Operational Amplifier + Diode

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

1. gianx80New Member

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I draw the current arrow, I suppose it's correct. Thanks for the help with LTSpice, if I have any other problem, I'll open a new thread in another section ;-).

Thanks again

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Hello again,

Well that is very close, but for the current arrows we like to draw them inside the wire itself, as part of the wire, to sort of show that the current is inside the wire. The voltage arrows though we like to draw outside the component so that we can see the polarity of the voltage across the components.

So the next exercise is to draw the voltage arrows. The arrows will look like the arrow you have drawn in your picture there this time, outside the components (and wires). The head of the arrows simply align with the most positive part of the component, and the tail of the arrow aligns with the most negative part of the component. We have three components here, so there has to be three arrows. Note also that since the sum of voltage drops in a series circuit is zero there will be some arrows pointing with the direction of the current and some against it.
Refer to one of my previous drawings to refresh on how to draw the arrows.

I want to remind you that as i said before, we are moving fairly slow for now and covering some material that may seem redundant but after these few basic details we'll get to the equations and solve these circuits. It might also seem that there are other ways to do these things, and there are, but we need to establish some ground work to form a common base for discussion. In other words, this will all make a lot more sense very shortly

3. gianx80New Member

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Hi, excuse me for the late reply, but yesterday I took an examination and it went very well. Now I'm back to electronics ;-). Ok, I modified the image ... is it now correct?

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4. DaveNew Member

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Hi again,

Oh yes that's very good. I see you got the arrows correct for the voltage source, which looks opposite to the resistor arrows. That's a good way to think of it.

Next all we want to do is label each node so we can start writing equations. I'll assume you can do this because im sure you can, so we'll start with the top node and work our way down. Before we do that though, it helps to identify a common node which we can call zero volts. This is usually known as 'ground'. The negative side of the source looks good, so we'll call that 0. That makes the lowest node (dot) become the zero (0) node.
Ok back to the top, we'll all that upper most node v1, and the center node v2. We also label the current (the arrow inside the wire) as i1.
Thus so far we have voltage nodes v1, v2, and 0, and we have current i1.

Next all we do is use Ohm's Law for each voltage arrow. When we 'know' the voltage at each side of a resistor we know the voltage difference and since we 'know' the current is i1 we can write equations that relate all of this together.
The arrow to the left is obviously equal to V so that's labeled V, but for the other two arrows we equate the current though the element to the voltage across the element.
Using Ohm's Law I=V/R we get different equations for each resistor as shown in the first attachment.

Does this make sense so far?

If it does, take a look at fig 3e in the second attachment. As you can see, i've added a resistor R3. This brought in a new node too.
What you should do now is take that drawing and label the new node in the same way the other nodes are labeled, then write the equation for that voltage arrow in the same way the others are done.
The voltage arrow equation is written by dividing the voltage difference across the resistor by it's resistance and since that equals the current fllowing through it we simply equate that to the assumed current i1.
Notice you'll also have to fix the second equation because that other voltage changed (it's no longer 0).

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6. gianx80New Member

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I think I understood what you wrote ;-). I modified the image, I hope it's correct.

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Hello again,

Oh yes, very good. You seem to be catching on quick.
Now we jump back to the 2 resistor circuit, but now i've added two more resistors in the circuit, R3 and R4. This introduces another node so we have to label that, plus the connection is in parallel to the other two resistors so the current now splits into two currents which we are calling i1 and i2.
Can you figure out how to label the new node and the two new voltage arrows?
(Note that the second 'ground' 0v node is not really a new node, it's drawn for reference. The two nodes form was is sometimes called a "super node" but it's no big deal really, we didnt have to draw that node over again to the right).

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8. gianx80New Member

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Ok, I tried to do what you asked. Is it correct what I wrote?

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Hi again,

Oh yes, that's right, very good really. Now one more and then we take off What we are doing here is basic to all the circuits we will do so it's good to get this out of the way.

