CN1213215A - Bipolar operational transconductance amplifier based on inverse hyperbolic tangent-hyperbolic tangent transformation - Google Patents

Abstract

An OTA having a completely linear transconductance is provided. This OTA is comprised of a V-I converter, a first, a second and a third I-V converter, a first differential pair of first and second emitter-coupled bipolar transistors driven by a first constant current, and a second differential pair of third and fourth emitter-coupled bipolar transistors driven by a second constant current. The V-I converter converts a differential input voltage to first and second output currents linearly related to the differential input voltage and generates a third output current. The first I-V converter generates a first output voltage. The second I-V converter generates a second output voltage. The third I-V converter generates a third output voltage.

Description

Based on the bipolar operational transconductance amplifier from the inverse hyperbolic tangent to the Tanh Transform

The invention relates to operation transconductance amplifier and more specifically be about having complete linear transconductance and being applicable to bipolar operational transconductance amplifier based on the semiconductor integrated circuit (IC) of inverse hyperbolic tangent one Tanh Transform.

The differential amplifier circuit that has fabulous transconductance linearity in quite wide input voltage range is known as " operation transconductance amplifier (OTA) ".Therefore OTA can be called " transconductance linearity " circuit.

Figure 1 illustrates common ambipolar transconductance linearity multiplier, it is to be 2,512 by the Japanese Patent No. of announcing in April, 1996,385 is disclosed, and this Japan Patent correspondence the U.S. Patent number 5,331 of in July, 1994 issue, 289, this multiplier has the identical configuration of essence with OTA.

As shown in Figure 1, the npn type bipolar transistor Q109 of emitter-coupled and the differential pair of Q110 are subjected to constant current source C101 (electric current: driving IOA), and constituted voltage-to-current (V-I) transducer 11.Transistor Q109 and Q110 have emitter resistance R101 and R102 respectively.The base stage of transistor Q109 and Q110 constituted V-I transducer 11 inputs right and they be applied in difference input voltage Vi.The collector electrode of transistor Q109 and Q110 has constituted the pair of output of V-I transducer 11.

The end of diode pair 15 of diode D101 and D104 of being connected in series is connected to the collector electrode of transistor Q109.The other end of diode pair 15 is connected to and is applied with reference voltage V _REFReference voltage line.The end of diode pair 16 of diode D103 and D106 of being connected in series is connected to the collector electrode of transistor Q110, and the other end of diode pair 16 is connected to and applies reference voltage V _BEFReference voltage line.

And then, the additional diode pair 17 that be connected in series diode D102 and D105 are provided.One end of diode pair 17 is connected to constant current source C102 (electric current: I _OB).The other end of diode pair 17 is connected to and applies reference voltage V _REFReference voltage line.

The electric current I that the nPn type bipolar transistor Q101 of emitter-coupled and the differential pair 12 of Q102 are produced by current mirror circuit 13 _BODriving.The transistor Q101 of a pair of input of formation differential pair 12 and the base stage of Q102 are connected respectively to collector electrode and the diode pair 17 of transistor Q110.The base stage of transistor Q101 and Q102 applies difference input voltage Δ V ₁The transistor Q101 of the pair of output of formation differential pair 12 and the collector electrode of Q102 are connected respectively to the emitter of nPn type bipolar transistor Q105 and Q106.

The nPn type bipolar transistor Q103 of emitter-coupled and the differential pair 14 of Q104 are subjected to the electric current I that current mirroring circuit 13 produces _BODriving.The transistor Q103 of a pair of input of formation differential pair 14 and the base stage of Q104 are connected respectively to the collector electrode of diode pair 17 and transistor Q109.The base stage of transistor Q103 and Q104 applies difference input voltage Δ V ₂The transistor Q103 of the pair of output of formation differential pair 14 and the collector electrode of Q104 are connected respectively to the emitter of nPn type bipolar transistor Q107 and Q108.

Transistor Q105, Q106, the base stage of Q107 and Q108 is connected to the voltage V that applies jointly _CB, one on the collector electrode of transistor Q105 and Q107 is formed into the first output transistor Q106 of transconductance linearity multiplier and the collector electrode of Q108 is connected second output that constitutes the transconductance linearity multiplier.

Current mirroring circuit 13 is by nPn type bipolar transistor Q111, and Q112 and Q113 form, and their emitter is connected to jointly and applies current/voltage V _EEPressure-wire and current source C103 input current I is provided _BOTransistor Q111, the base stage of Q112 and Q113 is coupled in one.The collector electrode of transistor Q111 and Q112 is connected respectively to the emitter that is coupled of transistor Q101 and Q102 and the emitter that connects together of transistor Q103 and Q104.The base stage of transistor Q113 and collector electrode are connected in one and be connected to current source C103.

Subsequently, explain the operating principle of common transconductance linearity multiplier shown in Figure 1 below.

Suppose that base width modulation (being initial voltage) is left in the basket and the dc common-base current gain factor of bipolar transistor equals a unit, the collector current of bipolar transistor is provided by following expression formula (1). $I_C=I_S\left\{\exp \right(\frac{V_{\mathrm{BE}}}{V_T})-1\}---\left(1\right)$

In expression formula (1), V _BEBe bipolar transistor base-emitter voltage, I _SIt is the full electric current that closes wherein.Equally, V _TBe defined as V _T=KT/q, thermal voltage, K is a Boltzmann constant here, T is that the absolute temperature and the q of the degree Kelvin number of degrees is electron charges.

When bipolar transistor carries out normal running and base-emitter voltage V _BEBe approximately equal to 600mV, the exponential part exp (V of expression formula (1) _BE/ V _T) value that has is approximately e ¹⁰Therefore constant component " 1 " can be ignored.

Like this, above-mentioned expression formula (1) can be approximately as follows: $I_C=I_S\exp \left(\frac{V_{\mathrm{BE}}}{V_T}\right)---\left(1^{\′}\right)$

This equation (1 ') can be rewritten as following equation (2) $V_{\mathrm{BE}}=V_T{I}_n\left(\frac{I_C}{I_S}\right)---\left(2\right)$

V-I transducer 11 is according to difference input voltage V _iThe collector electrode of transistor Q109 and Q110 is exported a pair of difference output current I respectively _A1And I _A2

Suppose that V-I transducer 11 has linear transmission characteristic, difference output current I _A1And I _A2Each can be expressed as the constant current (I of the output that flows through respective transducer 11 _OA/ 2) and (being transistor Q109 and Q110) and variable-current (G _mV _i/ 2) and the difference input voltage V that applies _iProportional, (G here _m/ 2) be the mutual conductance of transducer 11.

Therefore, the difference output current I of transducer 11 _A1And I _A2Provide by following equation (3) and (4) respectively. $I_{A1}=\frac{1}{2}(I_{\mathrm{OA}}+G_m{V}_i)---\left(3\right)$ $I_{A2}=\frac{1}{2}(I_{\mathrm{OA}}-G_m{V}_i)---\left(4\right)$

Usually, the forward drop of diode equals the base-emitter voltage of bipolar transistor.Therefore, the base-emitter voltage V in above-mentioned equation (2) _BECan be rewritten as the voltage drop of each diode D1, D2, D3, D4, D5 and D6.On the other hand, difference input current I _A1And I _A2Flow through by diode pair 15 and 16 constant current I respectively _OBThe diode pair 17 of flowing through.

