
Philips Semiconductor Product data
NE5517/NE5517A/
AU5517
Dual operational transconductance amplifier
2002 Dec 06 9
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
qIn 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
qIn I5
I4[KT
qI5*I4
I4(3)
and I4 ≅ I5 ≅ IB
KT
qInI5
I4[KT
qI5*I4
1ń2IB+2KT
qI5*I4
IB+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:
VINǒIBq
2KTǓ+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
qIn
ID
2)IS
ID
2*IS
+KT
qIn 1ń2(IB)IO)
1ń2(IB*IO)(6)
IO+IS2IB
IDfor |IS|tID
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 @R3
R2)R3@gM;
VOUT +IOUT @RL;
A+VOUT
VIN +R3
R2)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.