This next one might seem a little stranger but all we have to remember is that when the current flows into one end of a passive element (resisitor) it induces a voltage that is most positive at that end of the element. There are two ways to show this though when we dont know the polarity of the voltages yet. Thus we assume on direction and then that shows us the way to draw the voltage arrow too. So we assume a current direction (and draw the current arrow), then draw the voltage arrow, then write the equation (as we have done with the other resistors already) noting which voltage is at the head of the arrow and which is at the tail of the arrow. Put another way, sometimes we will have two choices for the current arrow and in that case we simply pick one and go with that. We just keep in mind that once we decide which way we will indicate the current flow, the voltage arrow (polarity) must follow suite.

So given the above do you want to try this next one? If you have any problem with this one that's ok because once you see how this is done you'll then know how this is done for any circuit. After this we will be ready to write the equations for the entire circuit and then of course we can solve the circuit.

(One little note, i moved one of the equations up a little to make room for the new equation so i drew a line from that equation to it's associated voltage arrow as you can see in the diagram).

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10. gianx80New Member

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I tried to do what you asked, I hope it's correct!

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Hello again,

Oh yes, that's absolutely correct. The new resistor introduces a new current and a new voltage so another equation too. You got that perfect.

What *i* neglected to do however was to indicate that when i added another resistor that the two currents in R2 and R4 had to be labeled as new currents too. Not too much of a problem however, as i simply took your new diagram and corrected my own error by labeling the two resistors R2 and R4 with the new currents i4 and i5. You see how the current splits with the new resistor? That means we have different currents through those two resistors now. No big deal though, as i have attached the new diagram which includes your recent work with the new resistor R5.

Ok, now it's time to write the equations for the circuit. Up till now we've written equations for each circuit element. Now it's time to combine all those results.

First we compile a list of all the equations we already wrote for each element. We get this list:

i1=(v1-v2)/R1
i2=(v1-v3)/R3
i3=(v2-v3)/R5
i4=(v2-0)/R2
i5=(v3-0)/R4

Note that this is just the list of what we already found for each element, and some of these may simplify immediately (such as i5=v3/R4).

Next we need to write equations for each node (nodal analysis), and to accomplish this we use KCL (Kirchhoff's Current Law) which states that the algebraic sum of currents entering a node equals the sum of currents leaving a node. What this means in simple terms is that if we have a current entering a node of say 1 amp, then we have to have one or more currents that are leaving the node who's sum totals 1 amp too. Alternately, if we have two currents entering the node that sum to 1 amp then we have to have at least one node that has 1 amp leaving. There are many possibilities here, but all we have to do is simply add all the currents entering the node and subtract all of the currents leaving the node and equate that to zero. We do this for all of the nodes we dont yet know the voltage of.

(see new diagram)
Starting with the node labeled v1, we really already know that this node equals the source voltage, so we dont have to equate that one. We do have to do v2 and v3, so lets do that now.

Starting at node v2, there are three currents associated with that node, i1, i3, and i4. The current arrow for i1 points toward the node, so we add i1. The current arrow for i3 points away from the node, so we subtract i3. The current arrow for i4 also points away from the node, so we subtract i4 also. We equate these currents to zero like this:
+i1-i3-i4=0
and as you can see the ones we added got plus signs and the ones we subtracted got negative signs, and they all were equated to zero.
What you see here is the magic of nodal analysis. By simply equating the node currents like this we can solve the circuit.

We still have another node to do, the node v3. We do that one the same way. Since i2 points toward the node and i3 points toward the node and i5 points away from the node we end up with:
+i2+i3-i5=0

Now we are done with most of the work. Everything else is just substitution and then we solve the system of equations to get the answers.
The two simultaneous equations look like this so far:
i1-i3-i4=0
i2+i3-i5=0

but to solve them we need to do a few substitutions. We have to go back to our list of element equations and replace every instance of every current with it's element equation.

