T · ln (I 2 / I S2) ... (1) where, I _S1, I _S2 is the saturation current, V _T is (26 mV at 300 ° K) thermal voltage, as the current amplification factor (h _FE) is sufficiently high to ignore the I _B. When the emitter area of the transistor Q33 is two times the emitter area of the transistor _{Q34, I S1 = 2I S2 ...} (2) , and the potential V _R of the resistor R31 is _{V R = V BE (Q34)} -V BE (Q33) = V _T ｛ln (I ₂ / I _S2 ) −ln (I ₁ / I _S1 )｝ = V _T · ln (2I ₂ / I ₁ ) (3) I ₁ and I ₂ are PNP transistors Q31 and Q32. Since I ₁ = I ₂ ... (4) by the current mirror circuit, I ₂ = V _R / R ₃₁ = V _T / R ₃₁ .ln2 = I ₀₁ = I ₀₂ (5), and the output current I _{For 01} and I ₀₂ , the same value is obtained as the out current and the sink current, and since V _T and R ₃₁ are constants, a constant current is obtained. Note that _IST is a start-up current. Output current of the Widlar constant current circuit as described above, by only the resistor R _31, is determined, it can be seen that the absolute value variation, extremely effective circuit to the power supply voltage variation of the V _BE. [Problems to be Solved by the Invention] The above-described conventional constant current circuit is formed by ignoring the base current, that is, assuming that the current amplification factor (h _FE ) is large. By the way, in the integrated electronic circuit, the NPN transistor can control the current amplification factor _hFE relatively easily, and can set the center value at around 200. However, at present, it is difficult to control the PNP transistor because it is a lateral element using the surface, and it is difficult to increase the current amplification factor _hFE . Therefore, for example, the base current of an NPN transistor whose current amplification factor h _FE is at least about 10 to 15 can be neglected because h _FE can be made high, but the base current of a PNP transistor cannot be made high because h _FE cannot be made high. become unable. In that case, Widlar constant current circuit in FIG. 6 will become the I ₁ and I ₂ are different, there is a problem becomes an output current is large variations in it. That is, when the current amplification factor of the PNP transistor and beta _P, I ₁ is the sum of the collector current and Q31, Q32, Q35 base current of Q31. Here, the sum of the base currents of Q31, Q32, and Q35 is n
If I (n is the sum of the emitter area to Q31) ₂ / β _{P and, I 1 = I 2 + nI} 2 / β = (1 + n / β P) I 2 ... (6) next, I ₁ is (3), (6) from the _{equation, I 1 = V R / R} 31 = V T / R 31 · ln (2I 2 / I 1) = V T / R 31 · ln {2 / (1 + n / β P)} ... (7 ) And I ₂ (6) and (7), I ₂ = 1 / (1 + n / β _P ) · I ₁ = V _T / ｛(1 + n / β _P ) R ₃₁ ｝ · ln ｛2 / ( 1 + n / β _P )｝
... (8) Here, 1 + n / β _P = x and differentiated by x. ∂I ₂ / ∂x = −V _T / x ² R · ln (2 / x) −V _T / xR · 1 / x = −V _T / x ² R · ｛(1 + ln (2 / x)｝ V _T / x ² R is positive at any x and 1 + ln2 / x is x
Is as follows. x <2e Positive x = 2e 0 x> 2e Negative 1 + n / β _P ≧ 2e, then n / β _P ≧ 4.42. For example, if β _P = 5 and n> 22, this does not occur, and in actual use, This is unlikely to occur and 1 + ln2 / x is also considered positive. Therefore, ∂I ₂ / ∂x <0. This indicates that when β _P decreases, x increases, and that β ₂ decreases monotonically with the decrease of β _P. Further, since I ₂ = I ₀₁ = I ₀₂ , it can be seen that the output current also decreases as a result. That is, the output current is greatly affected by the current amplification factor of the PNP transistor, and this is a serious problem particularly in an integrated electronic circuit having a large variation in the absolute value of the current amplification factor. An object of the present invention is to provide a constant current circuit having an integrated electronic configuration with a small output current change. [Means for Solving the Problems] A constant current circuit according to the present invention includes a current source circuit including first and second transistors and a resistor, and a third current source circuit connected to the current source circuit and having bases connected to each other.
