CN102455727B - Current control circuit with rang of 100pA-1muA - Google Patents

Abstract

The invention provides a current control circuit which comprises a first power transistor, a second power transistor and a controller, wherein the first power transistor and the second power transistor are connected between a first power source and the output end of the current control circuit, and the width-to-length ratio of the first power transistor is smaller than that of the second power transistor; the controller responds to a received enabling signal, so that the first power transistor and the second power transistor are conducted in sequence; when receiving the enabling signal, the controller conducts the first power transistor, so that power supplied from the first power source is supplied to the output end through the first power transistor; and when the voltage level of the output end is equal to the level of the preset output voltage of the current control circuit, the controller conducts the second power transistor, so that power supplied from the first power source is supplied to the output end through the second power transistor which operates in a saturation region. Therefore, the current control circuit can perform the function of soft startup, and restricts the inrush current.

Description

The weak current source of 100pA-1 μ A range

Technical field

The present invention relates to a kind of numerical control wide-range current source, particularly a kind of faint voltage-controlled current source at 100pA-1 μ A range.

Background technology

The weak current source is crucial instrument in many research and development and product measurement application, especially has a wide range of applications at aspects such as semiconductor, nanometer technology and superconductors.Current source mainly contains three kinds of condenser types, resistance-type, ionization type, for the resistance type current source, although the circuit form of current source is various, but its circuit thought is substantially the same, pass through resistance, adopt voltage-current transformation that the voltage signal of standard is transformed into it proportional current signal, in the situation that input voltage is constant, the electric current that conversion obtains is also constant.This mode is simple, and it is convenient to realize, but since in the application circuit of constant current source the impedance of load uncertain, and load is very large on the impact of constant current source, particularly in the high occasion of precision, this problem is more outstanding.

In order to address this problem, a kind of technical scheme based on feedback thought has appearred, and shown in Figure 1 namely is a kind of this type of constant-current source circuit, reference voltage V _RefSend into the in the same way input end of operational amplifier, the output termination load of amplifier, current sampling resistor R is voltage with the current conversion of output terminal, the reverse input end that enters amplifier consists of negative feedback.If operational amplifier is individual desirable, its open-loop gain is enough large, i.e. the current potential approximately equal of two input ends.Establish simultaneously the input base current I of operational amplifier _B=0; Then by the character of operational amplifier, can get

V _ref=V _out=I _load·R （1）

Namely at V _RefIn the constant situation, flow through load R _LoadElectric current I _LoadRelevant with R, thus realized the constant output of electric current.

Yet there are two problems in constant-current source circuit shown in Figure 1:

(1) bias current of amplifier reverse input end and offset voltage can affect the precision of circuit, especially affect very large for microampere order and following circuit;

(2) in this circuit, load is not altogether.Owing to having sealed in resistance at ground wire, the electric current of inflow place level will produce voltage at sample resistance, and this voltage will appear at the form of ground level noise on all ground nodes of system, will have a strong impact on like this precision of circuit.

Therefore, the output current of this class constant current source can not be used for carrying out the output of weak current more than the mA level.

Summary of the invention

Technical matters to be solved by this invention is for the defective of above-mentioned background technology and deficiency, a kind of weak current source of 100pA-1 μ A range is provided, it can solve the shortcoming of existing current source circuit, eliminate load to the impact of output current, overcome the existing little shortcoming of current source output current scope, realize the adjustable weak current output of electric current 100pA-1 μ A range.

The present invention is for solving the problems of the technologies described above, and the technical scheme of employing is:

A kind of weak current source of 100pA-1 μ A range, comprise control module, display unit, input block, D/A converting unit, reference voltage source, differential motion amplifying unit, voltage current transformation unit and negative feedback unit, wherein, the input end of control module connects the input block that is used for input preset current value, output terminal connects display unit, and the preset current value is sent into display unit shows; The output terminal of control module also connects the input end of D/A converting unit, and the current value analog quantity that input block is inputted is converted to digital quantity, and sends into the D/A converting unit; The D/A converting unit will be converted to analog quantity from the preset current value digital quantity of control unit, send into the input end of differential motion amplifying unit, the input end of described D/A converting unit also connects reference voltage source, by the comparison of reference voltage source realization with the input quantity of control module; The output terminal of differential motion amplifying unit connects the input end of voltage current transformation unit, the input end of voltage current transformation unit is connection control unit also, realize multiple range current conversion by control voltage current transformation unit, and the output terminal of voltage current transformation unit connects the input end of differential motion amplifying unit via the negative feedback unit, and output voltage and preset voltage are superposeed.

