# OP-AMPS and hell lot similar!

Okay folks, this assign. is due tomorrow and I wasn't able to work on it due to tests and I really need to complete this assignment.
So please if anyone knows any of the problems in the posted pdf file, please let me know as to how to go about solving it, 'cause that would save me a lot of time and will give me some sleep.

Thanks a lot...!

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IS anyone going to even try on the problems posted???

Or is everyone "scared" of giving it a try?

enigma
Staff Emeritus
Gold Member

I'm bumping this thread over to homework help.

Curious3141
Homework Helper
I don't know if this is too late to be helpful, if so, I apologise. But really, you should be posting your own ideas before we can help you, that's our policy.

Anyway, that's a lot of problems. I only tried the first (because it looked simple, and it was), and the last (because it looked interesting and hard, and it was both).

If all the opamps are assumed to be ideal (infinite gain, zero output impedance and infinite input impedance) the problems become doable.

For question 1, you know that the voltage applied to the + terminal is Vin. From this you should infer that the voltage experienced at the - terminal is also Vin because the gain is infinite and only a zero volt difference at the input can produce a finite Vout. Look at the bottom half of the circuit, does it look like a voltage divider with 2Rs in series with a supply of Vout ? In fact it can be treated as such because the current drawn by the inputs of the op-amp are zero. The voltage in the middle of the symmetric voltage divider is Vin.

Hence we have :

$$V_{in} = \frac{R}{R + R}V_{out}$$

giving $$V_{in} = \frac{1}{2}V_{out}$$

and hence $$A_V = 2$$

For the input impedance we need to find the ratio of Vin to Iin, where Iin is the input current. Since the inputs draw zero current,

$$V_{out} - V_{in} = I_{in}.R$$

Since $$V_{out} = 2V_{in}$$

$$\frac{V_{in}}{I_{in}} = R$$

giving the input impedance as R.

The last problem is challenging, but go back to first principles. I worked with the capacitance as a complex reactance, giving me a neat expression for the gain at first (note that R1 does not affect the gain), but it quickly became unwieldy when I took the magnitude. To really find out the behaviour when $\omega$ varies, I would need to differentiate it, but I haven't done it yet. Nevertheless just "eyeballing" it, it's a low pass filter, but it could be called a bandpass because I think the maximum gain occurs at a frequency higher than zero, and tapers off at lower and higher frequencies. But take note this is a tentative conclusion, I'll have to differentiate the expression to be sure.