Since the current i1 had equation:
i1=(v1-v2)/R1

we substitute that into the two system equations, and we get:
(v1-v2)/R1-i3-i4=0
i2+i3-i5=0

Next we take that set and do the same for i2, and we get:
(v1-v2)/R1-i3-i4=0
(v1-v3)/R3+i3-i5=0

Of course we do this for i3 also and we get:
(v1-v2)/R1-(v2-v3)/R5-i4=0
(v1-v3)/R3+(v2-v3)/R5-i5=0

Following that procedure with i4 and i5 we end up with:
(v1-v2)/R1-(v2-v3)/R5-(v2-0)/R2=0
(v1-v3)/R3+(v2-v3)/R5- (v3-0)/R4=0

and all we do now is simplify those a little:
v3/R5-v2/R5-v2/R2-v2/R1+v1/R1=0
-v3/R5+v2/R5-v3/R4-v3/R3+v1/R3=0

and group terms according to the variables for the nodes v2 and v3, and we get:
v3/R5-v2*(1/R5+1/R2+1/R1)+v1/R1=0
-v3*(1/R5+1/R4+1/R3)+v2/R5+v1/R3=0

We can immediately simplify because we know what v1 is, so we replace v1 with the known voltage V and we get:
v3/R5-v2*(1/R5+1/R2+1/R1)+V/R1=0
-v3*(1/R5+1/R4+1/R3)+v2/R5+V/R3=0

and this is just a system of two equations in the two unknowns v2 and v3. All we have to do now is solve the system for v2 and v3 and we will have our solution this this circuit. We can do this algebraically but the result will come out a little long, but if we substitute the values for all the resistors and for the voltage source V we can do it a bit more straightforward.

Ok so next you can substitute the value for V as 2 volts, and the values for the resistors as follows:
R1=1
R2=2
R3=3
R4=4
R5=5
(each resistor has a value in ohms equal to it's enumeration).

So take the last result of our system of equations and using those values above solve for v2 and v3. We'll compare results next.

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12. gianx80New Member

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Ok, so, if I correctly solved the system, v2 = 1.32v and v3 = 1.19v

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Hello again,

Well before i reply to that question, let me ask one if you dont mind. Did you check your results using the circuit simulator? What were the results of the simulation?
The reason i ask this is because when no one is around to ask you'll still have a way to verify your analysis results if you use the simulator in combination with your circuit works.
Also referring to the equations we already wrote for the currents, can you calculate the total current flowing from the voltage source to the circuit?
One more, can you calculate the current flowing through R1?
I think you'll get these two but just in case you have a problem we'll talk about it.

Ok, we now move to another circuit which is just slightly more complicated. In the new diagram i have added two more resistors, so the current splits again at the two nodes it is connected to.
We are given the voltage source voltage and the value of all the resistors again. The voltage will again be 2v and the resistors will again have values the same as their enumerations (R1=1 ohm, R2=2 ohms, R3=3 ohms, etc., as before). You are asked to solve for the unknown node voltages.

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14. gianx80New Member

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I didn't do the simulation. However I've just done it and it confirms my results . However the current flowing through R1 is 0.68mA (verified with spice). I guess that the total current is i1+i2+i3+i4+i5 = 0.68A + 0.27A + 0.026A + 0.66A + 0.2975 A = 1.93 A.

I solved the last proposed exercise. The nodes have been already marked by you. I drew the current and voltage arrows as I did in the previous exercise. Then I labelled the nodes and the arrows. After that I wrote the current equations, then I wrote the equations at the nodes. I solved the system of three equations in three variables. Then I calculated the currents flowing through all the components. At the end I verified my results with SPICE.

Equations at the nodes:
i1 = (v1-v2)/R1
i2 = (v1-v3)/R3
i3 = (v2-v3)/R5
i4 = (v2-v4)/R2
i5 = (v3- 0)/R4
i6 = (v4-v3)/R7
i7 = (v4- 0)/R6

System:
+i1-i3-i4 = 0
+i2+i3+i6-i5 = 0
+i4-i6-i7 = 0

System with substitutions:
+(v1-v2)/R1-(v2-v3)/R5-(v2-v4)/R2 = 0
+(v1-v3)/R3+(v2-v3)/R5+(v4-v3)/R7-(v3- 0)/R4 = 0
+(v2-v4)/R2-(v4-v3)/R7-(v4- 0)/R6 = 0

