LT1228(Rev_A) 查看數據表（PDF） - Linear Technology
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LT1228 Datasheet PDF : 20 Pages
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LT1228
APPLICATI S I FOR ATIO
Substituting into the equation for transconductance gives:
gm
=
a
1.94R
=
10
R
The temperature variation in the term “a” can be ignored
since it is much less than that of the term “T” in the
equation for Vbe. Using a 2.5V source this way will main-
tain the gain constant within 1% over the full temperature
range of –55°C to 125°C. If the 2.5V source is off by 10%,
the gain will vary only about ±6% over the same tempera-
ture range.
We can also temperature compensate the transconduc-
tance without using a 2.5V reference if the negative power
supply is regulated. A Thevenin equivalent of 2.5V is
generated from two resistors to replace the reference. The
two resistors also determine the maximum set current,
approximately 1.1V/RTH. By rearranging the Thevenin
equations to solve for R4 and R6 we get the following
equations in terms of RTH and the negative supply, VEE.
R4 = R TH and R6 = R THVEE
1
–
2.5V
VEE
2.5V
Temperature Compensation of gm with a Thevenin Voltage
R6
6.19kΩ
R'
R4
1.24kΩ
gm
4
5
ISET
–15V
1.03k R'
ISET
Vbe
VTH = 2.5V
Vbe
LT1228 • TA05
Voltage Controlled Gain
To use a voltage to control the gain of the transconduc-
tance amplifier requires converting the voltage into a
current that flows into pin 5. Because the voltage at pin 5
is two diode drops above the negative supply, a single
resistor from the control voltage source to pin 5 will suffice
in many applications. The control voltage is referenced to
the negative supply and has an offset of about 900mV. The
conversion will be monotonic, but the linearity is deter-
mined by the change in the voltage at pin 5 (120mV per
decade of current). The characteristic is very repeatable
since the voltage at pin 5 will vary less than ± 5% from part
to part. The voltage at pin 5 also has a negative tempera-
ture coefficient as described in the previous section. When
the gain of several LT1228s are to be varied together, the
current can be split equally by using equal value resistors
to each pin 5.
For more accurate (and linear) control, a voltage-to-
current converter circuit using one op amp can be used.
The following circuit has several advantages. The input no
longer has to be referenced to the negative supply and the
input can be either polarity (or differential). This circuit
works on both single and split supplies since the input
voltage and the pin 5 voltage are independent of each
other. The temperature coefficient of the output current is
set by R5.
R3
1M
R1
1M
V1
R2
1M
V2
+
LT1006
–
R4
1M
50pF
R1 = R2
R3 = R4
IOUT =
(V1 – V2)
R5
× R3
R1
= 1mA/V
R5
1k
IOUT
TO PIN 5
OF LT1228
LT1228 • TA19
Digital control of the transconductance amplifier gain is
done by converting the output of a DAC to a current flowing
into pin 5. Unfortunately most current output DACs
sink rather than source current and do not have output
10
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