GB2067374A - Stable integrated transconductance amplifier - Google Patents

Abstract

The amplifier comprises a current source Q1 to Q4 and an amplifying section Q5 to Q7. All transistors, plus the 3 resistors, are made in a single integrated circuit, so that sizes and other properties are matched. Q1 and Q2 form a current mirror, and Q3 and Q4 form a second current mirror; the current in Q2 is mirrored into Q1, and that in Q3 is mirrored into Q4. The output current of this current source is thus determined by R1. This sets the base voltage of Q7, which is the "tail" impedance for the transistor pair Q5 and Q6. The gain of the amplifier section is determined by the load resistors RL and the conditions at Q7. It can be shown that all variables (temperature, processing constants, etc.) cancel out, leaving the gain dependent only on the geometrical size ratios of the transistors and the ratio of the resistors, which are fixed. CMOS transistors can be used instead of junction resistors. <IMAGE>

Description

SPECIFICATION Stable integrated transconductance amplifier The present invention relates to stable integrated transconductance amplifiers for use where the amplifier gain must be controlled over temperature, voltage, and manufacturing processing in order to maintain a specific and stable gain. Known transconductance amplifiers fail to provide adequate stability or predictability.

The object of the present invention is to provide a totally integrated gain stage amplifier the gain of which is predictable and stable over a wide temperature range and processing parameters.

Accordingly the present invention provides an integrated transconductance amplifier comprising a current source and an amplifying section, the current source comprising: first and second transistors of a first conductivity type with the emitter of the first connected through a reference resistor and the emitter of the second connected directly to the first side of a voltage supply, and the two bases and the collector of the second being connected together; third and fourth transistors, of the opposite type, with their emitters connected to the second side of the voltage supply, their bases connected together and to the collectors of the first and third transistors, and the collector of the fourth transistor being connected to the base of the first: and the amplifying section comprising:: fifth and sixth transistors of the first type, with their collectors connected through respective load resistors to the second side of the voltage supply, the input being applied between their bases and the output being taken from between their collectors, and a seventh transistor having its emitter connected to the first side of the voltage supply, its base to the base of the first transistor, and its collector to the emitters of the fifth and sixth transistors.

The transconductance amplifier of this present invention comprises a compensating current source and an amplifier section. The current source establishes a reference current which offsets the temperature and processing dependencies of the amplifier. The current source is comprised of two current mirrors and a reference resistor. Current mirrors establish a current through the reference resistor which is dependent of the supply voltage. By matching the reference resistor to the load resistors in the amplifier, the processing and temperature variations are cancelled.

Two embodiments of the invention will now be described by way of example with reference to the accompanying drawings, in which Figures 1 and 2 are circuit diagrams of junction transistor and CMOS transistor embodiments respectively.

Figure 1 shows a transconductance amplifier using ordinary junction transistors, and having a current source comprised of transistors Q1, 02, Q3 and Q4 and a resistor R1, and an amplifier comprised of transistors Q5, Q6 and 07 and a pair of resistors RL.

Transistor Q1 in the current source is connected as shown, with a reference resistor R1 connected between Ol and an earth (relatively negative) supply line 15. Transistor Q2 is connected as shown, with its base connected directly to its collector, thus forcing the collector-base voltage to zero and causing the transistor to behave as if it were in the forward-active region. Connected in this way, transistor Q2 acts as a diode.

Transistor Q1 and Q2 form a first current mirror.

Transistor Q3 of the current source also has its base and collector electrodes connected directly to each other, thereby causing transistor Q3 to act as a diode, and is connected as shown to a positive supply line 25.

Transistor Q4 is connected as shown. Transistors Q3 and Q4 form a second current mirror.

The amplifier section of the transconductance amplifier is comprised of transistors OS and Q6 with their emitters connected to each other and to the collector of transistor 07. The base electrodes of transistors Q5 and Q6 are fed from input terminals 33 and 32, and the collector electrodes feed output terminals 42 and 43 and a pair of load resistors R. Transistor Q7 has its base connected to the current source 01-Q4 as shown.

Transistor 03 is a PNP transistor which has its emitter-junction area matched to that of PNP transistor 04.

Transistors Q1 and 02 are NPN transistors with matched emitter junction areas such that the emitter junction area of transistor Ol is N times as large as the emitter junction area of transistor 02. The current through a diode is approximately as follows:

where J,B iS the leakage term determined by the processing hfe is the current gain A1 is the emitter junction area Vbe is the base-emitter voltage q is the constant of electronic charge K is Boitzman's constant T is the temperature in Kelvin MES is the manufacturer's constant.

For device on a monolithic IC, the diode equation becomes I = C1 Aj exp (OVbe) where C1 and H are constants for all devices in the IC. The current 11 appears across R1, so Ii = VR1/R1 and VR1 = Vbe2 - Vbel.

Remembering that Aj of 01 is equal to NAj of 02: I1 = (Vbe2 - Vbe1)/R1 = In(N Aj/Aj) /Ra13 = In(N)/RtO The operation of the current source is initiated by the presence of residual leakage current in, or the introduction of external current to, one of the current mirrors. If the emitter junction area of transistor Q1 is twice as large as the emitter junction area of transistor 02, it can be shown that the voltage across reference resistor R1 will be approximately 18 mV.

