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Chapter5

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‫ﻓﺼﻞ ﭘﻨﺠﻢ‬ ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﺍﻳﻦ ﻣﺒﺤﺚ ﺩﺭ ﮐﺘﺎﺏ ﻧﻴﺴﺖ . ﺍﺯ ‪ Cooper‬ﻳﺎ ﺳﺎﻭﻧﻲ ﺑﺨﻮﺍﻧﻴﺪ.‬ ‫1‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ)ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫2‬
‫ﻣﻘﺪﻣﻪ :‬ ‫ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻳﻚ ﺑﺎﺭ ﻣﻘﺎﻭﻣﺘﻲ ﺩﺭﻳﻚ ﺷﺒﻜﺔ ‪ DC‬ﻣﻲ ﺗﻮﺍﻧﻴﻢ‬ ‫ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﻳﻢ . ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺩﻭ ﺭﻭﺵ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ‬ ‫‪Ic‬‬ ‫ﺑﮑﺎﺭ ﺑﺮﺩ .‬ ‫‪Ic‬‬ ‫‪IR‬‬ ‫+ _‬ ‫‪A‬‬ ‫‪Ip‬‬ ‫‪+ V‬‬ ‫‪c‬‬ ‫‪VR‬‬ ‫+‬ ‫‪V‬‬ ‫_‬ ‫2‬ ‫‪Vp‬‬ ‫_‬ ‫+ ‪IR‬‬ ‫‪R‬‬ ‫_‬ ‫‪V‬‬ ‫‪Ip‬‬ ‫+‬ ‫‪V‬‬ ‫‪A‬‬ ‫‪Vp‬‬ ‫_‬ ‫1‬ ‫2‬ ‫‪VR‬‬ ‫= ‪PP‬‬ ‫‪RV‬‬ ‫3‬ ‫2 ‪PC = R A I R‬‬ ‫‪Pm = I C V P = ( I R + I P ) V R = PR + PP where‬‬ ‫)1‬ ‫‪Pm = I C V P = ( V R + V C ) I R = PR + PC where‬‬ ‫)2‬
‫ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﻣﻼﺣﻈﻪ ﻣﻲ ﺷﻮﺩ ﺩﺭ ﻫﺮ ﺩﻭ ﺭﻭﺵ ‪) Pm‬ﺗﻮﺍﻧﻲ ﮐﻪ ﻣﻲ ﺧﻮﺍﻧﻴﻢ(‬ ‫ﺑﻴﺶ ﺍﺯ ‪) PR‬ﺗﻮﺍﻧﻲ ﮐﻪ ﻭﺍﻗﻌﺎ ﺩﺭ ﺑﺎﺭ ﻣﺼﺮﻑ ﻣﻲ ﺷﻮﺩ( ﺍﺳﺖ .‬ ‫ﺍﮔﺮﺟﺮﻳﺎﻥ ﺑﺎﺭ ﺯﻳﺎﺩ ﺑﺎﺷﺪ ﻭ ﻭﻟﺘﺎﮊ ﺁﻥ ﻛﻢ ﺭﻭﺵ ﺍﻭﻝ ﺑﻬﺘﺮ ﺍﺳﺖ.‬ ‫ﺍﮔﺮ ﺟﺮﻳﺎﻥ ﺑﺎﺭ ﻛﻢ ﺑﺎﺷﺪ ﻭ ﻭﻟﺘﺎﮊ ﺁﻥ ﺯﻳﺎﺩ ﺭﻭﺵ ﺩﻭﻡ ﺑﻬﺘﺮ ﺍﺳﺖ.‬ ‫4‬
‫ﻣﻌﻤﻮﻻ ﺭﻭﺵ )1( ﺭﺍ ﺗﺮﺟﻴﺢ ﻣﻲ ﺩﻫﻴﻢ :‬ ‫ﹰ‬ ‫ﺍﻟﻒ( ﻭﻟﺘﻤﺘﺮﻫﺎﻱ ﺍﻣﺮﻭﺯﻱ ﻣﻘﺎﻭﻣﺖ ﻭﺭﻭﺩﻱ ﺧﻴﻠﻲ ﺯﻳﺎﺩﻱ ﺩﺍﺭﻧﺪ .‬ ‫ﺏ( ﺍﮔﺮ ﻭﻟﺘﺎﮊ ﺛﺎﺑﺖ ﺑﺎﺷﺪ ﺍﻣﺎ ﺑﺎﺭ ﻣﺘﻐﻴﺮ ﺑﺎﺷﺪ ﻣﻘﺪﺍﺭ ‪ PP = V 2 / R V‬ﺛﺎﺑﺖ‬ ‫ﺍﺳﺖ ﻭ ﻣﻲ ﺗﻮﺍﻥ ﻣﻘﺪﺍﺭ ﺁﻧﺮﺍ ﺩﺭ ﺣﺎﻟﺖ ﺑﻲ ﺑﺎﺭﻱ ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ ﻭ ﺍﺯ ﺗﻤﺎﻡ‬ ‫ﻗﺮﺍﺋﺘﻬﺎ ﮐﻢ ﮐﺮﺩ.‬ ‫ﻣﻌﻤﻮﻻ ﺑﺠﺎﻱ ﺍﻳﻦ ﮐﺎﺭ ﺍﺯ ﻭﺍﺗﻤﺘﺮ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﻨﻴﻢ ﮐﻪ ﺩﻳﮕﺮ ﺍﺣﺘﻴﺎﺝ ﺑﻪ‬ ‫ﺿﺮﺏ ﺍﻋﺪﺍﺩ ﺩﺭ ﻫﻢ ﻧﺪﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ ﻭ ﺻﺤﺖ ﻫﻢ ﺑﻬﺘﺮ ﺍﺳﺖ.‬ ‫5‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫6‬
‫ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ:‬ ‫ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ ﻧﻮﻉ ﻭﺍﺗﻤﺘﺮ ﺍﺳﺖ ﮐﻪ ﻣﻲ ﺗﻮﺍﻧﺪ ﺗﻮﺍﻥ ‪ DC‬ﻳﺎ ‪) AC‬ﺳﻴﻨﻮﺳﻲ ﻳﺎ‬ ‫ﻏﻴﺮ ﺳﻴﻨﻮﺳﻲ( ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﺩ . )ﻧﻮﻋﻬﺎﻱ ﺩﻳﮕﺮ: ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ،‬ ‫ﺗﺮﻣﻮﮐﻮﭘﻠﻲ، ﺍﻟﮑﺘﺮﻭﺍﺳﺘﺎﺗﻴﮑﻲ، ﻓﺮﻭﻣﻐﻨﺎﻃﻴﺴﻲ(‬ ‫ﺩﻳﺪﻳﻢ ﻛﻪ ﺩﺭﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮ :‬ ‫‪dM‬‬ ‫‪Ic‬‬ ‫+‬ ‫‪VR‬‬ ‫_‬ ‫7‬ ‫‪Ip‬‬ ‫‪R‬‬ ‫‪p‬‬ ‫‪IC IP‬‬ ‫=‪T‬‬ ‫‪dθ‬‬ ‫‪1 dM‬‬ ‫‪IC IP‬‬ ‫=‪θ‬‬ ‫‪S dθ‬‬
‫ﺣﺎﻝ ﺍﮔﺮ ‪ ، I p = VR / R p ، I c = I R‬ﺑﺎﺷﺪ ﺩﺍﺭﻳﻢ:‬ ‫‪θ‬‬ ‫‪RP‬‬ ‫=‪θ‬‬ ‫=‪⇒P‬‬ ‫‪K‬‬ ‫′‪K‬‬ ‫ﻭﻟﻲ ﺩﺭ ﻭﺍﻗﻊ ،‬ ‫‪1 dM VR‬‬ ‫‪P‬‬ ‫=‪θ‬‬ ‫‪IR‬‬ ‫‪=K‬‬ ‫‪S dθ R P‬‬ ‫‪RP‬‬ ‫‪VR‬‬ ‫‪I R VR‬‬ ‫‪θ = KI C I P = KI C‬‬ ‫‪≈K‬‬ ‫‪= K′P‬‬ ‫‪RP‬‬ ‫‪RP‬‬ ‫ﻳﻌﻨﻲ ‪ θ ≈ K ′P‬ﺩﺭ ﺣﺎﻟﻲ ﮐﻪ ‪) θ = K ′P‬ﻳﻌﻨﻲ ﻭﺍﺗﻤﺘﺮ ﺑﺮ ﺍﺳﺎﺱ ﺍﻳﻦ ﺿﺮﻳﺐ‬ ‫ﻣﺪﺭﺝ ﻣﻲ ﺷﻮﺩ.(‬ ‫‪m‬‬ ‫8‬
‫ﺍﮔﺮ ﺟﺮﻳﺎﻧﻬﺎ ‪ ac‬ﺑﺎﺷﻨﺪ ﻧﻴﺰ‬ ‫ﻭ ﺍﮔﺮ ‪ i = i C‬ﻭ ‪ i P = v‬ﺑﺎﺷﺪ ﺩﺍﺭﻳﻢ :‬ ‫‪i‬‬ ‫‪RP‬‬ ‫‪ic‬‬ ‫+‬ ‫‪V‬‬ ‫_‬ ‫‪Z‬‬ ‫‪ip‬‬ ‫‪1T‬‬ ‫‪θ = k ∫ i Ci P dt‬‬ ‫0‪T‬‬ ‫‪T‬‬ ‫1 ‪k‬‬ ‫‪k‬‬ ‫=‪θ‬‬ ‫‪P = k′P‬‬ ‫= ‪∫ ivdt‬‬ ‫0 ‪RP T‬‬ ‫‪RP‬‬ ‫‪R‬‬ ‫‪p‬‬ ‫‪1T‬‬ ‫ﻫﻤﺎﻥ ﺗﻌﺮﻳﻒ ﺗﻮﺍﻥ ﺍﮐﺘﻴﻮ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﻣﻮﺭﺩ ﺳﻴﻨﻮﺳﻲ )‪VICos(ϕ‬‬ ‫ﮐﻪ ‪∫ ivdt‬‬ ‫0‪T‬‬ ‫9‬ ‫ﻣﻲ ﺷﻮﺩ .‬
: ‫ﻭ ﻳﺎ‬ RP RP T Pm = θ= ∫ i Ci P dt k T 0 : ‫ﺍﻟﺒﺘﻪ ﺧﻄﺎﻳﻲ ﮐﻪ ﻗﺒﻼ ﺍﺷﺎﺭﻩ ﮐﺮﺩﻳﻢ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻭ ﺩﺭ ﻭﺍﻗﻊ‬ k 1T k 1T kT θ = ∫ i Ci P dt ≈ ∫ i C vdt ≈ ∫ ivdt = k ′P RP T 0 RP T 0 T0 P≈ Rp k θ ‫ ﻭﻟﻲ‬Pm = Rp k θ ‫ﺩﺭ ﺍﻳﻨﺠﺎ ﻧﻴﺰ‬ 10
± ± 1 11
‫ﺩﺭ ﺍﻳﻨﺠﺎ ﻫﻢ ﺩﻭ ﺟﻮﺭ ﻣﻲ ﺗﻮﺍﻥ ﻭﺍﺗﻤﺘﺮ ﺭﺍ ﺑﺴﺖ. ﻣﻌﻤﻮﻻ ﻳﮏ ﺳﺮ ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﺟﺮﻳﺎﻥ ﻭ ﻳﮏ ﺳﺮ ﺳﻴﻢ ﭘﻴﭻ ﻭﻟﺘﺎﮊ ﺑﺎ ﻋﻼﻣﺖ ± ﻣﺸﺨﺺ ﺷﺪﻩ ﺍﺳﺖ.‬ ‫ﺟﺮﻳﺎﻥ ﻭﺍﺭﺩ ﻗﻄﺐ ± ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﺷﺪﻩ ﻭ ﻗﻄﺐ ± ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﻭﻟﺘﺎﮊ ﺑﻪ ﻳﮏ ﻃﺮﻑ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﻭﺻﻞ ﻣﻲ ﺷﻮﺩ. )ﺍﮔﺮ ﺩﻳﺪﻳﺪ ﻋﻘﺮﺑﻪ‬ ‫ﺑﺮﻋﮑﺲ ﻣﻨﺤﺮﻑ ﻣﻲ ﺷﻮﺩ ﺟﻬﺖ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﻌﻮﻳﺾ ﮐﻨﻴﺪ(.‬ ‫±‬ ‫±‬ ‫±‬ ‫±‬ ‫21‬ ‫2‬ ‫1‬
‫ﮐﻪ ﮔﻔﺘﻴﻢ ﻣﻌﻤﻮﻻ ﺭﻭﺵ ﺍﻭﻝ ﺗﺮﺟﻴﺢ ﺩﺍﺭﺩ ﭼﻮﻥ ﻭﻟﺘﺎﮊ ﺑﺮﻕ ﺷﻬﺮ ﺛﺎﺑﺖ ﻓﺮﺽ ﺷﺪ‬ ‫ﭘﺲ ‪ Pp‬ﺛﺎﺑﺖ ﺍﺳﺖ ﮐﻪ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎﺯﺍﻱ ‪ ZL‬ﺑﻴﻨﻬﺎﻳﺖ ﺁﻥ ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﮐﺮﺩ ﻭ‬ ‫ﻫﻤﻴﺸﻪ ﺍﺯ ﻗﺮﺍﺋﺘﻬﺎ ﮐﻢ ﮐﺮﺩ.‬ ‫ﺑﺮﺍﻱ ﺣﺬﻑ ﺍﺛﺮ ‪ i p‬ﺩﺭ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺟﺮﻳﺎﻥ ﺗﻮﺳﻂ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ ﺳﻴﻢ‬ ‫ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺩﺭ ﺟﻬﺖ ﺧﻼﻑ ﺩﻭﺭ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﭘﻴﭽﻴﺪ ﺗﺎ ﻓﻠﻮﻱ ﺍﻳﺠﺎﺩ‬ ‫ﺷﺪﻩ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ i = i − i‬ﺑﺎﺷﺪ )ﻭﺍﺗﻤﺘﺮ ﺟﺒﺮﺍﻥ ﺷﺪﻩ(.‬ ‫ﻋﻼﻭﻩ ﺑﺮ ﻋﺎﻣﻞ ﻓﻮﻕ ﻋﺎﻣﻞ ﺩﻳﮕﺮﻱ ﮐﻪ ﺑﺮﺍﻱ ﺧﻄﺎ ﺩﺭ ﺣﺎﻟﺖ ‪ ac‬ﺩﺍﺭﻳﻢ ﺍﻳﻦ ﺍﺳﺖ ﮐﻪ‬ ‫‪ i p = V / R p‬ﻧﻴﺴﺖ ﻭ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻠﻒ ﺭﺍ ﻫﻢ ﺩﺍﺭﻳﻢ . ﺩﺭ ﺣﺎﻟﺖ ﺳﻴﻨﻮﺳﻲ ﮐﻪ‬ ‫ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ ﺣﺎﻟﺖ ﺍﺳﺖ ﺍﻳﻦ ﻣﺴﺎﻟﻪ ﺭﺍ ﺑﺮﺭﺳﻲ ﻣﻲ ﮐﻨﻴﻢ .‬ ‫‪p‬‬ ‫31‬ ‫‪c‬‬
‫ﺁﻧﺎﻟﻴﺰ ﻭﺍﺗﻤﺘﺮ ﺩﺭ ﺣﺎﻟﺖ ﺩﺍﺋﻤﻲ ﺳﻴﻨﻮﺳﻲ‬ ~ iC = I C Cos (ωt + ϕC ) I C = I C e jϕ ~ iP = I P Cos (ωt + ϕ P ) I P = I P e jϕ ~ v(t ) = VCos (ωt ) V =V < 0 C P ~ iC = Re{I C e jωt } ~ iP = Re{I P e jωt } ~ IC R p ~ I LP ~ IP ~ Z 14
RP T 1T ‫ ﺍﻣﺎ‬Pm = ∫ iC iP dt ‫ ﻭ‬θ = k ∫ i Ci P dt ‫ﺩﻳﺪﻳﻢ ﮐﻪ‬ T 0 T0 1T 1 ~~ (ϕc − ϕ P ) = 1 Re I C I *P ∫ iC iP dt = I C I PCos 2 2 T0 { { RP ~~ Pm = Re I C I * P 2 } } : ‫ﻳﻌﻨﻲ‬ 15
:‫ﺣﺎﻝ ﺑﺮﺍﻱ ﺑﺪﺳﺖ ﺁﻭﺭﺩﻥ ﺧﻄﺎﻱ ﻭﺍﺗﻤﺘﺮ‬ ~ = ze jϕ z ~ = R + j ωL = z e j β zp p p p ‫ﻳﺎ‬ ~ ~ ~ ~ ~ v V − jϕ I C = I + I P I = ~ = e = Ie − jϕ z z ~ V − jβ v ~ Ip = ~ = e zp zp 16
