Greetings, Can anyone let me know the equation for Vout(t) if I were to place an inductor over the negative feedback of an op amp? More specifically in the circuit below, throw an inductor (L1) in place of the 100kΩ resistor, R1 for the Vin resistor, and R2 for the Vout resistor, and the voltage source is AC. I think the equation is something like Vout(t)= -(R1)(L1) dvi/dt... That of a differentiating amplifier. But I'm not sure. I've read through the book here online and all it mentions is that any reactive component can be used, but it seems the effects of capacitors and inductors would be different, if not opposite. Anyone know for sure? Thanks, Bill
Do you know about the virtual ground concept in inverting op amp circuits? If you do, the calculation of Vout is simple. If you don't know about virtual ground, you need to understand the concept before the circuit will cease to be a mystery.
Ok. The that means the Vout is simply equal to - VL, or the negative of the voltage over the inductor. And I can find that because the current in is Vin/R1, meaning the voltage over the inductor is... hmm di_in/dt? Am I on the right track here? I'm not well practiced in AC power yet.
I think you've got it. Negative, of course, as you mentioned in your previous post. Post your answer when you have it finalized.
As I'm working through this I realized I forgot the second resistor. Do I include this resistor in my Z total for calculating current (R1 + R2= real component of Z)? And once I have the voltage over the inductor do I use a voltage divider to find Vout?
Don't take this as a putdown. I'm surprised that you are aware of the virtual ground concept, but you apparently don't know that the output impedance of an op amp with feedback is very low (≈zero), so the output looks like a voltage source. The other resistor is just another load on that voltage source, and does not enter into the transfer function of your differentiator.
Bill, you brought up an interesting concept -- I've never seen an inductor used in the feedback path of an op amp. It got me to thinking about the problem and I surmised that the gain would be a function of frequency. Think of the inductor as a variable resistor with a resistance that increases with frequency. It stands to reason that when the reactance X=2*pi*f*L equaled the value of the sampling resister you'd get unity gain. I got so curious I set a circuit up with Rs=220 ohms and used a feedback inductor of 84uH. I thought that at 450kHz -- value of f for 220 ohms -- I'd get unity gain. What I got surprised me. It acted like a bandpass filter with a center frequency at 450kHz. I could vary Rs with resistors higher than 220 ohms and calculate the center frequency, but not for smaller values. My guess is that Rs dictates the roll on frequency and the transfer function for the op amp (741) caused the roll off. Hope this helped. There's nothing like practical experimentation! Incidently, this circuit could be used to demodulate FM by feeding its output to a simple AM detector!
You might get different results with a more modern op amp. 741's have very slow slew rates - something like .5 V/uS.
A real inductor has stray capacitance, so you might get some unexpected results. Also, the 90 deg lag from output to input makes the phase margin worse, which can lead to oscillation unless a shunt resistor is added to put a zero in the feedback loop.
Oh man, I kind of understand what you're talking about there, but I don't know enough yet to see it as exciting. Perhaps I'll have my "Ah Ha!" moment sometime later as I'm trying to get to sleep. I got the correct answer for my problem, it was just the negative voltage over the inductor. I didn't put it in terms of frequency. I was making this much more difficult than it needed to be. I appreciate all the help I received here, especially from you Ron. Thanks. I'll be returning frequently as I'm just beginning my ECE degree and will surely have many more questions.
Beenthere, I think your right. Especially if I were to use a 318 wide bandwidth op amp. My guess is that when the reactance of the inductor equaled Rs I'd get unity gain. Then the gain should go up directly proportional to the frequency. This thought comes from X = 2*pi*f*L. After all the gain is Rf/Rs and in this case Rf is an inductor. What are your thoughts?
Ron, as I said above, I had theorized that the roll on frequency would be at about unity gain. This is because for XL (inductive reactance) smaller than Rs the amp would attenuate the signal. Above Rs the gain should be directly proportional to the frequency, from XL/Rs where XL is directly proportional to frequency. And so, as I said to been there, I think a high-bandwidth LM318 op amp would give a reponse closer to my original prediction. Thoughts? Soon I'll be ordering more parts. I'll include an LM318 and see what happens.
No, I'm not holding my breath. Also, I don't need the answer anymore. I completed my Master of Science degree in Electrical and Computer Engineering in March. Thanks again for the help during the fundamentals