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LT1228CJ8 Datasheet(PDF) 10 Page - Linear Technology |
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LT1228CJ8 Datasheet(HTML) 10 Page - Linear Technology |
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10 / 20 page  10 LT1228 1228fc S APPLICATI I FOR ATIO Substituting into the equation for transconductance gives: g a RR m == 194 10 . 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. R R V V and R RV V TH EE TH EE 4 1 25 6 25 = ⎛ ⎝⎜ ⎞ ⎠⎟ = – . . Temperature Compensation of gm with a Thevenin Voltage LT1228 ¥ TA05 R4 1.24k Ω –15V gm 5 4 R' ISET ISET 1.03k Vbe Vbe VTH = 2.5V R' R6 6.19k Ω 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 refer- enced to the negative supply and has an offset of about 900mV. The conversion will be monotonic, but the linear- ity is determined 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 temperature coefficient as described in the pre- vious 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. LT1228 ¥ TA19 R5 1k R1 1M V1 V2 IOUT TO PIN 5 OF LT1228 50pF R1 = R2 R3 = R4 IOUT = • = 1mA/V – + R2 1M R3 1M R4 1M (V1 – V2) R5 R3 R1 LT1006 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 |
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