[ ] ( 1 1 ~ ~ ~* 1 ~~ 2 Pm = R p Re (I + I p )I p = RP I P + RP Re I I p* 2 2 2 ⎛ − jϕ V + jβ ⎞ 1 = PP + RP Re⎜ Ie e ⎟ ⎜ ⎟ 2 zp ⎝ ⎠ 1 RP = PP + VI cos(ϕ − β ) 2 zP 1 = PP + VI cos(β ) cos(ϕ − β ) 2 ωLP ‫ ﻭ‬RP = cos(β) = sin(β ) zP zP ) : ‫ﺯﻳﺮﺍ‬ 17
‫ﺑﺮﺍﻱ 0 = ‪ ωLP‬ﺩﺍﺭﻳﻢ ‪ ~P = z P = RP‬ﻭ 0 = ‪ β‬ﻭﻟﺬﺍ:‬ ‫‪z‬‬ ‫1‬ ‫‪Pm = PP + VI cos(ϕ) = PP + P‬‬ ‫2‬ ‫ﮐﻪ ﺍﻳﻦ ﻫﻤﺎﻥ ﭼﻴﺰﻱ ﺍﺳﺖ ﮐﻪ ﻣﺒﺤﺚ ﻗﺒﻞ ﺩﻳﺪﻳﻢ ﻭ ﺍﺛﺮ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ‪ LP‬ﺭﺍ‬ ‫ﺩﺭ ﻧﻈﺮ ﻧﮕﺮﻓﺘﻪ ﺑﻮﺩﻳﻢ ﻭﮔﺮﻧﻪ ﺩﺭ ﺻﻮﺭﺕ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ‪ PP‬ﺩﺍﺭﻳﻢ:)ﺿﺮﻳﺐ‬ ‫ﺗﺼﺤﻴﺢ(‬ ‫)‪cos(ϕ‬‬ ‫‪P‬‬ ‫=‬ ‫) ‪Pm cos(β ) cos(ϕ − β‬‬ ‫81‬
‫ﻣﻌﻤﻮﻻ ﺍﮔﺮﭼﻪ 0 ≠ ‪ LP‬ﻭﻟﻲ ‪ ωLP << RP‬ﻭ ﺍﺯ ﺗﻘﺮﻳﺒﻲ ﮐﻪ ﺩﺭ ﺍﺩﺍﻣﻪ‬ ‫ﺧﻮﺍﻫﺪ ﺁﻣﺪ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﻨﻴﻢ:‬ ‫1‬ ‫)‪Pm = PP + VI cos β(cos β cos ϕ + sin β sin ϕ‬‬ ‫2‬ ‫1‬ ‫)‪= PP + VI cos 2 β(cos ϕ + tgβ sin ϕ‬‬ ‫2‬ ‫ﺑﺎ ﻓﺮﺽ ‪ ωLP << RP‬ﺑﺘﺎ ﺧﻴﻠﻲ ﮐﻮﭼﮏ ﺧﻮﺍﻫﺪ ﺑﻮﺩ )ﻣﻌﻤﻮﻻ ﭼﻨﺪ ﺩﻗﻴﻘﻪ(‬ ‫ﻭ ﻟﺬﺍ 1 ≈ ‪ cos 2 β‬ﭘﺲ:‬ ‫‪ωLP‬‬ ‫= ‪tgβ‬‬ ‫‪RP‬‬ ‫91‬ ‫1‬ ‫‪Pm ≈ PP + VI (cos ϕ + tgβ sin ϕ) where‬‬ ‫2‬
‫ﮐﻪ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ :‬ ‫1‬ ‫‪Pm ≈ Pp + P + VI tgβ sin ϕ‬‬ ‫2‬ ‫ﭘﺲ ﻋﻼﻭﻩ ﺑﺮ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺩﺭ ﺷﺎﺧﻪ ﻭﻟﺘﻤﺘﺮ ، ﺧﻄﺎﻱ ﺩﻳﮕﺮ ﻧﺎﺷﻲ ﺍﺯ‬ ‫ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ‪ LP‬ﻣﻲ ﺑﺎﺷﺪ ﮐﻪ ﻣﻘﺪﺍﺭ ﺧﻄﺎ ﺑﻪ ﺯﺍﻭﻳﻪ ﻓﺎﺯ ﺑﺎﺭ ‪ ϕ‬ﻫﻢ ﺑﺴﺘﮕﻲ‬ ‫ﺩﺍﺭﺩ. ﻫﺮ ﭼﻪ ﺑﺎﺭ ﻣﻘﺎﻭﻣﺘﻲ ﺗﺮ ﺑﺎﺷﺪﺍﻳﻦ ﺧﻄﺎ ﮐﻤﺘﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ )ﺑﺮﺍﻱ ﺑﺎﺭ‬ ‫ﺳﻠﻔﻲ ﻣﺜﺒﺖ ﻭ ﺑﺮﺍﻱ ﺑﺎﺭ ﺧﺎﺯﻧﻲ ﻣﻨﻔﻲ ، ﺳﻠﻒ ‪ LP‬ﺳﺒﺐ ﻫﻤﻔﺎﺯﺗﺮ ﺷﺪﻥ ‪I C‬‬ ‫ﻭ ‪ I P‬ﻣﻲ ﺷﻮﺩ(.‬ ‫ﺑﺮﺍﻱ ﺟﺒﺮﺍﻥ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﮐﺎﺭﻱ ﻣﻲ ﮐﻨﻴﻢ ﮐﻪ‬ ‫0∠ ‪z ≈ R‬‬ ‫‪p‬‬ ‫02‬ ‫‪p‬‬
LP RP r ~ = jωL + ( R − r ) + r (1 − rjωc ) zp P P 1 + r 2c 2ω 2 c ( z p ≈ RP − r + r + jω LP − r 2c ) ، ‫ ﺁﻧﮕﺎﻩ‬r 2c 2ω 2 << 1 ‫ﺍﮔﺮ‬ 21
‫ﺩﺭ ﻧﺘﻴﺠﻪ :‬ ‫‪LP = r 2C‬‬ ‫ﺑﺎ ﺍﻳﻦ ﺭﻭﺵ ﺗﺎ ﺣﺪﻭﺩ ﻓﺮﮐﺎﻧﺲ ‪ 10kHz‬ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻴﻢ ﭘﻴﭻ ﺟﺒﺮﺍﻥ ﻣﻲ ﺷﻮﺩ .‬ ‫22‬
‫ﻣﺜﺎﻝ :‬ ‫ﻳﮏ ﻭﺍﺕ ﺳﻨﺞ ﺍﻟﮑﺘﺮﻭﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ ﻣﻄﺎﺑﻖ ﺷﮑﻞ ﺯﻳﺮ ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﻳﮏ‬ ‫ﻣﺪﺍﺭ ﺗﮏ ﻓﺎﺯ ﺑﮑﺎﺭ ﺭﻓﺘﻪ ﺍﺳﺖ . ﻭﻟﺘﺎﮊ ﺑﺎﺭ ‪ 100V‬ﻭ ﺟﺮﻳﺎﻥ ﺑﺎﺭ‪ 9A‬ﻭ ﺿﺮﻳﺐ ﺗﻮﺍﻥ‬ ‫1.0 ﭘﺲ ﻓﺎﺯ ﻣﻲ ﺑﺎﺷﺪ . ﻣﺪﺍﺭ ﻭﻟﺘﺎﮊ ﻭﺍﺗﻤﺘﺮ ﺩﺍﺭﺍﻱ ﻣﻘﺎﻭﻣﺖ 0003 ﺍﻫﻢ ﻭ‬ ‫ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ‪ 30mH‬ﻣﻲ ﺑﺎﺷﺪ . )‪(f=50Hz‬‬ ‫~‬ ‫‪IC‬‬ ‫~‬ ‫‪I‬‬ ‫~‬ ‫‪Z‬‬ ‫32‬ ‫~‬ ‫‪IP‬‬ ‫‪LP‬‬ ‫‪R‬‬ ‫‪p‬‬
‫ﺍﻟﻒ( ﺩﺭﺻﺪ ﺧﻄﺎﻱ ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ ﺭﺍﭘﻴﺪﺍ ﮐﻨﻴﺪ .‬ ‫ﺏ( ﺩﺭ ﺻﻮﺭﺕ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ‪ ، PP‬ﺑﺮﺍﻱ ﭼﻪ ﺿﺮﻳﺐ ﺗﻮﺍﻧﻲ ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ‬ ‫ﺻﻔﺮ ﺧﻮﺍﻫﺪ ﺷﺪ ؟‬ ‫ﺝ( ﭼﮕﻮﻧﻪ ﻣﻲ ﺗﻮﺍﻥ ﺧﻄﺎﻱ ﻧﺎﺷﻲ ﺍﺯ ‪ LP‬ﺭﺍ ﺣﺬﻑ ﮐﺮﺩ؟‬ ‫42‬
P = VI cos( ϕ ) = 100 × 9 × 0.1 = 90W Pm = PP + VI cos( β ) cos( ϕ − β ) cos( ϕ ) = 0.1 ⇒ ϕ = 84.26° X P = 2π( 50 )( 30 × 10 − 3 ) = 9.42Ω , 9.42 β = tg ( ) = 0.18° 3000 V 2 100 2 PP ≈ = = 3.33W RP 3000 RP = 3000Ω : ‫ﺣﻞ‬ (‫ﺍﻟﻒ‬ −1 ( X P << RP ) Pm = 3.33 + 100 × 9 × cos( 0.18 ) cos( 84.26 − 0.18 ) = 96.16W ‫8.6% = 001 * 09 −61.69 = ﺩﺭ ﺻﺪ ﺧﻄﺎ‬ 90 25
(‫ﺏ‬ cos( ϕ − β ) = 0 ⇒ ϕ − β = ±90° ⇒ ϕ − 0.18 = −90° ⇒ ϕ = −89.82 (‫ﺝ‬ LP = 0.03H = r 2C : ‫ﺑﺮﺍﻱ ﻣﺜﺎﻝ‬ C = 30nf , r = 1kΩ , R1 = 2kΩ 26
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫72‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ﺍﮔﺮ ﺳﻴﺴﺘﻢ ﺳﻪ ﻓﺎﺯ ﻣﺘﻌﺎﺩﻝ 4 ﺳﻴﻤﻪ ﺑﺎﺷﺪ ﻣﻲ ﺗﻮﺍﻥ ﺗﻮﺍﻥ ﻳﻚ ﻓﺎﺯ ﺭﺍ ﺑﺎ‬ ‫ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﻭﺍﺗﻤﺘﺮ ﺑﺸﮑﻞ ﺯﻳﺮ ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ ﻭ ﺩﺭ 3 ﺿﺮﺏ ﻛﺮﺩ . ﻭﻟﻲ‬ ‫ﺍﻏﻠﺐ ﺑﺎﺭ ﻧﺎﻣﺘﻌﺎﺩﻝ ﺍﺳﺖ ﻭ ﺍﺯ ﺍﻳﻦ ﺭﻭﺵ ﻧﻤﻲ ﺗﻮﺍﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ ﻭ ﺑﺎﻳﺪ‬ ‫3 ﻭﺍﺗﻤﺘﺮ ﻳﺎ ﻻﺍﻗﻞ ﺩﻭ ﻭﺍﺗﻤﺘﺮ)ﺑﺮﺍﻱ ﺳﻴﺴﺘﻢ ﺳﻪ ﺳﻴﻤﻪ( ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ .‬ ‫‪R‬‬ ‫‪S‬‬ ‫‪T‬‬ ‫‪N‬‬ ‫82‬
‫ﻗﻀﻴﻪ ﺑﻠﻮﻧﺪﻝ: ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺳﻴﺴﺘﻢ ‪ N‬ﺳﻴﻤﻪ 1-‪ N‬ﻭﺍﺗﻤﺘﺮ‬ ‫ﻛﺎﻓﻲ ﺍﺳﺖ. )ﻳﻚ ﺳﻴﻢ ﺭﺍ ﻣﺒﻨﺎ ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻴﻢ( .‬ ‫92‬
‫ﺭﻭﺵ 3 ﻭﺍﺗﻤﺘﺮﻱ‬ 1 + i1 V 1 _ v C + _ 3 + ' v3 v'2 V 3 V 2 + _ 2 + + i3 O _ _ _ + ' v1 i2 _ 30
‫ﺳﺮﻫﺎﻱ ﭘﺘﺎﻧﺴﻴﻞ ﻭﺍﺗﻤﺘﺮﻫﺎ ﺭﺍ ﺑﻪ ﻧﻘﻄﻪ ﻣﺸﺘﺮﻙ )‪ (c‬ﻭﺻﻞ ﻣﻲ ﻛﻨﻴﻢ. ﮐﻪ‬ ‫ﻣﺜﻼ ﻣﻲ ﺗﻮﺍﻧﺪ ﻫﻤﺎﻥ ‪ O‬ﺑﺎﺷﺪ.‬ ‫′‬ ‫1‪v1 = v + v‬‬ ‫′‬ ‫2‪v2 = v + v‬‬ ‫′‬ ‫3‪v3 = v + v‬‬ ‫13‬ ‫‪1T‬‬ ‫′‬ ‫‪P = ∫ v1i1dt‬‬ ‫1‬ ‫0‪T‬‬ ‫‪1T‬‬ ‫′‬ ‫‪P2 = ∫ v2i2 dt‬‬ ‫0‪T‬‬ ‫‪1T‬‬ ‫′‬ ‫‪P3 = ∫ v3i3dt‬‬ ‫0‪T‬‬
⎞⎞ ⎛ 1 T⎛ P + P2 + P3 = ∫ ⎜ v1i1 + v2i2 + v3i3 + v⎜ i1 + i2 + i3 ⎟ ⎟ dt 1 4 3 ⎜ 1 24 ⎟ ⎟ T 0⎜ ⎝ ⎠⎠ 0 ⎝ 1T = ∫ (v1i1 + v2i2 + v3i3 )dt = P = ‫ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺑﺎﺭ‬ T0 32
‫ﺭﻭﺵ 2 ﻭﺍﺗﻤﺘﺮﻱ‬ ‫ﺍﮔﺮ ﻧﻘﻄﻪ ﻣﺸﺘﺮﮎ ﺳﻴﻢ ﭘﻴﭽﻬﺎﻱ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺭﻭﻱ ﻳﮑﻲ ﺍﺯ ﺧﻄﻮﻁ ﺑﮕﻴﺮﻳﻢ ﻳﮏ‬ ‫ﻭﺍﺗﻤﺘﺮ ﮐﻤﺘﺮ ﻻﺯﻡ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.‬ ‫1‬ ‫+‬ ‫‪V‬‬ ‫1‬ ‫_‬ ‫_‬ ‫33‬ ‫‪V‬‬ ‫3‬ ‫+‬ ‫+‬ ‫‪V‬‬ ‫2‬ ‫_‬ ‫3‬ ‫2‬
1T P = ∫ i1( v1 − v3 )dt 1 T0 1T P2 = ∫ i2 ( v2 − v3 )dt T0 1T P + P2 = ∫ (i1( v1 − v3 ) + i2 ( v2 − v3 ))dt 1 T0 1T = ∫ (i1v1 + i2v2 + i3v3 ))dt = P T0 i3 = −( i1 + i2 ) : ‫ﺯﻳﺮﺍ‬ 34
‫ﺩﺭ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚ ﻫﻢ :‬ ‫1‬ ‫+‬ ‫_‬ ‫1‪V‬‬ ‫‪V‬‬ ‫3‬ ‫_‬ ‫+‬ ‫1‪i‬‬ ‫3‪i‬‬ ‫3‬ ‫_‬ ‫2‪i‬‬ ‫+‬ ‫‪V‬‬ ‫2‬ ‫2‬ ‫53‬
1T P = ∫ − v3 ( i1 − i3 )dt 1 T0 1T P2 = ∫ v2 ( i2 − i1 )dt T0 1T( P + P2 = ∫ v3i3 + v2i2 − i1 (v2 + v3 ))dt 1 T0 1T = ∫ (i1v1 + i2 v2 + i3v3 ))dt = P T0 v1 + v2 + v3 = 0 36
‫ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﺩﺭ ﺣﺎﻟﺖ ﮐﻠﻲ ﺑﺮﻗﺮﺍﺭﻧﺪ ﺍﺯ ﺟﻤﻠﻪ ﺩﺭ ﺣﺎﻟﺖ ﺩﺍﺋﻤﻲ ﺳﻴﻨﻮﺳﻲ .‬ ‫ﭘﺲ ﺑﺮﺍﻱ ﺳﻴﻨﻮﺳﻲ ﺑﺎ ﺑﺎﺭ ﻧﺎﻣﺘﻌﺎﺩﻝ ﻧﻴﺰ ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﺑﺮﻗﺮﺍﺭ ﺑﻮﺩﻩ ﻭ ‪P + P2 = P‬‬ ‫1‬ ‫ﻣﺜﻼ ﺩﺭ ﺣﺎﻟﺖ ﺳﺘﺎﺭﻩ ﺩﺍﺭﻳﻢ :‬ ‫3‪P = V1I1 cos ϕ1 + V2 I 2 cos ϕ2 + V3 I 3 cos ϕ‬‬ ‫32‪,V‬‬ ‫ﮐﻪ ‪ Vi‬ﻭ ‪ I i‬ﻫﺎ ﻣﻘﺎﺩﻳﺮ ‪ rms‬ﻫﺴﺘﻨﺪ .‬ ‫73‬ ‫2‬ ‫‪= I1V13 cos ϕ I ,V + I 2V23 cos ϕ I‬‬ ‫31‬ ‫1‬