According to this, diode pair 15,16 and 17 voltage drop V ₁, V ₂And V ₃The equation that is expressed as respectively (5), (6) and (7). ${\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1=2V_T\mathrm{In}\left(\frac{I_{A1}}{I_S}\right)=2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{2I_S}\right)---\left(5\right)$ ${\mathrm{<figure-callout\; id="2"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_2=2V_T\mathrm{In}\left(\frac{I_{A2}}{I_S}\right)-2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_S}\right)---\left(6\right)$ ${\mathrm{<figure-callout\; id="3"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_3=2V_T\mathrm{In}\left(\frac{I_{\mathrm{OB}}}{I_S}\right)---\left(7\right)$

According to this, use above-mentioned equation (5), (6) and (7), differential pair 12 and 14 difference input voltage Δ V ₁With Δ V ₂Provide by following equation (8) and (9) respectively. $\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1=V_2-{\mathrm{<figure-callout\; id="3"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_3=2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_{\mathrm{OB}}}\right)---\left(8\right)$ $\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="2"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_2=V_3-{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1=2V_T\mathrm{In}\left(\frac{2I_{\mathrm{OB}}}{I_{\mathrm{OA}}+G_m{V}_i}\right)\&NotEqual;\mathrm{\&Delta;}{V}_1--\left(9\right)$

See difference input voltage Δ V as knowing from equation (8) and (9) ₁With Δ V ₂Be unequal.This means, voltage V ₃Be not equal to voltage V ₁And V ₂And middle point voltage (or half); That is,

V ₃≠(1/2)(V ₁+V ₂),

By constant current I _BOThat drive and respectively by difference input voltage Δ V ₁With Δ V ₂The intersection that applies is coupled the difference output current Δ V=(I of differential pair 12 and 14 _O1-I _O2) be represented as following expression formula (10a) $\mathrm{\&Delta;I}=I_{O1}-{\mathrm{<figure-callout\; id="2"\; label="I\; O"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">I</figure-callout>}}_O=I_{\mathrm{BO}}\{\tanh \left(\frac{\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1}{2V_T}\right)+\tanh \left(\frac{\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="2"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_2}{2V_T}\right)---\left(10a\right)$

Above-mentioned equation (8) and (9) are updated to equation (10a) provide following equation (10b). $\mathrm{\&Delta;I}=I_{\mathrm{BO}}\left[\tanh \right\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_{\mathrm{OB}}}\right)\}+\tanh \{\mathrm{In}\left(\frac{2I_{\mathrm{OB}}}{I_{\mathrm{OA}}+G_m{V}_i}\right)\left\}\right]---\left(10b\right)$ $=I_{\mathrm{BO}}[\frac{\sinh \left\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_{\mathrm{OB}}}\right)\right\}}{\cosh \left\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_{\mathrm{OB}}}\right)\right\}}+\frac{\sinh \left\{\mathrm{In}\left(\frac{2I_{\mathrm{OB}}}{I_{\mathrm{OA}}+G_{m{V}_i}}\right)\right\}}{\cosh \{\mathrm{In}\left(\frac{2I_{\mathrm{OB}}}{I_{\mathrm{OA}}+G_m{V}_i}\right)}]$

If hyperbolic sine (sinh) in equation (10b) and hyperbolic cosine (cosh) function are cancelled by known sign, equation (10b) can be rewritten as following equation (11) $\mathrm{\&Delta;I}=I_{\mathrm{BO}}[\frac{{(I_{\mathrm{OA}}-G_m{V}_i)}^2-4{I_{\mathrm{OB}}}^2}{{(I_{\mathrm{OA}}-G_m{V}_t)}^2+4{{\mathrm{<figure-callout\; id="2"\; label="I\; OB"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">I</figure-callout>}}_{\mathrm{OB}}}^2}+\frac{4{I_{\mathrm{OB}}}^2-{(I_{\mathrm{OA}}+G_m{V}_i)}^2}{{(I_{\mathrm{OA}}+G_m{V}_i)}^2+4{{I_{\mathrm{OB}}}^2}_{}}]$ $=I_{\mathrm{BO}}\{\frac{{I_{\mathrm{OA}}}^2-{\left(G_m{V}_i\right)}^2-4{I_{\mathrm{OB}}}^2-2I_{\mathrm{OA}}{G}_m{V}_i}{{I_{\mathrm{OA}}}^2-{\left(G_m{V}_i\right)}^2+4{I_{\mathrm{OB}}}^2-2I_{\mathrm{OA}}{G}_m{V}_i}---\left(11\right)$ $+\frac{4{I_{\mathrm{OB}}}^2-{I_{\mathrm{OA}}}^2-{\left(G_m{V}_i\right)}^2-2I_{\mathrm{OA}}{G}_m{V}_i}{{{\mathrm{<figure-callout\; id="2"\; label="I\; OA"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">I</figure-callout>}}_{\mathrm{OA}}}^2+{\left(G_m{V}_i\right)}^2+4{I_{\mathrm{<figure-callout\; id="2"\; label="OB"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">OB</figure-callout>}}}^2+2I_{\mathrm{OA}}{G}_m{V}_i}\}$ $=-32I_{\mathrm{OA}}{I_{\mathrm{BO}}}^3\left[\frac{G_m{V}_i}{{({I_{\mathrm{<figure-callout\; id="2"\; label="OA"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">OA</figure-callout>}}}^2+{G_m}^2{V_i}^2+4{I_{\mathrm{OB}}}^2)}^2-4{I_{\mathrm{OA}}}^2{G_m}^2{V_i}^2}\right]$

The following fact derives from equation (11).

At first, except difference input voltage V _iOutside, when the electric current I of current source C103 _BOWhen being used, circuit arrangement shown in Figure 1 can be used as analog multiplier.Yet, this clearly, difference output current Δ I be not accurately with difference input voltage V _iWith input current I _BOProduct (V ₁* I _BO) proportional.

The second, when the electric current I of current source C103 _BOWhen being set to steady state value, the circuit arrangement shown in Fig. 1 can be as OTA.Yet, this clearly, difference output current Δ I is not accurately and difference input voltage V _iProportional.

OTA and analog multiplier are the bases that analog signal is used, the basic function piece.OTA is required to have the mutual conductance of complete linearity.Require multiplier to have multiplier characteristic completely.

Yet common multiplier shown in Figure 1 does not have multiplier characteristic completely.If this multiplier is used as OTA, it does not have linear transconductance completely.

According to this, target of the present invention provides the OTA with complete linear transconductance.

Another target of the present invention provides the OTA of the multiplier that can be used as transconductance linearity.

By following description, the above-mentioned target that does not have specific description is very clearly for the those skilled in the art.

OTA according to first aspect present invention comprises the V-I transducer, first current-voltage (I-V) transducer, second I-V transducer, the 3rd I-V transducer, by first differential pair of the first and second emitter-coupled bipolar transistors of first constant current driven with by second differential pair of the third and fourth emitter-coupled bipolar transistor of second constant current driven.

This V-I transducer conversion difference input voltage be first and second output currents and generation the 3rd output current with difference input voltage linear correlation.

It is first output voltage that is applied to the 4th transistor base that first I-V transducer is changed first output current.

It is second output voltage that is applied to the first transistor base stage that second I-V transducer is changed second output current.

It is the 3rd output voltage that is applied to the second and the 3rd transistor base that the 3rd I-V transducer is changed the 3rd output current, the 3rd output voltage equal first and second output voltages and mid-point voltage (being half).

The first and the 3rd transistorized collector electrode of first differential pair is connected in first output that has constituted OTA, and the second and the 4th transistorized collector electrode of second differential pair is connected in second output that has constituted OTA.

