Ok I'm doing my electronics homework and I'm too lazy to think (it's Friday) so I was wondering if someone could help me out. Here's the problem: Design an inverting op amp (ideal) with a gain of 46dB that can deliver +/- 15V to a load resistance of 5kohms. The maximum current that can be delivered is 4mA. So, how should I go about doing this? I know the gain is -R2/R1 = Vo/Vs = 200. But does it matter what values I choose for R1 and R2?

Man, I've already forgotten all that stuff. Do you mean design the op-amp external circuit? I'm assuming you do from the ideal part and the references to R1 and R2. 46dB = votlage gain of 39810. So, R2/R1 = 39810, and like you say, you have to kind of just pick R1. If you pick 100 ohms for R1, then R2 = 4Meg. (4,000,000 / 100 = 40,000) Since you said Ideal op-amp, the power supply doesn't matter, nor does the output load, nor does the output current capability. An ideal op amp has infinite input impedance, 0 output impedance, and infinite gain. I think that's it. I graduated, but my grades sucked. Chris

I just realize the 15V supply, 5K load, and 4mA output limits help determine R2. 15V across 5K = 3 mA, so you have 1mA to send back around the feedback loop. Since this is an Ideal op amp with + input grounded, the - input will have 0V on it (infinite gain), and if the output is at +15V, and you want 1mA to flow, you need 15K minimum in the feedback loop. Then back calculate R2/R1 = 40,000 where R2 = 15K. R1 = 2.67 ohms. Kind of freaky, but still works out. I'd rather use the 4Meg resistor and 100 ohm resistor myself. Chris

I don't see where you got the voltage gain of 39810. We learned that Vout/Vsource = R2/R1 = Av. And 20*log(Av) = gain in dB. Working backwards from 46 dB I get Av = 200. In the back of the book where all the answers are it says one possibility is R1 = 1k and R2 = 20k. But with those numbers R2/R1 = 20, not 200, so either I'm missing something or the book is wrong (which isn't all that uncommon). I get what you're saying about how the feedback loop needs to get 1mA. That helps me out a lot. If you use R2/R1 = 20 like the book apparently did and arbitrarily choose 20k for R2 (which is > the minimum of 15k) you get 1k for R1. If I use R2/R1 = 200, with R2 = 20k then R1 = 100 ohms. hmm. I wonder who's right?

My bad on the dB calculation. I was doing power calculation which is (10)Log10 ratio instead of (20)Log10 ratio. Now, using (20) Log10; 46/20 = 2.3, and 10**2.3 = 199.5 => 200. Cool. That's better. Now, work it out like before, picking 20K for R2, gives 100 ohms for R1. I'd guess that the book is wrong. Go for 20K and 100R, or go for 200K and 1K. I told you, 5 years after graduating you've forgotten all this stuff. Chris

The input resistor is R1. You must choose that value so that it doesn't load the stage preceeding the op amp. The junction between the input and the feedback resistor is a virtual ground point in an inverting circuit: that means the Zin of the op amp stage will be virtually equal to R1. If the previous stage / signal generator or whatever can't drive 100 Ohms, you can't use 100 Ohms as an input resistor. The actual values used will depend on what type of op amp you use. Really large values of feedback res. for bipolar op amps (eg 741) are a no-no. However, 2Meg Ohm shouldn't be a problem which means you could use a 10K for R1: no worries for the opamp or preceeding stage. Calculate the load resistor from Ohms law. In reality a (bipolar) opamp will deliver o/p voltage to within approx. 2 Volts of either supply rail: that is, +- 13 Volts. So, 13/.004 = 3250 Ohms; 3K3 for the next prefered value. The amp should then sink or source slightly less than 4 milliamps with 3K3 so be within your design tollerances so driving 5K is no problem. Does that make sense? John

The real answer to your problem is, of course, that you should have taken up a profession that pays loads of $$$ - lawyer, accountant, etc and leave electronics to the idiots like me. RJ