‫ﺣﺎﻝ ﺍﮔﺮ ﺩﺭ ﺣﺎﻟﺖ ﺩﺍﺋﻤﻲ ﺳﻴﻨﻮﺳﻲ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ ﺑﺎﺷﺪ ، ﺍﮔﺮ ﻭﻟﺘﺎﮊ ‪rms‬‬ ‫ﻓﺎﺯﻫﺎ ﺭﺍ ﺑﺎ 1‪ V‬ﻭ 2‪ V‬ﻭ 3‪ V‬ﻧﺸﺎﻥ ﺩﻫﻴﻢ ﻭ ﺟﺮﻳﺎﻥ ‪ rms‬ﺁﻧﻬﺎ ﺭﺍ ﺑﺎ 1‪ I‬ﻭ 2 ‪ I‬ﻭ‬ ‫3 ‪ I‬ﻧﺸﺎﻥ ﺩﻫﻴﻢ ، ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ :‬ ‫1‬ ‫3‬ ‫83‬ ‫2‬
V 13 V 1 I I V 1 V 23 3 I 3 V 2 2 V 31 39
V = V 1 = V 2 = V 3 ‫ﺑﺎ ﻓﺮﺽ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ‬ I = I = I = I 1 V 12 2 = V 3 = V 13 23 = 3V P = V I cos( 30 − ϕ ) = 1 13 P = V 2 23 1 I 2 c os( 30 3VI cos( 30 − ϕ ) +ϕ) = 3VI cos( 30 + ϕ ) P + P = 3VI [cos( 30 + ϕ ) + cos( 30 − ϕ ) ] = 3 VI cos( ϕ ) = P P − P = 3VI sin( ϕ ) = 1 1 ⇒ 2 2 Q 3 P1 − P2 tg ( ϕ ) 3VI sin( ϕ ) = = P1 + P2 3 VI cos( ϕ ) 3 40
‫ﻳﻌﻨﻲ ﺍﮔﺮ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ ﺑﺎﺷﺪ ﺩﺭﺣﺎﻟﺖ ﺩﺍﺋﻤﯽ ﺳﻴﻨﻮﺳﯽ ﺑﺎ ﺭﻭﺵ ﺩﻭ ﻭﺍﺗﻤﺘﺮﻱ‬ ‫ﻣﻲ ﺗﻮﺍﻥ ﻋﻼﻭﻩ ﺑﺮ ﺗﻮﺍﻥ ﺍﮐﺘﻴﻮ ، ﺗﻮﺍﻥ ﺭﺍﮐﺘﻴﻮ ﻳﺎ ‪ ϕ‬ﺭﺍ ﻧﻴﺰ ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ.‬ ‫3‬ ‫‪ϕ = 0 , cos ϕ = 1 ⇒ P = 3VI , P = P2 = VI‬‬ ‫1‬ ‫2‬ ‫3‬ ‫3‬ ‫0 = 2‪ϕ = 60 , cos ϕ = 0.5 ⇒ P = VI , P = VI , P‬‬ ‫1‬ ‫2‬ ‫2‬ ‫0 < 2‪ϕ > 60 , cos ϕ < 0.5 ⇒ P‬‬ ‫14‬ ‫3‬ ‫3‬ ‫= ‪ϕ = 90 , cos ϕ = 0 ⇒ P‬‬ ‫0 = ‪VI , P2 = − VI ⇒ P‬‬ ‫1‬ ‫2‬ ‫2‬ ‫3‬ ‫3‬ ‫‪ϕ = −60 ⇒ P = VI , P = 0 , P2 = VI‬‬ ‫1‬ ‫2‬ ‫2‬ ‫0 < ‪ϕ < −60 ⇒ P‬‬ ‫1‬
‫ﺍﮔﺮ ﻭﺍﺗﻤﺘﺮﻱ ﻣﻨﻔﻲ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ، ﺑﺎﻳﺪ ﺟﻬﺖ ﻳﻜﻲ ﺍﺯ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﺭﺍ ﻋﻮﺽ‬ ‫ﻛﺮﺩ ﺗﺎ ﻗﺮﺍﺋﺖ ﻣﺜﺒﺖ ﺷﻮﺩ ﻭﻟﻲ ﺑﺎﻳﺪ ﻋﻼﻣﺖ ﻣﻨﻔﻲ ﺭﺍ ﺧﻮﺩﻣﺎﻥ ﺩﺭﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ.‬ ‫ﻣﺜﺎﻝ:‬ ‫ﺗﻮﺍﻥ ﻳﻚ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ 3 ﺳﻴﻤﻪ ﻣﺘﻌﺎﺩﻝ ﺑﻪ ﺭﻭﺵ ﺩﻭ ﻭﺍﺗﻤﺘﺮﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ‬ ‫ﺷﺪﻩ ﺍﺳﺖ. ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ 1 ، ‪ 7500w‬ﻭ ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ 2، ‪ -1500w‬ﺍﺳﺖ.‬ ‫ﺍﻟﻒ( ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﺑﺎﺭ؟‬ ‫ﺏ( ﺍﮔﺮﻭﻟﺘﺎﮊ ﺧﻂ ‪ 400v‬ﺑﺎﺷﺪ ﭼﻪ ﻇﺮﻓﻴﺘﻲ)ﺧﺎﺯﻧﻲ( ﺩﺭﻫﺮ ﻓﺎﺯ ﺑﺎﻳﺪ ﺍﺿﺎﻓﻪ ﻛﺮﺩ ﺗﺎ‬ ‫ﻛﻞ ﺗﻮﺍﻥ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺷﺪﻩ ﺭﻭﻱ ﻭﺍﺗﻤﺘﺮ 1 ﻇﺎﻫﺮ ﺷﻮﺩ ؟‬ ‫24‬
‫ﺍﻟﻒ(‬ ‫‪P = 7500 w, P2 = −1500 w‬‬ ‫1‬ ‫‪P = P + P2 = 6000 w‬‬ ‫1‬ ‫2‪P1 − P‬‬ ‫3‬ ‫953.0 = ) ‪= 68.9° ⇒ cos( ϕ‬‬ ‫2‪P1 + P‬‬ ‫1−‬ ‫‪ϕ = tg‬‬ ‫ﺏ(‬ ‫ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺑﺎﻳﺪ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ :‬ ‫34‬ ‫°06 = ‪ϕ‬‬
‫ ﻭﻟﺘﺎﮊ ﻓﺎﺯ‬V = 400 = 231v 3 ‫ ﻫﺮ ﻓﺎﺯ‬P = 6000 = 2000 w 3 2000 ⇒I= = 24.11A 231×0.359 231v = 9.58Ω Z= 24.1A R = Z cos( ϕ ) = 3.44Ω : ‫ﻗﺒﻞ ﺍﺯ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ﺧﺎﺯﻥ ﺩﺍﺭﻳﻢ‬ X = Z sin( ϕ ) = 8.94Ω tg ( ϕ′ ) = 3 = 1.73 X = Rtg ( ϕ′ ) = 3.44 × 1.73 = 5.96 ⇒ Capacitor' s _ Re ac tan ce = 8.94 − 5.96 = 2.98Ω 1 = 1068µF C= 2π( 50 )( 2.98 ) 44
‫ﻭﺍﺗﻤﺘﺮ 3 ﻓﺎﺯ:‬ ‫ﻳﻚ ﻭﺍﺗﻤﺘﺮ ﺍﺳﺖ ﻛﻪ ﺩﺍﺭﺍﻱ ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﻭﻟﺘﺎﮊ ﻭ ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ‬ ‫ﺍﺳﺖ ﻭ ﻛﺎﺭ 2 ﻭﺍﺗﻤﺘﺮ 1‪ P‬ﻭ 2‪ P‬ﺭﺍ ﻳﻜﺠﺎ ﺍﻧﺠﺎﻡ ﻣﻲ ﺩﻫﺪ ﻭ ﺣﺎﺻﻞ ﺟﻤﻊ‬ ‫ﺭﺍ ﺑﻪ ﻣﺎ ﻣﻲ ﺩﻫﺪ.‬ ‫54‬
‫ﻭﺍﺭﻣﺘﺮ‬ ‫ﺍﻳﻦ ﻭﺳﻴﻠﻪ ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺭﺍﻛﺘﻴﻮ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ. ﺩﺭ ﻭﺍﺗﻤﺘﺮ‬ ‫ﺗﮑﻔﺎﺯ ﺍﮔﺮ ﺑﺠﺎﻱ ﺳﺮﻱ ﻛﺮﺩﻥ ﻣﻘﺎﻭﻣﺖ ، ﺳﻠﻒ ﺑﺰﺭﮔﻲ ﺑﺎ ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﭘﺘﺎﻧﺴﻴﻞ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮ ﺳﺮﻱ ﻛﻨﻴﻢ ﺟﺮﻳﺎﻥ ﮔﺬﺭﻧﺪﻩ ﺍﺯ ﺳﻴﻢ ﭘﻴﭻ ﺑﺎ‬ ‫ﻭﻟﺘﺎﮊ ﺁﻥ ﺗﻘﺮﻳﺒﺎ 09 ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ .‬ ‫ﹰ‬ ‫‪V‬‬ ‫‪I‬‬ ‫‪ϕ‬‬ ‫‪Ip‬‬ ‫64‬ ‫) ‪Qm = VI cos( 90 − ϕ ) = VI sin( ϕ‬‬
‫ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺭﺍﻛﺘﻴﻮ ﺩﺭ ﻣﺪﺍﺭ ﺳﻪ ﻓﺎﺯ ﻣﺘﻌﺎﺩﻝ:‬ ‫) ‪Pm = V31 I 2Cos( 90 − ϕ ) = 3VISin( ϕ‬‬ ‫1‬ ‫2‬ ‫3‬ ‫74‬
‫ﺗﺮﺍﻧﺴﻔﻮﺭﻣﺮﻫﺎﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ :‬ ‫ﺩﺭ ﺍﺩﻭﺍﺗﻲ ﺍﺯ ﻗﺒﻴﻞ ﺍﻟﮑﺘﺮﻭﺩﻳﻨﺎﻣﻮ ﻣﺘﺮ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﺷﻨﺖ ﻳﺎ ﺿﺮﺏ‬ ‫ﮐﻨﻨﺪﻩ ﺑﻪ ﺩﻟﻴﻞ ﻭﺟﻮﺩ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ، ﻧﺴﺒﺖ ﺛﺎﺑﺘﻲ ﺭﺍ ﺑﺎ‬ ‫ﺗﻐﻴﻴﺮ ﻓﺮﮐﺎﻧﺲ ﻧﻤﻲ ﺩﻫﺪ . ﺭﺍﻩ ﺩﻳﮕﺮ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﺮﺍﻧﺲ ﺍﺳﺖ ﮐﻪ ﻋﻼﻭﻩ‬ ‫ﺑﺮ ﻧﺴﺒﺖ ﺛﺎﺑﺖ ، ﺍﻳﺰﻭﻻﺳﻴﻮﻥ ﻫﻢ ﺍﻳﺠﺎﺩ ﻣﻲ ﮐﻨﺪ ﻭﮔﺮﻧﻪ ﺁﻭﺭﺩﻥ ﻳﮏ‬ ‫ﻭﻟﺘﺎﮊ ﺧﻴﻠﻲ ﺯﻳﺎﺩ ﺑﻪ ﺩﺳﺘﮕﺎﻩ ﻫﻢ ﺧﻄﺮﻧﺎﮎ ﺍﺳﺖ ﻭ ﻫﻢ ﺩﺭ ﺍﺑﻌﺎﺩ ﮐﻮﭼﮏ‬ ‫ﺩﺳﺘﮕﺎﻩ ﻋﺎﻳﻘﻬﺎ ﺗﺤﻤﻞ ﻧﻤﻲ ﮐﻨﻨﺪ.‬ ‫84‬
49 1:n n:1
50
‫ﺗﺮﺍﻧﺲ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻧﺴﻲ ﺍﺳﺖ ﮐﻪ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺛﺎﻧﻮﻳﻪ ﺑﻴﺸﺘﺮ ﺍﺯ ﺍﻭﻟﻴﻪ ﺍﺳﺖ ﻭ ﻃﻮﺭﻱ‬ ‫ﻃﺮﺍﺣﻲ ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﺣﺎﻟﺖ ﺍﺗﺼﺎﻝ ﮐﻮﺗﺎﻩ ﺛﺎﻧﻮﻳﻪ )ﺁﻣﭙﺮﻣﺘﺮ ﻳﺎ ﺭﻟﻪ ﻳﺎ ...(‬ ‫ﺑﺘﻮﺍﻧﺪ ﮐﺎﺭ ﮐﻨﺪ )ﺟﺮﻳﺎﻥ ﺯﻳﺎﺩﻱ ﺭﺩ ﻣﻲ ﺷﻮﺩ(. ﻫﺮﮔﺰ ﻧﺒﺎﻳﺪ ﺛﺎﻧﻮﻳﻪ ﺗﺮﺍﻧﺲ‬ ‫ﺟﺮﻳﺎﻥ ﺭﺍ ﺯﻳﺮ ﺑﺎﺭ ﻣﺪﺍﺭ ﺑﺎﺯ ﮐﺮﺩ ﻭﺍﻻ ‪ emf‬ﺑﺰﺭﮔﻲ ﮐﻪ ﺍﻳﺠﺎﺩ ﻣﻲ ﺷﻮﺩ ﻣﻲ‬ ‫ﺗﻮﺍﻧﺪ ﻋﺎﻳﻖ ﺭﺍ ﺑﻪ ﺷﮑﺴﺖ ﺑﺒﺮﺩ. ﺍﺻﻮﻻ ﺍﮔﺮ ﻣﻘﺎﻭﻣﺖ ﺑﺎﺭ ﺗﺮﺍﻧﺲ ﺟﺮﻳﺎﻥ ، ﺑﺰﺭﮒ‬ ‫ﺑﺎﺷﺪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺟﺮﻳﺎﻧﻲ ﮐﻪ ﺍﺯ ﺛﺎﻧﻮﻳﻪ ﻣﻲ ﮔﺬﺭﺩ )‪ 1/n‬ﺟﺮﻳﺎﻥ ﺍﻭﻟﻴﻪ( 2 ‪RI‬‬ ‫ﺯﻳﺎﺩ ﺷﺪﻩ ﻭ ﻭﻟﺘﺎﮊ ﺯﻳﺎﺩ ﺣﺎﺻﻠﻪ ﺗﺮﺍﻧﺲ ﺭﺍ ﻣﻲ ﺳﻮﺯﺍﻧﺪ . ﻣﺜﻼ ﺍﮔﺮ ﻧﺴﺒﺖ 1 ﺑﻪ‬ ‫001 ﺑﺎﺷﺪ ﻭ ﺍﻭﻟﻴﻪ ‪ 5A‬ﺑﺎﺷﺪ ، ﺟﺮﻳﺎﻥ ﺛﺎﻧﻮﻳﻪ ‪ 0.05A‬ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﮐﻪ ﺑﺮﺍﻱ‬ ‫‪ , R=100K‬ﺗﻮﺍﻥ 052 ﻭﺍﺕ ﻭ ﻭﻟﺘﺎﮊ ‪ 5KV‬ﻣﻲ ﺷﻮﺩ.‬ ‫15‬