After adopting such scheme, the present invention can eliminate load to the influence of peak current, and noise and leakage current are weakened the impact of output current in the circuit, and simultaneously, equivalent internal resistance can reach 2.5 * 10 ¹⁴Ω.

Description of drawings

Fig. 1 is the connection layout of existing constant-current source circuit;

Fig. 2 is integrated stand composition of the present invention;

Fig. 3 is the practical circuit figure of control module among the present invention;

Fig. 4 is the internal circuit diagram of weak current generation unit among the present invention;

Fig. 5 is the equivalent internal resistance circuit diagram of Fig. 4;

Fig. 6 be normal voltage when 1V, the corresponding relation figure of resistance and output current;

Actual and the theoretical graph of a relation of exporting of electric current when Fig. 7 is 1nA-10nA.

Embodiment

Below with reference to drawings and the specific embodiments, structure of the present invention and principle of work are elaborated.

As shown in Figure 2, the invention provides a kind of weak current source of 100pA-1 μ A range, comprise control module 1, display unit 2, input block 3, D/A converting unit 4, reference voltage source 5, differential motion amplifying unit 6, voltage current transformation unit 7 and negative feedback unit 8, the below introduces respectively.

In the present embodiment, control module 1 adopts single-chip microcomputer, adopt keyboard in its input end connection input block 3(the present embodiment), can be by input block 3 input preset current values, and control module 1 is converted to digital quantity with the preset current value by analog quantity and exports, and generates control command according to the preset current value; The output terminal of control module 1 connects display unit 2, can by these display unit 2 simultaneous displays, carry out reference for the operator by the preset current value of input block 3 inputs.

D/A converting unit 4 adopts the AD558 of direct voltage output types, and the output terminal of its input end connection control unit 1 receives the preset current value of digital quantity form, be converted to analog quantity after, send into again differential motion amplifying unit 6; The input end of D/A converting unit 4 also connects reference voltage source 5, utilizes this reference voltage source 5 to produce benchmark numerical value, and the current value that is used for transmitting with control module 1 compares.

The differential motion amplifying unit 6 of high common mode adopts dynamic auto school zero operational amplifier ICL7650, its output terminal connects the input end of voltage current transformation unit 7, to send into voltage current transformation unit 7 through amplifying the analog quantity of processing, carry out the electric current and voltage conversion by voltage current transformation unit 7 again, produce corresponding current signal.

The input end of voltage current transformation unit 7 is connection control unit 1 also, selected the range output of voltage current transformation unit 7 by control module 1 control, specifically, cooperate shown in Figure 4, control command according to control module 1, control electronic switch ADG508(also is the electronic switch S1-S5 among Fig. 4) select the high value resistor of voltage current transformation, realize the electric current output of different ranges; The output terminal of voltage current transformation unit 7 is used for carrying out constant current output on the one hand, also connects simultaneously the input end of negative feedback unit 8.

Negative feedback unit 8 is used for the voltage of sampled voltage current transformation unit 7 output terminals, and is superimposed upon the input end of the differential motion amplifying unit 6 of high common mode, thereby reaches the purpose of constant current.

Specifically, cooperate shown in Figure 3, it is the control module 1 that provides of the present embodiment and a circuit connection structure of D/A converting unit 4, voltage current transformation unit 7, input block 3 and display unit 2, its working method is the current value that control module 1 control inputs unit, 3 inputs will be exported, control simultaneously the demonstration of its current value, and the range of control D/A converting unit 4 and voltage current transformation unit 7.Thereby realize numerical control and the output of multiple range electric current of electric current.