The output of OTA from the first and second output difference guide.

Use is according to the OTA of a first aspect of the present invention, and V-I transducer conversion difference input voltage be first and second output currents and generation the 3rd output current with difference input voltage linear correlation.

It is first output voltage that is applied to the 4th transistorized base stage that first I-V transducer is changed first output current.It is second output voltage that is applied to the first crystal base stage that second I-V transducer is changed second output current.It is the 3rd output voltage that is applied to the second and the 3rd transistor base that the 3rd I-V transducer is changed the 3rd output current.The 3rd output voltage equal first and second output voltages and mid-point voltage, (being half).

Therefore, be applied to poor between the first and the 3rd output voltage that difference between the second and the 3rd output voltage of first differential pair always equals to be applied to second differential pair.This means, and the output of OTA equals the twice of one output of first and second differential pairs.

The tanh of the difference that applies between each of the output of first and second differential pairs and the second and the 3rd output voltage or the first and the 3rd output voltage is proportional.

According to this, the output of OTA and difference input linear correlation, in other words has linear transconductance completely according to OTA of the present invention.

According to this, if driving first and second constant currents of first and second differential pairs is set to equate and be used as input current, OTA according to a first aspect of the present invention can be used as the transconductance linearity multiplier, is used for difference input voltage and input current and multiplies each other.

In the preferred embodiment according to the OTA of first aspect present invention, the long-pending square root of first and second output currents of the 3rd output current of the 3rd I-V transducer and V-I transducer is proportional.

In this case, be worth preferably, the first, the second and each of the 3rd I-V transducer form by two serial connection diodes.Each diode can be made of p-n junction diode or the bipolar transistor that connects into diode that is connected together by its base stage and collector electrode.

Composition according to the OTA of second aspect present invention comprises V-I transducer, first I-V transducer, second I-V transducer, the 3rd I-V transducer, first differential pair that constituted by the bipolar transistor of first and second emitter-coupled of first constant current driven, second differential pair that the bipolar transistor of third and fourth emitter-coupled of second constant current driven constitutes.

This V-I transducer conversion difference input voltage is first and second output currents of difference input voltage linear correlation.

It is first output voltage that is applied to the 4th transistor base that this first I-V transducer is changed first output current.

It is second output voltage that is applied to the first transistor base stage that this second I-V transducer is changed second output current.

It is the 3rd output voltage that is applied to the second and the 3rd transistor base that the 3rd I-V transducer is changed first and second output currents.The 3rd output voltage equal first and second output voltages and mid-point voltage (being half).

The first and the 3rd transistorized collector electrode of first differential pair connects together and has constituted first output of OTA.The second and the 4th transistorized collector electrode of second differential pair connects together and has constituted second output of OTA.

Derive the difference output of OTA by first and second outputs.

Use is according to the OTA of second aspect present invention, and only the configuration of the 3rd I-V transducer is different with OTA according to first aspect present invention, but still is according to the advantage of the OTA of first aspect here.

In preferred embodiment according to the OTA of a second aspect of the present invention, first I-V transducer is to be made of first and second diodes that are connected in series, second I-V transducer be constitute by third and fourth diode that is connected in series and the 3rd I-V transducer constitute by the 5th diode and the 5th bipolar transistor.

First output current flows through the 5th diode.Second output current flows through the 5th transistor.The 3rd output voltage by the forward drop of the 5th diode and the 5th transistorized base-emitter voltage and provide.

OTA according to third aspect present invention comprises V-I transducer, first I-V transducer; Second I-V transducer, the 3rd I-V transducer, four end units of the bipolar transistor that the first, the second and the 4th emitter that is driven by public constant-current source is coupled.

The second and the 3rd transistorized emitter area is than the big K of the first and the 4th transistorized emitter area ₁Doubly, K here ₁It is the constant of growing up in 1.

This V-I transducer conversion difference input voltage is the first, the second, third and fourth output current with difference input voltage linear correlation.The 3rd output current equals first output current and takes advantage of a, and a is a constant here.The 4th output current equals second output current and takes advantage of b, and b is a constant here.

It is first output voltage that is applied to the 4th transistorized base stage that first I-V transducer is changed first output current, and first I-V transducer is to be made of the 5th and the 6th bipolar transistor that is connected in series mutually with diode connection.The 6th transistorized emitter region K ₂Doubly to the 5th transistorized emitter region, K here ₂It is constant greater than 1.

It is second output voltage that applies the first transistor base stage that second I-V transducer is changed second output current.Second I-V transducer is to be made of the 7th and the 8th bipolar transistor that is connected in series with diode connection.The 8th transistorized emitter area K ₂Greater than the 7th transistorized emitter area.

It is the 3rd output voltage that is applied to the second and the 3rd transistor base that the 3rd I-V transducer is changed the 4th output current.The 3rd I-V transducer is made of the 9th and the tenth bipolar transistor.The 9th transistor has diode and connects.The 3rd output current flows through the 9th transistor.The 4th output current the tenth transistor of flowing through.The 9th transistorized emitter area K ₃Doubly to the 5th and the 7th transistorized emitter area, K here ₃It is constant greater than unit 1.The tenth transistorized emitter area K ₄Doubly to the 5th and the 7th transistorized emitter area, K here ₄It is constant greater than unit 1.

The 3rd output voltage by the forward drop of the 5th diode and the 5th transistorized base-emitter voltage and provide.

Constant a, b, K ₁, K ₂, K ₃, and K ₄Relation below satisfying. $\frac{\mathrm{ab}{K}_2}{K_1{K}_3{\mathrm{<figure-callout\; id="4"\; label="K"\; filenames="A9810946700311.png,A9810946700321.png"\; state="\{\{state\}\}">K</figure-callout>}}_4}=1$

The first and the 3rd transistorized collector electrode of four end units connects together and has constituted first output of OTA.The second and the 4th transistorized collector electrode of four end units connects together and has constituted second output of OTA.

Derive from first and second outputs on the output difference ground of OTA.

Use is according to the OTA of a third aspect of the present invention, because the configuration of this OTA is identical with the configuration according to the OTA of first aspect, also has here and according to those advantages of the OTA of first aspect.

According to invention is described with reference to the drawings, the present invention has various effects really.

The circuit diagram of Fig. 1 shows common ambipolar transconductance linearity multiplier.

The circuit diagram of Fig. 2 shows the ambipolar OTA according to first embodiment of the invention.

The circuit diagram of Fig. 3 shows the ambipolar OTA according to second embodiment of the invention.

The circuit diagram of Fig. 4 shows the ambipolar OTA according to second embodiment of the invention, and wherein configuration of the physical circuit of V-I transducer and constant current source are illustrated.

The circuit diagram of Fig. 5 shows the ambipolar OTA according to third embodiment of the invention.

The circuit diagram of Fig. 6 shows the V-I transducer that uses in the OTA according to third embodiment of the invention.

The circuit diagram of Fig. 7 shows the V-I transducer that uses in the OTA according to first embodiment of the invention.

The preferred embodiments of the present invention are described with reference to the accompanying drawing in front.

As shown in Figure 2, OTA according to first embodiment of the invention comprises V-I transducer 1a, the first diode pair 5a of diode D1 and D4 is connected in series, the second diode pair 6a of diode D2 and D5 is connected in series, the 3rd diode pair 7a of diode D3 and D6 is connected in series, npn type bipolar transistor Q1 that is coupled by emitter and first differential pair 2 of Q2 are coupled second differential pair 4 of npn type bipolar transistor Q3 and Q4 by emitter.