‫ﺗﺮﺍﻧﺲ ﻭﻟﺘﺎﮊ ﺗﺮﺍﻧﺴﻲ ﺍﺳﺖ ﺑﺎ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺍﻭﻟﻴﻪ ﺑﻴﺸﺘﺮ ﺍﺯ ﺛﺎﻧﻮﻳﻪ ﮐﻪ ﻃﻮﺭﻱ ﻃﺮﺍﺣﻲ‬ ‫ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﺣﺎﻟﺖ ﻣﺪﺍﺭﺑﺎﺯ ﺑﻮﺩﻥ ﺛﺎﻧﻮﻳﻪ ﺑﺘﻮﺍﻧﺪ ﮐﺎﺭ ﮐﻨﺪ)ﺟﺮﻳﺎﻥ ﭼﻨﺪﺍﻥ ﺭﺩ‬ ‫ﻧﻤﻲ ﺷﻮﺩ ﻭﻟﻲ ﺍﺧﺘﻼﻑ ﭘﺘﺎﻧﺴﻴﻞ ﺯﻳﺎﺩ ﺍﺳﺖ(.ﺿﻤﻨﺄ ﻭﺟﻮﺩ ﺗﺮﺍﻧﺴﻬﺎﯼ ﻓﻮﻕ ﺳﺒﺐ‬ ‫ﮐﺎﻫﺶ ﺍﺛﺮ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺷﺪﻩ ﺍﺳﺖ.‬ ‫ﺑﻌﻀﺎ ﺩﺭ ﺗﺮﺍﻧﺲ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﺍﻳﻨﮑﻪ ﻣﺠﺒﻮﺭ ﺑﻪ ﻗﻄﻊ ﺳﻴﻤﻲ ﮐﻪ ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺟﺮﻳﺎﻥ‬ ‫ﺁﻧﺮﺍ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﻳﻢ ﻧﺸﻮﻳﻢ ﺍﺯ ﺧﻮﺩ ﺁﻥ ﺳﻴﻢ ﺑﻌﻨﻮﺍﻥ ﺍﻭﻟﻴﻪ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﻨﻨﺪ ﺍﻣﺎ‬ ‫ﺗﺮﺍﻧﺲ ، ‪ dc‬ﺭﺍ ﻋﺒﻮﺭ ﻧﻤﻲ ﺩﻫﺪ ﻭ ﺣﺘﻲ ﻭﺟﻮﺩ ‪ dc‬ﺑﺎﻋﺚ ﺍﺷﺒﺎﻉ ﻣﻐﻨﺎﻃﻴﺴﻲ ﻣﻲ‬ ‫ﺷﻮﺩ ﻟﺬﺍ ﺩﺭ ﭘﺮﻭﺑﻬﺎﻱ ﺟﺮﻳﺎﻥ ﻭ ﻫﺮﺟﺎ ﺍﺣﺘﻴﺎﺝ ﺑﻪ ﺗﺒﺪﻳﻞ ﺟﺮﻳﺎﻥ ﺑﺎ ﭘﺎﺳﺦ‬ ‫ﻓﺮﮐﺎﻧﺴﻲ ﺷﺎﻣﻞ ‪ dc‬ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ ﺍﺯ ﺍﺛﺮ ﻫﺎﻝ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ. ﺍﻭﻟﻴﻦ -‪clip‬‬ ‫‪ on milliammeter‬ﺍﺛﺮ ﻫﺎﻝ ﺭﺍ ‪ HP‬ﺩﺭ ﺳﺎﻝ 8591 ﺑﻪ ﺑﺎﺯﺍﺭ ﺩﺍﺩ .‬ ‫25‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫35‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ)ﮐﻨﺘﻮﺭ(‬ ‫‪t‬‬ ‫′ ‪W ( t ) = ∫ P( t ′ )dt‬‬ ‫ﺍﻧﺮﮊﻱ، ﺍﻧﺘﮕﺮﺍﻝ ﺗﻮﺍﻥ ﺑﺮﺣﺴﺐ ﺯﻣﺎﻥ ﺍﺳﺖ.‬ ‫0‬ ‫ﻣﺒﻨﺎﻱ ﻣﺤﺎﺳﺒﻪ ﻗﻴﻤﺖ ، ﻣﻴﺰﺍﻥ ﻣﺼﺮﻑ ﺍﻧﺮﮊﻱ ﺑﺮ ﺣﺴﺐ ‪ kwh‬ﺍﺳﺖ. ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ‬ ‫ﻭﺳﻴﻠﻪ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ ﺍﻟﮑﺘﺮﻳﮑﻲ ، ﻛﻴﻠﻮﻭﺍﺕ ﺳﺎﻋﺖ ﻣﺘﺮ ﺍﻧﺪﻭﻛﺴﻴﻮﻧﻲ )ﮐﻨﺘﻮﺭ‬ ‫ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ( ﺍﺳﺖ .‬ ‫ﮐﻴﻠﻮﻭﺍﺕ ﺳﺎﻋﺖ ﻣﺘﺮ ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ :‬ ‫ﺩﺭﺍﻳﻦ ﺩﺳﺘﮕﺎﻩ ﻳﻚ ﺻﻔﺤﺔ ﭼﺮﺧﻨﺪﻩ ﺩﺍﺭﻳﻢ ﻛﻪ ﺳﺮﻋﺖ ﭼﺮﺧﺶ ﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ‬ ‫ﺑﺎ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺯﺩﻩ ﺷﺪﻩ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺍﻧﺮﮊﻱ ﻣﺼﺮﻓﻲ ﺧﻮﺍﻫﺪ‬ ‫ﺑﻮﺩ . ﺩﺭﺍﻳﻨﺠﺎ ﺑﺮﺧﻼﻑ ﻭﺍﺗﻤﺘﺮ ﻛﻪ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺩﺭﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ θ‬ﺑﻮﺩ،‬ ‫ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ dθ/dt‬ﺍﺳﺖ.‬ ‫45‬
‫ﺍﺻﻮﻝ ﻛﺎﺭ ﻣﻮﺗﻮﺭ ﺍﻧﺪﻭﻛﺴﻴﻮﻧﻲ :‬ ‫ﺍﮔﺮ ﺣﻠﻘﻪ ﺍﻱ ﺣﺎﻭﻱ ﺟﺮﻳﺎﻥ ‪ i‬ﺩﺍﺧﻞ ﻣﻴﺪﺍﻥ ﻣﻐﻨﺎﻃﻴﺴﻲ ‪ B‬ﻗﺮﺍﺭ ﺑﮕﻴﺮﺩ، ﮔﺸﺘﺎﻭﺭﻱ‬ ‫ﺑﺮﺍﺑﺮ ﺑﺎ ‪ BAi‬ﺑﺮﺁﻥ ﻭﺍﺭﺩ ﻣﻲ ﺷﻮﺩ. ﺑﻌﺒﺎﺭﺕ ﺩﻳﮕﺮ ﮔﺸﺘﺎﻭﺭ ﻭﺍﺭﺩﻩ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ ﺑﺎ ‪ϕi‬‬ ‫ﻭ ﺍﮔﺮ ‪ ϕ‬ﻭ ‪ i‬ﻫﺮ ﺩﻭ ‪ ac‬ﺑﺎﺷﻨﺪ ﻭ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ:‬ ‫‪ϕ = Φ sin ωt‬‬ ‫) ‪i = I sin(ωt − α‬‬ ‫‪1T‬‬ ‫) ‪ ∝ ∫ ϕ(t )i(t )dt = 1 ΦI cos( α‬ﮔﺸﺘﺎﻭﺭ ﻣﺘﻮﺳﻂ‬ ‫0‪T‬‬ ‫2‬ ‫55‬ ‫⇒‬
‫* ﺩﺭ ﻛﻨﺘﻮﺭ ﺍﺯ ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﻭ ﻳﻚ ﺻﻔﺤﺔ ﺩﻭﺍﺭ ﺁﻟﻮﻣﻴﻨﻴﻮﻣﻲ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ)ﻗﺮﺍﺭﺩﺍﺩﻫﺎﻱ ﺟﻬﺖ ﻣﺜﺒﺖ ‪ ϕ‬ﻭ ‪:( i‬‬ ‫1‪i‬‬ ‫2‪i‬‬ ‫2‪ϕ‬‬ ‫65‬ ‫1‪ϕ‬‬
‫ﭼﻮﻥ 1‪ ϕ‬ﻭ 2‪ ϕ‬ﻣﺘﻐﻴﺮ ﺑﺎ ﺯﻣﺎﻥ ﻫﺴﺘﻨﺪ، ﺍﻳﺠﺎﺩ ‪ emf‬ﺍﻟﻘﺎﻳﻲ ﻭ ﻧﺘﻴﺠﺘﺎ ﺟﺮﻳﺎﻥ ﻓﻮﻛﻮ‬ ‫ﹰ‬ ‫ﺩﺭ ﺻﻔﺤﺔ ﺁﻟﻮﻣﻴﻨﻴﻮﻣﻲ ﻣﻲ ﻛﻨﻨﺪ . ﺑﺎ ﻓﺮﺽ ﮐﺎﻣﻼ ﻣﻘﺎﻭﻣﺘﻲ ﺑﻮﺩﻥ ﻣﺴﻴﺮ ﺟﺮﻳﺎﻥ ﻭ‬ ‫ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﻣﺴﻴﺮ :‬ ‫1‪dϕ‬‬ ‫1‪e‬‬ ‫ﮐﻪ 1‪ R‬ﻣﻘﺎﻭﻣﺖ ﻣﺴﻴﺮ ﺍﺳﺖ.‬ ‫75‬ ‫1‪R‬‬ ‫= ‪if‬‬ ‫1‬ ‫‪dt‬‬ ‫− = 1‪e‬‬
‫) ‪ϕ1 = Φ1 sin( ωt‬‬ ‫ﺍﮔﺮ:‬ ‫1 ‪ωΦ‬‬ ‫1 ‪ωΦ‬‬ ‫− = 1‪⇒ i f‬‬ ‫= ‪cos ωt‬‬ ‫) 09 − ‪sin(ωt − 90 ) = I f1 sin( ωt‬‬ ‫1‪R‬‬ ‫1‪R‬‬ ‫~‬ ‫1‪Φ‬‬ ‫~‬ ‫‪If‬‬ ‫ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ~ ، 09 ﺩﺭﺟﻪ ﭘﺲ ﻓﺎﺯ ﺍﺯ ~‬ ‫‪If‬‬ ‫2‪ ϕ‬ﺍﺳﺖ. ﻣﻼﺣﻈﻪ ﻣﻲ ﮐﻨﻴﺪ ﮐﻪ ﺑﺎ‬ ‫‪~ I‬‬ ‫ﺗﻮﺟﻪ ﺑﻪ ﺯﺍﻭﻳﻪ 09 ﺩﺭﺟﻪ ﺍﻱ ﺑﻴﻦ ‪ ~f‬ﻭ 2‪ ϕ‬ﮔﺸﺘﺎﻭﺭ ﺣﺎﺻﻞ ﺍﺯ ﺁﻧﻬﺎ ﺻﻔﺮ ﺍﺳﺖ )ﺩﺭ‬ ‫1‬ ‫2‬ ‫2‬ ‫ﻭﺍﻗﻊ ﻣﺴﻴﺮﻫﺎﻱ ﻓﻮﮐﻮ ﮐﺎﻣﻼ ﻣﻘﺎﻭﻣﺘﻲ ﻧﻴﺴﺘﻨﺪ ﻭ ﺍﻳﻨﻬﺎ ﻗﺪﺭﻱ ﺍﻳﺠﺎﺩ ﺗﺮﻣﺰ ﻣﻲ ﮐﻨﻨﺪ(.‬ ‫~‬ ‫~ ~‬ ‫~‬ ‫ﺍﻣﺎ ﺍﮔﺮ 1‪ ϕ‬ﻭ 2‪ ϕ‬ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﺍﺛﺮ ﻣﺘﻘﺎﺑﻞ ‪ ~f‬ﻭ 1‪ ϕ‬ﻭ ﻧﻴﺰ ‪ ~f‬ﻭ 2‪ϕ‬‬ ‫‪I‬‬ ‫‪I‬‬ ‫ﺍﻳﺠﺎﺩ ﮔﺸﺘﺎﻭﺭ ﻣﺤﺮﮎ ﻣﻲ ﮐﻨﺪ .‬ ‫2‬ ‫85‬ ‫1‬
ϕ1 ϕ2 ϕ1 > 0 ,i1 < 0 ,ϕ2 < 0 ,i2 < 0 ⇒ F ‫ ﻫﻢ ﺟﻬﺖ‬T12 ,T21 ϕ1 ϕ2 i2 i1 F 59
~ ~ :‫ ﺩﺭﺟﻪ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ، ﺩﺍﺭﻳﻢ‬β ‫ ﺑﺎﻧﺪﺍﺯﻩ‬ϕ2 ‫ ﻭ‬ϕ1 ‫ﺍﮔﺮ‬ β β ~ If2 ~ I f1 ~ ϕ1 ~ ϕ2 T12 ∝ Φ1I f cos( 90 + β ) ∝ Φ1Φ 2 cos( 90 + β ) 2 T21 ∝ Φ 2 I f cos( 90 − β ) ∝ Φ1Φ 2 cos( 90 − β ) 1 60
‫ﺑﺎ ﻧﻮﺷﺘﻦ ﺭﻭﺍﺑﻂ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ :‬ ‫) ‪T ∝ T21 − T12 ∝ Φ1Φ 2 sin( β‬‬ ‫* ﺍﮔﺮ 0 = ‪ β‬ﺑﺎﺷﺪ 0=‪ T‬ﺍﺳﺖ. ﺑﻴﺸﺘﺮﻳﻦ ﮔﺸﺘﺎﻭﺭ ﻭﻗﺘﻲ ﺍﺳﺖ ﻛﻪ °09 = ‪. β‬‬ ‫16‬
‫ﺍﺻﻮﻝ ﮐﺎﺭ ﮐﻨﺘﻮﺭ :‬ ‫ﺩﺭ ﮐﻨﺘﻮﺭ ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ ﻳﮏ ﺳﻴﻢ ﭘﻴﭻ 1‪ ϕ‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺟﺮﻳﺎﻥ ﻭ 2‪ ϕ‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ‬ ‫ﻭﻟﺘﺎﮊ ﻭﻟﻲ ﺑﺎ ﺍﺧﺘﻼﻑ ﻓﺎﺯ 09 ﺩﺭﺟﻪ ﺍﺳﺖ )ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺯﻳﺎﺩﻱ‬ ‫ﺩﺍﺭﺩ(. ﻣﻲ ﺧﻮﺍﻫﻴﻢ ‪ T‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺑﺎﺷﺪ ﭘﺲ ﻛﺎﻓﻴﺴﺖ 1‪ Φ‬ﻣﺘﻨﺎﺳﺐ‬ ‫ﺑﺎ ﺟﺮﻳﺎﻥ ﻭ 2 ‪ Φ‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﻭﻟﺘﺎﮊ ﺑﺎﺭ ﺑﺎﺷﺪ ﻭﻟﻲ ﺑﺎ 09 ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﻧﺴﺒﺖ ﺑﻪ‬ ‫ﻭﻟﺘﺎﮊ ﺑﺎﺭ ﺍﺳﺖ. ﻟﺬﺍ :‬ ‫‪T ∝ IV sin( 90 − ϕ ) = IV cos( ϕ ) = P‬‬ ‫~‬ ‫‪V‬‬ ‫‪ϕ‬‬ ‫~‬ ‫1‪ϕ‬‬ ‫26‬ ‫~‬ ‫‪I‬‬ ‫‪β‬‬ ‫~‬ ‫2‪ϕ‬‬
‫ﺩﺭ ﻭﺍﻗﻊ ﺩﺭ ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﻗﺪﺭﻱ ﮐﻤﺘﺮ ﺍﺯ‬ ‫09 ﺩﺭﺟﻪ ﻣﻲ ﺑﺎﺷﺪ ﮐﻪ ﺑﺮﺍﻱ ﺟﺒﺮﺍﻥ ﺁﻥ ﺗﻤﻬﻴﺪﺍﺗﻲ ﺷﺪﻩ ﺍﺳﺖ.)ﺳﻴﻢ‬ ‫ﭘﻴﭻ ﺳﺎﻳﻪ ﺍﻧﺪﺍﺯ ﺩﺭ ﮐﻨﺎﺭ ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ(.‬ ‫ﺑﺮﺍﻱ ﺍﻳﺠﺎﺩ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺍﺯ ﺁﻫﻨﺮﺑﺎﻱ ﺩﺍﺋﻤﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ .‬ ‫36‬
‫ﺷﺒﻴﻪ ﺭﺍﺑﻄﻪ ﺍﻱ ﮐﻪ ﺑﺮﺍﻱ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺣﺎﺻﻞ ﺍﺯ ﮔﺮﺩﺵ ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﻣﺘﺤﺮﮎ ﮔﺎﻟﻮﺍﻧﻮﻣﺘﺮ ﺩﺭ ﻣﻴﺪﺍﻥ ﻣﻐﻨﺎﻃﻴﺴﻲ ﺩﺍﺷﺘﻴﻢ) ‪، ( ( NBA )2 dθ‬‬ ‫‪R‬‬ ‫‪dt‬‬ ‫ﺩﺭ ﺍﻳﻨﺠﺎ ﻧﻴﺰ :‬ ‫ﺳﺮﻋﺖ ﭼﺮﺧﺶ‬ ‫‪dθ‬‬ ‫= ‪ ∝ n‬ﮔﺸﺘﺎﻭﺭﻣﻘﺎﻭﻡ‬ ‫‪dt‬‬ ‫ﻟﺬﺍ :‬ ‫‪ ⇒ n ∝ P‬ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ=ﮔﺸﺘﺎﻭﺭ ﻣﺤﺮﮎ‬ ‫46‬
