Transformers and Power: How Can Impedance Matching Improve Power Transfer?

[Aristotle]

Homework Statement

Homework Equations

Z_Impedance = Z_Load / n^2

The Attempt at a Solution

Hello everyone,
I understand that in order to provide maximum power to load, the load resistance (R_L) has to match the source resistance R_S. My main confusion is from the question itself on what my teacher meant by using "one additional component".
What I'm thinking is to change the load on the right circuit to frequency domain and add the two impedance. From there I would be able to add a impedance component on the left circuit to match the two, correct?
Can somebody possibly lead me to the right direction in this problem? Thank you :)[/B]

 

[gneill]

Your load has both resistance and reactance (an inductor) so the load is a complex impedance. The component you add to the primary will have to deal with matching the imaginary (reactive) part.

Take a look at the topic: "transformer impedance matching" or "transformer impedance reflection".

Hint: Start by choosing a transformer turns ratio to match the resistance in the source to that of the load.

 

[Aristotle]

Your load has both resistance and reactance (an inductor) so the load is a complex impedance. The component you add to the primary will have to deal with matching the imaginary (reactive) part.

Take a look at the topic: "transformer impedance matching" or "transformer impedance reflection".

Hint: Start by choosing a transformer turns ratio to match the resistance in the source to that of the load.

Okay so I did some reading on that topic of impedance reflection and from doing some math acknowledging that the Z_load = v2/i2 I came to realize that the impedance seen by the primary coil is equivalent to Z_L/(n^2). So what I did is I first found the impedance of the load by adding them (in series) & got Z_L= 100 + j2.05.
How would I choose a transformer turns ratio arbitrary if you don't mind me asking? Thank you for your reply!

 

[gneill]

How would I choose a transformer turns ratio arbitrary if you don't mind me asking?

Take a look at the hint I gave. The transformer ratio that you choose should match the resistances according to the maximum power transfer theorem.

 

[Aristotle]

Take a look at the hint I gave. The transformer ratio that you choose should match the resistances according to the maximum power transfer theorem.

Hmm, well the transformer ratio (n) relies on the voltage across the secondary coil to voltage across primary..how would this correlate with the resistances?
Maximum power transfer theorem states that the load impedance is equivalent to the conjugate of thevenin impedance.

 

[gneill]

How is the load impedance reflected into the primary? What's the relationship?

 

[Aristotle]

How is the load impedance reflected into the primary? What's the relationship?

Z_Imp= Z_L/(n^2) soo Z_imp / Z_L = 1 / n^2
We know Z impedance and Z load...so we could potentially solve for n.

 

[gneill]

Yes. As I've mentioned, start by considering just the real part of the impedances in order to fix the ratio.

 

[Aristotle]

Yes. As I've mentioned, start by considering just the real part of the impedances in order to fix the ratio.

So the component is 92 - j2.05 ohms then ?

 

[gneill]

So the component is 92 - j2.05 ohms then ?

I'm not sure what you're showing me there.

You want the reflected load resistance to match the source's resistance. What turns ratio will make the 100 Ohm load resistance "look like" 8 Ohms from the primary's viewpoint?

 

[Aristotle]

I'm not sure what you're showing me there.

You want the reflected load resistance to match the source's resistance. What turns ratio will make the 100 Ohm load resistance "look like" 8 Ohms from the primary's viewpoint?

If we only consider the real parts, then the turn ratio is just 1

 

[gneill]

No, then the load's 100 Ohms would look like 100 Ohms from the primary. That would not match the source's 8 Ohms.

 

[Aristotle]

No, then the load's 100 Ohms would look like 100 Ohms from the primary. That would not match the source's 8 Ohms.

Solving for n I got an answer of 1.04.

Work:
Zimp/Zload = 1/(n^2). n= sqrt(100/92) = 1.04

So turn ratio is 1: 1.04

 

[gneill]

Where does the 92 come from? The source resistance is 8 Ohms, the load impedance is 100 Ohms. You want the load resistance to look like 8 Ohms when "viewed" from the primary. Then it will satisfy the maximum power transfer theorem.

 

[Aristotle]

Where does the 92 come from? The source resistance is 8 Ohms, the load impedance is 100 Ohms. You want the load resistance to look like 8 Ohms when "viewed" from the primary. Then it will satisfy the maximum power transfer theorem.

Ooh right. 1:3.53

 

[gneill]

That's better

Now, using that same ratio, what does the load's inductive impedance look like from the primary?

