resistor mismatch

[meenakshi22]

(i) Obtain an expression for the CMRR of a difference amplifier, in terms of the tolerance of the resistances used in the amplifier. Hint: Assume that 𝑅1/𝑅2=(𝑅3/𝑅4)(1+𝛼), where 𝛼 is the degree of mismatch. Relate 𝛼 to the tolerance of the resistances (𝜎).
(ii) Plot the relationship between CMRR and 𝜎.
(iii) From the plot obtained in (ii) above, determine the tolerance required for a CMRR of 100 dB.

I know how to calculate gain via resistance but I don't know how to correlate resistance with tolerance of resistors and degree of mismatch. Can you please help with that. (We can take tolerance of each resistor as 1%.) I have added my work on finding gain

 

[KlausST]

Hi,

an 1k resistor (for example) with tolerance of 1% may have a value of 990 Ohms ... 1010 Ohms.
.. since 1% of 1k = 10 Ohms.

So if you use 2 resistors the one is 990 Ohms, the other 1010 Ohms. resulting in different gain and different offset.
(This is the extreme, but not expectable)

Klaus

 

[meenakshi22]

So how do i relate this with mismatch??
Can u explain that

 

[danadakk]

Another method use Matlab :

Regards, Dana.

 

[D.A.(Tony)Stewart]

This exercise is called computing the partial sensitivity derivative for tolerance error.

Some tools often apply Monte Carlo methods to show mean and min max errors in linear amps or filters.

The "Sensitivity" symbol is a capital S somewhat like the Integral Sign with sub & superscripts to define error % of each variable with respect to the function. However you are simply taking the partial derivative

However in the end with linear multipliers and resistor dividers most of the time it is just the sum of all the parts with tolerances in the equation. so 4x1% = 4% w.c. error.

So with four 1% R's the gain and offset or degradation in CMRR of a Diff. Amp are now 4% worst case or CMRR= est. -28 dB and now you wished you had an INA or chose that ratiometric R array with fixed dividers to 0.1%.
.
But if you found that you used parts like LEDs from the same batch # and they were verified to have the same Vf @ If within 3% instead of 30% of the mean value , your design would be better in some way. For R tolerances you have CMRR, gain and offset errors are now 20 dB better if you chose the more expensive binned parts with 0.1%

Anecdotal

Sometimes we need to test and bin parts to achieve a low cost design for tolerance errors and the errors are non-linear.

I recall a need for a $1 design of -40 to +70'C TCXO with 1ppm max, the designer made a diode varicap C(v2/v1) ratio bin tester. This was needed for the 928MHz ISM transmitter tolerance spec of 1 ppm. The other was the XTAL tester with a flex PCB R heater and thermistor to make the mini can into a mini oven with styrafoam using a 2oz flex PCB with R-0805 heaters and thermistor to regulate heat and I could ramp the temperature from 20'C to 40 to 70'C in seconds and measure the the ppm shift of the Xtal in a simple oscillator circuit to a frequency counter. From that linear slope of the 3rd order from 40'C to 70'C for the common "AT-cut" XTAL I had already computed the set of all 3rd order equations for every "minute" then the equation to generate all curves using an equation generator from a DOS program from Microwave Journal. (early 90's). I could predict the entire Xtal error from -40'C to 70'C Xtal from this 3rd order equation using only 2 Xtal frequency shift measurements in a 20 second micro-oven test to less than 0.5 ppm. This way I could get a 20 cent xtal with $10 performance.

Courtesy to CTS

 

[danadakk]

The long way of doing it, attached.


Regards, Dana.

 

[D.A.(Tony)Stewart]

The long way of doing it, attached.


Regards, Dana.

wow.. So the net result is CMRR is only 50% of the sum of tolerance errors.

 

[FvM]

Even if you follow the consideration in post #2 that worst case +/- deviation isn't expectable, you need at least 10 ppm matching to achieve 100 dB CMRR for G=1 amplifier. Worst case 5 ppm.

 

[starlord1607]

So what do you conclude? Whats the final expression? @meenakshi22