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Sci.Electronics.Basics -> How to solve this circuit
There are 4 messages in this thread.
You are currently looking at messages 1 to 4.
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Author: Dan CobyDate: 18:14 21-04-08
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Author: Dan CobyDate: 01:24 24-04-08
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<lionelgreenstreet@gmail.com> wrote in message
news:abcb5809-a004-4fa6-a744-44f0512dd88c@a23g2000hsc.googlegroups.com...
>> Set Rs to 0 and then do your calculations.
>
> ok, i tried but my procerure is wrong....Can you say me how obtain
> Pout?
Yes, I also get a different result.
If Rs is 0 then it is a simple short. Note: The problem stated that Rs is much
less that RL. To really assume that Rs is negligible then we also need to
assume that Rs is less than 1 / (w * CL) for range of frequencies that are being
considered. At very high frequencies, this may not be true.
Then the circuit reduces to a simple parallel combination of the current source
Ip(w) with Rj, RL, Cj, and CL. Define Req = Rj * RL / (Rj + RL) and Ct = Cj = CL.
Then the impedance of the capacitors is Zc = 1 / s (Cj + CL).
(Where s is the Laplacian operator which is commonly treated as sqrt(-1) * w.)
The impedance of the parallel combination of Req and Zc is:
Zt = Req * Zc / (Req + Zc)
or Zt = Req / (Req/Zc + 1)
The voltage across this impedance (which is also the voltage across RL) is:
Vt = VL = Ip(w) * Zt
The power in RL is: Pout(w) = VL^2 / RL
or Pout(w) = Ip(w)^2 *Zt^2 / RL
It looks like they are simply using the magnitude of the impedance, i.e.
mag(Zt) = sqrt((Req^2 / (w * Ct))^2 / (Req^2 + 1 / (w * Ct)^2))
or mag(Zt) = sqrt(Req^2 / (Req^2 * (w * Ct)^2 + 1))
Using the magnitude of Zt we get the following for Pout:
or Pout(w) = Ip(w)^2 * Req^2 / (RL * (Req * (w * CT)^2 + 1)
Which differs from their equation by a factor of (Req^2 / RL)
I think that my result is correct. If Rj is infinity then the two equations are
equal. If Rj is zero then Ip is shorted and there should be no power in RL
however their equation gives Ip^2 * RL.
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