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AU5517D

Dual operational transconductance amplifier

Philips Electronics 

Philips Semiconductor

Dual operational transconductance amplifier

Product data

NE5517/NE5517A/

AU5517

CIRCUIT DESCRIPTION

The circuit schematic diagram of one-half of the AU5517/NE5517, a

dual operational transconductance amplifier with linearizing diodes

and impedance buffers, is shown in Figure 8.

1. Transconductance Amplifier

The transistor pair, Q4 and Q5, forms a transconductance stage. The

ratio of their collector currents (I4 and I5, respectively) is defined by

the differential input voltage, VIN, which is shown in equation 1.

VIN

+

KT

q

In

I5

I4

(1)

Where VIN is the difference of the two input voltages

KT ≅ 26 mV at room temperature (300 °k).

Transistors Q1, Q2 and diode D1 form a current mirror which focuses

the sum of current I4 and I5 to be equal to amplifier bias current IB:

I4 + I5 = IB

(2)

If VIN is small, the ratio of I5 and I4 will approach unity and the Taylor

series of In function can be approximated as

KT

q

In

I5

I4

[

KT

q

I5 * I4

I4

(3)

and I4 ≅ I5 ≅ IB

KT

q

In

I5

I4

[

KT I5 * I4

q 1ń2IB

+

2KT I5

q

*

IB

I4

+

VIN

(4)

I5

*

I4

+

VIN

ǒIBqǓ

2KT

The remaining transistors (Q6 to Q11) and diodes (D4 to D6) form

three current mirrors that produce an output current equal to I5

minus I4. Thus:

ǒ Ǔq

VIN IB 2KT + IO

(5)

The term

ǒIBqǓ

2KT

is

then

the

transconductance

of

the

amplifier

and

is

proportional to IB.

2. Linearizing Diodes

For VIN greater than a few millivolts, equation 3 becomes invalid and

the transconductance increases non-linearly. Figure 9 shows how

the internal diodes can linearize the transfer function of the

operational amplifier. Assume D2 and D3 are biased with current

sources and the input signal current is IS. Since

I4 + I5 = IB and I5 – I4 = I0, that is:

I4 = (IB – I0), I5 = (IB + I0)

For the diodes and the input transistors that have identical

geometries and are subject to similar voltages and temperatures,

the following equation is true:

T

q

In

ID

2

)

IS

ID

2

*

IS

+

KT

q

In

1ń2(IB ) IO)

1ń2(IB * IO)

(6)

IO

+

IS

2IB

ID

for

|IS|

t

ID

2

The only limitation is that the signal current should not exceed ID.

3. Impedance Buffer

The upper limit of transconductance is defined by the maximum

value of IB (2 mA). The lowest value of IB for which the amplifier will

function therefore determines the overall dynamic range. At low

values of IB, a buffer with very low input bias current is desired. A

Darlington amplifier with constant-current source (Q14, Q15, Q16, D7,

D8, and R1) suits the need.

APPLICATIONS

Voltage-Controlled Amplifier

In Figure 10, the voltage divider R2, R3 divides the input-voltage into

small values (mV range) so the amplifier operates in a linear

manner.

It is:

IOUT

+

*

VIN

@

R2

R3

)

R3

@

gM;

VOUT + IOUT @ RL;

A

+

VOUT

VIN

+

R2

R3

)

R3

@

gM

@

RL

(3) gM = 19.2 IABC

(gM in µmhos for IABC in mA)

Since gM is directly proportional to IABC, the amplification is

controlled by the voltage VC in a simple way.

When VC is taken relative to –VCC the following formula is valid:

IABC

+

(VC

* 1.2V)

R1

The 1.2 V is the voltage across two base-emitter baths in the current

mirrors. This circuit is the base for many applications of the

AU5517/NE5517.

2002 Dec 06

9

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