‫ﭼﻮﻥ ﺳﺮﻋﺖ ﭼﺮﺧﺶ ﺻﻔﺤﻪ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺍﺳﺖ ، ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺯﺩﻩ‬ ‫ﺷﺪﻩ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺍﻧﺮﮊﻱ ﻣﺼﺮﻓﻲ ﺧﻮﺍﻫﺪ ﺑﻮﺩ .‬ ‫ﺍﮔﺮ ﺍﺯ ﻓﻨﺮ ﺑﺠﺎﻱ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﺮﺩﻳﻢ ، ﻭﺍﺗﻤﺘﺮ ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ‬ ‫ﺩﺍﺷﺘﻴﻢ.‬ ‫ﺑﺎ ﺗﻐﻴﻴﺮ ﻣﺤﻞ ﺁﻫﻨﺮﺑﺎﻱ ﺩﺍﺋﻤﻲ ﻣﻲ ﺗﻮﺍﻥ ﻣﻴﺰﺍﻥ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺭﺍ ﺗﻐﻴﻴﺮ ﺩﺍﺩ.‬ ‫ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ R‬ﻓﺎﺻﻠﻪ ﺁﻫﻨﺮﺑﺎ ﺍﺯ ﻣﺮﮐﺰ ﺍﺳﺖ ﻭ ﺍﮔﺮ ﻓﺎﺻﻠﻪ ﺍﺯ‬ ‫ﻣﺮﻛﺰ ﺯﻳﺎﺩ ﮔﺮﺩﺩ ، ‪T‬ﻣﻘﺎﻭﻡ ﺑﻴﺸﺘﺮ ﺷﺪﻩ ، ﻭ ﺻﻔﺤﻪ ﻛﻨﺪﺗﺮ ﻣﻲ ﭼﺮﺧﺪ . ﺑﻪ ﺍﻳﻦ‬ ‫ﺗﺮﺗﻴﺐ ﻣﻲ ﺗﻮﺍﻥ ﮐﻨﺘﻮﺭ ﺭﺍ ﮐﺎﻟﻴﺒﺮﻩ ﮐﺮﺩ )ﮐﺎﻟﻴﺒﺮﺍﺳﻴﻮﻥ ﺩﺭ ﺑﺎﺭ ﮐﺎﻣﻞ(‬ ‫)ﻫﻤﭽﻨﻴﻦ ﮐﺎﻟﻴﺒﺮﺍﺳﻴﻮﻥ ﺩﺭ %01 ﺑﺎﺭ ﮐﺎﻣﻞ ﺗﻮﺳﻂ ﺗﻨﻈﻴﻢ ﺳﻴﻢ ﭘﻴﭻ ﺳﺎﻳﻪ‬ ‫ﺍﻧﺪﺍﺯ ﺻﻮﺭﺕ ﻣﻲ ﮔﻴﺮﺩ( .‬ ‫56‬
‫ﺟﻨﺲ ﺁﻫﻨﺮﺑﺎ ﺑﺎﻳﺪ ﺑﺴﻴﺎﺭ ﺧﻮﺏ ﺑﺎﺷﺪ ﺗﺎ ﺩﺭ ﺍﺛﺮ ﮔﺬﺷﺖ ﺯﻣﺎﻥ ﺿﻌﻴﻒ ﻧﺸﻮﺩ.‬ ‫ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﺩﺭﺟﻪ ﺣﺮﺍﺭﺕ ﺧﺎﺻﻴﺖ ﺁﻫﻨﺮﺑﺎﻳﻲ ﮐﻢ ﻭ ﻣﻘﺎﻭﻣﺖ ﻣﺴﻴﺮ‬ ‫ﺍﻓﺰﺍﻳﺶ ﻣﻲ ﻳﺎﺑﺪ ، ﻳﻌﻨﻲ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﮐﻢ ﻣﻴﺸﻮﺩ .‬ ‫* ﺩﺭﮐﻨﺘﻮﺭ ﺳﻪ ﻓﺎﺯ ﻫﺮ ﻓﺎﺯ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﻭ ﺻﻔﺤﻪ ﺁﻟﻮﻣﻴﻨﻴﻮﻣﻲ ﻣﺨﺼﻮﺹ ﺑﻪ‬ ‫ﺧﻮﺩ ﺭﺍ ﺩﺍﺭﺩ ﻭﻟﻲ ﻫﻤﺔ ﮔﺸﺘﺎﻭﺭﻫﺎﻱ ﺍﻳﻦ ﺳﻪ، ﺑﻪ ﻳﻚ ﺷﻔﺖ ﻣﺘﺼﻞ ﻣﻲ‬ ‫ﺷﻮﻧﺪ ﮐﻪ ﺩﺭ ﻧﻬﺎﻳﺖ ﺳﺮﻋﺖ ﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﮐﻞ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺳﻪ ﻓﺎﺯ‬ ‫ﺍﺳﺖ .‬ ‫66‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫76‬
‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺿﺮﻳﺐ ﺗﻮﺍﻥ‬ ‫ﺩﺳﺘﮕﺎﻩ ‪ Cosφ‬ﻣﺘﺮ، ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ ﺍﺳﺖ ﻛﻪ ﺑﺠﺎﻱ ﻳﻚ ﺳﻴﻢ ﭘﻴﺞ‬ ‫ﻣﺘﺤﺮﻙ، ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﻣﺘﺤﺮﻙ ﻣﺘﻘﺎﻃﻊ )ﻋﻤﻮﺩ ﺑﺮﻫﻢ( ﺩﺍﺭﺩ. ﻣﻲ ﺩﺍﻧﻴﻢ‬ ‫ﺩﺭ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮ:‬ ‫‪dM‬‬ ‫2 ‪I1 I‬‬ ‫‪dθ‬‬ ‫86‬ ‫=‪T‬‬
‫ﺍﮔﺮ ﻣﻴﺪﺍﻥ ﺑﺼﻮﺭﺕ ﺑﺎﻻ ﺑﺎﺷﺪ:‬ ‫‪dM‬‬ ‫→ ‪M ∝ cos θ‬‬ ‫‪∝ sin θ‬‬ ‫‪dθ‬‬ ‫ﺟﻬﺖ ﺳﻴﻢ ﭘﻴﭽﻬﺎﻱ ‪ A‬ﻭ ‪ B‬ﻃﻮﺭﻱ ﺑﺴﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﮔﺸﺘﺎﻭﺭﻫﺎﻱ ﺁﻧﻬﺎ‬ ‫ﺑﺮﺧﻼﻑ ﻫﻢ ﺑﺎﺷﺪ . ﺩﺭ ﻭﺍﻗﻊ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺑﻪ ﺍﻳﻦ ﺻﻮﺭﺕ ﺍﻳﺠﺎﺩ ﻣﻴﺸﻮﺩ ﻭ‬ ‫ﺩﺳﺘﮕﺎﻩ ﻓﻨﺮﻱ ﻧﺪﺍﺭﺩ ﺑﺨﺎﻃﺮ ﻫﻤﻴﻦ ﻫﻢ ﻫﺴﺖ ﮐﻪ ﭘﺲ ﺍﺯ ﻗﺮﺍﺋﺖ ، ﻋﻘﺮﺑﻪ ﺳﺮ‬ ‫ﺟﺎﻳﺶ ﻣﻴﻤﺎﻧﺪ .‬ ‫96‬
TA = KVI cos( ϕ ) sin( θ ) TB = KVI cos( 90 − ϕ ) sin( 90 + θ )= KVI sin( ϕ ) cos( θ ) :‫ﺩﺭﺣﺎﻟﺖ ﺗﻌﺎﺩﻝ‬ TA = TB ⇒ cos θ sin ϕ = cos ϕ sin θ ⇒ tgθ = tgϕ ⇒ θ = ϕ 70
‫ﺻﻔﺤﺔ ﺩﺳﺘﮕﺎﻩ ﺭﺍ ﺑﺮﺣﺴﺐ ‪ Cosφ‬ﻣﺪﺭﺝ ﻣﻲ ﻛﻨﻨﺪ . ﺑﺮﺍﻱ ‪ 50Hz‬ﮐﺎﻟﻴﺒﺮﻩ‬ ‫ﺷﺪﻩ ﺍﺳﺖ ﻭ ﺩﺭ ﻓﺮﮐﺎﻧﺴﻬﺎﻱ ﺩﻳﮕﺮ ﺧﻄﺎ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ.‬ ‫ﺩﺭ ‪ Cosφ‬ﻣﺘﺮ ﺳﻪ ﻓﺎﺯ، ﻓﺎﺯ 1 ﺭﺍ ﺍﺯ ﺳﻴﻢ ﭘﻴﭻ ﺛﺎﺑﺖ ﻋﺒﻮﺭ ﻣﻴﺪﻫﻨﺪ ﻭﺩﻭ ﺳﻴﻢ‬ ‫ﭘﻴﭻ ﺩﻳﮕﺮ ﺭﺍ ﺑﺎ ﻣﻘﺎﻭﻣﺖ ﺑﻪ 21‪ V‬ﻭ 31‪ V‬ﻭﺻﻞ ﻣﻲ ﮐﻨﻨﺪ. ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ‬ ‫ﻣﻴﺘﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﮐﻪ ﺑﺎ ﻓﺮﺽ ﻣﺘﻌﺎﺩﻝ ﺑﻮﺩﻥ ﻓﺎﺯﻫﺎ ‪ϕ = θ‬ﺧﻮﺍﻫﺪ ﺑﻮﺩ .‬ ‫ﻣﺮﺍﺟﻌﻪ ﮐﻨﻴﺪ ﺑﻪ ﺳﺎﻭﻧﻲ ﺻﻔﺤﻪ 226.‬ ‫17‬

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Chapter5

  • 1.
    ‫ﻓﺼﻞ ﭘﻨﺠﻢ‬ ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱﺗﻮﺍﻥ ﻭ ﺍﻧﺮﮊﻱ‬ ‫ﺍﻳﻦ ﻣﺒﺤﺚ ﺩﺭ ﮐﺘﺎﺏ ﻧﻴﺴﺖ . ﺍﺯ ‪ Cooper‬ﻳﺎ ﺳﺎﻭﻧﻲ ﺑﺨﻮﺍﻧﻴﺪ.‬ ‫1‬
  • 2.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ)ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫2‬
  • 3.
    ‫ﻣﻘﺪﻣﻪ :‬ ‫ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩﮔﻴﺮﻱ ﺗﻮﺍﻥ ﻳﻚ ﺑﺎﺭ ﻣﻘﺎﻭﻣﺘﻲ ﺩﺭﻳﻚ ﺷﺒﻜﺔ ‪ DC‬ﻣﻲ ﺗﻮﺍﻧﻴﻢ‬ ‫ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﻳﻢ . ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺩﻭ ﺭﻭﺵ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ‬ ‫‪Ic‬‬ ‫ﺑﮑﺎﺭ ﺑﺮﺩ .‬ ‫‪Ic‬‬ ‫‪IR‬‬ ‫+ _‬ ‫‪A‬‬ ‫‪Ip‬‬ ‫‪+ V‬‬ ‫‪c‬‬ ‫‪VR‬‬ ‫+‬ ‫‪V‬‬ ‫_‬ ‫2‬ ‫‪Vp‬‬ ‫_‬ ‫+ ‪IR‬‬ ‫‪R‬‬ ‫_‬ ‫‪V‬‬ ‫‪Ip‬‬ ‫+‬ ‫‪V‬‬ ‫‪A‬‬ ‫‪Vp‬‬ ‫_‬ ‫1‬ ‫2‬ ‫‪VR‬‬ ‫= ‪PP‬‬ ‫‪RV‬‬ ‫3‬ ‫2 ‪PC = R A I R‬‬ ‫‪Pm = I C V P = ( I R + I P ) V R = PR + PP where‬‬ ‫)1‬ ‫‪Pm = I C V P = ( V R + V C ) I R = PR + PC where‬‬ ‫)2‬
  • 4.
    ‫ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﻣﻼﺣﻈﻪﻣﻲ ﺷﻮﺩ ﺩﺭ ﻫﺮ ﺩﻭ ﺭﻭﺵ ‪) Pm‬ﺗﻮﺍﻧﻲ ﮐﻪ ﻣﻲ ﺧﻮﺍﻧﻴﻢ(‬ ‫ﺑﻴﺶ ﺍﺯ ‪) PR‬ﺗﻮﺍﻧﻲ ﮐﻪ ﻭﺍﻗﻌﺎ ﺩﺭ ﺑﺎﺭ ﻣﺼﺮﻑ ﻣﻲ ﺷﻮﺩ( ﺍﺳﺖ .‬ ‫ﺍﮔﺮﺟﺮﻳﺎﻥ ﺑﺎﺭ ﺯﻳﺎﺩ ﺑﺎﺷﺪ ﻭ ﻭﻟﺘﺎﮊ ﺁﻥ ﻛﻢ ﺭﻭﺵ ﺍﻭﻝ ﺑﻬﺘﺮ ﺍﺳﺖ.‬ ‫ﺍﮔﺮ ﺟﺮﻳﺎﻥ ﺑﺎﺭ ﻛﻢ ﺑﺎﺷﺪ ﻭ ﻭﻟﺘﺎﮊ ﺁﻥ ﺯﻳﺎﺩ ﺭﻭﺵ ﺩﻭﻡ ﺑﻬﺘﺮ ﺍﺳﺖ.‬ ‫4‬
  • 5.
    ‫ﻣﻌﻤﻮﻻ ﺭﻭﺵ )1(ﺭﺍ ﺗﺮﺟﻴﺢ ﻣﻲ ﺩﻫﻴﻢ :‬ ‫ﹰ‬ ‫ﺍﻟﻒ( ﻭﻟﺘﻤﺘﺮﻫﺎﻱ ﺍﻣﺮﻭﺯﻱ ﻣﻘﺎﻭﻣﺖ ﻭﺭﻭﺩﻱ ﺧﻴﻠﻲ ﺯﻳﺎﺩﻱ ﺩﺍﺭﻧﺪ .‬ ‫ﺏ( ﺍﮔﺮ ﻭﻟﺘﺎﮊ ﺛﺎﺑﺖ ﺑﺎﺷﺪ ﺍﻣﺎ ﺑﺎﺭ ﻣﺘﻐﻴﺮ ﺑﺎﺷﺪ ﻣﻘﺪﺍﺭ ‪ PP = V 2 / R V‬ﺛﺎﺑﺖ‬ ‫ﺍﺳﺖ ﻭ ﻣﻲ ﺗﻮﺍﻥ ﻣﻘﺪﺍﺭ ﺁﻧﺮﺍ ﺩﺭ ﺣﺎﻟﺖ ﺑﻲ ﺑﺎﺭﻱ ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ ﻭ ﺍﺯ ﺗﻤﺎﻡ‬ ‫ﻗﺮﺍﺋﺘﻬﺎ ﮐﻢ ﮐﺮﺩ.‬ ‫ﻣﻌﻤﻮﻻ ﺑﺠﺎﻱ ﺍﻳﻦ ﮐﺎﺭ ﺍﺯ ﻭﺍﺗﻤﺘﺮ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﻨﻴﻢ ﮐﻪ ﺩﻳﮕﺮ ﺍﺣﺘﻴﺎﺝ ﺑﻪ‬ ‫ﺿﺮﺏ ﺍﻋﺪﺍﺩ ﺩﺭ ﻫﻢ ﻧﺪﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ ﻭ ﺻﺤﺖ ﻫﻢ ﺑﻬﺘﺮ ﺍﺳﺖ.‬ ‫5‬
  • 6.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫6‬
  • 7.
    ‫ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ:‬ ‫ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦﻧﻮﻉ ﻭﺍﺗﻤﺘﺮ ﺍﺳﺖ ﮐﻪ ﻣﻲ ﺗﻮﺍﻧﺪ ﺗﻮﺍﻥ ‪ DC‬ﻳﺎ ‪) AC‬ﺳﻴﻨﻮﺳﻲ ﻳﺎ‬ ‫ﻏﻴﺮ ﺳﻴﻨﻮﺳﻲ( ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﺩ . )ﻧﻮﻋﻬﺎﻱ ﺩﻳﮕﺮ: ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ،‬ ‫ﺗﺮﻣﻮﮐﻮﭘﻠﻲ، ﺍﻟﮑﺘﺮﻭﺍﺳﺘﺎﺗﻴﮑﻲ، ﻓﺮﻭﻣﻐﻨﺎﻃﻴﺴﻲ(‬ ‫ﺩﻳﺪﻳﻢ ﻛﻪ ﺩﺭﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮ :‬ ‫‪dM‬‬ ‫‪Ic‬‬ ‫+‬ ‫‪VR‬‬ ‫_‬ ‫7‬ ‫‪Ip‬‬ ‫‪R‬‬ ‫‪p‬‬ ‫‪IC IP‬‬ ‫=‪T‬‬ ‫‪dθ‬‬ ‫‪1 dM‬‬ ‫‪IC IP‬‬ ‫=‪θ‬‬ ‫‪S dθ‬‬
  • 8.
    ‫ﺣﺎﻝ ﺍﮔﺮ ‪، I p = VR / R p ، I c = I R‬ﺑﺎﺷﺪ ﺩﺍﺭﻳﻢ:‬ ‫‪θ‬‬ ‫‪RP‬‬ ‫=‪θ‬‬ ‫=‪⇒P‬‬ ‫‪K‬‬ ‫′‪K‬‬ ‫ﻭﻟﻲ ﺩﺭ ﻭﺍﻗﻊ ،‬ ‫‪1 dM VR‬‬ ‫‪P‬‬ ‫=‪θ‬‬ ‫‪IR‬‬ ‫‪=K‬‬ ‫‪S dθ R P‬‬ ‫‪RP‬‬ ‫‪VR‬‬ ‫‪I R VR‬‬ ‫‪θ = KI C I P = KI C‬‬ ‫‪≈K‬‬ ‫‪= K′P‬‬ ‫‪RP‬‬ ‫‪RP‬‬ ‫ﻳﻌﻨﻲ ‪ θ ≈ K ′P‬ﺩﺭ ﺣﺎﻟﻲ ﮐﻪ ‪) θ = K ′P‬ﻳﻌﻨﻲ ﻭﺍﺗﻤﺘﺮ ﺑﺮ ﺍﺳﺎﺱ ﺍﻳﻦ ﺿﺮﻳﺐ‬ ‫ﻣﺪﺭﺝ ﻣﻲ ﺷﻮﺩ.(‬ ‫‪m‬‬ ‫8‬
  • 9.
