0
\$\begingroup\$

Estimate the gain tolerance (+/- X%) of the non-inverting amplifier in Fig. 3 assuming that the tolerance is solely determined by the errors in the resistor values. Only resistors involved in the gain definition contribute. Their tolerances are simply added up. You have 5%-tolerance resistors.

Fig. 3:

enter image description here

I understand that $$A = 1 + \frac{R_2}{R_1}$$ for non-inverting op-amps, but I don't understand what this question is asking.

\$\endgroup\$
  • \$\begingroup\$ What do you think the question is asking? \$\endgroup\$ – awjlogan Nov 10 '17 at 18:43
  • \$\begingroup\$ I think it's asking for gain tolerance (trivial). I don't understand what X represents in +/- X%, and how to use the given figure to estimate what's being asked for. Is it related at all to that equation for A? Is there some formula that's slipping my mind? \$\endgroup\$ – Hello Nov 10 '17 at 18:45
  • \$\begingroup\$ The X is the percentage variation from the calculated ideal value. Say I want my gain to be 100, that X% is the error after taking into account the components real values. If I require a max gain of 105, and min gain of 95, I could handle 5% gain tolerance. So, the question is an error handling question. Your resistors are +-5%, what will the error in your gain be? \$\endgroup\$ – awjlogan Nov 10 '17 at 18:49
  • \$\begingroup\$ Using the equation the gain is 11. Given the 5% tolerance it could range from 6 to 16? Is this what you mean? \$\endgroup\$ – Hello Nov 10 '17 at 18:53
  • \$\begingroup\$ No, it's a percentage error not an absolute figure you need to calculate. If you buy a 10K resistor, it will be 10K +/- 5% (for this question), so will actually be somewhere between 9.5K and 10.5K. The question is asking how you can take your calculated gain of 11 and work out in what range the real gain will be. Hint, it's not 5% error - search for how to combine errors, in this case the 5% errors in the resistor values. \$\endgroup\$ – awjlogan Nov 10 '17 at 19:01
1
\$\begingroup\$

Resistors cannot be manufactured perfectly, a 10k resistance can be bought with a certain tolerance (like 1% or 0.1%), and it is guaranteed by the manufacturer to be in this range. The more tolerance the more expensive the resistor. This means that every circuit you build in the real world will be different. Because of this when you buy said 10k resistor with 1% tolerance, you don't know if your getting a 10.1k or a 9.9k resistor or any value in between. So you need to see if this will affect circuit performance.

You need an equations like this:

$$ A = 1 + \frac{R_2±R_{2tol} }{R_1+ R_{1tol}}$$

Which gives you four equations:

$$ A = 1 + \frac{R_{2max} }{R_{1max}} $$ $$ A = 1 + \frac{R_{2min} }{R_{1max}} $$ $$ A = 1 + \frac{R_{2max} }{R_{1min}} $$ $$ A = 1 + \frac{R_{2min} }{R_{1min}} $$

You then take the min an max of the gains and find the highest gain and lowest gain, if gain tolerance is acceptable in your design move forward.

\$\endgroup\$
  • \$\begingroup\$ So 1 + (9.5K/1.05K) = 10.0476 for the smallest and 1 + (10.5K/0.95K) = 12.0526 for the largest? \$\endgroup\$ – Hello Nov 10 '17 at 20:05
  • 1
    \$\begingroup\$ Why in the world did this generate negative rep? I didn't solve the problem for them I also used latex. \$\endgroup\$ – laptop2d Nov 10 '17 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.