The current in transistor 02 sets a particular value of the base-emitter voltage in transistor 02. The is voltage is equal to the sum of the emitter-base voltage of transistor Q1 plus the voltage across reference resistor R1. If the voltage across R1 is not equal to 18 mV, the base-emitter voltage of Q1 will exceed the voltage required to conduct the current through R1. Transistor Q1 will therefore conduct more current than transistor 02. Transistor 03 will mirror this current into transistor Q4, which in turn will supply more current to transistor 02, raising its base-emitter voltage. This feedback cycle continues until the voltage across reference resistor R1 equais the value necessary to make Ol and 02 have equal currents. At this point the current source is stable.

The voltage gain in the transconductance amplifier is as follows: V2N1 = gm RL = Av where the transconductance gm = 0 I.

The two load resistors RL are equal and matched, and transistors Q5 and Q6 are matched and biased so that the current through transistor 05 is equal to the current through transistor 06 at V1 = 0. Transistor 07 is matched to transistors 01 and 02 and will therefore have the same base-emitter voltage as transistor 02.

Transistor 07 will thus mirror the current through transistor 02 as a ratio of emitter junction area of transistors 02 and Q7.

102 = Ii = M11 where M is the junction area of transistor Q7 divided by the emitter junction area of transistor 02.

The current through transistor Q7 is divided equally between transistors OS and 06. Thus, the current 1L through each of the load resistors RL is half the current through transistor 07.

lL=Mli/2 Substituting terms, Av=0IR= Mel1 RL/2 = M # In(N) R/2 # R1 = M In(N) R/2R1.

The final expression for the gain involves only M, N, and R1 and RL. M and N are geometrical constants, and since R1 and RL are matched, the ratio of R1 to RL is not affected by variations in processing of the circuit during its manufacture. The circuit operates at microcurrent levels and the design is totally integratable. It will thus produce a predictable, stable gain over processing, temperature, and voltage variations.

Figure 2 shows a corresponding circuit implemented with CMOS technology. The operation of the circuit of Figure 2 is essentially the same as that of the Figure 1 circuit. However, the equations underlying the operation are somewhat different.

P-channel MOSFETS Q3 and 04 and N-channel MOSFETS Q1 and 02 comprise the two current mirrors from which the current reference is established. Q3 and Q4 have equal gate areas, and mirror equal currents.

Q1 has a gate aspect ratio of N(W/L), and Q2 has a gate aspect ratio of (W/L). R1 is the current setting resistor.

The current in a diode-connected MOSFET transistor is given approximately by the Sa h equation: ID = k (W/L) (VGs - VT)2 where k = Ft t a processing constant Wand L are the gate width and length, respectively VGS is the gate to source voltage VT is the threshold voltage of the device, assumed to be identical for all devices on the integrated circuit.

Hence for transistors 01 and Q2, we have 1oi = k N (W/L) (VGs - VT)2 (1) lD2 = k (W/L (VGs2 - VT)2 (2) Assuming that these two currents are equal, VGS1 -VT= (VGS2 VT) #N The voltage in equations (1) and (2) are related thus: VGSZ = VGS1 + ID1 R1.

hence ID1 = 1D2 = ((Vcs - VT) - (VGS1 - VT)}/R1 Substituting from equation (3), we get ID2 = (Vcs2 -VT) (#N - 1)/R1 #N (4) Combining equations (2) and (4) and eliminating the voltages, we get lD2 = (W1 )2/{k (W/L) R12 N} (5) We now turn for the moment to the amplifier section of the circuit. The gain of the transconductance amplifier is given by Av= gm RL = 2 k K (W/L) (VGS5 - VT) RL (6) where the gate aspect ratio of OS is K(W/L).Also, from the Sah equation, ID5 = k K (W/L) (VGss - VT)2 (7) Eliminating the voltage from equations (6) and (7), we get ID5 = AV2/4 k K (W/L) RL2 (8) We now have to relate the current source and amplifier sections of the circuit. Taking the gate aspect ratio of Q7 as M(W/L), the fact that VGS is the same for 02 and 07 means that ID7 = M 102.

In addition, since the current through 07 is split equally between Q5 and Q6, we have os = ID7 / 2- Eliminating lD7 from these last 2 equations, we get lD2 = 21D5/ M This enables us to combine equations (5) and (8), eliminating the currents. The resulting equation can be manipulated to give an expression forAv:

In this expression, K, M, and N are constants relating to gate aspect ratios, and R1 and RL are resistors with equal temperature coefficients and matched properties and of fixed relative geometries. If follows that the gain Avis independent of voltage, temperature, and processing (manufacturing) parameters.

Claims (4)

1. An integrated transconductance amplifier comprising a current source and an amplifying section, the current source comprising: first and second transistors of a first conductivity type with the emitter of the first connected through a reference resistor and the emitter of the second connected directly to the first side of a voltage supply, and the two bases and the collector of the second being connected together; third and fourth transistors, of the opposite type, with their emitters connected to the second side of the voltage supply, their bases connected together and to the collectors of the first and third transistors, and the collector of the fourth transistor being connected to the base of the first: and the amplifying section comprising: : fifth and sixth transistors of the first type, with their collectors connected through respective load resistors to the second side of the voltage supply, the input being applied between their bases and the output being taken from between their collectors, and a seventh transistor having its emitter connected to the first side of the voltage supply, its base to the base of the first transistor, and its collector to the emitters of the fifth and sixth transistors.

2. An amplifier according to Claim 1, wherein the transistors are junction transistors.

3. An amplifier according to Claim 1, wherein the transistors and CMOS field-effect transistors, with the emitters, bases, and collectors being the sources, gates, and drains.

4. An integrated transconductance amplifier substantially as described with reference to Figure 1 or Figure 2.