Be applied to V-I transducer 1a and export the first, the second and the 3rd output current I respectively with difference input voltage V1 _A1, I _A2And (I _A1+ I _A2) ^1/2

The end of the first diode pair 5a is connected to first output of V-I transducer 1a.The other end of the first diode pair 5a is connected to and applies reference voltage V _REFReference voltage line.

The end of the second diode pair 6a is connected to first output of V-I transducer 1a.The other end of the second diode pair 6a is connected to and applies reference voltage V _REFReference voltage line.

The end of the 3rd diode pair 7a is connected to the 3rd output of V-I transducer 1a.The other end of the 3rd diode pair 7a is connected to and applies reference voltage V _REFReference voltage line on.

Supply voltage can be used as reference voltage V _REF

The first diode pair 5a is as first I-V transducer, and being used to change the electric current of flowing through to 5a is first output voltage V ₁(being forward drop).The second diode pair 6a is as second I-V transducer, and being used to change the electric current of flowing through to 6a is second output voltage V ₂(being forward drop).The 3rd diode pair 7a is as the 3rd I-V transducer, and being used to change by the electric current to 7a is the 3rd output voltage V ₂(being forward drop).

The transistor Q1 of first differential pair 2 and the emitter of Q2 are connected to constant current I are provided _BOThe constant-current source (not shown).First differential pair 2 is subjected to constant current I _BODriving.

The transistor Q1 of formation differential pair 2 a pair of inputs and the base stage of Q2 are connected respectively to the end points of the second and the 3rd diode pair 6a and 7a.Difference input voltage Δ V ₁Be applied to the base stage of transistor Q1 and Q2, here Δ V ₁=V ₂-V ₃

The transistor Q3 of second differential pair 4 and the emitter of Q4 are connected to constant current I are provided _BOThe constant-current source (not shown).Constant current I _BODrive second differential pair 4.

The transistor Q3 of a pair of input of formation differential pair 4 and the base stage of Q4 are connected respectively to the end points of the 3rd and first diode pair 7a and 5a.Apply difference input voltage Δ V ₂To the base stage of transistor Q3 and Q4, Δ V here ₂=V ₃-V ₁

The collector electrode of transistor Q1 and Q2 is at first output that constitutes OTA.The collector electrode of transistor Q2 and Q4 is connected in second output that has constituted OTA.

Difference output current Δ I according to the OTA of first embodiment promptly derives from first and second outputs from the collector electrode that is coupled that is coupled collector electrode and transistor Q2 and Q4 of transistor Q1 and Q3.

Secondly, the operation principle according to the OTA of first embodiment shown in Figure 2 will be described below.

Suppose that V-I transducer 1 has linear transmission characteristic, each difference output current I _A1And I _A2Can be expressed as the constant current (I of the corresponding output end of the transducer 1 of flowing through _OA/ 2) and and variable-current (G _mV ₁/ 2) and the difference input voltage V that applies ₁Proportional, (G here _m/ 2) be the mutual conductance of transducer 11.

Therefore, the difference output current I of transducer 1 _A1And I _A2Provide by above-mentioned equation (3) and (4) respectively.

Because the difference output current I of flow through respectively diode pair 15 and 16 is arranged _A1And I _A2, diode pair 5 and 6 voltage drop V ₁And V ₂Respectively by above-mentioned equation (5) and (6) expression.

The output current of diode pair 17 of flowing through is (I _A1I _A2) ^1/2, therefore, the voltage drop V of diode pair 17 ₃Available equation (12) is represented. ${\mathrm{<figure-callout\; id="3"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_3=2V_T\mathrm{In}\left(\frac{\sqrt{I_{A1}{I}_{A2}}}{I_S}\right)=2V_T\mathrm{In}\left(\frac{\sqrt{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}}{2I_S}\right)$ $=2V_T\mathrm{In}\left(\frac{\sqrt{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}}{2I_S}\right)$ $=V_T\mathrm{In}\left(\frac{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}{2I_S}\right)---\left(12\right)$ $=\frac{1}{2}[2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{2I_S}\right)+2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_S}\right)]$ $=\frac{1}{2}({\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1+V_2).$

Equation (12) means the voltage drop V of diode pair 17 ₃Equal the first and second voltage drop V ₁And V ₂And half.In other words, voltage drop V ₃Equal voltage drop V ₁And V ₂And mid-point voltage.

Equation (5) above using, (6) and (12), differential pair 12 and 14 difference input voltage Δ V ₁With Δ V ₂Provide by equation (13) and (14) respectively. $\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1=V_2-{\mathrm{<figure-callout\; id="3"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_3=V_2-\frac{{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1+{\mathrm{<figure-callout\; id="3"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_3}{2}=\frac{V_2-{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1}{2}$ $=V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_{\mathrm{OB}}}\right)-V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{2I_{\mathrm{OB}}}\right)---\left(13\right)$ $=V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)$ $\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="2"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_2=V_3-{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1=\frac{{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_2}{2}-{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1=\frac{V_2-{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1}{2}$ $=V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_{\mathrm{OB}}}\right)-V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{2I_{\mathrm{OB}}}\right)---\left(14\right)$ $=V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)=\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1$

From equation (13) and (14) as can be seen, difference input voltage Δ V ₁With Δ V ₂Equate.

Substitution equation (13) and (14) provide equation (15) to above-mentioned equation (10a). $\mathrm{\&Delta;I}=I_{\mathrm{BO}}\left[\tanh \right\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)\}+\tanh \{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)\left\}\right]$ $=2I_{\mathrm{BO}}\tanh \left\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)\right\}---\left(15\right)$

As a result of, difference output current Δ I is expressed as follows: $\mathrm{\&Delta;I}=2I_{\mathrm{BO}}\tanh \left\{\frac{1}{2}\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)\right\}$ $=2I_{\mathrm{BO}}\frac{\sinh \left\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)\right\}}{\cosh \left\{\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_{\mathrm{OA}}+G_m{V}_i}\right)\right\}+1}$ $=2I_{\mathrm{BO}}\left[\frac{{(I_{\mathrm{OA}}-G_m{V}_i)}^2-{(I_{\mathrm{OA}}+G_m{V}_i)}^2}{(I_{\mathrm{OA}}+G_m{V}_i)+{(I_{\mathrm{OA}}-G_m{V}_i)}^2}\right]$ $=2I_{\mathrm{BO}}\frac{\{(I_{\mathrm{OA}}-G_m{V}_i)+(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)-(I_{\mathrm{OA}}+G_m{V}_i)\}}{\{(I_{\mathrm{OA}}+G_m{V}_i)+(I_{\mathrm{OA}}-G_m{V}_i){\}}^2}$ $=2I_{\mathrm{BO}}\left[\frac{(I_{\mathrm{OA}}-G_m{V}_i)-(I_{\mathrm{OA}}+G_m{V}_i)}{(I_{\mathrm{OA}}+G_m{V}_i)+(I_{\mathrm{OA}}-G_m{V}_i)}\right]---\left(16\right)$ $=2I_{\mathrm{BO}}\frac{2G_m{V}_i}{2I_{\mathrm{OA}}}$ $=2I_{\mathrm{BO}}\left[\frac{G_m{V}_i}{I_{\mathrm{OA}}}\right]$

From equation (16) as can be seen, OTA difference output current shown in Fig. 2 and difference input voltage V ₁Linear correlation.