Cooperate Fig. 4 that the generation principle of weak current is elaborated, below will the generation principle of weak current be elaborated, wherein, operational amplifier A 1, A2 consist of high common mode and input differential proportional amplifier, A3 is voltage follower, A4 is sign-changing amplifier, when load increases or be detected the offset voltage increase of instrument, output current produces voltage drop at output terminal and increases, the electric current of output reduces, and at this moment, testing circuit detects the voltage drop of output terminal, by the circuit that A3, A4 consist of, adjust output voltage V according to this voltage drop _Out, guarantee that the voltage on the precision resistor R equals reference voltage, i.e. V all the time _Out-V _Load=V _Ref, so that output current equals V _Ref/ R, thus reach the purpose of constant current.

Very faint because producing in the circuit, the circuit that consists of the voltage current transformation unit may have leakage current on circuit board, will produce error, adopts the method for teflon binding post in the present embodiment, and is floating empty.Because adopting high resistance resistor, its stability directly affects the performance of circuit.A key factor that affects high value resistor stability is ambient humidity, this is because malaria can form water membrane on the high resistance resistor surface, particularly high resistance resistor surface contaminated after, the dust that adheres to, salt grade the substance dissolves of easily ionizable in moisture film, and moisture film has certain conductive characteristic, form leakage current, thereby significantly reduce the resistance value of high resistance resistor, therefore adopt in the present embodiment the high resistance resistor of agent of low hygroscopicity insullac, thereby ambient humidity reduces obviously on the impact of resistance value; Reducing humidity is that high resistance resistor is sealed in the glass tube with vacuum on the another kind of method that high resistance resistor affects; the more difficult burn into oxidation of resistive film and damage; resistance value is more stable; vacuum tube internal leakage electric current can be ignored; simultaneously; must adopt electric leakage protection technology in the circuit; force exactly low-resistance node in the circuit and the equipotential a kind of technology of high resistant input node myopia; add that in the outside of high value resistor vacuum tube the voltage identical with high resistant two ends size forms equipotential; thereby do not produce potential difference (PD), produce leakage current.

Have thermonoise in the circuit in the high value resistor, the noise voltage that resistance produces is determined by following formula:

$E_t=\sqrt{4\mathrm{kTRB}}---\left(2\right)$

In the formula, k is Boltzmann constant, and T is absolute temperature, and B is bandwidth, and R is resistance.

For this reason at electric capacity of resistance two ends parallel connection, E _tBe the thermal noise voltage of resistance R, then the noise voltage at electric capacity two ends is

E _To=E _t/(1+j2πRC) （3）

Equivalent heat noise bandwidth is

B=1/4RC （4）

The thermonoise effective value that substitution gets the electric capacity two ends is

$E_t=\sqrt{\mathrm{kT}/C}---\left(5\right)$

The output effective value of the thermonoise of RC parallel circuit and resistance are irrelevant, only depend on electric capacity and absolute temperature, increase capacitance and can reduce thermal noise voltage, thereby reduce resistance to the impact of generation current.

As follows to the circuit analysis among Fig. 4, according to the character of operational amplifier, ignore operational amplifier to the impact of circuit, can get

$V_{\mathrm{out}}=\frac{R_5{R}_2}{R_3{R}_1}{V}_{\mathrm{ref}}+\frac{R_8}{R_7}\frac{R_5}{R_4}{V}_{\mathrm{load}}---\left(6\right)$

$V_{\mathrm{out}}-V_{\mathrm{load}}=\frac{R_5{R}_2}{R_3{R}_1}{V}_{\mathrm{ref}}+\frac{R_8}{R_7}\frac{R_5}{R_4}{V}_{\mathrm{load}}-V_{\mathrm{load}}---\left(7\right)$

$I_{\mathrm{out}}R=\frac{R_5{R}_2}{R_3{R}_1}{V}_{\mathrm{ref}}+\frac{R_8}{R_7}\frac{R_5}{R_4}{V}_{\mathrm{load}}-V_{\mathrm{load}}---\left(8\right)$

$I_{\mathrm{out}}=\frac{\frac{R_5{R}_2}{R_3{R}_1}{V}_{\mathrm{ref}}}{R+(\frac{R_8}{R_7}\frac{R_5}{R_4}-1)R_{\mathrm{load}}}---\left(9\right)$

From the visible output current of formula (9) and resistance R, R ₁, R ₂, R ₃, R ₄, R ₅, R ₇, R ₈, reference voltage source V _Ref, pull-up resistor R _LoadRelevant, therefore, as long as select suitable resistance and reference source, just can realize constant current.