    ‫ﺍﮔﺮ ﺟﺮﻳﺎﻧﻬﺎ ‪ac‬ﺑﺎﺷﻨﺪ ﻧﻴﺰ‬ ‫ﻭ ﺍﮔﺮ ‪ i = i C‬ﻭ ‪ i P = v‬ﺑﺎﺷﺪ ﺩﺍﺭﻳﻢ :‬ ‫‪i‬‬ ‫‪RP‬‬ ‫‪ic‬‬ ‫+‬ ‫‪V‬‬ ‫_‬ ‫‪Z‬‬ ‫‪ip‬‬ ‫‪1T‬‬ ‫‪θ = k ∫ i Ci P dt‬‬ ‫0‪T‬‬ ‫‪T‬‬ ‫1 ‪k‬‬ ‫‪k‬‬ ‫=‪θ‬‬ ‫‪P = k′P‬‬ ‫= ‪∫ ivdt‬‬ ‫0 ‪RP T‬‬ ‫‪RP‬‬ ‫‪R‬‬ ‫‪p‬‬ ‫‪1T‬‬ ‫ﻫﻤﺎﻥ ﺗﻌﺮﻳﻒ ﺗﻮﺍﻥ ﺍﮐﺘﻴﻮ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﻣﻮﺭﺩ ﺳﻴﻨﻮﺳﻲ )‪VICos(ϕ‬‬ ‫ﮐﻪ ‪∫ ivdt‬‬ ‫0‪T‬‬ ‫9‬ ‫ﻣﻲ ﺷﻮﺩ .‬
  • 10.
    : ‫ﻭ ﻳﺎ‬ RP RPT Pm = θ= ∫ i Ci P dt k T 0 : ‫ﺍﻟﺒﺘﻪ ﺧﻄﺎﻳﻲ ﮐﻪ ﻗﺒﻼ ﺍﺷﺎﺭﻩ ﮐﺮﺩﻳﻢ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻭ ﺩﺭ ﻭﺍﻗﻊ‬ k 1T k 1T kT θ = ∫ i Ci P dt ≈ ∫ i C vdt ≈ ∫ ivdt = k ′P RP T 0 RP T 0 T0 P≈ Rp k θ ‫ ﻭﻟﻲ‬Pm = Rp k θ ‫ﺩﺭ ﺍﻳﻨﺠﺎ ﻧﻴﺰ‬ 10
  • 11.
  • 12.
    ‫ﺩﺭ ﺍﻳﻨﺠﺎ ﻫﻢﺩﻭ ﺟﻮﺭ ﻣﻲ ﺗﻮﺍﻥ ﻭﺍﺗﻤﺘﺮ ﺭﺍ ﺑﺴﺖ. ﻣﻌﻤﻮﻻ ﻳﮏ ﺳﺮ ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﺟﺮﻳﺎﻥ ﻭ ﻳﮏ ﺳﺮ ﺳﻴﻢ ﭘﻴﭻ ﻭﻟﺘﺎﮊ ﺑﺎ ﻋﻼﻣﺖ ± ﻣﺸﺨﺺ ﺷﺪﻩ ﺍﺳﺖ.‬ ‫ﺟﺮﻳﺎﻥ ﻭﺍﺭﺩ ﻗﻄﺐ ± ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﺷﺪﻩ ﻭ ﻗﻄﺐ ± ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﻭﻟﺘﺎﮊ ﺑﻪ ﻳﮏ ﻃﺮﻑ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﻭﺻﻞ ﻣﻲ ﺷﻮﺩ. )ﺍﮔﺮ ﺩﻳﺪﻳﺪ ﻋﻘﺮﺑﻪ‬ ‫ﺑﺮﻋﮑﺲ ﻣﻨﺤﺮﻑ ﻣﻲ ﺷﻮﺩ ﺟﻬﺖ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﻌﻮﻳﺾ ﮐﻨﻴﺪ(.‬ ‫±‬ ‫±‬ ‫±‬ ‫±‬ ‫21‬ ‫2‬ ‫1‬
  • 13.
    ‫ﮐﻪ ﮔﻔﺘﻴﻢ ﻣﻌﻤﻮﻻﺭﻭﺵ ﺍﻭﻝ ﺗﺮﺟﻴﺢ ﺩﺍﺭﺩ ﭼﻮﻥ ﻭﻟﺘﺎﮊ ﺑﺮﻕ ﺷﻬﺮ ﺛﺎﺑﺖ ﻓﺮﺽ ﺷﺪ‬ ‫ﭘﺲ ‪ Pp‬ﺛﺎﺑﺖ ﺍﺳﺖ ﮐﻪ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎﺯﺍﻱ ‪ ZL‬ﺑﻴﻨﻬﺎﻳﺖ ﺁﻥ ﺭﺍ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﮐﺮﺩ ﻭ‬ ‫ﻫﻤﻴﺸﻪ ﺍﺯ ﻗﺮﺍﺋﺘﻬﺎ ﮐﻢ ﮐﺮﺩ.‬ ‫ﺑﺮﺍﻱ ﺣﺬﻑ ﺍﺛﺮ ‪ i p‬ﺩﺭ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺟﺮﻳﺎﻥ ﺗﻮﺳﻂ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ ﺳﻴﻢ‬ ‫ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺩﺭ ﺟﻬﺖ ﺧﻼﻑ ﺩﻭﺭ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ ﭘﻴﭽﻴﺪ ﺗﺎ ﻓﻠﻮﻱ ﺍﻳﺠﺎﺩ‬ ‫ﺷﺪﻩ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ i = i − i‬ﺑﺎﺷﺪ )ﻭﺍﺗﻤﺘﺮ ﺟﺒﺮﺍﻥ ﺷﺪﻩ(.‬ ‫ﻋﻼﻭﻩ ﺑﺮ ﻋﺎﻣﻞ ﻓﻮﻕ ﻋﺎﻣﻞ ﺩﻳﮕﺮﻱ ﮐﻪ ﺑﺮﺍﻱ ﺧﻄﺎ ﺩﺭ ﺣﺎﻟﺖ ‪ ac‬ﺩﺍﺭﻳﻢ ﺍﻳﻦ ﺍﺳﺖ ﮐﻪ‬ ‫‪ i p = V / R p‬ﻧﻴﺴﺖ ﻭ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻠﻒ ﺭﺍ ﻫﻢ ﺩﺍﺭﻳﻢ . ﺩﺭ ﺣﺎﻟﺖ ﺳﻴﻨﻮﺳﻲ ﮐﻪ‬ ‫ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ ﺣﺎﻟﺖ ﺍﺳﺖ ﺍﻳﻦ ﻣﺴﺎﻟﻪ ﺭﺍ ﺑﺮﺭﺳﻲ ﻣﻲ ﮐﻨﻴﻢ .‬ ‫‪p‬‬ ‫31‬ ‫‪c‬‬
  • 14.
    ‫ﺁﻧﺎﻟﻴﺰ ﻭﺍﺗﻤﺘﺮ ﺩﺭﺣﺎﻟﺖ ﺩﺍﺋﻤﻲ ﺳﻴﻨﻮﺳﻲ‬ ~ iC = I C Cos (ωt + ϕC ) I C = I C e jϕ ~ iP = I P Cos (ωt + ϕ P ) I P = I P e jϕ ~ v(t ) = VCos (ωt ) V =V < 0 C P ~ iC = Re{I C e jωt } ~ iP = Re{I P e jωt } ~ IC R p ~ I LP ~ IP ~ Z 14
  • 15.
    RP T 1T ‫ ﺍﻣﺎ‬Pm= ∫ iC iP dt ‫ ﻭ‬θ = k ∫ i Ci P dt ‫ﺩﻳﺪﻳﻢ ﮐﻪ‬ T 0 T0 1T 1 ~~ (ϕc − ϕ P ) = 1 Re I C I *P ∫ iC iP dt = I C I PCos 2 2 T0 { { RP ~~ Pm = Re I C I * P 2 } } : ‫ﻳﻌﻨﻲ‬ 15
  • 16.
    :‫ﺣﺎﻝ ﺑﺮﺍﻱ ﺑﺪﺳﺖﺁﻭﺭﺩﻥ ﺧﻄﺎﻱ ﻭﺍﺗﻤﺘﺮ‬ ~ = ze jϕ z ~ = R + j ωL = z e j β zp p p p ‫ﻳﺎ‬ ~ ~ ~ ~ ~ v V − jϕ I C = I + I P I = ~ = e = Ie − jϕ z z ~ V − jβ v ~ Ip = ~ = e zp zp 16
  • 17.
    [ ] ( 1 1 ~ ~ ~*1 ~~ 2 Pm = R p Re (I + I p )I p = RP I P + RP Re I I p* 2 2 2 ⎛ − jϕ V + jβ ⎞ 1 = PP + RP Re⎜ Ie e ⎟ ⎜ ⎟ 2 zp ⎝ ⎠ 1 RP = PP + VI cos(ϕ − β ) 2 zP 1 = PP + VI cos(β ) cos(ϕ − β ) 2 ωLP ‫ ﻭ‬RP = cos(β) = sin(β ) zP zP ) : ‫ﺯﻳﺮﺍ‬ 17
  • 18.
    ‫ﺑﺮﺍﻱ 0 =‪ ωLP‬ﺩﺍﺭﻳﻢ ‪ ~P = z P = RP‬ﻭ 0 = ‪ β‬ﻭﻟﺬﺍ:‬ ‫‪z‬‬ ‫1‬ ‫‪Pm = PP + VI cos(ϕ) = PP + P‬‬ ‫2‬ ‫ﮐﻪ ﺍﻳﻦ ﻫﻤﺎﻥ ﭼﻴﺰﻱ ﺍﺳﺖ ﮐﻪ ﻣﺒﺤﺚ ﻗﺒﻞ ﺩﻳﺪﻳﻢ ﻭ ﺍﺛﺮ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ‪ LP‬ﺭﺍ‬ ‫ﺩﺭ ﻧﻈﺮ ﻧﮕﺮﻓﺘﻪ ﺑﻮﺩﻳﻢ ﻭﮔﺮﻧﻪ ﺩﺭ ﺻﻮﺭﺕ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ‪ PP‬ﺩﺍﺭﻳﻢ:)ﺿﺮﻳﺐ‬ ‫ﺗﺼﺤﻴﺢ(‬ ‫)‪cos(ϕ‬‬ ‫‪P‬‬ ‫=‬ ‫) ‪Pm cos(β ) cos(ϕ − β‬‬ ‫81‬
  • 19.
    ‫ﻣﻌﻤﻮﻻ ﺍﮔﺮﭼﻪ 0≠ ‪ LP‬ﻭﻟﻲ ‪ ωLP << RP‬ﻭ ﺍﺯ ﺗﻘﺮﻳﺒﻲ ﮐﻪ ﺩﺭ ﺍﺩﺍﻣﻪ‬ ‫ﺧﻮﺍﻫﺪ ﺁﻣﺪ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﻨﻴﻢ:‬ ‫1‬ ‫)‪Pm = PP + VI cos β(cos β cos ϕ + sin β sin ϕ‬‬ ‫2‬ ‫1‬ ‫)‪= PP + VI cos 2 β(cos ϕ + tgβ sin ϕ‬‬ ‫2‬ ‫ﺑﺎ ﻓﺮﺽ ‪ ωLP << RP‬ﺑﺘﺎ ﺧﻴﻠﻲ ﮐﻮﭼﮏ ﺧﻮﺍﻫﺪ ﺑﻮﺩ )ﻣﻌﻤﻮﻻ ﭼﻨﺪ ﺩﻗﻴﻘﻪ(‬ ‫ﻭ ﻟﺬﺍ 1 ≈ ‪ cos 2 β‬ﭘﺲ:‬ ‫‪ωLP‬‬ ‫= ‪tgβ‬‬ ‫‪RP‬‬ ‫91‬ ‫1‬ ‫‪Pm ≈ PP + VI (cos ϕ + tgβ sin ϕ) where‬‬ ‫2‬
  • 20.
    ‫ﮐﻪ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ:‬ ‫1‬ ‫‪Pm ≈ Pp + P + VI tgβ sin ϕ‬‬ ‫2‬ ‫ﭘﺲ ﻋﻼﻭﻩ ﺑﺮ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺩﺭ ﺷﺎﺧﻪ ﻭﻟﺘﻤﺘﺮ ، ﺧﻄﺎﻱ ﺩﻳﮕﺮ ﻧﺎﺷﻲ ﺍﺯ‬ ‫ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ‪ LP‬ﻣﻲ ﺑﺎﺷﺪ ﮐﻪ ﻣﻘﺪﺍﺭ ﺧﻄﺎ ﺑﻪ ﺯﺍﻭﻳﻪ ﻓﺎﺯ ﺑﺎﺭ ‪ ϕ‬ﻫﻢ ﺑﺴﺘﮕﻲ‬ ‫ﺩﺍﺭﺩ. ﻫﺮ ﭼﻪ ﺑﺎﺭ ﻣﻘﺎﻭﻣﺘﻲ ﺗﺮ ﺑﺎﺷﺪﺍﻳﻦ ﺧﻄﺎ ﮐﻤﺘﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ )ﺑﺮﺍﻱ ﺑﺎﺭ‬ ‫ﺳﻠﻔﻲ ﻣﺜﺒﺖ ﻭ ﺑﺮﺍﻱ ﺑﺎﺭ ﺧﺎﺯﻧﻲ ﻣﻨﻔﻲ ، ﺳﻠﻒ ‪ LP‬ﺳﺒﺐ ﻫﻤﻔﺎﺯﺗﺮ ﺷﺪﻥ ‪I C‬‬ ‫ﻭ ‪ I P‬ﻣﻲ ﺷﻮﺩ(.‬ ‫ﺑﺮﺍﻱ ﺟﺒﺮﺍﻥ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﮐﺎﺭﻱ ﻣﻲ ﮐﻨﻴﻢ ﮐﻪ‬ ‫0∠ ‪z ≈ R‬‬ ‫‪p‬‬ ‫02‬ ‫‪p‬‬
  • 21.
    LP RP r ~ = jωL+ ( R − r ) + r (1 − rjωc ) zp P P 1 + r 2c 2ω 2 c ( z p ≈ RP − r + r + jω LP − r 2c ) ، ‫ ﺁﻧﮕﺎﻩ‬r 2c 2ω 2 << 1 ‫ﺍﮔﺮ‬ 21
  • 22.
    ‫ﺩﺭ ﻧﺘﻴﺠﻪ :‬ ‫‪LP= r 2C‬‬ ‫ﺑﺎ ﺍﻳﻦ ﺭﻭﺵ ﺗﺎ ﺣﺪﻭﺩ ﻓﺮﮐﺎﻧﺲ ‪ 10kHz‬ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻴﻢ ﭘﻴﭻ ﺟﺒﺮﺍﻥ ﻣﻲ ﺷﻮﺩ .‬ ‫22‬
  • 23.
    ‫ﻣﺜﺎﻝ :‬ ‫ﻳﮏ ﻭﺍﺕﺳﻨﺞ ﺍﻟﮑﺘﺮﻭﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ ﻣﻄﺎﺑﻖ ﺷﮑﻞ ﺯﻳﺮ ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﻳﮏ‬ ‫ﻣﺪﺍﺭ ﺗﮏ ﻓﺎﺯ ﺑﮑﺎﺭ ﺭﻓﺘﻪ ﺍﺳﺖ . ﻭﻟﺘﺎﮊ ﺑﺎﺭ ‪ 100V‬ﻭ ﺟﺮﻳﺎﻥ ﺑﺎﺭ‪ 9A‬ﻭ ﺿﺮﻳﺐ ﺗﻮﺍﻥ‬ ‫1.0 ﭘﺲ ﻓﺎﺯ ﻣﻲ ﺑﺎﺷﺪ . ﻣﺪﺍﺭ ﻭﻟﺘﺎﮊ ﻭﺍﺗﻤﺘﺮ ﺩﺍﺭﺍﻱ ﻣﻘﺎﻭﻣﺖ 0003 ﺍﻫﻢ ﻭ‬ ‫ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ‪ 30mH‬ﻣﻲ ﺑﺎﺷﺪ . )‪(f=50Hz‬‬ ‫~‬ ‫‪IC‬‬ ‫~‬ ‫‪I‬‬ ‫~‬ ‫‪Z‬‬ ‫32‬ ‫~‬ ‫‪IP‬‬ ‫‪LP‬‬ ‫‪R‬‬ ‫‪p‬‬
  • 24.