And a means for generating first and second currents having a ratio equal to the collector current ratios of the first and second transistors, in a Widlar constant current circuit formed by a current mirror circuit including a fourth transistor and a fourth transistor. A second means for converting the first and second current differences into a voltage difference, a transconductance amplifier for generating an output current corresponding to the voltage difference and having an output terminal connected to the bases of the third and fourth transistors; And a negative feedback configuration including Embodiment Next, the present invention will be described with reference to the drawings. FIG. 1 is a circuit diagram of an embodiment of the present invention, in which Q11 and Q12 are PNP transistors, Q13 to Q16 are NPN transistors, and TCA is a transconductance amplifier. R11 to R14 are resistors, 101 is a power supply, and 102 is a constant current source. Said transistor Q11~Q14 and a resistor R11 constitute a Widlar constant current circuit, the startup current I _ST is passing flows to the constant current source 102. Also, an NPN transistor Q15 having twice the emitter area,
The bases of the one-time NPN transistor Q16 are connected, and the emitter of the former transistor Q15 is grounded through a resistor R12 having the same value as the resistance of the Widlar constant current circuit. The collectors of the transistors Q15 and Q16 are connected to the power supply 101 through resistors R13 and R14, respectively, and each of them is directly connected to the input of the transconductance amplifier TCA. The output of the transconductance amplifier TCA is connected to the bases of the PNP transistors Q11 and Q12 of the Widlar circuit, in other words, the collector of the NPN transistor Q13 whose emitter area is doubled. According to this circuit configuration, the reduction of the current amplification factor h _FE of the PNP transistor Q11, Q12, a current error of each collector current I _1, I ₂ are transistors Q15, Q16 and the resistor R12 collector currents I _1, I ₂ and The same currents I ₁ ′ and I ₂ ′ are generated and the resistances R13 and R14
It is converted into a voltage error [Delta] V _d by. Then, current I _F of the voltage error [Delta] V _d transconductance amplifier TCA
Into the transistor Q11 of the Widlar constant current circuit,
By adding to the collector connection point of Q13, a negative feedback loop is formed. That, (I _B current is high) → (I ₁ -I ₂ current _{Sadai) → (I 1 '-I 2} ' current Sadai) → ([Delta] V _d Univ) → (I _F Univ.)
→ This is a loop of (I ₁ −I ₂ current difference small). This negative feedback loop makes it possible to keep the output current change during the current amplification factor _hFE small. Next, this will be described using equations. The currents I ₁ and I ₂ is given by equation (6), I ₁ = I ₁ because it is _{', I 2 = I 2'} , I 1 '= (1 + n / β P) · I 2' ... (6 '), And I ₂ is given by I ₂ ′ = 1 / (1 + n / β _P ) · I ₁ ′ = V _T / (1 + n / β _P ) R ₁₁ · ln ｛2 / (1 + n / β _P) )｝ (8 ′) Therefore, the error current ΔI is given by ΔI = I ₁ ′ −I ₂ ′ = (1 + n / β _P ) · I ₂ ′ −I ₂ ′ = n / β _P · I ₂ ′ = nV _T / (β _P + n) R ₁₁ · ln {2 / (1 + n / β P)} ... (11) , and the error voltage [Delta] V _{d is, ΔV d = ΔI · R 14} = R 14 nV T / R 11 (β P + n) · ln {2 / ( 1 + n / β _P )｝ (12) Here, there is a circuit shown in FIG. 2 as an example of the transconductance amplifier TCA. In the figure, Q12 and Q22 are N
The PN transistors Q23 and Q24 are PNP transistors and are configured as a differential amplifier. 201 is a voltage source, 202
Is a constant current source. The input voltage V _d of this transconductance amplifier TCA
Output current I _F characteristics for is known to be given by the following equation. I _F = α _F I _EE tan h (V _d / 2V _T ) (13) where α _F is a common base current amplification factor, and I _EE is a differential pair emitter current. (13) By substituting (12) into equation, the output current I _F of the transconductance amplifier is as follows. I _F = α _F I _EE · tan h [{R ₁₄ h / 2R ₁₁ (β _P + n) · ln ｛2 / (1 + n / β _P )}] (14) That is, feedback as in equation (14) A current flows and negative feedback is applied. From the above results, FIG. 3 shows a configuration using the transconductance amplifier circuit of FIG. 2 for the transconductance amplifier TCA of FIG. In this figure, parts corresponding to FIGS. 1 and 2 are denoted by the same reference numerals, and in FIG. 3, Q25 is a PNP transistor. FIG. 4 shows the result of a simulation performed using SPICE-F in this circuit. In this figure, n = 1
