Ideal inverting amplifier with ideal diodes

awesommee

Joined Nov 18, 2019
13
In the image circuit, the operation amplifiers are ideal and operate in linear mode. The diodes are ideal with Vd = 0, and it is known that Vr = 2.5V and
R = 10kΩ. Determine and draw the characteristics Vi (Vg) for the input voltage range −3.5V ≤ Vg ≤ 3.5V.

I realized that this is inverting amplifier. I was thinking of changing these 2 parts with Thevenin equivalents.

My problem is that I don't know how to prove my assumption that both diodes are ON at Vg = -3.5V. Also, When I change circuit part circled with blue I will get Vt and Rt.In inverting amplifier there is only R there.

WBahn

Joined Mar 31, 2012
26,398
You are going to run into problems if you try to replace the portion you've circled in orange for a couple of reasons. First off, the two ports (the nodes that cross the boundary between what's inside the Thevenin equivalent and what's not) are shorted together. Second, if you rearrange things so that this isn't an issue, then you will possibly end up with a situation in which the op-amp constraint that you need to impose (the virtual short between op-amp inputs) involves a node inside the Thevenin equivalent, and that won't work.

Just walk the analysis up from one end of the input voltage range to the other. Start at Vg = -3.5 V and determine whether the diodes are on or off. Then replace them with their equivalents in those states and determine the output of the resulting circuit. Then determine at what input one of the diodes changes state and repeat the process from that point. You should only have at most three combinations of diode state that are accessible since once a diode changes state it shouldn't change state again.

awesommee

Joined Nov 18, 2019
13
Just walk the analysis up from one end of the input voltage range to the other. Start at Vg = -3.5 V and determine whether the diodes are on or off. Then replace them with their equivalents in those states and determine the output of the resulting circuit. Then determine at what input one of the diodes changes state and repeat the process from that point. You should only have at most three combinations of diode state that are accessible since once a diode changes state it shouldn't change state again.
I know all that, but what I don't know is how to actually do the analysis. Can you do the analysis for a situation D1 - OFF and D2 - OFF so I can get the idea how it is done. Is there any chance you can help me out on Discord?

WBahn

Joined Mar 31, 2012
26,398
No, I'm not going to do part of your homework for you, you need to show your best attempt and then I (and others) can review it and point out where your reasoning is going astray so that you can address your understanding of the material.

But going back to what you said at the end of your original post, namely that your problem is that you don't know how to prove your assumption that both diodes are ON at Vg = -3.5V, let's start with that.

Modify the circuit so that both diodes are on. Also, let's label all of the nodes and define all of the currents so that we can discuss them and stay on the same page.

Now, look over this circuit and see if there are any voltages or currents that you can determine because they are constants (maybe zero, maybe something other than zero) regardless of what Vg happens to be?

awesommee

Joined Nov 18, 2019
13
No, I'm not going to do part of your homework for you, you need to show your best attempt and then I (and others) can review it and point out where your reasoning is going astray so that you can address your understanding of the material.
Yeah, I don't want you to do it for me and not be able to understand it.

This is what I did and I am not sure if this is right. That is why I asked if you could do one part so I can see how it is done.

WBahn

Joined Mar 31, 2012
26,398
Could you try to get an image that has more contrast? My eyes aren't able to make out quite a bit of that.

awesommee

Joined Nov 18, 2019
13

WBahn

Joined Mar 31, 2012
26,398
You are off to a pretty good start. The first stumbling block you have is when you say that

I2 = I6.

You are overlooking the connection to the branches above it.

With the diodes on, nodes B, D, and E are the same node, so you need to apply KCL to that node as a whole.

You start going off the rails after this point, as well.

You say that

I6 = V_R * R

Consider the units very closely. Ohm's Law is

V = I·R, so a voltage times a resistance has units of current times resistance-squared, which whatever that might work out to be, it is NOT units of current. So you know that this is wrong and everything that follows from it is wrong.

Now ask yourself how it is that you multiplied 2.5 V by 10,000 Ω and got 25 mA.

Let's look to the basics. By Ohm's Law, I6 is the ratio of the voltage difference across that resistor divided by that resistance, with the voltage polarity consistent with the current polarity (in accordance with the passive sign convention).

So I6 = (V_B - V_BottomOfThatResistor) / R

What is V_B always going to be in this circuit, provided the op-amp is operating in its active region?

What is V_BottomOfThatResistor going to always be?

I