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MAT02 数据表(PDF) 9 Page - Analog Devices

部件名 MAT02
功能描述  Low Noise, Matched Dual Monolithic Transistor
PDF  12 Pages
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制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

MAT02 数据表(HTML) 9 Page - Analog Devices

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MAT02
–9–
REV. C
approximately 26 mV and the error due to an rBEIC term will be
rBEIC/26 mV. Using an rBE of 0.4
Ω for the MAT02 and assum-
ing a collector current range of up to 200
µA, then a peak error
of 0.3% could be expected for an rBEIC error term when using
the MAT02. Total error is dependent on the specific application
configuration (multiply, divide, square, etc.) and the required
dynamic range. An obvious way to reduce ICrBE error is to re-
duce the maximum collector current, but then op amp offsets
and leakage currents become a limiting factor at low input lev-
els. A design range of no greater than 10
µA to 1 mA is generally
recommended for most nonlinear function circuits.
A powerful technique for reducing error due to ICrBE is shown in
Figure 20. A small voltage equal to ICrBE is applied to the tran-
sistor base. For this circuit:
VB =
RC
R2
V1 and ICrBE =
rBE
R1
V1
(10)
The error from rBEIC is cancelled if RC/R2 is made equal to rBE/
R1. Since the MAT02 bulk resistance is approximately 0.39
Ω,
an RC of 3.9
Ω and R
2 of 10 R1 will give good error cancellation.
In more complex circuits, such as the circuit in Figure 19, it
may be inconvenient to apply a compensation voltage to each
individual base. A better approach is to sum all compensation to
the bases of Q1. The “A” side needs a base voltage of (VO/RO +
VZ/R3) rBE and the “B” side needs a base voltage of (VX/R1+VY/
R2) rBE. Linearity of better than
±0.1% is readily achievable with
this compensation technique.
Operational amplifier offsets are another source of error. In Fig-
ure 20, the input offset voltage and input bias current will cause
an error in collector current of (VOS/R1) + IB. A low offset op
amp, such as the OP07 with less than 75
µV of V
OS and IB of
less than
±3 nA, is recommended. The OP22/OP32, a program-
mable micropower op amp, should be considered if low power
consumption or single-supply operation is needed. The value of
frequency-compensating capacitor (CO) is dependent on the op
amp frequency response and peak collector current. Typical val-
ues for CO range from 30 pF to 300 pF.
. . .
FOUR-QUADRANT MULTIPLIER
A simplified schematic for a four-quadrant log/antilog multiplier
is shown in Figure 21. As with the previously discussed one-
quadrant multiplier, the circuit makes IO = I1 I2/I3. The two
input currents, I1 and I2, are each offset in the positive direction.
This positive offset is then subtracted out at the output stage.
Assuming ideal op amps, the currents are:
I1 =
V X
R1
+
V R
R2
, I2 =
VY
R1
+
V R
R2
(11)
IO =
V X
R1
+
VY
R1
+
V R
R2
+
VO
RO
, I3 =
V R
R2
From IO = I1 I2/I3, the output voltage will be:
VO =
ROR2
R1
2
V XVY
V R
(12)
Figure 20. Compensation of Bulk Resistance Error
Extrinsic resistive terms and the early effect cause departure
from the ideal logarithmic relationship. For small VCB, all of
these effects can be lumped together as a total effective bulk re-
sistance rBE. The rBEIC term causes departure from the desired
logarithmic relationship. The rBE term for the MAT02 is less
than 0.5
Ω and ∆r
BE between the two sides is negligible.
Returning to the multiplier/divider circuit of Figure 1 and using
Equation (4):
VBE1A + VBE2A – VBE2B –VBE1B + (I1 + I2 – IO – I3) rBE = 0
If the transistor pairs are held to the same temperature, then:
kT
q
In
I1I2
I3IO
=
kT
q
In
IS1AIS2A
IS1BIS2B
+ (I1 + I2 – IO – I3) rBE
(6)
If all the terms on the right-hand side were zero, then we would
have In (I1 I2/I3 IO) equal to zero which would lead directly to
the desired result:
IO =
I1I2
I3
, where I1, I2, I3, IO > 0
(7)
Note that this relationship is temperature independent. The
right-hand side of Equation (6) is near zero and the output cur-
rent IO will be approximately I1 I2/I3. To estimate error, define ø
as the right-hand side terms of Equation (6):
ø = In
IS1AIS2A
IS1BIS2B
+
q
kT
(I1 + I2 – IO – I3) rBE
(8)
For the MAT02, In (ISA/ISB) and ICrBE are very small. For small
ø,
εØ ~ 1 + ø and therefore:
I1I2
I3IO
= 1 + ø
(9)
IO ~
I1I2
I3
(1 – ø)
The In (ISA/ISB) terms in ø cause a fixed gain error of less than
±0.6% from each pair when using the MAT02, and this gain
error is easily trimmed out by varying RO. The ICrBE terms are
more troublesome because they vary with signal levels and
are multiplied by absolute temperature. At 25
°C, kT/q is



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