    ‫ﺍﻟﻒ( ﺩﺭﺻﺪ ﺧﻄﺎﻱﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ ﺭﺍﭘﻴﺪﺍ ﮐﻨﻴﺪ .‬ ‫ﺏ( ﺩﺭ ﺻﻮﺭﺕ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ‪ ، PP‬ﺑﺮﺍﻱ ﭼﻪ ﺿﺮﻳﺐ ﺗﻮﺍﻧﻲ ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ‬ ‫ﺻﻔﺮ ﺧﻮﺍﻫﺪ ﺷﺪ ؟‬ ‫ﺝ( ﭼﮕﻮﻧﻪ ﻣﻲ ﺗﻮﺍﻥ ﺧﻄﺎﻱ ﻧﺎﺷﻲ ﺍﺯ ‪ LP‬ﺭﺍ ﺣﺬﻑ ﮐﺮﺩ؟‬ ‫42‬
  • 25.
    P = VIcos( ϕ ) = 100 × 9 × 0.1 = 90W Pm = PP + VI cos( β ) cos( ϕ − β ) cos( ϕ ) = 0.1 ⇒ ϕ = 84.26° X P = 2π( 50 )( 30 × 10 − 3 ) = 9.42Ω , 9.42 β = tg ( ) = 0.18° 3000 V 2 100 2 PP ≈ = = 3.33W RP 3000 RP = 3000Ω : ‫ﺣﻞ‬ (‫ﺍﻟﻒ‬ −1 ( X P << RP ) Pm = 3.33 + 100 × 9 × cos( 0.18 ) cos( 84.26 − 0.18 ) = 96.16W ‫8.6% = 001 * 09 −61.69 = ﺩﺭ ﺻﺪ ﺧﻄﺎ‬ 90 25
  • 26.
    (‫ﺏ‬ cos( ϕ −β ) = 0 ⇒ ϕ − β = ±90° ⇒ ϕ − 0.18 = −90° ⇒ ϕ = −89.82 (‫ﺝ‬ LP = 0.03H = r 2C : ‫ﺑﺮﺍﻱ ﻣﺜﺎﻝ‬ C = 30nf , r = 1kΩ , R1 = 2kΩ 26
  • 27.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫72‬
  • 28.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ﺍﮔﺮ ﺳﻴﺴﺘﻢ ﺳﻪ ﻓﺎﺯ ﻣﺘﻌﺎﺩﻝ 4 ﺳﻴﻤﻪ ﺑﺎﺷﺪ ﻣﻲ ﺗﻮﺍﻥ ﺗﻮﺍﻥ ﻳﻚ ﻓﺎﺯ ﺭﺍ ﺑﺎ‬ ‫ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﻭﺍﺗﻤﺘﺮ ﺑﺸﮑﻞ ﺯﻳﺮ ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ ﻭ ﺩﺭ 3 ﺿﺮﺏ ﻛﺮﺩ . ﻭﻟﻲ‬ ‫ﺍﻏﻠﺐ ﺑﺎﺭ ﻧﺎﻣﺘﻌﺎﺩﻝ ﺍﺳﺖ ﻭ ﺍﺯ ﺍﻳﻦ ﺭﻭﺵ ﻧﻤﻲ ﺗﻮﺍﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ ﻭ ﺑﺎﻳﺪ‬ ‫3 ﻭﺍﺗﻤﺘﺮ ﻳﺎ ﻻﺍﻗﻞ ﺩﻭ ﻭﺍﺗﻤﺘﺮ)ﺑﺮﺍﻱ ﺳﻴﺴﺘﻢ ﺳﻪ ﺳﻴﻤﻪ( ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ .‬ ‫‪R‬‬ ‫‪S‬‬ ‫‪T‬‬ ‫‪N‬‬ ‫82‬
  • 29.
    ‫ﻗﻀﻴﻪ ﺑﻠﻮﻧﺪﻝ: ﺑﺮﺍﻱﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺳﻴﺴﺘﻢ ‪ N‬ﺳﻴﻤﻪ 1-‪ N‬ﻭﺍﺗﻤﺘﺮ‬ ‫ﻛﺎﻓﻲ ﺍﺳﺖ. )ﻳﻚ ﺳﻴﻢ ﺭﺍ ﻣﺒﻨﺎ ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻴﻢ( .‬ ‫92‬
  • 30.
  • 31.
    ‫ﺳﺮﻫﺎﻱ ﭘﺘﺎﻧﺴﻴﻞ ﻭﺍﺗﻤﺘﺮﻫﺎﺭﺍ ﺑﻪ ﻧﻘﻄﻪ ﻣﺸﺘﺮﻙ )‪ (c‬ﻭﺻﻞ ﻣﻲ ﻛﻨﻴﻢ. ﮐﻪ‬ ‫ﻣﺜﻼ ﻣﻲ ﺗﻮﺍﻧﺪ ﻫﻤﺎﻥ ‪ O‬ﺑﺎﺷﺪ.‬ ‫′‬ ‫1‪v1 = v + v‬‬ ‫′‬ ‫2‪v2 = v + v‬‬ ‫′‬ ‫3‪v3 = v + v‬‬ ‫13‬ ‫‪1T‬‬ ‫′‬ ‫‪P = ∫ v1i1dt‬‬ ‫1‬ ‫0‪T‬‬ ‫‪1T‬‬ ‫′‬ ‫‪P2 = ∫ v2i2 dt‬‬ ‫0‪T‬‬ ‫‪1T‬‬ ‫′‬ ‫‪P3 = ∫ v3i3dt‬‬ ‫0‪T‬‬
  • 32.
    ⎞⎞ ⎛ 1 T⎛ P +P2 + P3 = ∫ ⎜ v1i1 + v2i2 + v3i3 + v⎜ i1 + i2 + i3 ⎟ ⎟ dt 1 4 3 ⎜ 1 24 ⎟ ⎟ T 0⎜ ⎝ ⎠⎠ 0 ⎝ 1T = ∫ (v1i1 + v2i2 + v3i3 )dt = P = ‫ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺑﺎﺭ‬ T0 32
  • 33.
    ‫ﺭﻭﺵ 2 ﻭﺍﺗﻤﺘﺮﻱ‬ ‫ﺍﮔﺮﻧﻘﻄﻪ ﻣﺸﺘﺮﮎ ﺳﻴﻢ ﭘﻴﭽﻬﺎﻱ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺭﻭﻱ ﻳﮑﻲ ﺍﺯ ﺧﻄﻮﻁ ﺑﮕﻴﺮﻳﻢ ﻳﮏ‬ ‫ﻭﺍﺗﻤﺘﺮ ﮐﻤﺘﺮ ﻻﺯﻡ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.‬ ‫1‬ ‫+‬ ‫‪V‬‬ ‫1‬ ‫_‬ ‫_‬ ‫33‬ ‫‪V‬‬ ‫3‬ ‫+‬ ‫+‬ ‫‪V‬‬ ‫2‬ ‫_‬ ‫3‬ ‫2‬
  • 34.
    1T P = ∫i1( v1 − v3 )dt 1 T0 1T P2 = ∫ i2 ( v2 − v3 )dt T0 1T P + P2 = ∫ (i1( v1 − v3 ) + i2 ( v2 − v3 ))dt 1 T0 1T = ∫ (i1v1 + i2v2 + i3v3 ))dt = P T0 i3 = −( i1 + i2 ) : ‫ﺯﻳﺮﺍ‬ 34
  • 35.
    ‫ﺩﺭ ﺍﺗﺼﺎﻝ ﻣﺜﻠﺚﻫﻢ :‬ ‫1‬ ‫+‬ ‫_‬ ‫1‪V‬‬ ‫‪V‬‬ ‫3‬ ‫_‬ ‫+‬ ‫1‪i‬‬ ‫3‪i‬‬ ‫3‬ ‫_‬ ‫2‪i‬‬ ‫+‬ ‫‪V‬‬ ‫2‬ ‫2‬ ‫53‬
  • 36.
    1T P = ∫− v3 ( i1 − i3 )dt 1 T0 1T P2 = ∫ v2 ( i2 − i1 )dt T0 1T( P + P2 = ∫ v3i3 + v2i2 − i1 (v2 + v3 ))dt 1 T0 1T = ∫ (i1v1 + i2 v2 + i3v3 ))dt = P T0 v1 + v2 + v3 = 0 36
  • 37.
    ‫ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﺩﺭﺣﺎﻟﺖ ﮐﻠﻲ ﺑﺮﻗﺮﺍﺭﻧﺪ ﺍﺯ ﺟﻤﻠﻪ ﺩﺭ ﺣﺎﻟﺖ ﺩﺍﺋﻤﻲ ﺳﻴﻨﻮﺳﻲ .‬ ‫ﭘﺲ ﺑﺮﺍﻱ ﺳﻴﻨﻮﺳﻲ ﺑﺎ ﺑﺎﺭ ﻧﺎﻣﺘﻌﺎﺩﻝ ﻧﻴﺰ ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﺑﺮﻗﺮﺍﺭ ﺑﻮﺩﻩ ﻭ ‪P + P2 = P‬‬ ‫1‬ ‫ﻣﺜﻼ ﺩﺭ ﺣﺎﻟﺖ ﺳﺘﺎﺭﻩ ﺩﺍﺭﻳﻢ :‬ ‫3‪P = V1I1 cos ϕ1 + V2 I 2 cos ϕ2 + V3 I 3 cos ϕ‬‬ ‫32‪,V‬‬ ‫ﮐﻪ ‪ Vi‬ﻭ ‪ I i‬ﻫﺎ ﻣﻘﺎﺩﻳﺮ ‪ rms‬ﻫﺴﺘﻨﺪ .‬ ‫73‬ ‫2‬ ‫‪= I1V13 cos ϕ I ,V + I 2V23 cos ϕ I‬‬ ‫31‬ ‫1‬
  • 38.
    ‫ﺣﺎﻝ ﺍﮔﺮ ﺩﺭﺣﺎﻟﺖ ﺩﺍﺋﻤﻲ ﺳﻴﻨﻮﺳﻲ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ ﺑﺎﺷﺪ ، ﺍﮔﺮ ﻭﻟﺘﺎﮊ ‪rms‬‬ ‫ﻓﺎﺯﻫﺎ ﺭﺍ ﺑﺎ 1‪ V‬ﻭ 2‪ V‬ﻭ 3‪ V‬ﻧﺸﺎﻥ ﺩﻫﻴﻢ ﻭ ﺟﺮﻳﺎﻥ ‪ rms‬ﺁﻧﻬﺎ ﺭﺍ ﺑﺎ 1‪ I‬ﻭ 2 ‪ I‬ﻭ‬ ‫3 ‪ I‬ﻧﺸﺎﻥ ﺩﻫﻴﻢ ، ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ :‬ ‫1‬ ‫3‬ ‫83‬ ‫2‬
  • 39.
  • 40.
    V = V 1 =V 2 = V 3 ‫ﺑﺎ ﻓﺮﺽ ﺑﺎﺭ ﻣﺘﻌﺎﺩﻝ‬ I = I = I = I 1 V 12 2 = V 3 = V 13 23 = 3V P = V I cos( 30 − ϕ ) = 1 13 P = V 2 23 1 I 2 c os( 30 3VI cos( 30 − ϕ ) +ϕ) = 3VI cos( 30 + ϕ ) P + P = 3VI [cos( 30 + ϕ ) + cos( 30 − ϕ ) ] = 3 VI cos( ϕ ) = P P − P = 3VI sin( ϕ ) = 1 1 ⇒ 2 2 Q 3 P1 − P2 tg ( ϕ ) 3VI sin( ϕ ) = = P1 + P2 3 VI cos( ϕ ) 3 40
  • 41.
    ‫ﻳﻌﻨﻲ ﺍﮔﺮ ﺑﺎﺭﻣﺘﻌﺎﺩﻝ ﺑﺎﺷﺪ ﺩﺭﺣﺎﻟﺖ ﺩﺍﺋﻤﯽ ﺳﻴﻨﻮﺳﯽ ﺑﺎ ﺭﻭﺵ ﺩﻭ ﻭﺍﺗﻤﺘﺮﻱ‬ ‫ﻣﻲ ﺗﻮﺍﻥ ﻋﻼﻭﻩ ﺑﺮ ﺗﻮﺍﻥ ﺍﮐﺘﻴﻮ ، ﺗﻮﺍﻥ ﺭﺍﮐﺘﻴﻮ ﻳﺎ ‪ ϕ‬ﺭﺍ ﻧﻴﺰ ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ.‬ ‫3‬ ‫‪ϕ = 0 , cos ϕ = 1 ⇒ P = 3VI , P = P2 = VI‬‬ ‫1‬ ‫2‬ ‫3‬ ‫3‬ ‫0 = 2‪ϕ = 60 , cos ϕ = 0.5 ⇒ P = VI , P = VI , P‬‬ ‫1‬ ‫2‬ ‫2‬ ‫0 < 2‪ϕ > 60 , cos ϕ < 0.5 ⇒ P‬‬ ‫14‬ ‫3‬ ‫3‬ ‫= ‪ϕ = 90 , cos ϕ = 0 ⇒ P‬‬ ‫0 = ‪VI , P2 = − VI ⇒ P‬‬ ‫1‬ ‫2‬ ‫2‬ ‫3‬ ‫3‬ ‫‪ϕ = −60 ⇒ P = VI , P = 0 , P2 = VI‬‬ ‫1‬ ‫2‬ ‫2‬ ‫0 < ‪ϕ < −60 ⇒ P‬‬ ‫1‬
  • 42.
    ‫ﺍﮔﺮ ﻭﺍﺗﻤﺘﺮﻱ ﻣﻨﻔﻲﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ، ﺑﺎﻳﺪ ﺟﻬﺖ ﻳﻜﻲ ﺍﺯ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﺭﺍ ﻋﻮﺽ‬ ‫ﻛﺮﺩ ﺗﺎ ﻗﺮﺍﺋﺖ ﻣﺜﺒﺖ ﺷﻮﺩ ﻭﻟﻲ ﺑﺎﻳﺪ ﻋﻼﻣﺖ ﻣﻨﻔﻲ ﺭﺍ ﺧﻮﺩﻣﺎﻥ ﺩﺭﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ.‬ ‫ﻣﺜﺎﻝ:‬ ‫ﺗﻮﺍﻥ ﻳﻚ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ 3 ﺳﻴﻤﻪ ﻣﺘﻌﺎﺩﻝ ﺑﻪ ﺭﻭﺵ ﺩﻭ ﻭﺍﺗﻤﺘﺮﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ‬ ‫ﺷﺪﻩ ﺍﺳﺖ. ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ 1 ، ‪ 7500w‬ﻭ ﻗﺮﺍﺋﺖ ﻭﺍﺗﻤﺘﺮ 2، ‪ -1500w‬ﺍﺳﺖ.‬ ‫ﺍﻟﻒ( ﺿﺮﻳﺐ ﺗﻮﺍﻥ ﺑﺎﺭ؟‬ ‫ﺏ( ﺍﮔﺮﻭﻟﺘﺎﮊ ﺧﻂ ‪ 400v‬ﺑﺎﺷﺪ ﭼﻪ ﻇﺮﻓﻴﺘﻲ)ﺧﺎﺯﻧﻲ( ﺩﺭﻫﺮ ﻓﺎﺯ ﺑﺎﻳﺪ ﺍﺿﺎﻓﻪ ﻛﺮﺩ ﺗﺎ‬ ‫ﻛﻞ ﺗﻮﺍﻥ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺷﺪﻩ ﺭﻭﻱ ﻭﺍﺗﻤﺘﺮ 1 ﻇﺎﻫﺮ ﺷﻮﺩ ؟‬ ‫24‬
  • 43.
    ‫ﺍﻟﻒ(‬ ‫‪P = 7500w, P2 = −1500 w‬‬ ‫1‬ ‫‪P = P + P2 = 6000 w‬‬ ‫1‬ ‫2‪P1 − P‬‬ ‫3‬ ‫953.0 = ) ‪= 68.9° ⇒ cos( ϕ‬‬ ‫2‪P1 + P‬‬ ‫1−‬ ‫‪ϕ = tg‬‬ ‫ﺏ(‬ ‫ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻨﻈﻮﺭ ﺑﺎﻳﺪ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ :‬ ‫34‬ ‫°06 = ‪ϕ‬‬
  • 44.
    ‫ ﻭﻟﺘﺎﮊ ﻓﺎﺯ‬V= 400 = 231v 3 ‫ ﻫﺮ ﻓﺎﺯ‬P = 6000 = 2000 w 3 2000 ⇒I= = 24.11A 231×0.359 231v = 9.58Ω Z= 24.1A R = Z cos( ϕ ) = 3.44Ω : ‫ﻗﺒﻞ ﺍﺯ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ﺧﺎﺯﻥ ﺩﺍﺭﻳﻢ‬ X = Z sin( ϕ ) = 8.94Ω tg ( ϕ′ ) = 3 = 1.73 X = Rtg ( ϕ′ ) = 3.44 × 1.73 = 5.96 ⇒ Capacitor' s _ Re ac tan ce = 8.94 − 5.96 = 2.98Ω 1 = 1068µF C= 2π( 50 )( 2.98 ) 44
  • 45.
