Can someone figure this out for me? I really don't know where to start. The 1K resistor going up top has me very confused.

 

Can someone figure this out for me? I really don't know where to start. The 1K resistor going up top has me very confused.

The 1k resistor can be considered part of the second stage. Think of the second stage as an inverting summing circuit. Hopefully, this hint will be enough to let you figure this problem out.

One question I have is, are you sure that the first stage has the inverting terminal grounded? If this is drawn correctly, then the first stage has positive feedback.

 

Yea that's they way our teacher gave it to us. That hint makes it a little easier to understand. So how does the 1K resistor affect the output of the first amplifier?

 

Yea that's they way our teacher gave it to us.

OK, so keep the fact that there is positive feedback in mind when you analyze the response of the first stage.

So how does the 1K resistor affect the output of the first amplifier?

You should be able to figure this out. Look at the two nodes that the 1k resistor connects to. One end is a voltage source and the other node is a virtual ground. Ask yourself if the 1k resistor can change these nodes. If so, under what conditions will that happen? Then ask yourself if changes in these nodes can affect the output of the first stage. If so, which node will affect it?

 

It will affect the node going into the positive terminal of the first amp, right?
You said to think of the second like a summing amplifier. So it would be adding the output of the first amplifier (2*Vin) and whatever is coming through that 1K resistor (just Vin)?
The equation I came up with is Vout = (2k/5k)*Vin*2 + 2k/1k*Vin

You said it would be inverting too, so I assume that whole thing is negative.

Something tells me that that isn't even close. I find that on these types of problems my first guess is not even close.

 

But that 1K resistor would have to take up some of the current that would otherwise be going to the first amplifier, so wouldn't that change the equation I have?

As you can probably tell, I'm not very confident when it comes to this stuff.

 

As you can probably tell, I'm not very confident when it comes to this stuff.

Yes, you need to become more familiar with the common op-amp configurations and you also need to be able to calculate the response when you encounter configurations you are not familiar with.

You have some of it right, and some of your analysis is not right.

The first stage is not an amplifier, but is a comparitor circuit with hysteresis. You can google this to get more information. The positive feedback prevents this first stage from acting like an amplifier with gain of 2, as you wrote. The comparitor will go to the positive rail if the input voltage goes sufficiently above ground, and it will go to the negative rail if it is sufficiently below ground.

The 1K resistor will not really affect the output of the first stage under most circumstances. I would go into details on when it could do so, but you are not at the stage to understand this yet. The input voltage Vin is an ideal voltage source, so it won't change with the presence of the 1K resistor.

 

So if the first stage is a comparator with Hysteresis, does the 1k resistor up top affect the theta of the comparator? In our lecture notes it says this type of comparator has a theta of R1/R2. So I'm thinking it's just 1, which means the width of the hysteresis window is 32 volts, since the saturation voltages are 16 and -16. Is that correct?

So if I make V1 the output of the first stage, whatever that might be, is the equation:

Vout = -2/5(V1) + -2(Vin)

correct?

 

LOL, I just stumbled upon this thread... I'm in the same class as you...

 

I think there are a number of us on here. Did you figure it out? I still haven't