In above-mentioned equation (16), following equation (17) is utilized. $\tanh \left(\frac{x}{2}\right)=\frac{\sinh \left(x\right)}{\cosh \left(x\right)+1}---\left(17\right)$

The aforementioned calculation method and the known tanh that in according to the OTA among first embodiment, use ^-1Tanh (inverse hyperbolic tangent-tanh) conversion has similarity, and a pair of output current of linear here V-I transducer is by (tanh ^-1) conversion, producing first and second output voltages, the difference between the one the second output voltages puts on difference channel, is used for the conversion at the tanh of known Gilbert gain unit.

Yet, at this (tanh ^-1In the)-tanh conversion method, following equation (18) is utilized. $x=\tanh \left\{\frac{1}{2}\mathrm{In}\left(\frac{1+x}{1-x}\right)\right\}---\left(18\right)$

As can be seen, the X in equation (17) equals half of in equation (18) X from equation (17) and (18).In addition, two diodes are connected in series among 6a and the 7a at each the first, the second and the 3rd transducer 5a.

In complying with the OTA of first embodiment, necessary supply voltage is by the base-emitter voltage V of bipolar transistor _BEBecome and be higher than (tanh ^-1The voltage of)-tanh conversion method.Yet, base-emitter voltage V _BEBiasing because the fluctuation of transistor characteristic can be reduced to (1/2) ^1/2(=0.707).This is because the biasing V of the base-emitter voltage of two diodes that are connected in series _BEBe restricted to (2) of single diode voltage ^1/2Doubly.

Fig. 3 shows the OTA according to the second embodiment of the present invention

OTA according to second embodiment has and the identical structure of first embodiment shown in Fig. 2.Except producing four output currents and diode D3 and npn type bipolar transistor Q5, V-I transducer 16 formed the 3rd I-V transducer 7b.

Therefore, by to Fig. 3 identical or corresponding part or element added beyond the identical label, the explanation of identical configuration is omitted, to reach the simplification purpose of description.

The difference input voltage V that V-16 conversions of I transducer apply ₁Be the first, the second, the third and fourth difference output current I _A1And I _A2And I _A1And I _A2The third and fourth difference output current I _A1And I _A2Diode D3 and transistor Q5 flow through respectively.

V falls in tertiary voltage ₃Equal base-emitter voltage and the diode D of transistor Q5 ₃The forward drop sum, the voltage drop of the 3rd I-V transducer 7b is provided by equation (19) by using above-mentioned equation (2). ${\mathrm{<figure-callout\; id="3"\; label="V"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_3=V_T\mathrm{In}\left(\frac{I_{A1}}{I_S}\right)+V_T\mathrm{In}\left(\frac{I_{A2}}{I_S}\right)$ $=V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{2I_S}\right)+V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_S}\right)$ $=V_T\mathrm{In}\left(\frac{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}{4{I_S}^2}\right)---\left(19\right)$ $=\frac{1}{2}[2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{2I_S}\right)+2V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{2I_S}\right)]$ $=\frac{1}{2}({\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_1+V_2)$

The voltage drop V of the 3rd I-V transducer 7b ₃Equal the first and second voltage drop V ₁And V ₂And half.As a result of, the identical advantage that has an embodiment here.

Linear V-I transducer 1b is for example illustrated by Fig. 4 that circuit arrangement realizes.Drive the constant current I of first and second differential pairs 2 and 4 _BORealized simply by the current mirroring circuit shown in Fig. 4.

In Fig. 4, npn type bipolar transistor Q6 and Q7 have constituted differential pair and npn type bipolar transistor Q8 and Q9 and have constituted another differential pair.Transistor Q6, Q7, Q8 and Q9 have emitter resistance R1 respectively, R2, R3 and R4.The emitter of transistor Q6 and Q7 is associated in together by emitter resistance R1 and R2 coupling, and the emitter of transistor Q8 and Q9 is coupled in one by emitter resistance R3 and R4.

The tie point of resistance R 1 and R2 is connected to provides constant current I _ORThe end of constant-current source C1.The other end of constant-current source C1 is connected to ground.

The tie point of resistance R 3 and R4 is connected to provides constant current I _OAThe end of constant-current source C2.The other end of constant-current source C2 is connected to ground.

Difference input voltage V ₁Be applied to the base stage of transistor Q6 and Q7 and the base stage of transistor Q8 and Q9.

Transistor Q6, Q7, the collector electrode of Q8 and Q9 is connected to diode pair 5a, diode pair 6a, diode D3 and transistor Q5.

The positive pole of diode D7 is connected to diode D1, the collector electrode of the negative pole of D2 and D3 and transistor Q5.The negative pole of diode D7 is connected to and applies supply voltage V _CCPower line.Diode D7 is as definition reference voltage V _REF

V-I transducer the 1b that has Fig. 4 configuration does not have linear completely.Yet it has realized actual linear V-I transducer characteristic with simple configuration.

Current mirroring circuit 3 is by npn type bipolar transistor Q10, and Q11 and Q12 form, and their emitter is connected to ground jointly and current source C3 provides constant current I _BOTransistor Q10, the base stage of Q11 and Q12 is connected one.The collector electrode of transistor Q11 and Q12 is connected respectively to the emitter of transistor Q1 and Q2 and the emitter of transistor Q3 and Q4.The base stage of transistor Q10 and collector electrode are linked one and be connected to current source C3.

Load resistance R5 is connected collector electrode and the supply voltage V of transistor Q1 and Q3 _CCBetween.Another load resistance R6 is connected the collector electrode and the supply voltage V of the coupling connection of transistor Q2 and Q4 _CCBetween.

The differential output voltage Δ V of OTA ₀Be from load resistance R6, to draw.

Fig. 5 shows the OTA according to third embodiment of the invention.

OTA according to the 3rd embodiment is by V-I transducer 1C, first I-V transducer 5b, and second I-V transducer 6b, the 3rd I-V transducer 7c and four end units 8 are formed.

V-I transducer is applied in difference input voltage V1 and exports the the first, the second, the 3rd and the 4th output current I respectively _A1, I _A2, aI _A1And bI _A2, a and b are constants here.

The one I-V transducer 5b changes the first output current I _A1First output voltage for the base stage that is applied to transistor Q4.First I-V transducer 5b is made of the npn type bipolar transistor Q13 and the Q14 that are connected in series with diode connection.The emitter area K of transistor Q14 ₂The area of times transistor Q13, K here ₂It is constant greater than unit 1.

Second I-V transducer 6b changes the second output current I _A2For being applied to second output voltage of transistor Q1 base stage.Second I-V transducer 6b is made of the npn type bipolar transistor Q15 and the Q16 that are connected in series with diode connection.The emitter area K of transistor Q16 ₂Doubly to the emitter area of transistor Q15.

The 3rd I-V transducer 7C conversion output current aI _A1For being applied to the 3rd output voltage of transistor Q2 and Q3 base stage.The 3rd I-V transducer 7C is made of npn type bipolar transistor Q17 and Q18.Transistor Q17 has diode and connects.The 3rd output current aI _A1Transistor Q17 flows through.The 4th output current bI ₂Flow through transistor Q16.The emitter area K of transistor Q17 ₃Doubly to the emitter area of transistor Q13 and Q15, K here ₃It is constant greater than unit 1.The emitter area K of transistor Q18 ₄Doubly to the emitter area of transistor Q13 and Q15, K here ₄It is constant greater than unit 1.

The 3rd output voltage V ₃Fall by diode Q17 forward voltage transistor Q18 base-emitter voltage and provide.

