Datasheet LT1228 (Analog Devices) - 10

LT1228
APPLICATIONS INFORMATION
Substituting into the equation for transconductance gives: diode drops above the negative supply, a single resistor a 10 from the control voltage source to Pin 5 will suffice in gm = = many applications. The control voltage is referenced to 1.94R R the negative supply and has an offset of about 900mV. The temperature variation in the term “a” can be ignored The conversion will be monotonic, but the linearity is since it is much less than that of the term “T” in the equa- determined by the change in the voltage at Pin 5 (120mV tion for V per decade of current). The characteristic is very repeat- be. Using a 2.5V source this way will maintain the gain constant within 1% over the full temperature range of able since the voltage at Pin 5 will vary less than ±5% –55°C to 125°C. If the 2.5V source is off by 10%, the gain from part to part. The voltage at Pin 5 also has a negative will vary only about ±6% over the same temperature range. temperature coefficient as described in the previous sec- tion. When the gain of several LT1228s are to be varied We can also temperature compensate the transconductance together, the current can be split equally by using equal without using a 2.5V reference if the negative power supply value resistors to each Pin 5. is regulated. A Thevenin equivalent of 2.5V is generated from two resistors to replace the reference. The two resis- For more accurate (and linear) control, a voltage-to-current tors also determine the maximum set current, approxi- converter circuit using one op amp can be used. The fol- mately 1.1V/R lowing circuit has several advantages. The input no longer TH. By rearranging the Thevenin equations to solve for R4 and R6 we get the following equations in has to be referenced to the negative supply and the input terms of R can be either polarity (or differential). This circuit works TH and the negative supply, VEE. on both single and split supplies since the input voltage R R R4 = TH and R6 = THVEE and the Pin 5 voltage are independent of each other. The  2.5V  2.5V 1– temperature coefficient of the output current is set by R5.  VEE 
Temperature Compensation of g
R3
m with a Thevenin Voltage
1M 1.03k R' R1 1M V1 R5 + ISET 1k IOUT R2 LT1006 TO PIN 5 gm V 1M be OF LT1228 V2 – 4 R6 V R4 TH = 2.5V 6.19kΩ 1M 5 R' ISET Vbe 50pF R4 R1 = R2 1.24kΩ R3 = R4 –15V (V1 – V2) R3 IOUT = • = 1mA/V LT1228 • TA05 R5 R1 LT1228 • TA19
Voltage Controlled Gain
Digital control of the transconductance amplifier gain is To use a voltage to control the gain of the transconductance done by converting the output of a DAC to a current flow- amplifier requires converting the voltage into a current ing into Pin 5. Unfortunately most current output DACs that flows into Pin 5. Because the voltage at Pin 5 is two sink rather than source current and do not have output 1228fd 10 Document Outline Features Applications Description Typical Application Absolute Maximum Ratings Pin Configuration Order Information Electrical Characteristics Typical Performance Characteristics Simplified Schematic Applications Information Typical Applications Package Description Revision History Typical Applications Related Parts