 

[Aristotle]

That's better

Now, using that same ratio, what does the load's inductive impedance look like from the primary?

Since Zimp/ZL = 1/3.53, solving for Zimp, we get:
Zimp = 100-j2.05 / 3.53
= 28.33 - j .5807

This is what the load impedance look like from primary

 

[gneill]

Okay, a couple of things. First, note that your 100 Ohms is now 28.3 rather than 8. This is because you forgot to square the turns ratio.
Second, the impedance of an inductor is a positive imaginary value: Z_L = jωL. It's a capacitor that ends up with a negative imaginary value.

So, fix the sign of the inductor impedance and square the turns ratio and give it another go.

 

[Aristotle]

Okay, a couple of things. First, note that your 100 Ohms is now 28.3 rather than 8. This is because you forgot to square the turns ratio.
Second, the impedance of an inductor is a positive imaginary value: Z_L = jωL. It's a capacitor that ends up with a negative imaginary value.

So, fix the sign of the inductor impedance and square the turns ratio and give it another go.

Woops pretend you did not see that
8.025 - j .1654

Oh that makes sense.. Capacitor impedance is -j*1/(wc)..
So if we equate -j.1654 to -j*1/(wc) we can find the capacitor value!

 

[gneill]

Yes!

 

[Aristotle]

Yes!

You're a great teacher thank you!
So the question that the problem wants is to draw the circuit. Would I have to draw it as an ideal transformer circuit with the core in the middle or equivalent circuit?

 

[gneill]

You're a great teacher thank you!
So the question that the problem wants is to draw the circuit. Would I have to draw it as an ideal transformer circuit with the core in the middle or equivalent circuit?

It sounds like they want a diagram showing all the components. That would be your original sketch with the transformer and capacitor inserted and labeled appropriately.

 

[Aristotle]

It sounds like they want a diagram showing all the components. That would be your original sketch with the transformer and capacitor inserted and labeled appropriately.

That's the part that makes me scratch my head..we are also adding 2 coils..or 2 inductors side by side to create this transformer circuit. Would these inductors need a value?

 

[gneill]

That's the part that makes me scratch my head..we are also adding 2 coils..or 2 inductors side by side to create this transformer circuit. Would these inductors need a value?

No, you need only specify the turns ratio.

 

[Aristotle]

I know I didn't label the capacitor value yet. But here is what I have so far in terms of drawing it
I'm assuming we can also assign dots and current directions if we wanted to, but that wasn't specify.Update my capacitor value is -.148 F

 

[gneill]

View attachment 96922 I know I didn't label the capacitor value yet. But here is what I have so far in terms of drawing it
I'm assuming we can also assign dots and current directions if we wanted to, but that wasn't specify.

You could, but it's not necessary. Your diagram looks fine (save for the missing capacitor value).

Update my capacitor value is -.148 F

Note that the capacitance is positive quantity. It's the impedance of it that's negative.

 

[Aristotle]

You could, but it's not necessary. Your diagram looks fine (save for the missing capacitor value).

Note that the capacitance is positive quantity. It's the impedance of it that's negative.

Well I knew that the impedance looking through the primary got us 8.00 + j.16400.
So I equated j.16400 to 1/(jwC) and found that C was equal to -0.148 F.-0.16400 = 1/(41C)
C = negative value

 

[gneill]

Given: $Z_C = -j~0.164~Ω$

Then:
$Z_C = \frac{1}{j ω C}$

$C = \frac{1}{j ω Z_C}$

$~~~ = \frac{1}{j ω (-j~0.614~Ω)}$

$~~~ = \frac{1}{ω (0.614~Ω)}$

 

[gneill]

Note that a real capacitor constructed from real materials cannot have a negative capacity. Take a look at the equation for the parallel plate capacitor: In order for the capacitance to be negative the permittivity of space ε_o would have to be negative; it's the only part of the expression that isn't a simple geometric quantity. Since ε_o is not negative, neither is the capacitance of a real capacitor.

 

[Aristotle]

Note that a real capacitor constructed from real materials cannot have a negative capacity. Take a look at the equation for the parallel plate capacitor: In order for the capacitance to be negative the permittivity of space ε_o would have to be negative; it's the only part of the expression that isn't a simple geometric quantity. Since ε_o is not negative, neither is the capacitance of a real capacitor.[/SUB]

I see..I appreciate the big picture. So the reason why you changed the imaginary part seen from primary to load to "-j.16400" is due to the maximum power theorem correct?