Actual operational amplifier, because its input offset current and voltage are non-vanishing, the condition of zero output when not satisfying zero input.Consider the impact of input offset voltage and the input current of operational amplifier, offset voltage on the impact of output voltage is

$V_{\mathrm{outos}}=V_{\mathrm{os}3}\left(\frac{R_8}{R_7}\frac{R_5}{R_4}\right)-(1+\frac{R_8}{R_7})V_{\mathrm{os}4}+(1+\frac{R_5}{R_3//R_4})V_{\mathrm{os}2}-(1+\frac{R_2}{R_1})\left(\frac{R_5}{R_4}\right)V_{\mathrm{os}1}---\left(10\right)$

Get R1=R2=R3=R4=R5, R7=R8

Then can get the input offset voltage of operational amplifier and input current on the impact of output current is:

$I_{\mathrm{outos}}=\frac{V_{\mathrm{os}3}-2V_{\mathrm{os}4}+3V_{\mathrm{os}2}-2V_{\mathrm{os}1}}{R}+I_{\mathrm{os}3}---\left(11\right)$

Then total output current is

$I_{\mathrm{out}}=\frac{\frac{R_5{R}_2}{R_3{R}_1}{V}_{\mathrm{ref}}}{R+(\frac{R_8}{R_7}\frac{R_5}{R_4}-1)R_{\mathrm{load}}}+\frac{V_{\mathrm{os}3}-2V_{\mathrm{os}4}+3V_{\mathrm{os}2}-2V_{\mathrm{os}1}}{R}+I_{\mathrm{os}3}---\left(12\right)$

R wherein ₁, R ₂, R ₃, R ₄, R ₅, R ₇, R ₈Corresponding resistor for each resistor among Fig. 4; V _Os1, V _Os2, V _Os3, V _Os4Be respectively the offset voltage of operational amplifier A 1, A2, A3, A4; I _Os3Be the input offset current of operational amplifier A 3, V _RefBe reference data voltage source, R _LoadBe pull-up resistor, I _OutBe output current.

In the formula (12), its every physical significance is: first is the relation function of electric current and voltage, second be the offset voltage of each operational amplifier on the impact of output current, the 3rd is that the offset current of operational amplifier A 3 is on the impact of output current.After tested the calibration after in the ideal case, resistance R ₁, R ₂, R ₃, R ₄, R ₅, R ₇, R ₈All be invariable, R ₁=R ₂=R ₃=R ₄=R ₅, R ₇=R ₈, ignore the impact of operational amplifier, have

$I_{\mathrm{out}}=\frac{V_{\mathrm{ref}}}{R}---\left(13\right)$

Be not difficult to find out by formula (13), output current and R are linear, and it doesn't matter to output current in load.As long as R is certain, output current is steady.For obtaining other electric current, can by changing resistance or the reference voltage level of R, just can obtain different current values simultaneously.

In actual conditions, the temperature coefficient of resistance and the temperature coefficient of voltage-reference will affect the stability of output current, and there are error in resistance and operational amplifier, suppose that the absolute error of resistance is respectively Δ R ₁, Δ R ₂, Δ R ₃, Δ R ₄, Δ R ₅, Δ R ₇, Δ R ₈, Δ R, relative error is δ R ₁, δ R ₂, δ R ₃, δ R ₄, δ R ₆, δ R ₇, δ R ₈, δ R; The offset voltage of operational amplifier A i is V _Osi, then for the constant-current source circuit among Fig. 4, from formula (5), can obtain because the output current error that resistance error brings is:

$\mathrm{\&Delta;}{I}_{\mathrm{out}}\text{=}\frac{\frac{({\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_5+\mathrm{\&Delta;}{R}_5)({\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_2+\mathrm{\&Delta;}{R}_2)}{({\mathrm{<figure-callout\; id="3"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_3+\mathrm{\&Delta;}{R}_3)({\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_1+\mathrm{\&Delta;}{R}_1)}\frac{V_{\mathrm{ref}}}{R+\mathrm{\&Delta;<figure-callout\; id="1"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}}{1+[\frac{{\mathrm{<figure-callout\; id="8"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_8+\mathrm{\&Delta;}{R}_8}{{\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_7+\mathrm{\&Delta;}{R}_7}\frac{{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_5+\mathrm{\&Delta;}{R}_5}{{\mathrm{<figure-callout\; id="4"\; label="R"\; filenames="101027095806.png"\; state="\{\{state\}\}">R</figure-callout>}}_4+\mathrm{\&Delta;}{R}_4}-1]\frac{R_{\mathrm{load}}}{R}}-I_{\mathrm{out}}$

$=\frac{\frac{(1+{\mathrm{\&delta;}}_{R5})(1+{\mathrm{\&delta;}}_{R2})}{(1+{\mathrm{\&delta;}}_{R3})(1+{\mathrm{\&delta;}}_{R1})}\frac{I_{\mathrm{out}}}{1+{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}}}{1+[\frac{(1+{\mathrm{\&delta;}}_{R8})(1+{\mathrm{\&delta;}}_{R5})R}{(1+{\mathrm{\&delta;}}_{R7})(1+{\mathrm{\&delta;}}_{R_4})}-1]\frac{R_{\mathrm{load}}}{R}}-I_{\mathrm{out}}$

$\&ap;(1+{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R2}-{\mathrm{\&delta;}}_{R3}-{\mathrm{\&delta;}}_{R1}-{\mathrm{\&delta;}}_R)\{1-[(1+{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="8"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}8}+{\mathrm{\&delta;}}_{R5}-{\mathrm{\&delta;}}_{R_7}-{\mathrm{\&delta;}}_{R4})-1\left]\frac{R_{\mathrm{load}}}{R}\right\}{I}_{\mathrm{out}}-I_{\mathrm{out}}$

$\&ap;(1+{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R2}-{\mathrm{\&delta;}}_{R3}-{\mathrm{\&delta;}}_{R1}-{\mathrm{\&delta;}}_R)\{1+[(-{\mathrm{\&delta;}}_{R8}-{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{{\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_7}+{\mathrm{\&delta;}}_{R4})\left]\frac{R_{\mathrm{load}}}{R}\right\}{I}_{\mathrm{out}}-I_{\mathrm{out}}$

$\&ap;[(1+{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R2}-{\mathrm{\&delta;}}_{R3}-{\mathrm{\&delta;}}_{R1}-{\mathrm{\&delta;}}_R)+(-{\mathrm{\&delta;}}_{R8}-{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{{\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_7}+{\mathrm{\&delta;}}_{R4})\frac{R_{\mathrm{load}}}{R}]I_{\mathrm{out}}-I_{\mathrm{out}}$

$\&ap;[({\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R2}-{\mathrm{\&delta;}}_{R3}-{\mathrm{\&delta;}}_{R1}-{\mathrm{\&delta;}}_R)+(-{\mathrm{\&delta;}}_{R8}-{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{{\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_7}+{\mathrm{\&delta;}}_{R4})\frac{R_{\mathrm{load}}}{R}]I_{\mathrm{out}}---\left(14\right)$

The relative error of output current is

${\mathrm{\&delta;}}_{\mathrm{Iout}}=\frac{{\mathrm{\&Delta;I}}_{\mathrm{out}}}{I_{\mathrm{out}}}$

$\&ap;({\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R2}-{\mathrm{\&delta;}}_{R3}-{\mathrm{\&delta;}}_{R1}-{\mathrm{\&delta;}}_R)-({\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R8}-{\mathrm{\&delta;}}_{R4}-{\mathrm{\&delta;}}_{R7})\frac{R_{\mathrm{load}}}{R}---\left(15\right)$

In the formula (15), its physical significance is respectively: front four is R ₁, R ₂, R ₃And R ₅The relative error to output current that differential scale operation error causes; The 5th is the relative error to output current that the resistance R relative error causes; Last is R ₈And R ₇Proportional error and R ₄And R ₅The impact on constant current source that proportional error causes.