The transistor Q1 of four-terminal network 8 and the collector electrode of Q3 are connected in first output that constitutes OTA.The collector electrode of tetrapolar transistor Q2 and Q4 is connected in second output that constitutes OTA.

Derive from first and second outputs on the output difference ground of OTA.

Secondly, the operation principle according to the OTA of the 3rd embodiment shown in Fig. 5 is explained as follows.

In Fig. 5, the transistor Q1 of four-terminal network 8, Q2, the collector current of Q3 and Q4 is respectively defined as I _C1, I _C2, I _C3, and I _C4By using above-mentioned equation (1), they are represented as following equation (20) respectively, (21), (22) and (23). ${\mathrm{<figure-callout\; id="1"\; label="I\; C"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">I</figure-callout>}}_C=I_S\exp \left(\frac{V_{B1}-V_E}{V_T}\right)---\left(20\right)$ ${\mathrm{<figure-callout\; id="3"\; label="I\; C"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">I</figure-callout>}}_C=K_1{I}_S\exp \left(\frac{V_{B2}-V_E}{V_T}\right)---\left(21\right)$ ${\mathrm{<figure-callout\; id="3"\; label="I\; C"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">I</figure-callout>}}_C=K_1{I}_S\exp \left(\frac{V_{B3}-V_E}{V_T}\right)=I_{C2}---\left(22\right)$ ${\mathrm{<figure-callout\; id="4"\; label="I\; C"\; filenames="A9810946700311.png,A9810946700321.png"\; state="\{\{state\}\}">I</figure-callout>}}_C=I_S\exp \left(\frac{V_{B4}-V_E}{V_T}\right)---\left(23\right)$

Here V _EBe transistor Q1, Q2, the common emitter voltage of Q3 and Q4, and V _B1, V _B2, V _B3And V _B4Be their base voltages separately.

Equally, four-terminal network 8 is subjected to single current I _BO, driving, following equation (24) is set up,

I _C1+I _C2+I _C3+I _C4=I _BO (24)

Here I _C2=I _C3

Separate equation (20) to (24) and provide following equation (25). $I_S\exp \left(\frac{V_E}{V_T}\right)=\frac{I_{\mathrm{BO}}}{\exp \left(\frac{V_{B1}}{V_T}\right)+2{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">K</figure-callout>}}_1\exp \left(\frac{V_{B2}}{V_T}\right)+\exp \left(\frac{V_{B4}}{V_T}\right)}---\left(25\right)$

According to this, the difference output current Δ I of OTA shown in Figure 5 is expressed as follows.

ΔI=I ₀₁-I ₀₂ $=\frac{I_{\mathrm{BO}}\{\exp \left(\frac{V_{B1}}{V_T}\right)-\exp \left(\frac{V_{B4}}{V_T}\right)\}}{\exp \left(\frac{V_{B1}}{V_T}\right)+2{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">K</figure-callout>}}_1\exp \left(\frac{V_{B2}}{V_T}\right)+\exp \left(\frac{V_{B4}}{V_T}\right)}---\left(26\right)$

If V-I transducer 1C and difference input voltage V _iLinear correlation, output current I so _A1And I _A2Can be by above-mentioned equation (3) and (4) expression.

According to this, the transistor Q1 of four-terminal network 8, Q2, the base voltage V of Q3 and Q4 _B1, V _B2, V _B3And V _B4Represented by following equation (27), (28) and (29) respectively. ${\mathrm{<figure-callout\; id="1"\; label="V\; B"\; filenames="A9810946700331.png"\; state="\{\{state\}\}">V</figure-callout>}}_B=V_{\mathrm{CC}}-V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{I_S}\right)-V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}-G_m{V}_i}{K_2{I}_S}\right)$ $=V_{\mathrm{CC}}-V_T\mathrm{In}\left\{\frac{(I_{\mathrm{OA}}-G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}{K_2{I_S}^2}\right\}---\left(27\right)$ ${\mathrm{<figure-callout\; id="2"\; label="V\; B"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_B={\mathrm{<figure-callout\; id="3"\; label="V\; B"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_B=V_{\mathrm{CC}}-V_T\mathrm{In}\left\{\frac{a(I_{\mathrm{OA}}+G_m{V}_1)}{K_3{I}_S}\right\}-V_T\mathrm{In}\left\{\frac{b(I_{\mathrm{OA}}-G_m{V}_1)}{K_4{I}_S}\right\}---\left(28\right)$ $=V_{\mathrm{CC}}-V_T\mathrm{In}\left\{\frac{\mathrm{ab}(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}{K_3{K}_4{I_S}^2}\right\}$ ${\mathrm{<figure-callout\; id="4"\; label="V\; B"\; filenames="A9810946700311.png,A9810946700321.png"\; state="\{\{state\}\}">V</figure-callout>}}_B=V_{\mathrm{CC}}-V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{I_S}\right)-V_T\mathrm{In}\left(\frac{I_{\mathrm{OA}}+G_m{V}_i}{K_2{I}_S}\right)$ $=V_{\mathrm{CC}}-V_T\mathrm{In}\left\{\frac{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}+G_m{V}_i)}{K_2{I_S}^2}\right\}---\left(29\right)$

For equation (27, (28) and (29) obtain following equation (30a) to (26).ΔI=I _BO× $\frac{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}+G_m{V}_i)-(I_{\mathrm{OA}}-G_m{V}_i)(I_{\mathrm{OA}}-G_m{V_i}_{})}{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}+G_m{V}_i)-(I_{\mathrm{OA}}-G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)+\frac{2K_1{K}_3{\mathrm{<figure-callout\; id="4"\; label="K"\; filenames="A9810946700311.png,A9810946700321.png"\; state="\{\{state\}\}">K</figure-callout>}}_4}{\mathrm{ab}{K}_2}(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i)}---\left(30a\right)$

In order to make equation (30a) be the form of above-mentioned equation (1b), following relation (30b) needs to satisfy. $\mathrm{\&Delta;I}=I_{\mathrm{BO}}\frac{\{(I_{\mathrm{OA}}+G_m{V}_i)+(I_{\mathrm{OA}}+G_m{V}_i)\}\{(I_{\mathrm{OA}}-G_m{V}_i)-(I_{\mathrm{OA}}-G_m{V}_i)\}}{\{(I_{\mathrm{OA}}+G_m{V}_i)(I_{\mathrm{OA}}+G_m{V}_i)+(I_{\mathrm{OA}}-G_m{V}_i)(I_{\mathrm{OA}}-G_m{V}_i){\}}^2}---\left(30b\right)$ $=I_{\mathrm{BO}}\frac{(I_{\mathrm{OA}}+G_m{V}_i)-(I_{\mathrm{OA}}+G_m{V}_i)}{(I_{\mathrm{OA}}+G_m{V}_i)+(I_{\mathrm{OA}}-G_m{V}_i)}$

According to this, following relation (30c) needs to satisfy. $\frac{2K_1{K}_3{\mathrm{<figure-callout\; id="4"\; label="K"\; filenames="A9810946700311.png,A9810946700321.png"\; state="\{\{state\}\}">K</figure-callout>}}_4}{\mathrm{<figure-callout\; id="2"\; label="ab\; K"\; filenames="A9810946700321.png"\; state="\{\{state\}\}">ab</figure-callout>}{K}_{}{}_2}=2---\left(30c\right)$

As a result of, following relation of plane (31) needs to satisfy. $\frac{\mathrm{ab}{K}_2}{K_1{K}_3{\mathrm{<figure-callout\; id="4"\; label="K"\; filenames="A9810946700311.png,A9810946700321.png"\; state="\{\{state\}\}">K</figure-callout>}}_4}=1---\left(31\right)$

When concerning (31) when being satisfied, difference output current Δ I is represented as follows. $\mathrm{\&Delta;I}=\frac{I_{\mathrm{BO}}{G}_m{V}_i}{I_{\mathrm{OA}}}---\left(32\right)$

From expression (32) as can be seen, comply with the output current difference input voltage V of the OTA of the 3rd embodiment _iLinear correlation.

The V of using among the 3rd embodiment-I transducer 1C for example can be realized by the current arrangements shown in Fig. 6.This current arrangements in fact and the inventor, Kimura, the known OTA of exploitation is identical and at inthe Japanese Non-Examined Patent Publication No.9-23802 Published in september1997, and in IEEE Transaction On Circuit and System, PartI, Vol, 45, No.1, pp 108-113 discloses among the January 1998.

As shown in Figure 6, this V-I transducer 1C comprises the differential pair of the balance of npn bipolar transistor Q31 that its emitter area is equal to each other and Q32.

The emitter of transistor Q31 and Q32 is that the emitter resistance R31 of R is connected to one by resistance.The emitter of transistor Q31 further increases current mirroring circuit 31 by emitter follower and is connected to ground.The emitter of transistor Q32 further increases current mirroring circuit 32 by emitter follower and is connected to ground.

Current mirroring circuit 31 and 32 is respectively as the active load of transistor Q31 and Q32, the output current I of V-I transducer 1C _A1And aI _A1, I _A2And bI _A2From current mirror 31 and 32, draw respectively.

Supply voltage V _CCBy constant current I is provided _OAConstant-current source be applied to the collector electrode of transistor Q31.Transistor Q31 is by constant current I _OADrive.

Identical supply voltage V _CCBy identical constant current I is provided _OAConstant-current source be applied to the collector electrode of transistor Q32.This transistor Q32 is subjected to constant current I _OADriving.

Difference input voltage V _iBe applied to the base stage of transistor Q31 and Q32.Current i is according to the difference input voltage V that applies _iValue will flow through emitter resistance R31.

Use the V shown in Fig. 6-I conversion 1C, because transistor Q31 is subjected to identical constant current I with Q32 _OADriving, base-emitter voltage V _BE31And V _BE32Be equal to each other.Therefore, following equation (33) is set up.

V _i=R _i (33)

Electric current I is expressed as according to this $i=\frac{V_i}{R}---\left(34\right)$

Like this, from current mirroring circuit 31 and the 32 output current I that export _A1And I _A2Respectively by providing in following equation (35a) and (35). $I_{A1}=I_{\mathrm{OA}}+1=I_{\mathrm{OA}}+\frac{V_i}{R}--\left(35a\right)$ $I_{A2}=I_{\mathrm{OA}}-1=I_{\mathrm{OA}}-\frac{V_i}{R}---\left(35\right)$

Current mirroring circuit 31 is connected one npn bipolar transistor Q33 by its base stage, and Q34 and Q46 form, and npn bipolar transistor Q35 provides constant voltage V as emitter-follower transistor and constant voltage source V31 _LS6 times of emitter area that transistor Q46 has are to the area of transistor Q33 and Q34.

The collector electrode of transistor Q33 is connected to the emitter of transistor Q31.The emitter of transistor Q33 is connected to ground.Transistor Q33, the base stage of the coupling connection of Q34 and Q46 is connected to the negative electrode of voltage source V 33.The emitter of transistor Q34 is connected to ground.The base stage of transistor Q35 is connected to the collector electrode of transistor Q31.The collector electrode of transistor Q35 is applied with supply voltage V _CCThe emitter of transistor Q35 is connected to the negative pole of voltage source V 31.

Output current I _A2=I _OA-(V _i/ R) draw from the collector electrode of transistor Q34.Output current bI _A2Draw from the collector electrode of transistor Q46.

Constant pressure source V31 is as mobile transistor Q33, the voltage level of the coupling symbasis utmost point of Q34 and Q46.

Similarly, current mirror 32 is connected in three npn bipolar transistor Q36 of one by base stage, and Q37 and Q47 form, and npn bipolar transistor Q38 is as emitter-follower transistor, and constant pressure source V32 provides identical constant voltage V _LSWith voltage source V 31 provide the same.The emitter area a that transistor Q47 has is doubly to the emitter area of transistor Q36 and Q37.

The collector electrode of transistor Q36 is connected to the emitter of transistor Q32.The emitter of transistor Q36 is connected to ground.The base stage that transistor Q36, Q37 are coupled is connected to the negative pole of voltage source V 32.The emitter of transistor Q37 is connected to ground.The base stage of transistor Q38 is connected to the collector electrode of transistor Q32.The collector electrode of transistor Q38 is added with supply voltage V _CCThe emitter of transistor Q38 connects the positive pole of voltage source V 32.

Output current I _A1=I ₀+ (V _i/ R) draw from the collector electrode of transistor Q34.Output current aI _A1From the collector electrode of transistor Q47, draw.

Constant voltage source V32 is servo as mobile transistor Q36, the voltage level that is coupled base stage of Q37 and Q47.

Can know from equation (35a) with (35b) and to find out that the V shown in Fig. 6-I transducer 1C has complete or fabulous linear operation.

Fig. 7 shows at the example according to the V-I transducer 1a that uses among the OTA of first embodiment shown in Figure 2.

This circuit arrangement correspondence by adding npn type bipolar transistor Q39 and Q40 and npn type bipolar transistor Q41, Q42, Q43, Q44, the configuration that Q45 obtains to above-mentioned V shown in Figure 6-I transducer 1C is except bipolar transistor Q46 and Q47 are omitted.

As shown in Figure 7, transistor Q39 and Q40 have constituted current mirroring circuit.The base stage of transistor Q39 and Q40 is connected in one, and the emitter of transistor Q39 and Q40 is applied in supply voltage V _CCThe base stage of crystal Q39 and collector electrode are linked one.The collector electrode of transistor Q39 is connected to the collector electrode of transistor Q37.Reference current from transistor Q37 is applied to the collector electrode of transistor Q39, image electric current I _A1Collector electrode output from transistor Q40.

Transistor Q41 and Q43 have constituted current mirroring circuit.The emitter of transistor Q41 and Q43 is connected to ground.The base stage of transistor Q41 and Q43 is connected to the collector electrode of transistor Q34 jointly.The collector electrode of transistor Q41 is connected to the collector electrode of transistor Q40.From the next reference current I of transistor Q40 _A1Be applied to the collector electrode of transistor Q41, image electric current I _A1Collector electrode output from transistor Q43.

The base stage of transistor Q42 and Q44 is connected to the collector electrode of transistor Q40 jointly.The emitter of transistor Q42 is connected to the base stage of transistor Q41 and Q43 and the collector electrode of transistor Q34.The emitter of transistor Q44 is connected to base stage and the collector electrode of transistor Q45.The base stage of transistor Q45 and collector electrode are connected in one.The emitter of transistor Q45 is connected to ground.