System results:
v2 = 1.7039V
v3 = 1.2850V
v4 = 1.2792V

Currents flowing through the components:
i1 = 0.2961A
i2 = 0.2383A
i3 = 0.0838A
i4 = 0.2123A
i5 = 0.3212A
i6 = -0.0008A
i7 = 0.2132A

Now, look at i6 current. It is negative. However I think that it has to be positive. I verified with SPICE and the correct value seems to be 837,696 uA (micro Ampere) ... a positive value. The equation for i6 current is:

i6 = (v4-v3)/R7

so, substituting the correct values, I have:

i6 = (1.2792 - 1.2850)/7 = -0.0008A

that is a negative current. Could you explain me where is the error? Is this current so small that has not to be considered? I was thinking: if I round off my voltage values to 2 decimal places, I find that there is no current flowing through R7.

P.S.
Thanks again for your irreplaceable help

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Hello again,

Oh yes, very good! You've gotten all the right results and you see you verified this yourself so you have a way to check your answers when you do a circuit. And about i6 being negative and small, that is something we should talk about.

If you look at this next diagram (Fig. 5) you'll see that [A] shows the current i6, and that it worked to to be a negative current equal to -I. In , i've labeled it as a negative current as you can see, and that would mean the voltage arrow would be indicating a negative voltage. In [C], i turned the current arrow around so it now points to the left, and that allows us to change the polarity of the current I to positive. Since the current is now positive that means we should turn the arrow around if we want it to indicate a positive voltage (as shown in [D] ).
So you see our original assumed direction for the current was wrong so the current came out negative, and since the current arrow shows direction that means that we have the following equivalency:
"A positive current flowing in the positive direction is the same as a negative current flowing in the negative direction".
Remember that the arrow shows direction and the polarity of the current can be plus or minus, but it can all be shown as positive current in the positive direction. When we do end up with a negative current like this all we have to do is flip the arrow and change the polarity and then flip the voltage arrow too. This then also means that a negative current flowing in the positive direction creates a negative voltage drop across a resistor, but we can always adjust to our liking.
If this is not clear we can do a couple more examples to illustrate this concept again. If you have any questions about this, now is the time to ask.

Back to figure 3g, where we have the 5 resistors in the circuit and one voltage source...
When we talk about the 'total' current from the source we talk not about the sum of all the currents, but only that current flowing through the source itself. If you look at the top node of the circuit you will see three currents associated with that node. We have i0 from the source, and i1 and i2 flowing through resistors. Since i0 is entering the node and i1 and i2 are leaving the node we end up with:
i0-i1-i2=0
and if we rearrange that just a bit we get:
i0=i1+i2
Thus, the i0 current is equal to simply the sum of i1 and i2. When we solve the circuit we can solve for i1 and i2 and then we can simply add them and get the value of i0. This would be called the total current, but i should have told you that this is really i0 as the phrase "total current" can be misleading.
Ok, so with that in mind, you should next calculate i0 for that five resistor circuit, and also i0 for the new seven resistor circuit. This current is often of great importance because we often have to know the power that the source is delivering to the circuit. This brings us to the subject of power while we are at it...
Knowing this current i0 for both circuits and the source voltage (as given as 2v for both) can you calculate the power that the source delivers to all the resistors for each circuit?

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16. gianx80New Member

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Ok, I understood what you wrote. You said that we can adjust the polarity because "A positive current flowing in the positive direction is the same as a negative current flowing in the negative direction". But you also said that we'll end up with a negative voltage drop across the resistor. So, with "negative voltage drop" do you mean the fact that the voltage arrow is now flipped over?

For the five resistors circuit the total current is (keeping in mind i1=(v1-v2)/R1 and i2=(v1-v3)/R3):
i0 = i1+i2 = 0.68A + 0.27A = 0.95A (SPICE tells me that this current should be negative ... Do I have to change any polarity?)

For the seven resistors circuit the total current is (keeping in mind i1 = (v1-v2)/R1 and i2 = (v1-v3)/R3):
i0 = i1+i2 = 0.2961A + 0.2383A = 0.5344A

I verified with SPICE and these results should be right.