The below will be according to side circuit model shown in Figure 5, the equivalent output impedance characteristic of labor voltage-current converter circuit.Voltage-current converter circuit equivalence output resistance is defined as Thevenin equivalence small-signal resistance, namely from circuit output end see into resistance.In equivalent electrical circuit shown in Figure 5, A ₀₁, A ₀₂, A ₀₃, A ₀₄Respectively the limited open-loop gain of operational amplifier A 1, A2, A3, A4, R ₀₁, R ₀₂, R ₀₃, R ₀₄Be respectively the internal output impedance of operational amplifier, for calculating output impedance, with the zero setting of circuit input voltage, add electric current I at output terminal _O, output end voltage is V _O

According to the empty short and virtual earth characteristics of operational amplifier work and linear zone, can get the voltage input/output relation of circuit among Fig. 5:

$\frac{V_{o4}}{{\mathrm{<figure-callout\; id="3"\; label="V\; o"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">V</figure-callout>}}_o}=-\frac{R_8}{{\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_7}\frac{1}{1+\frac{(1+{\mathrm{<figure-callout\; id="4"\; label="R\; o"\; filenames="101027095806.png"\; state="\{\{state\}\}">R</figure-callout>}}_o/R_8)(1+{\mathrm{<figure-callout\; id="8"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_8/{\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_7+{\mathrm{<figure-callout\; id="8"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_8/R_{\mathrm{in}})}{A_{o4}-{\mathrm{<figure-callout\; id="4"\; label="R\; o"\; filenames="101027095806.png"\; state="\{\{state\}\}">R</figure-callout>}}_o/{\mathrm{<figure-callout\; id="8"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_8}}$

$\frac{V_{o2}}{{\mathrm{<figure-callout\; id="4"\; label="V\; o"\; filenames="101027095806.png"\; state="\{\{state\}\}">V</figure-callout>}}_o}=-\frac{R_5}{{\mathrm{<figure-callout\; id="4"\; label="R"\; filenames="101027095806.png"\; state="\{\{state\}\}">R</figure-callout>}}_4}\frac{1}{1+\frac{(1+{\mathrm{<figure-callout\; id="2"\; label="R\; o"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_o/R_5)(1+{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_5/{\mathrm{<figure-callout\; id="4"\; label="R"\; filenames="101027095806.png"\; state="\{\{state\}\}">R</figure-callout>}}_4+{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_5/R_{\mathrm{in}})}{A_{o2}-{\mathrm{<figure-callout\; id="2"\; label="R\; o"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_o/R_5}}---\left(16\right)$

In the top formula, A _O2, A _O3, A ₀₄Respectively the open-loop gain of operational amplifier A 2, A3, A4, V _O2, V _O3, V _O4Be respectively the output voltage of operational amplifier A 2, A3, A4, R _O2, R _O3, R _O4Be respectively the internal output impedance of operational amplifier, R _InBe the input impedance of operational amplifier, for discussing conveniently, establishing operational amplifier is ideal operational amplifier, i.e. A _O2, A _O3, A ₀₄Respectively the open-loop gain of operational amplifier A 2, A3, A4 and for infinitely great, R _O2, R _O3, R _O4Be respectively the internal output impedance 0 of operational amplifier, output impedance R _InBe infinity, obtain

$V_O={\mathrm{<figure-callout\; id="3"\; label="V\; O"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">V</figure-callout>}}_O;{\mathrm{<figure-callout\; id="4"\; label="V\; O"\; filenames="101027095806.png"\; state="\{\{state\}\}">V</figure-callout>}}_O=-\frac{R_8}{R_7}{V}_O;{\mathrm{<figure-callout\; id="2"\; label="V\; O"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">V</figure-callout>}}_O=-\frac{R_5}{R_4}{V}_{O4}---\left(17\right)$

Output contact in operational amplifier A 2 satisfies the KCL equation:

$I_O=\frac{V_{O2}-V_O}{R}+\frac{V_O-V_{O3}}{R_{\mathrm{in}}+R_6}---\left(18\right)$