From the next electric current I of the collector electrode of transistor Q34 _A2Collector electrode output from transistor Q42.Electric current (I _A1I _A2) ^1/2Collector electrode output from transistor Q44.It is the reasons are as follows:

If transistor Q41, Q42, the base-emitter voltage of Q44 and Q45 is defined as V _BE41, V _BE42, V _BE44, and V _BE45, use above-mentioned equation (2), substrate-emitter voltage V _BE41, V _BE42, V _BE44Be represented as following equation (36), (37) and (38). $V_{\mathrm{BE}41}=V_T\mathrm{In}\left(\frac{I_{A1}}{I_S}\right)---\left(36\right)$ $V_{\mathrm{BE}42}=V_T\mathrm{In}\left(\frac{I_{A2}}{I_S}\right)---\left(37\right)$ $V_{\mathrm{BE}44}=V_{\mathrm{BE}45}=V_T\mathrm{In}\left(\frac{I_{\mathrm{OUT}}}{I_S}\right)---\left(38\right)$

On the other hand, because being connected in an emitter with transistor Q41 and Q45, the base stage of transistor Q42 and Q44 is connected to ground, from base-emitter voltage V _BE41, V _BE42, V _BE44And V _BE45In can set up following equation (39).

V _BE41+V _BE42=V _BE44+V _BE45 (39)

Substitution (36), (37) and (38) draw following equation (40) to equation (39)

I _OUT ²=I _A1*I _A2 (40)

As a result of, can obtain following equation (41) $I_{\mathrm{OUT}}=\sqrt{I_{A1}\&CenterDot;I_{A2}}---\left(41\right)$

Like this, output current I _OUTAnd electric current I _A1And I _A2The square root of taking advantage of proportional.

Now described the preferred embodiment of the invention, should be appreciated that, under the situation that does not break away from spirit of the present invention, can make various modifications the those skilled in the art.Therefore scope of invention only has following claim definition.

Claims (7)

1. bipolar operational transconductance amplifier comprises:

(a) V-I transducer is used to change the difference input voltage for first and second output currents of difference input voltage linear correlation with produce the 3rd output current;

(b) first I-V transducer, being used to change said first output current is first output voltage;

(c) second I-V transducer, being used to change said second output current is second output voltage;

(d) the 3rd I-V transducer, being used to change said the 3rd output current is the 3rd output voltage;

Said the 3rd output voltage equal said first and second output voltages and mid-point voltage;

(e) by first differential pair of the first and second emitter-coupled bipolar transistors of first constant current driven; With

(f) by second differential pair of the third and fourth emitter-coupled bipolar transistor of second constant current driven;

Wherein, said first output voltage is applied to the said the 4th transistorized base stage, and base stage and said the 3rd output voltage that said second output voltage is applied to said the first transistor are applied to the said second and the 3rd transistorized base stage jointly;

The first and the 3rd transistorized collector electrode of wherein said first differential pair connects together to constitute first output of said OTA;

The second and the 4th transistorized collector electrode of wherein said second differential pair is connected in one to constitute second output of said OTA;

Wherein the output of said OTA is derived from said first and second output difference ground.

2. the amplifier of claim 1, wherein, the long-pending square root of said the 3rd output current of said the 3rd I-V transducer and said first and second output currents of said V-I transducer is proportional.

3. the amplifier of claim 1, wherein, each said the first, the second and the 3rd I-V transducer be to constitute by two diodes that are connected in series.

4. the amplifier of claim 1, wherein, each said diode is to be made of the bipolar transistor that its base stage and collector electrode are coupled together.

5. operation transconductance amplifier comprises:

(a) V-I transducer, be used to change the difference input voltage for the first, the second, the third and fourth output current of difference input voltage linear correlation;

(b) first I-V transducer, being used to change said first output current is first output voltage;

(c) second I-V transducer, being used to change said second output current is second output voltage;

(d) the 3rd I-V transducer, being used to change said the 3rd output current is the 3rd output voltage;

Said the 3rd output voltage equal said first and second output voltages and mid-point voltage;

(e) by first differential pair of the bipolar transistor of first and second emitter-coupled of first constant current driven; With

(f) by second differential pair of the bipolar transistor of third and fourth emitter-coupled of second constant current driven;

Wherein, said first output voltage is applied to the said the 4th transistorized base stage, and base stage and said the 3rd output voltage that said second output voltage is applied to said the first transistor are applied to the said second and the 3rd transistorized base stage;

The said first and the 3rd transistorized collector electrode of wherein said first differential pair is connected in one to constitute first output of said OTA;

The said second and the 4th transistorized collector electrode of said second differential pair wherein is connected in one to constitute second output of said OTA;

Wherein the output of said OTA is drawn from the said first and second output difference.

6. the amplifier of claim 5, wherein, said first I-V transducer is to be made of first and second diodes that are connected in series, said second I-V transducer be constitute by third and fourth diode that is connected in series and said the 3rd I-V transducer constitute by the 5th diode and the 5th bipolar transistor;

Wherein said first output current flows through said the 5th diode, flow through said the 5th transistor and said the 3rd output voltage of said second output current provided by the forward drop and the said the 5th transistorized base-emitter voltage sum of said the 5th diode.

7. operation transconductance amplifier comprises:

(a) V-I transducer, be used to change the difference input voltage for the first, the second, the third and fourth output current of said difference input voltage linear correlation;

Said the 3rd output current equals said first output current and multiply by a, and a is a constant here;

Said the 4th output current equals said first output current and multiply by b, and b is a constant here;

(b) first I-V transducer, being used to change said first output current is first output voltage;

Said first I-V transducer is made of the 5th and the 6th bipolar transistor that is connected in series with diode connection;

The said the 6th transistorized emitter area K ₂Doubly to the said the 5th transistorized emitter area, K is the constant greater than unit 1 here;

(c) second I-V transducer, said second output current of transducer is second voltage;

Said second I-V transducer is to be made of the 7th and the 8th bipolar transistor that is connected in series with diode connection;

The said the 8th transistorized emitter area K ₂Doubly to the said the 7th transistorized emitter area;

(d) the 3rd I-V transducer, being used to change said the 3rd output current is the 3rd output voltage;

Said the 3rd I-V transducer is to be made of the 9th and the tenth bipolar transistor;

Said the 9th transistor has diode and connects;

Said the 3rd output current said the 9th transistor of flowing through;

Said the 4th output current said the tenth transistor of flowing through;

The said the 9th transistorized emitter area K ₃Doubly to the said the 5th and the 7th transistorized emitter area, K here ₃It is constant greater than unit one;

The said the tenth transistorized emitter area K ₄Doubly to the said the 5th and the 7th transistorized emitter area, K here ₄It is constant greater than unit 1;

Said the 3rd output voltage is to be provided by the forward voltage drop of the 5th diode and the said the 5th transistorized base-emitter voltage sum.

(e) be coupled four end units of bipolar transistor by common constant current driven the first, the second, third and fourth emitter;

The said second and the 3rd emitter area K ₁Doubly to the said first and the 4th transistorized emitter area, K here ₁It is constant greater than unit 1;

Wherein said first output voltage puts on the said the 4th transistorized base stage, and said second output voltage puts on the base stage of said the first transistor; Said the 3rd output voltage jointly puts on the said second and the 3rd transistorized base stage;

The said first and the 3rd transistorized collector electrode of wherein said four end units connects together to constitute said OTA first output, and the second and the 4th transistorized collector electrode of said four end units is connected in one to constitute second output of said OTA;

Here said constant a, b, K ₁, K ₂, K ₃And K ₄Relation below satisfying $\frac{\mathrm{ab}{K}_2}{K_1{K}_3{K}_4}=1$ Here draw to the output difference of said OTA from said first and second outputs.

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