# Impedance matching w/ multiple secondary loads

Discussion in 'Homework Help' started by RogerMc, Feb 6, 2015.

1. ### RogerMc Thread Starter New Member

Feb 19, 2014
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0
Trying to get my head around this but making no progress. I've tried calculating the power on secondary with hope that I can solve by having the same power on the primary side. But without voltage it seems like I am missing a piece to the puzzle.

Of course I know that Zp = (Zs) x (Np / Ns) ** 2.

Perhaps I am confused because I don't know what Zs is in this case.

Can someone please point me in the right direction?

Thanks!

2. ### RogerMc Thread Starter New Member

Feb 19, 2014
12
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Here's my attempt ...

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3. ### crutschow Expert

Mar 14, 2008
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4,478
Roger, this is a Homework Help forum, not a We'll Do Your Homework For You forum. The OP needs to do the problem with our help.

You calculate Zp for each secondary winding separately (where R1 and R2 are Zs). Since the load appears as a parallel load to the primary, you then calculate the parallel value of the two values calculated, to give the final value of Zp.

4. ### RogerMc Thread Starter New Member

Feb 19, 2014
12
0
OK, I just thought you might want to see how I was attempting to solve the problem. Thanks for helping, I will try and solve in the manner you suggested.

5. ### RogerMc Thread Starter New Member

Feb 19, 2014
12
0
So

Zp for s1 = (10)(100/50)**2 = (10)(.5)**2 = 2.5 ohms
Zp for s2 = (25)(100/25)**2 = (10)(4)**2 = 400 ohms

Zp = 2.5 ohm in parallel with 400 ohm = 2.48 ohms

But if I choose a hypothetical voltage for Vp then the power calculations don't jive. Doesn't seem like 2.48 ohms is correct.

Assume Vp = 150V; then Vs1 = 75.5V & Vs2 = 37.5V
Ps1 = ((75.5)**2) / 10 = 570W
Ps2 = ((37.5)**2) / 25 = 56.25W
Pp = 570 + 56.25W = 626.25W

But (V*V)/R would be ((150)(150))/2.48 = 9072W

Seems like a Zp of 36 ohms would be closer than 2.48 ohms.

What am I missing?

Thanks!

6. ### RogerMc Thread Starter New Member

Feb 19, 2014
12
0
Decided it would be easier to select an arbitrary value for Vp and then perform power calcs on secondary side and then trace back to primary side (assuming it is desired for max power xfer). Result Zp was 36.36 ohms. Built a spreadsheet model and confirmed resistance of 36.36 ohm is consistent for all choices of Vp.

Thanks for helping ...

Mar 14, 2008
16,576
4,478

8. ### crutschow Expert

Mar 14, 2008
16,576
4,478
Zp for s1 = 10 (100/50)^2 = 10 (2^2) = 40Ω
40Ω in parallel with 400Ω is 36.36Ω, the value you calculated by other means.

RogerMc likes this.
9. ### RogerMc Thread Starter New Member

Feb 19, 2014
12
0
Thanks for catching that!

Now I have verification that my answer is correct ... Always like that.

Thanks for the assist ...