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Hello again,

Yes that is very good indeed. That's both currents accurate enough for many purposes. I think you have grasped the equation writing very very well so you wont find it very difficult to do the more complicated circuits we will look at because we just follow those basic rules.

Yes, some spice will show a negative current because the software assumes electron current flow, which is the flow of electrons from the negative battery terminal, though the circuit, and back to the positive terminal of the battery.
We are using conventional current flow which is from the positive battery terminal through the circuit and back to the negative terminal.

If we draw a current arrow pointing to the right we draw the voltage arrow pointing to the left. If the current is positive then the voltage arrow indicates a positive voltage. Without changing the current arrow if we still draw it pointing to the right but it is a negative current, then the voltage arrow indicates a negative voltage. So the voltage arrow always takes on the polarity of the current.
Because of this it is sometimes best to flip the voltage arrow if you flip the current arrow and that keeps everything the same. For the negative current pointing to the right we end up with the negative voltage pointing to the left, but when we flip the current arrow the current is then positive, so we flip the voltage arrow too so it keeps the same polarity as that of the current.
If this is not clear that's ok. I find that this is one of the hardest things for people to grasp when they first encounter it, but once they do a few examples it starts to make sense. We can do some more exercises too where we focus on this concept and that would help im sure.
As i said, if you still have questions, that's ok because that's why we are here today
Once we get past this we start to get into some interesting circuits.

BTW, in your study of math did you cover complex numbers yet (like X=a+b*j, or X=3+7*j)?

18. gianx80New Member

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I think I understood this concept, however I'll master it with more exercises.
Yes, I've already studied complex numbers in that form, however I'm not so familiar with them.

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Hello again,

Oh ok, well would you like to do a couple more circuits then?

We can refresh on complex numbers, that's no problem. We'll need them when we move to capacitors. All that we have done so far will be found to apply to capacitors as well, except we need complex numbers to help us.

Quick review:

Complex numbers are said to have a 'real' part and an 'imaginary part'. The abbreviations for real and imaginary are varied, and here are a few:
real, imag
re,im
re,img
realpart,imagpart

I like to use 'real' and 'imag' when talking about these because that makes it pretty clear yet still a little shorter.

Complex numbers have two parts, 'real' and 'imag'. This is often written as the sum of a real part and an imag part like this:
3+2j
or
3+2*j
where we can see that this is not just an addition but the little 'j' is multiplying the 2 there (in other pure math writings the letter used is 'i', but that gets confusing in electrical work so the lower case 'j' is used instead).

It helps to know what j really is. j is just the square root of minus 1, which could be written as "sqrt(-1)". Note that the square root of a negative number is not defined for real numbers, but it's interesting that if you take that 'j' and square it, you get a real number again. We know that if you square the square root of a number you get that number back, so if we do this with minus 1 (-1) we simply get that number back. Thus, (sqrt(-1))^2=-1 as expected. In this way we say that j^2=-1.
There is of course physical interpretation, but we'll get to that later sometime. For now a little table can help us here:
j^1 = sqrt(-1)
j^2 = -1
j^3 = -1*j=-j
j^4 = 1
Note that if you read down the right side of all these, you'll read
sqrt(-1), -1, -j, 1
which are four different results. This pattern repeats for powers above 4, so that j^5 for example is again equal to -1.

Ok, so the general form for a complex number is a+b*j, where 'a' is the real part and 'b' is the imag part.
Since we are calling this combination of terms a single entity, we can handle it as if it was a single number in some ways so we can define math operations with complex numbers just like we do with real numbers.
If we let X=a+b*j and Y=c+d*j, then we can define these:
SUM: X+Y=(a+c)+(b+d)*j and here a+c became the real part and b+d became the imag part.
PRODUCT: X*Y=(a*c)+(a*d*j)+(b*j*c)+(b*j*d*j)=(a*c)+(a*d*j)+(b*j*c)+(b*d*j^2)=(a*c)+(a*d*j)+(b*j*c)-(b*d)=(a*c-b*d)+(a*d+b*c)*j
Note that with the sum all we did was add the 'components' of the first complex number to the second complex number.
With the product, we had to cross multiply.

