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#11
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The impedance should be R + jwL + 1/(jwC)
You sure?, how do the j parts cancel at resonance if they are both added? |
#13
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Which voltage and which current? Clearly the voltage across the
capacitor will always be pi/2 relative to the current in that capacitor, and the same (but opposite sign) for an inductor. Also, are you doing an AC or a transient analysis? If transient, are the transients settled, are you really on the resonance frequency, and are you simulating with sufficient time resolution? Oh, and I'm not quite sure what you mean by "lambda/4" phase shift. Is that degrees or radians, and what's lambda? Cheers, Tom (Diego Stutzer) wrote in message . com... Hi, Well, I'm really confused. I simulate a simple serial R-C-L-Network (all in series). As far as I know the total (input-)Impedance of the network is: Z = R + jwL - j/(wC) resp. the resonance frequency (where Zin=R) is 1/sqrt(L*C). At resonance frequency, the Impedance should be real and therefore in my hummel opinion Voltage and Current schould be in phase. The funny thing is, when i build up such a network in Schematics (Cadence PSD 14.1/Orcad 9.2) and simulate it with the PSpice A/D Simulator, the current is displaced (relative to the voltage) about lambda/4 - obviously not in phase!? Can anyone tell my where I made a mistake? Or why this Problem is showing up? Thanks to anyone reading this and especially to those who post answers. D. Stutzer |
#14
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![]() Don Pearce wrote: On 14 Oct 2003 11:38:29 -0700, (Diego Stutzer) wrote: Hi, Well, I'm really confused. I simulate a simple serial R-C-L-Network (all in series). As far as I know the total (input-)Impedance of the network is: Z = R + jwL - j/(wC) resp. -j/wc= 1/jwc |
#15
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![]() Don Pearce wrote: On 14 Oct 2003 11:38:29 -0700, (Diego Stutzer) wrote: Hi, Well, I'm really confused. I simulate a simple serial R-C-L-Network (all in series). As far as I know the total (input-)Impedance of the network is: Z = R + jwL - j/(wC) resp. -j/wc= 1/jwc |
#16
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Multiply top and bottom of 1/jwC by j (This does not change its value) and
you get 1/jwC = minus j/wC. Back to school with your algebra. |
#17
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Multiply top and bottom of 1/jwC by j (This does not change its value) and
you get 1/jwC = minus j/wC. Back to school with your algebra. |
#18
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On Wed, 15 Oct 2003 02:03:05 +0000 (UTC), "Reg Edwards"
wrote: Multiply top and bottom of 1/jwC by j (This does not change its value) and you get 1/jwC = minus j/wC. Back to school with your algebra. His algebra looks perfectly fine to me. But as others have pointed out, he's left the 2pi out. |
#19
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On Wed, 15 Oct 2003 02:03:05 +0000 (UTC), "Reg Edwards"
wrote: Multiply top and bottom of 1/jwC by j (This does not change its value) and you get 1/jwC = minus j/wC. Back to school with your algebra. His algebra looks perfectly fine to me. But as others have pointed out, he's left the 2pi out. |
#20
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![]() "budgie" wrote in message ... On Wed, 15 Oct 2003 02:03:05 +0000 (UTC), "Reg Edwards" wrote: Multiply top and bottom of 1/jwC by j (This does not change its value) and you get 1/jwC = minus j/wC. Back to school with your algebra. His algebra looks perfectly fine to me. But as others have pointed out, he's left the 2pi out. ------------------------------------------ Yes. I apologise for my remark about school. I gained the incorrect impression from the previous replies. The w in wC stands for omega = 2*Pi*F, the angular frequency. |
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