$R_{\mathrm{out}}=\frac{V_O}{I_O}$

$=\frac{V_O}{\frac{V_{O2}-V_O}{R}+\frac{V_O-V_{O3}}{R_{\mathrm{in}}+{\mathrm{<figure-callout\; id="6"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_6}}=\frac{V_O}{\frac{\frac{R_5{R}_8}{R_4{R}_7}{V}_O-V_O}{R}+\frac{V_O-V_O}{R_{\mathrm{in}}+{\mathrm{<figure-callout\; id="6"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}}_6}}=\frac{R}{\frac{R_5{R}_8}{R_4{R}_7}-1}---\left(19\right)$

Work as R ₇=R ₈, R ₄=R ₅The time, in the stable situation of the resistance of resistance, it is infinitely great that output resistance is tending towards; But in practice, consider deviation and the temperature coefficient of resistance, then become corresponding relation by output resistance with the relative error of resistance.

In the situation that circuit is the Voltage-controlled Current Source characteristic, should satisfy the condition that equivalent output resistance equals pull-up resistor, namely the equivalent output resistance of constant-current source circuit is the resistance value when causing output current 50% relative error.The equivalent output resistance that can get circuit is

$R_{\mathrm{out}}=\frac{\frac{1}{2}R}{{\mathrm{\&delta;}}_{\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="101027095806.png,101027100124.png"\; state="\{\{state\}\}">R</figure-callout>}5}+{\mathrm{\&delta;}}_{R8}-{\mathrm{\&delta;}}_{R4}-{\mathrm{\&delta;}}_{R7}}---\left(20\right)$

Resistance R ₁-R ₈Adopt metalfilmresistor, temperature coefficient is less than 5ppm/ ℃; Wherein the relative error of ratio tracker is less than 0.002%, and the equivalent output resistance that obtains constant-current source circuit is

R _out=0.25×10 ⁵R （21）

According to top analysis as can be known, the principal element that affects constant current source stability is reference voltage V _RefAnd resistance R, according to present components and parts manufacture level, resistance R adopts metalfilmresistor, and its temperature coefficient is less than 5ppm/ ℃; Operational amplifier A 1, A2 and A4, can select from steady nucleus amplifier or the certainly steady nucleus amplifier of copped wave, the drift of offset voltage is less than 0.01 μ V/ ℃, offset current is less than 50pA, operational amplifier A 3 is larger because of the impact on output current, answer the more superior Low-bias Current operational amplifier of selectivity, offset voltage drifts about less than 5 μ V/ ℃, and offset current is less than 150fA; Calculate the B phase error of each range according to the formula (9) of front, at R _LoadUnder the condition of≤10R, calculate A item error by formula (12), calculate equivalent output resistance according to formula (18), the Circuit theory parameter is as shown in table 1.

Table 1

Can cooperate Fig. 6 and shown in Figure 7, wherein, when Fig. 6 is illustrated in normal voltage 1V, the corresponding relation figure of resistance and output current; The actual graph of a relation with theoretical output of electric current when Fig. 7 then represents 1nA-10nA, solid line representation theory output quantity wherein, and dotted line represents actual output quantity.

Above embodiment only for explanation technological thought of the present invention, can not limit protection scope of the present invention with this, every technological thought that proposes according to the present invention, and any change of doing on the technical scheme basis all falls within the protection domain of the present invention.

Claims (1)

1. the weak current source of a 100pA-1 μ A range, it is characterized in that: comprise control module, display unit, input block, D/A converting unit, reference voltage source, differential motion amplifying unit, voltage current transformation unit and negative feedback unit, wherein, the input end of control module connects the input block that is used for input preset current value, output terminal connects display unit, and the preset current value is sent into display unit shows; The output terminal of control module also connects the input end of D/A converting unit, and the current value analog quantity that input block is inputted is converted to digital quantity, and sends into the D/A converting unit; The D/A converting unit will be converted to analog quantity from the preset current value digital quantity of control unit, send into the input end of differential motion amplifying unit, and the input end of described D/A converting unit also connects reference voltage source; The output terminal of differential motion amplifying unit connects the input end of voltage current transformation unit, the input end of voltage current transformation unit is connection control unit also, select range by control module control, and the output terminal of voltage current transformation unit connects the input end of differential motion amplifying unit via the negative feedback unit, and output voltage and preset voltage are superposeed.

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