Quote of the Day
Who is that woman? She can't be the woman who raised me.
— I heard a coworker say this about his mother. He was shocked that his mother acted differently in her role as a grandmother than she acted as a mother. Personally, I plan on being an extremely doting grandfather. I can't wait to get the talk about my behavior from my son.
I have an existing circuit for which I need to modify the front-end gain. The gain is provided by an LT1101, which is a common instrumentation amplifier. This part is normally used with one of its two fixed gain settings (10x, 100x). As commonly happens, I need to find a way to resolve an issue without making major changes to a circuit. The designers of the LT1101 provided you a way to modify the amplifiers gain by adding two resistors to the circuit. Figure 1 shows the modified circuit, with the added resistors marked with red ovals and labeled Rx.
The LT1101 allows you to configure its fixed gains as follows:
- 10x Gain: leave pins 2 and 7 open.
- 100x Gain: (1) short pin 2 to pin 1, (2) short pin 7 to pin 8.
For gains G such that 10 ≤ G ≤ 100, the LT1101 datasheet provides us Equation 1 for adjusting the gain to values above 10 by putting identical resistors between pins 2 and 1, and pins 7 and 8.
- G is the desired gain.
- Rx is the external resistor being added.
- R is the internal resistor that has a nominal value of 9.2 kΩ.
What I am doing in this post works fine for a small number of test circuit, but long term it would be better to use an instrumentation amplifier whose gain can be programmed using a single resistor (e.g. LT1789).
Normally, I would not post an analysis this simple, but this solution does illustrate some interesting techniques:
- Use of an infix operator to represent the resistance of parallel resistors
- Setting up and solving circuit equations
- Using the partial fraction keyword to simplify an expression
- Using a datasheet's block diagram to modify a circuit's performance
Figure 2 shows my analysis of the LT1101's gain circuit. I applied Kirchhoff's voltage law to the internal opamp circuit. Note that VA is the output of the opamp labeled A in Figure 1.
I should mention that I wanted to convert the input voltages at the inverting (VN) and non-inverting pins (VP) into a common-mode (VC) and differential-mode (ΔV) representation. I did this through the substitutions and . With this substitution, the VC term was cancelled out and only the ΔV term was left.
I simulated this circuit modification using LTSpice and it worked perfectly. I now need to go into the lab and try it out.
This post just provides a simple demonstration of basic resistor calculations using Mathcad. While there is nothing complicated here, this example does show how easy it is to document your calculations using this type of tool. I can return to this circuit analysis years later, and I will easily be able to determine what I was doing.