    ‫ﻭﺍﺗﻤﺘﺮ 3 ﻓﺎﺯ:‬ ‫ﻳﻚﻭﺍﺗﻤﺘﺮ ﺍﺳﺖ ﻛﻪ ﺩﺍﺭﺍﻱ ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﻭﻟﺘﺎﮊ ﻭ ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﺟﺮﻳﺎﻥ‬ ‫ﺍﺳﺖ ﻭ ﻛﺎﺭ 2 ﻭﺍﺗﻤﺘﺮ 1‪ P‬ﻭ 2‪ P‬ﺭﺍ ﻳﻜﺠﺎ ﺍﻧﺠﺎﻡ ﻣﻲ ﺩﻫﺪ ﻭ ﺣﺎﺻﻞ ﺟﻤﻊ‬ ‫ﺭﺍ ﺑﻪ ﻣﺎ ﻣﻲ ﺩﻫﺪ.‬ ‫54‬
  • 46.
    ‫ﻭﺍﺭﻣﺘﺮ‬ ‫ﺍﻳﻦ ﻭﺳﻴﻠﻪ ﺑﺮﺍﻱﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺭﺍﻛﺘﻴﻮ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ. ﺩﺭ ﻭﺍﺗﻤﺘﺮ‬ ‫ﺗﮑﻔﺎﺯ ﺍﮔﺮ ﺑﺠﺎﻱ ﺳﺮﻱ ﻛﺮﺩﻥ ﻣﻘﺎﻭﻣﺖ ، ﺳﻠﻒ ﺑﺰﺭﮔﻲ ﺑﺎ ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﭘﺘﺎﻧﺴﻴﻞ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮ ﺳﺮﻱ ﻛﻨﻴﻢ ﺟﺮﻳﺎﻥ ﮔﺬﺭﻧﺪﻩ ﺍﺯ ﺳﻴﻢ ﭘﻴﭻ ﺑﺎ‬ ‫ﻭﻟﺘﺎﮊ ﺁﻥ ﺗﻘﺮﻳﺒﺎ 09 ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ .‬ ‫ﹰ‬ ‫‪V‬‬ ‫‪I‬‬ ‫‪ϕ‬‬ ‫‪Ip‬‬ ‫64‬ ‫) ‪Qm = VI cos( 90 − ϕ ) = VI sin( ϕ‬‬
  • 47.
    ‫ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱﺗﻮﺍﻥ ﺭﺍﻛﺘﻴﻮ ﺩﺭ ﻣﺪﺍﺭ ﺳﻪ ﻓﺎﺯ ﻣﺘﻌﺎﺩﻝ:‬ ‫) ‪Pm = V31 I 2Cos( 90 − ϕ ) = 3VISin( ϕ‬‬ ‫1‬ ‫2‬ ‫3‬ ‫74‬
  • 48.
    ‫ﺗﺮﺍﻧﺴﻔﻮﺭﻣﺮﻫﺎﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ:‬ ‫ﺩﺭ ﺍﺩﻭﺍﺗﻲ ﺍﺯ ﻗﺒﻴﻞ ﺍﻟﮑﺘﺮﻭﺩﻳﻨﺎﻣﻮ ﻣﺘﺮ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻘﺎﻭﻣﺖ ﺷﻨﺖ ﻳﺎ ﺿﺮﺏ‬ ‫ﮐﻨﻨﺪﻩ ﺑﻪ ﺩﻟﻴﻞ ﻭﺟﻮﺩ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ، ﻧﺴﺒﺖ ﺛﺎﺑﺘﻲ ﺭﺍ ﺑﺎ‬ ‫ﺗﻐﻴﻴﺮ ﻓﺮﮐﺎﻧﺲ ﻧﻤﻲ ﺩﻫﺪ . ﺭﺍﻩ ﺩﻳﮕﺮ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﺮﺍﻧﺲ ﺍﺳﺖ ﮐﻪ ﻋﻼﻭﻩ‬ ‫ﺑﺮ ﻧﺴﺒﺖ ﺛﺎﺑﺖ ، ﺍﻳﺰﻭﻻﺳﻴﻮﻥ ﻫﻢ ﺍﻳﺠﺎﺩ ﻣﻲ ﮐﻨﺪ ﻭﮔﺮﻧﻪ ﺁﻭﺭﺩﻥ ﻳﮏ‬ ‫ﻭﻟﺘﺎﮊ ﺧﻴﻠﻲ ﺯﻳﺎﺩ ﺑﻪ ﺩﺳﺘﮕﺎﻩ ﻫﻢ ﺧﻄﺮﻧﺎﮎ ﺍﺳﺖ ﻭ ﻫﻢ ﺩﺭ ﺍﺑﻌﺎﺩ ﮐﻮﭼﮏ‬ ‫ﺩﺳﺘﮕﺎﻩ ﻋﺎﻳﻘﻬﺎ ﺗﺤﻤﻞ ﻧﻤﻲ ﮐﻨﻨﺪ.‬ ‫84‬
  • 49.
  • 50.
  • 51.
    ‫ﺗﺮﺍﻧﺲ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻧﺴﻲﺍﺳﺖ ﮐﻪ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺛﺎﻧﻮﻳﻪ ﺑﻴﺸﺘﺮ ﺍﺯ ﺍﻭﻟﻴﻪ ﺍﺳﺖ ﻭ ﻃﻮﺭﻱ‬ ‫ﻃﺮﺍﺣﻲ ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﺣﺎﻟﺖ ﺍﺗﺼﺎﻝ ﮐﻮﺗﺎﻩ ﺛﺎﻧﻮﻳﻪ )ﺁﻣﭙﺮﻣﺘﺮ ﻳﺎ ﺭﻟﻪ ﻳﺎ ...(‬ ‫ﺑﺘﻮﺍﻧﺪ ﮐﺎﺭ ﮐﻨﺪ )ﺟﺮﻳﺎﻥ ﺯﻳﺎﺩﻱ ﺭﺩ ﻣﻲ ﺷﻮﺩ(. ﻫﺮﮔﺰ ﻧﺒﺎﻳﺪ ﺛﺎﻧﻮﻳﻪ ﺗﺮﺍﻧﺲ‬ ‫ﺟﺮﻳﺎﻥ ﺭﺍ ﺯﻳﺮ ﺑﺎﺭ ﻣﺪﺍﺭ ﺑﺎﺯ ﮐﺮﺩ ﻭﺍﻻ ‪ emf‬ﺑﺰﺭﮔﻲ ﮐﻪ ﺍﻳﺠﺎﺩ ﻣﻲ ﺷﻮﺩ ﻣﻲ‬ ‫ﺗﻮﺍﻧﺪ ﻋﺎﻳﻖ ﺭﺍ ﺑﻪ ﺷﮑﺴﺖ ﺑﺒﺮﺩ. ﺍﺻﻮﻻ ﺍﮔﺮ ﻣﻘﺎﻭﻣﺖ ﺑﺎﺭ ﺗﺮﺍﻧﺲ ﺟﺮﻳﺎﻥ ، ﺑﺰﺭﮒ‬ ‫ﺑﺎﺷﺪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺟﺮﻳﺎﻧﻲ ﮐﻪ ﺍﺯ ﺛﺎﻧﻮﻳﻪ ﻣﻲ ﮔﺬﺭﺩ )‪ 1/n‬ﺟﺮﻳﺎﻥ ﺍﻭﻟﻴﻪ( 2 ‪RI‬‬ ‫ﺯﻳﺎﺩ ﺷﺪﻩ ﻭ ﻭﻟﺘﺎﮊ ﺯﻳﺎﺩ ﺣﺎﺻﻠﻪ ﺗﺮﺍﻧﺲ ﺭﺍ ﻣﻲ ﺳﻮﺯﺍﻧﺪ . ﻣﺜﻼ ﺍﮔﺮ ﻧﺴﺒﺖ 1 ﺑﻪ‬ ‫001 ﺑﺎﺷﺪ ﻭ ﺍﻭﻟﻴﻪ ‪ 5A‬ﺑﺎﺷﺪ ، ﺟﺮﻳﺎﻥ ﺛﺎﻧﻮﻳﻪ ‪ 0.05A‬ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﮐﻪ ﺑﺮﺍﻱ‬ ‫‪ , R=100K‬ﺗﻮﺍﻥ 052 ﻭﺍﺕ ﻭ ﻭﻟﺘﺎﮊ ‪ 5KV‬ﻣﻲ ﺷﻮﺩ.‬ ‫15‬
  • 52.
    ‫ﺗﺮﺍﻧﺲ ﻭﻟﺘﺎﮊ ﺗﺮﺍﻧﺴﻲﺍﺳﺖ ﺑﺎ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺍﻭﻟﻴﻪ ﺑﻴﺸﺘﺮ ﺍﺯ ﺛﺎﻧﻮﻳﻪ ﮐﻪ ﻃﻮﺭﻱ ﻃﺮﺍﺣﻲ‬ ‫ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﺣﺎﻟﺖ ﻣﺪﺍﺭﺑﺎﺯ ﺑﻮﺩﻥ ﺛﺎﻧﻮﻳﻪ ﺑﺘﻮﺍﻧﺪ ﮐﺎﺭ ﮐﻨﺪ)ﺟﺮﻳﺎﻥ ﭼﻨﺪﺍﻥ ﺭﺩ‬ ‫ﻧﻤﻲ ﺷﻮﺩ ﻭﻟﻲ ﺍﺧﺘﻼﻑ ﭘﺘﺎﻧﺴﻴﻞ ﺯﻳﺎﺩ ﺍﺳﺖ(.ﺿﻤﻨﺄ ﻭﺟﻮﺩ ﺗﺮﺍﻧﺴﻬﺎﯼ ﻓﻮﻕ ﺳﺒﺐ‬ ‫ﮐﺎﻫﺶ ﺍﺛﺮ ﺑﺎﺭﮔﺬﺍﺭﻱ ﺷﺪﻩ ﺍﺳﺖ.‬ ‫ﺑﻌﻀﺎ ﺩﺭ ﺗﺮﺍﻧﺲ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﺍﻳﻨﮑﻪ ﻣﺠﺒﻮﺭ ﺑﻪ ﻗﻄﻊ ﺳﻴﻤﻲ ﮐﻪ ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺟﺮﻳﺎﻥ‬ ‫ﺁﻧﺮﺍ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﻳﻢ ﻧﺸﻮﻳﻢ ﺍﺯ ﺧﻮﺩ ﺁﻥ ﺳﻴﻢ ﺑﻌﻨﻮﺍﻥ ﺍﻭﻟﻴﻪ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﻨﻨﺪ ﺍﻣﺎ‬ ‫ﺗﺮﺍﻧﺲ ، ‪ dc‬ﺭﺍ ﻋﺒﻮﺭ ﻧﻤﻲ ﺩﻫﺪ ﻭ ﺣﺘﻲ ﻭﺟﻮﺩ ‪ dc‬ﺑﺎﻋﺚ ﺍﺷﺒﺎﻉ ﻣﻐﻨﺎﻃﻴﺴﻲ ﻣﻲ‬ ‫ﺷﻮﺩ ﻟﺬﺍ ﺩﺭ ﭘﺮﻭﺑﻬﺎﻱ ﺟﺮﻳﺎﻥ ﻭ ﻫﺮﺟﺎ ﺍﺣﺘﻴﺎﺝ ﺑﻪ ﺗﺒﺪﻳﻞ ﺟﺮﻳﺎﻥ ﺑﺎ ﭘﺎﺳﺦ‬ ‫ﻓﺮﮐﺎﻧﺴﻲ ﺷﺎﻣﻞ ‪ dc‬ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ ﺍﺯ ﺍﺛﺮ ﻫﺎﻝ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ. ﺍﻭﻟﻴﻦ -‪clip‬‬ ‫‪ on milliammeter‬ﺍﺛﺮ ﻫﺎﻝ ﺭﺍ ‪ HP‬ﺩﺭ ﺳﺎﻝ 8591 ﺑﻪ ﺑﺎﺯﺍﺭ ﺩﺍﺩ .‬ ‫25‬
  • 53.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫35‬
  • 54.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ)ﮐﻨﺘﻮﺭ(‬ ‫‪t‬‬ ‫′‪W ( t ) = ∫ P( t ′ )dt‬‬ ‫ﺍﻧﺮﮊﻱ، ﺍﻧﺘﮕﺮﺍﻝ ﺗﻮﺍﻥ ﺑﺮﺣﺴﺐ ﺯﻣﺎﻥ ﺍﺳﺖ.‬ ‫0‬ ‫ﻣﺒﻨﺎﻱ ﻣﺤﺎﺳﺒﻪ ﻗﻴﻤﺖ ، ﻣﻴﺰﺍﻥ ﻣﺼﺮﻑ ﺍﻧﺮﮊﻱ ﺑﺮ ﺣﺴﺐ ‪ kwh‬ﺍﺳﺖ. ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ‬ ‫ﻭﺳﻴﻠﻪ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ ﺍﻟﮑﺘﺮﻳﮑﻲ ، ﻛﻴﻠﻮﻭﺍﺕ ﺳﺎﻋﺖ ﻣﺘﺮ ﺍﻧﺪﻭﻛﺴﻴﻮﻧﻲ )ﮐﻨﺘﻮﺭ‬ ‫ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ( ﺍﺳﺖ .‬ ‫ﮐﻴﻠﻮﻭﺍﺕ ﺳﺎﻋﺖ ﻣﺘﺮ ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ :‬ ‫ﺩﺭﺍﻳﻦ ﺩﺳﺘﮕﺎﻩ ﻳﻚ ﺻﻔﺤﺔ ﭼﺮﺧﻨﺪﻩ ﺩﺍﺭﻳﻢ ﻛﻪ ﺳﺮﻋﺖ ﭼﺮﺧﺶ ﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ‬ ‫ﺑﺎ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺯﺩﻩ ﺷﺪﻩ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺍﻧﺮﮊﻱ ﻣﺼﺮﻓﻲ ﺧﻮﺍﻫﺪ‬ ‫ﺑﻮﺩ . ﺩﺭﺍﻳﻨﺠﺎ ﺑﺮﺧﻼﻑ ﻭﺍﺗﻤﺘﺮ ﻛﻪ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺩﺭﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ θ‬ﺑﻮﺩ،‬ ‫ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ dθ/dt‬ﺍﺳﺖ.‬ ‫45‬
  • 55.
    ‫ﺍﺻﻮﻝ ﻛﺎﺭ ﻣﻮﺗﻮﺭﺍﻧﺪﻭﻛﺴﻴﻮﻧﻲ :‬ ‫ﺍﮔﺮ ﺣﻠﻘﻪ ﺍﻱ ﺣﺎﻭﻱ ﺟﺮﻳﺎﻥ ‪ i‬ﺩﺍﺧﻞ ﻣﻴﺪﺍﻥ ﻣﻐﻨﺎﻃﻴﺴﻲ ‪ B‬ﻗﺮﺍﺭ ﺑﮕﻴﺮﺩ، ﮔﺸﺘﺎﻭﺭﻱ‬ ‫ﺑﺮﺍﺑﺮ ﺑﺎ ‪ BAi‬ﺑﺮﺁﻥ ﻭﺍﺭﺩ ﻣﻲ ﺷﻮﺩ. ﺑﻌﺒﺎﺭﺕ ﺩﻳﮕﺮ ﮔﺸﺘﺎﻭﺭ ﻭﺍﺭﺩﻩ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ ﺑﺎ ‪ϕi‬‬ ‫ﻭ ﺍﮔﺮ ‪ ϕ‬ﻭ ‪ i‬ﻫﺮ ﺩﻭ ‪ ac‬ﺑﺎﺷﻨﺪ ﻭ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ:‬ ‫‪ϕ = Φ sin ωt‬‬ ‫) ‪i = I sin(ωt − α‬‬ ‫‪1T‬‬ ‫) ‪ ∝ ∫ ϕ(t )i(t )dt = 1 ΦI cos( α‬ﮔﺸﺘﺎﻭﺭ ﻣﺘﻮﺳﻂ‬ ‫0‪T‬‬ ‫2‬ ‫55‬ ‫⇒‬
  • 56.
    ‫* ﺩﺭ ﻛﻨﺘﻮﺭﺍﺯ ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﻭ ﻳﻚ ﺻﻔﺤﺔ ﺩﻭﺍﺭ ﺁﻟﻮﻣﻴﻨﻴﻮﻣﻲ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ‬ ‫ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ)ﻗﺮﺍﺭﺩﺍﺩﻫﺎﻱ ﺟﻬﺖ ﻣﺜﺒﺖ ‪ ϕ‬ﻭ ‪:( i‬‬ ‫1‪i‬‬ ‫2‪i‬‬ ‫2‪ϕ‬‬ ‫65‬ ‫1‪ϕ‬‬
  • 57.
    ‫ﭼﻮﻥ 1‪ ϕ‬ﻭ2‪ ϕ‬ﻣﺘﻐﻴﺮ ﺑﺎ ﺯﻣﺎﻥ ﻫﺴﺘﻨﺪ، ﺍﻳﺠﺎﺩ ‪ emf‬ﺍﻟﻘﺎﻳﻲ ﻭ ﻧﺘﻴﺠﺘﺎ ﺟﺮﻳﺎﻥ ﻓﻮﻛﻮ‬ ