# Operational Amplifier + Diode

Discussion in 'Homework Help' started by gianx80, Feb 22, 2011.

1. ### gianx80New Member

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1. 9-2*j
2. 3+4*j
3. (19+42*j)/25
4. (19-42*j)/85
5. 1

If you intendend for point 5: ((9+2*j)/(3-4*j))*(25/(19+42*j))

Last edited: Mar 11, 2011
2. ### MrAlWell-Known MemberMost Helpful Member

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Hi there,

Well you did an exceedingly great job there i think. It's nice to be able to convey this kind of information to someone else who is as interested as you are. I hope you are enjoying the subject matter we are looking at here, and i think you are.

For #5 the idea was to calculate the reciprocal of X/Y, so that would be 1/(X/Y) which you should do X/Y first (as in #3) and then actually calculate the reciprocal. You should get another complex number.

Next, we really should look at a circuit with at least one current source in it before we get to the other circuits with more types of elements in them.

In the drawing Fig 6a, we see a current source with the arrow pointing up (i0= 2 amps) and three resistors. We have two nodes and we want to solve for the voltage of those two nodes v1 and v2. The trick is to write the element equations and then try to equate the known current i0 to other unknown currents to come up with a system of two equations (two unknowns means two equations).
If you have any difficulty that's ok really, we'll go over it here.

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3. ### gianx80New Member

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I finally managed to solve the exercise. Here my resolution (checked with SPICE):

i0 = 2A;
i1 = v1/R1
i2 = (v1-v2)/R2
i3 = v2/R3

i0-i1-i2 = 0
i2-i3 = 0

2-v1-(v1-v2)/2=0
(v1-v2)/2-v2/3=0

v1 = 1.6
v2 = 1

However, I have still a doubt ... a stupid doubt indeed. Can you explain, step by step, why i3 is v2/R3 and not (v1-v2)/R3? I know it's more or less what I did with i1, but I'd like to have a more in depth explanation

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

i3 (current through R3) is v2/R3 because it's really (v2-0)/R3 simplified, remember the ground is the 0 node and is 0v?

Oh yeah one more little detail...in the future we'll have to use at least 3 digits of precision because the exact v1 is 1.66666... volts and since you only wrote 1.6 i thought at first you didnt get the right answer. 1.66666...would round to 1.7v and that would be a little misleading too, so i think we should use a min of 3 digits like 1.67 or 1.667 even. Makes sense right?

Next we should take a quick look at lumped circuit elements. This is because we can often find significant circuit simplifications by simply combining two or more elements into one. This is usually known as lumping, and you may have heard of this before.
This circuit will make a good illustration of lumping, because we can simplify this circuit quite a bit by simply combining two elements together before we start. Can you spot the two? If you can that's good, but if you cant that's ok too because you'll get used to doing this thing pretty quick anyway.
If we look at R2 and R3, we'll note that they are in series. The only difference here is that we want to solve for v2 as well as v1. But also note that if we already knew what v1 was, we could solve this circuit almost by just looking at it.
This brings us to the idea of lumping two of the elements to convert those into only one element, then solving for v1, then using voltage division to solve for v2. This is a very handy concept.
With our previous circuit, if we add R2 and R3 in series, we get a single resistor of value 5 ohms.

So your next problem will be to calculate the voltage v1 using R2 and R3 as lumped into one resistor of 5 ohms, then use voltage division to calculate v2. You can actually do this on a hand calculator without writing any equations, that's the beauty of lumping.
You'll also note that once you calculate the lumped value of R2 plus R3, you can use current division to calculate the two currents, but an even simpler way is to note that another lumping is possible. Can you spot that other lumping we can do here (only after R2 and R3 are lumped)?
I'll ask that you calculate the new resistor values too, such as R2+R3=5 ohms (but there's at least one more lumping too as mentioned).

If you have any problems with this that's no big deal as once you see this you'll get it right away as i think you catch on pretty quick

Last edited: Mar 14, 2011
6. ### gianx80New Member

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Ok, I get it. And ok, I'll use 3 digits of precision in the future.

Ok, R2 and R3 are in series. So we can do the same with R1 and R2, am I right? However, as you said, I consider a 5 ohms resistor instead of R2 and R3 resistors. Doing this I obtain a voltage value of 1.667v. Then I apply the voltage divider (R3/(R2+R3))*v1 = 1v (1.0002v).

I tried to apply the current divider, but I didn't manage to do it. I know that is similar to the voltage divider and I understood the equation that describes it [(R2/(R1+R2))*I and the resistors have to be in parallel], however I can't figure out how to apply it to this circuit.

And I'm sorry but I can't see another lumping

P.S.
SPICE tells me that the current flowing through R3 is negative, however I found a positive current through R3 and I think that voltage arrows are oriented in the right directions. So, SPICE negative current could be related to the "conventional current flow" thing?

Last edited: Mar 15, 2011
7. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Yes R2 and R3 are in series, and once we put them in series we get a value of 5 ohms which we can now consider one single resistor. Lets call it R4. Now we have R1 in parallel with R4. You may want to draw up a quick schematic of this. You'll now see R1 in parallel with R4 (not in series). This means you can use current division to calculate the two currents.

Current division is simply as follows:

iRa=i*Rb/(Ra+Rb)
iRb=i*Ra/(Ra+Rb)
this is where
iRa is the current through Ra, and
iRb is the current through Rb, and
i is the total current flowing through both.

Note all we do here is divide the opposite resistor value by the sum of both resistors. We just have to remember that we use the other resistor value in the numerator (and the sum of both in the denominator). So to state this another way, we have both resistor values in the denominator, but we only have in the numerator the resistor that is NOT the resistor who's current we want to solve for. If this itself seems confusing we can look at this in more detail.

Ok, so you should go over that and calculate the two currents that way first. If you dont understand it you might still be able to use the two formulas above, but this is something you should understand because it will come in handy for future circuits.

Now we turn again to the second lumping...

For the second method, you can lump R4 (from above) with R1 because they are in parallel, and one formula for two parallel resistors is:
Rp=Ra*Rb/(Ra+Rb)
and then you can calculate the parallel resistance and the voltage across those two and then go from there to get v2.

So the first lumping was possible because R2 and R3 were in series (and then we can call this one resistor R4), and the second lumping is possible because R4 and R1 ended up in parallel. So one lumping was due to a series circuit of two elements, and the second lumping was due to two elements in parallel.

Dont worry if this isnt too clear, we can always go over it more, no problem.

These basics might still seem like we are going a little slow, but dont worry because what we cover here will help us go quickly through even much more complicated circuits later. All of this will be applicable to the more involved circuits and will make them much easier to solve. So once we get through some of this basic stuff you'll find that you can do a ton of circuits just knowing these few basic principles.

Almost forgot to mention:
spice often shows a negative current because it would use electron flow rather than 'conventional' current flow. You'll have to remember that.

Last edited: Mar 15, 2011
8. ### gianx80New Member

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Ok, I managed to correctly use the current divider, it's all clear now. So:

iR1 = i*R4/(R1+R4) = 2*(5/6) = 1.667A
iR4 = i*R1/(R1+R4) = 2*(1/6) = 0.333A

Ok, you're obviously right, I didn't notice that I can lump these two parallel resistors. So:

Rp = (R1*R4)/(R1+R4) = 0.833 ohms
Vp = Rp*i = 0.833*2 = 1.666v (1.667v)

Ok, for the current sign I remembered correctly what you told me, thanks for the reminder.

New circuital schemes attached.

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9. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Oh yes ok, very good. This is the process of lumping where we simplify the circuit using some pretty simple ideas and so we make the whole thing easier to solve. In this circuit we were able to eventually reduce the circuit down to 1 single resistor driven by a current source, so that makes the calculation for the voltage a lot easier.

Now we move to another circuit, but this circuit has two sources in it (see attachment). The two sources are both voltage sources, drawn as batteries E1 and E2. The voltages are shown as 1v and 2v respectively.
Examining this circuit we can see that we have three nodes, one at the top of E1, one at the top of E2, and one at the junction of the three resistors. This means if we use nodal analysis we have two of the three nodes already solved for because we already know E1 and E2, so the only node left is the center node v1. So you're next puzzle is to solve for the voltage v1 using nodal analysis as we discussed earlier. This will give you the voltage at v1.
There's also a second approach to this circuit which comes in handy with more complex circuits which can simplify the circuit like lumping did. This second approach uses the principle of "superposition". The basic idea here is that for this circuit the two sources each have a contribution to the voltage v1, and those contributions add up algebraically. What we do then to simplify the circuit is to calculate the individual contributions separately, then add the two to get the final result. The final voltage v1 is incomplete considering only one source acting independently, but once we consider both sources independently we can sum the results and get the total voltage of v1.
To use superposition we have to introduce the concept of "killing" a source, and "unkilling" a source. To kill a voltage source we short it out completely, and to kill a current source we open circuit it completely. To unkill a voltage source we simply connect the original source back up normally, and same with a current source.
Ok, so with our circuit in Fig. 7a we have two voltage sources so what we do is first kill one source, compute v1 due to the other source, then unkill that source and kill the other one, then compute v1 due to the opposite source. We end up with two partial voltages which we might call v1A and v1B. We then add the two to get the total v1:
v1=v1A+v1B
and that's the voltage at the node v1.
What you can do with this circuit is first kill E1, compute v1A, then kill E2 and computer v1B, then compute the total sum. If you like you can keep track of the partial voltages using the notation like v1(E1) and v1(E2), which tells you the voltage at v1 with respect to a given source. The notation is up to you.
So your next puzzle is to use superposition to calculate v1A, v1B, and the total sum v1.

Just one small note here. If we had more than two sources in the circuit we would start by killing all of the sources first, then unkill one source at a time and find v1A, then kill that one again and unkill the second and find v1B, then kill that one again and unkill the next and find v1C, and do this for all of the sources. Once this is done, we would add all of the partial voltages up to get the total voltage. So this works with any number of sources.

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10. ### gianx80New Member

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NODAL ANALYSIS solution:

i1 = (E1-v1)/R1
i2 = (E2-v1)/R2
i3 = v1/R3

i1+12-i3=0

(E1-v1)/R1+(E2-v1)/R2-v1/R3 = 0

(1-v1)/1+(2-v1)/2-v1/3 = 0

v1 = 1.091

i1 = 1-1.091 = -0.091A
i2 = (2-1.091)/2 = 0.454A

but v1 > E1, so

i1 = (v1 - E1)/R1 = 1.091-1 = 0.091A

SUPERPOSITION method solution:

[E1 OFF]
i1 = v1A/R1
i2 = (E2-v1)/R2
i3 = v1A/R3

i2-i1-i3 = 0

(E2-v1A)/R2 - v1A/R1 - v1A/R3

(2-v1A)/2-v1A-v1A/3 = 0

v1A = 0.545v

i1 = 0.545A
i2 = (2-0.545)/2 = 0.727A
i3 = 0.545/3 = 0.182A

------------------------------

[E2 OFF]
i1 = (E1-v1B)/R1
i2 = v1B/R2
i3 = v1B/R3

i1-12-i3 = 0

(E1-v1B)/R1-v1B/R2-v1B/R3 = 0

(1-v1B)-v1B/2-v1B/3 = 0

v1B = 0.545v

i1 = 1-0.545 = 0.455A
i2 = 0.545/2 = 0.272A
i3 = 0.545/3 = 0.182A

----------------------

v1 = v1A + v1B = 1.09v

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11. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Oh ok you did very good here again, i think you are picking this up pretty quick so we can move fast.

Just one question though, why did you decide to change the sign (polarity) of i1?
When it came out negative, that meant that in relationship to the current arrow you drew in your revised Fig. 7a drawing the current was really negative. You can change it to positive if you like, but then you have to change the direction of the arrow in your drawing. Does this make sense to you?

12. ### gianx80New Member

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Yes, I know, I changed the direction of the arrow but I didn't upload the correct drawing

13. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Oh ok, ha ha, i thought you probably figured that out but i just wanted to make sure. You're really doing good with this.

The next circuit we want to solve will be exactly like Fig. 7a, but we will replace the voltage source on the left (1v) with a current source of 1 amp, and the current arrow will point exactly how you have it in your revised Fig. 7a drawing. That means the current source will be a circle with an arrow inside it pointing up (the bottom terminal can be considered to connect to ground). The current from I1 (we'll call that source I1 for now) will thus flow through R1 from left to right (conventional current flow as usual).
You should solve this new circuit (we'll call it 8a) by superposition, as you did before except this time we have one current source and one voltage source.

Little question: Up to now we have only used independent sources, which are called independent because they dont depend on any other conditions in the circuit. Another class of sources are called "dependent" sources, because they do depend on other quantities in the circuit. Have you used any of these kinds of sources in the past yet? No problem if you didnt but i thought i would ask.

14. ### gianx80New Member

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Here my solution:

[I1 OFF]

There's no current flowing through R1.

i2 = (E2-v1a)/R2
i3 = v1a/R3

i2-i3 = 0
(E2-v1a)/R2 - v1a/R3 = 0
(2-v1a)/2 - v1a/3 = 0

v1a = 1.2v
i2 = (2-1.2)/2 = 0.4A
i3 = 1.2/3 = 0.4A

[E2 OFF]

i1 = 1A
i2 = v1b/R2
i3 = v1b/R3

(I change arrows direction for R2)
i1-i2-i3 = 0
1-v1b/R2-v1b/R3=0
1-v1b/2-v1b/3 = 0

v1b = 1.2v
i2 = 0.6A
i3 = 0.4A

[BOTH ON]

v1 = v1a+v1b = 1.2+1.2 = 2.4v

-----------------------------------

I think I've never used dependent sources.

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15. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Oh yes, very good, as usual

You're doing so good so today we'll dig right in. We are going to take a quick look at dependent sources because they are needed in your original circuit you posted, plus lots of other circuits.

You'll have to refer to the attached drawing for this discussion.

In Fig. 8a, you'll see a new circuit symbol. It's the diamond shape with a plus on top
and minus on bottom. That's a dependent voltage source. The polarity is the same as
a regular voltage source as shown with the plus and minus. Notice the "2Va" next to it
though. Also notice to the left of it there is another "Va" with a plus on top of it
and a minus on the bottom. That Va and associated polarity acts as the "input" for
the dependent voltage source. Since the dependent source has a "2" in front of the "Va"
that means take the value of Va from the left and using that polarity, multiply it
times 2 and that will give you the voltage across the dependent source which here works
out to 10v because the input is 5v and so Va=5 and 2*Va=10. Note R1 doesnt do too much
here except draw power from V1.

In Fig. 8b, a variation in the drawing of the dependent voltage source is shown. It works
exactly the same way except the two input lines are actually drawn too. The polarity is
shown on the source itself along with the usual source polarity markins. This is the
same circuit so the output is the same.

Ok, so in Fig. 8c try to solve for the output voltage. There isnt much to do here but it's
a simple look at how a dependent source works.

If you get all this (dont worry if you dont right now) then you may try the circuit in
Fig. 8d. It's a little different because the dependent source works in series with the
input voltage source. It's just a regular series circuit though so you can use Thev as
the sum of voltage drops around a closed circuit equals zero. You are to solve for the
current out of the input 120v source (same as current through the 30 ohm resistor.

In Fig. 8e there are shown a number of different representations of a dependent
current source. Note that there may be an input sense line that senses current or voltage.

In Fig. 8f there is one representation of a current controlled voltage source. That is,
it senses current and outputs voltage.

Note that none of these dependent sources have no internal feedback of any kind. They simply
sense the input and multiply it times some factor and that provides the output.

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16. ### gianx80New Member

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Thanks again for you help .

No problems with 8c, Vout = 60v. I had some problems with 8d. However I made this reasoning: the "+" control terminal is connected to ground, so we don't have any "gain", but we simply obtain -v1. In other words:

i1 = (v1-E1)/R1 = (120-(-120))/30 = (120+120)/30 = 240/30 = 8A

I checked the results with SPICE, however it's not all clear ... could you explain this concept more in depth?

Last edited: Mar 21, 2011
17. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Somehow you got the right answer anyway
In any case, we have a general procedure that can be applied to any circuit.

Since the polarity of Va is flipped compared to the way we usually calculate
the voltage across a resistor, we can flip Va if we also flip the source 2*Va.
That keeps everything the same except now the plus for Va is on top, and the
plus for 2*Va source is on the right. We dont have to flip anything but it
makes it a little nicer looking with the Va voltage the same as we have been
doing before. If we didnt flip anything it would just mean using different
polarities when we write the equation(s).
We can write an equation by assuming a current in the 30 ohm resistor flowing
from left to right. Starting with the 120v source and moving clockwise we get:
120-30*i+2*Va-i*15=0
Note all we did was follow the loop around and write a term according to what
element we encountered and what the polarity of that element was. It's a series
circuit so that's all that is required really.
Now we have an equation there with two variables, Va and i, but we notice that
Va is the voltage across the 15 ohm resistor, and the current through it is i,
so we can also write (using Ohm's Law):
Va=i*15
Now we can go back and substitute that in for Va in the former equation:
120-30*i+2*Va-i*15=0
we get:
120-30*i+2*(i*15)-i*15=0
or
120-30*i+2*i*15-i*15=0
which reduces to:
120-30*i+i*15=0
which reduces to:
120-i*15=0
which can be written:
120=i*15
and solving for i we get:
i=120/15=8 amps

That's a fairly general procedure where we just take Va as simply another variable to solve for
even if just temporary.

You can try another one similar to 8d now:
Take 8d and change the input source from 120v to 12v and solve for the current through the 15 ohm resistor.
Note that the polarities of dependent sources are twice as complicated as with a regular source because we
have two sets to think about. It's very important to get the polarities right or the results wont work out at
all. In fact, that could be an exercise to illustrate the difference.
Take 8d again only this time only flip Va across the 15 ohm resistor. That leaves the 2*Va source the
same with the plus on the left side. Solve for i this time and see the difference.

We should probably do one more like this, and the next circuit we do after that you will find very interesting i assure you

Last edited: Mar 21, 2011
18. ### gianx80New Member

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Thanks again .
You wrote:

but I can't completely understand this point. Why 30*i and i*15 are negative? I know, it's a stupid question but you know, sometimes the most obvious things are the most hard to understand.

For the 12v version of this exercise, the current i is 0.8A.

I think that in order to solve the last exercise proposed by you, I need to completely understand what I asked to you in the first part of this post

19. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

No it's not a stupid question. We never really did it this way before. You probably need a way to look at this based on what we already did before.

If you look at the 30 ohm resistor connected to the 120v source, if we were to write a nodal equation we would have to declare the terminal to the right of the 30 ohm resistor to be some voltage say v2. If we call the 120v source v1, then the voltage across the resistor we would call (v1-v2) and so the voltage arrow would be drawn from right to left (head left, tail right) across the resistor. This means when we used it in a nodal equation we would start with:
i=(v1-v2)/30
which equals
i=v1/30-v2/30
or:
30*i=v1-v2
or
0=v1-v2-30*i
or
v2=v1-30*i
So it's a positive current but it develops a voltage vR which gets subtracted from 120.

Another way to look at this:
Say we have 120v as before but we just have two resistors of 10 ohms each in series across that source.. The total current develops a voltage across both resistors, but the arrow of the 120v source points up (in the direction of the current) yet the voltage arrows of the two resistors points up too (in the direction opposite the assumed current flow direction) so the voltages of the two resistors get subtracted while the source is additive:
0=120-i*R1-i*R2
Note the units of each term are all in volts.

Does this help? We can do another circuit if you wish, no problem.

20. ### gianx80New Member

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I understood what you wrote, it was especially useful your comparison with nodal analysis .
Yes, I think it's useful to do another circuit so we could pass to that interesting circuit, the one that you mentioned earlier

P.S.
Thanks again for your irreplaceable patience

Last edited: Mar 24, 2011
21. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Oh you are welcome...it's nice to see your great interest in this subject area.

Next circuit we'll do is another series circuit with just plain old constant voltage sources and some resistors.

In the attached drawing, Fig. 9a, write the equations starting with the 14v voltage source, similar to how we did the last circuit, by following the loop around and writing each element equation. As we did in the last circuit, we dont really have to label the node voltages because it's a simple series circuit. We use the shorter method when possible.
The current arrow will be pointing out of the top of the 14v source. You can draw voltage arrows if you want or just do it mentally. Solve for the current out of the top of the voltage source in the direction pointing out of the top too. Be sure to watch the polarities of all the sources and the drawn or mental voltage arrows should be according to those individual polarities of course.
As you write each small equation (just one or two constants or variables) always keep in mind that the assumed direction of current is clockwise again. Follow this path of the current. When you reach the first terminal of a passive element (resistor) the voltage arrow will be such that the tip of the arrow is at that first terminal and the tail is at the other terminal. When you reach a voltage source however the voltage arrow is according to the polarity of the element: if the first terminal reached is positive then the tip of the arrow is at that first terminal too, but if the first terminal is negative then the tail of the arrow is at that first terminal. I hope this sounds clear but i think you get the picture.
As before, all of the element equations get summed to form one equation and this can be solved for the current 'i'.
What we are doing here is the application of, "The sum of the voltage drops around a closed path equals zero".
We are simply calculating each element voltage drop and summing them and equating that to zero. This allows us to solve for the current.

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