Does the resistor values used for opamp circuits come from the equations?

Question

I've the following circuit:

I used the LM741 as the opamp. The input resistance is \$2\cdot10^6\space\Omega\$, the output resistance is \$75\space\Omega\$ and the voltage gain is \$2\cdot10^5\$.

Question: I choose values for the resistors \$R_{in}\$ and \$R_f\$ that are between \$10\space k\Omega\$ and \$100\space k\Omega\$. But how does that range follow from the input resistance and output resistance equations?

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My work: from this page (slides 31/32/33) I found that the equations are:

• $$R_{input}\approx R_{in}+\frac{R_f+75}{1+2\cdot10^5}\to\infty_{ideal}\tag1$$
• $$R_{output}\approx\frac{75}{2\cdot10^5}\cdot\frac{R_{in}+R_f}{R_{in}}\to0_{ideal}\tag2$$

Now, I don't see how the values follow.

Answer 1

An ideal operational amplifier has infinite input impedance, zero parasitic leakage current and offsets, and zero output impedance. Therefore in this case ANY resistors are good, mathematically.

Real amplifiers do have finite output impedance (~30 Ohms and up to 1k), and noticeable input impedance (down to 300k for old designs like LM741). To get approximate functionality of OA to the best degree possible, the resistors must be much bigger than output impedance, but much smaller than input impedance. This gives you some choice to make. Sometimes the choice is narrow, like in your homework case, between 10k and 100k.

Answer 2

Welcome to the world of circuit synthesis.

Suppose you want a 20 bit system, with +-5 volts into the ADC. The resistors have 5 PPM temperature coefficients (perhaps these are Vishay metal-film resistors). The resistors, plus the PCB traces and the PCB FR-4 dielectric and the various VDD and GND planes and the metal chassis of the shielding case, provide 100 degree Centigrade per watt thermal resistance. The resistor thermal time constant is 11 milliSeconds; the PCB time constant is several seconds. Can we achieve 20 bits SINAD (signal to noise + distortion)? Can we keep the nonlinearity below 1 bit? Can we keep the self-heating of the resistors below 1 bit, or 1PPM?

For 1PPM, we need 0.2 degree Cent heating. At 100 degree C per watt, and we only budget 0.2 degree, we can only dissipate 2 milliWatts in the resistors.

With 5 volts across the resistors, what value is required?

P = V^2 / R; R = V^2/P = 5*5 / 0.002 = 25 * 500 = 12,500 ohms.

Now................are you able to achieve the Johnson Noise floor needed for 20 bits?

1K ohm in 1 Hz bandwidth is 4 nanoVolts RMS; in 1MHz Bandwidth, expect 4 microVolts.

That 12,500 ohm resistor will generate sqrt(12,500 / 1,000) = sqrt(12.5) ~~3.5X more noise,

or 4uV * 3.5 = 14 uVolts RMS.

Yet the random noise budget for 20 bit system, with 5 volt fullscale, is what?

5uV RMS?

Thus we are boxed in between nonlinearity of thermal heating, and the random noise floor.