It shows a change in the output current _I01 when the PNP transistor _hFE is changed under the conditions of 0, V _CC = 1.05V, I _EE = 20 μA, and α _F = 1. As is clear from this, h of the PNP transistor
_It can be seen that an extremely stable output current can be obtained with respect to the _FE change as compared with the Widlar constant current honeydew. For example, with respect to h _FE of the PNP transistor is 15～30～60 and the center value, 1/2-fold, when changed twice, Widlar a constant current circuit, though there is a change in -25% ~ + 28% for,
In the circuit of the present invention, a change of -2% to + 1% is suppressed. Also, compared to n = 3 and n = 10, 30%
It can be seen that the change of the ratio is suppressed to 3% in the circuit of the present invention. FIG. 5 shows another embodiment of the present invention, and the same or equivalent parts as those in FIG. 1 are denoted by the same reference numerals. Also in this embodiment, like the embodiment of Figure 1 converts the error of I ₁ and I ₂ into a voltage, transconductance amplifier TCA
A similar result can be obtained by applying current feedback. The present invention is also effective when a large number of PNP transistors are connected in parallel to increase the base current and decrease the output current. As it has been described [Effect of the Invention The present invention is of the PNP transistor h _FE
In Widlar constant current circuit such that I _B decreases increased output current by a change, which was converted into an error voltage by detecting the error current generated by I _B in Widlar constant current circuit, the transconductance amplifier this voltage And the output is fed back to the Widlar constant current circuit, so that the output current change can be reduced.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a diagram of a constant current circuit according to an embodiment of the present invention, FIG. 2 is a circuit diagram of a transconductance amplifier, and FIG. 3 is a circuit diagram of FIG. The applied circuit diagram, FIG.
Characteristic diagram showing a simulation result using the circuit of the figure,
FIG. 5 is a circuit diagram of another embodiment of the present invention, and FIG. 6 is a circuit diagram of a conventional Widlar constant current source. Q11, Q12: PNP transistor, Q13 to Q16: NPN transistor, Q21, Q22: NPN transistor, Q23, Q24, Q25 ...
… PNP transistors, Q31, Q32, Q35 …… PNP transistors, Q33, Q34, Q36 …… NPN transistors, R11-R14, R31…
... resistors, 101,201,301 ... power, 102,202,302 ... constant current sources, TCA ... transconductance amplifiers.

Claims (1)

(57) [Claims] In a Widlar constant current circuit including a current source circuit including first and second transistors and a resistor, and a current mirror circuit including third and fourth transistors connected to the current source circuit and having their bases connected to each other. , The first and second transistors having a ratio equal to the collector current ratio of the first and second transistors.
Means for generating a current, a second means for converting the first and second current differences into a voltage difference, and an output terminal for generating an output current corresponding to the voltage difference, and the output end of which generates the third and fourth currents. A constant current circuit comprising: a transconductance amplifier connected to a base of the transistor. 2. 2. The constant current circuit according to claim 1, wherein the first and second transistors have their bases connected to each other. 3. 2. The constant current circuit according to claim 1, wherein the first transistor has a resistor connected between the base and the collector, and the second transistor has a base connected to the collector of the first transistor.