Here's a few exercises:
1. Add and multiply 2+3*j and 6+7*j
2. Add and multiply 9+2*j and 7-1*j (note the negative sign)
3. Add and multiply 9+2*j and -1+3*j (again the sign)
4. Add and multiply -3-2*j and -4-8*j (note all the signs)

Note of course that you will get two answers for these four so that will be eight answers total.

A few words on notation...
Sometimes a number like 2+3*j is written as 2+3j or even as 2+j3 where the j is placed before the imag part (the 3 here).
Another form is 2+j*3. The 'j' is just used as an algebraic multiplier who's value is not yet determined completely.

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20. gianx80New Member

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Yes, if it's not a problem for you, I'd be happy to do a couple more circuits .
However, here there are my solutions for your exercises about complex numbers:

1. SUM: 8+10*j
PRO: -9+32*j

2. SUM: 16+1*j
PRO: 65+5*j

3. SUM: 8+5*j
PRO: -15+25*j

4. SUM: -7-10*j
PRO: -4+32*j

I hope that these results are correct.

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Hi again,

Ok very good for sure. Since you got all of those correct then i assume you understand multiplication and addition and thus subtraction (because subtraction is just a multiplication of -1 times a complex number before adding to to another one).
Now division is a little different, as you may know.

For the two complex numbers X and Y:
X=2+3*j
Y=5+2*j

if we want to divided X by Y we end up with X in the numerator but we have a complex number in the denominator. This presents a slight problem because the fraction does not simply break into two fractions like it could with the numerator. This means we have to do something special.

First we define the complex conjugate of a complex number, as simply the same number only with the sign of the imag part changed (ie multiplied by -1). This is only the imag part though, we leave the real part alone. Thus the complex number X above comes out to:
again X=2+3*j, and the conjugate of X is: 2-3*j. If X was 2-3*j, then the conjugate would have been simply 2+3*j as all we do is change the sign of the imag part.

There is a very interesting property of the complex conjugate too. This property is that when we multiply the conjugate times the original complex number we get a number that is purely real (ie no imag part). To see how this works we only need to multiply a number X by its conjugate and note the result. We will denote the complex conjugate of the number X by "conj(X)". In much literature, you'll find the complex conjugate denoted with an asterisk such as X*, but that gets a little confusing because we use that for multiplication. That would mean X*X* or X**X would mean the multiplication of X by conj(X) so the notation would look sloppy.
Ok, so we take a number X and multiply it by conj(X) and we get a purely real number. Since it is a purely real number, it simplifies the division so now we have a way to solve the division problem. We simply multiply the numerator and denominator by the conj(denominator) and after we simplify we have a single complex number again that is not a complex fraction.
Lets do an example.
We want to solve X/Y for X and Y given above as:
X=2+3*j
Y=5+2*j
Since Y is the denominator, we calcualte conj(Y)=5-2*j. Next we multiply that times the numerator X and denominator Y and we get:
Numer=X*conj(Y)
Denom=Y*conj(Y)
First the Denominator:
(5+2*j)*(5-2*j)=29 (note this is purely real as mentioned above)
Now the Numerator:
(2+3*j)*(5-2*j)=16+11*j
and now finally we perform the division:
(16+11*j)/29=16/29+11/29*j
and so the number X/Y comes out to one single complex number.
The answer expressed as floats looks like this approximately:
0.552+0.379*j

I suspect that you will pick this up quick or else you are just refreshing anyway, so here's a few problems to solve:
For X=9+2*j and Y=3-4*j, find:
1. The complex conjugate of X.
2. The complex conjugate of Y
3. The division, X/Y
4. The division, Y/X
5. The division of X/Y followed by the reciprocal of that result (you can use the result of #3 above).

I'll draw up a few more circuits you can test your knowledge with. Keep in mind that if you see a configuration that looks confusing to you somewhere, you can always try to solve it yourself and test your results out in the spice environment too. That's really good practice. Assign resistor values that are not the same to test your results, or if you feel up to it use a random number generator to generate the resistor values to use for the test circuit. It's also a good idea to test with more than one set of resistor values too to help reduce coincidental solutions that are not true solutions.

Last edited: Mar 10, 2011