Instrumental Amplifier Design Sheet

Electronic System Design 3 As our CMRR is defined as:

$$\text{CMRR}=\frac{\text{Differential Gain}}{2\rho /R}$$

And simply getting higher “precision” resistors is quite expensive, we attempt to achieve higher CMRR using our differential gain, and then using better resistors if needed. We load the gain into the first stage

Note:

$2\ast \rho R$ for a 1% for example is $2\ast 1\ast 0.01=0.02$ for 0.5% $2\ast 0.5\ast 0.01=0.01$

Total error for an instrum amp =

Assuming for a total accepted error of 0.0175 Gauss and $V_{input}=12.5mV/\text{Gauss}$ Therefore Error (RTI) = $0.0175\ast 12.5mV=218\mu V$ Gain = 80 So with an error budget of $218\mu V$ We first calculate the Resistor CMRR as this is the most expensive

$$\text{Resistor CMRR}=\frac{2.5V\ast (2\ast 1\%)}{80}=625\mu V$$

WAY TOO LARGE! Lets say we have 0.1%, expensive, but the best we can practically get

$$\text{Resistor CMRR}=\frac{2.5V\ast (2\ast 0.1\%)}{80}=62.5\mu V$$

Which is workable

Our usable Error Budget is now $218-62.5=155\mu V$ So we can create a figure of merit equal to our error

$$\text{FOM}=2\ast V_{os}+(R_{In(+)}+R_{In(+)})\ast I_B+\frac{1/2\ast V_s}{{\text{CMRR}}_{opamp}}\le 155uV$$ $$\text{FOM}=2\ast V_{os}+850\ast I_B+\frac{2.5V}{{\text{CMRR}}_{opamp}}\le 155uV$$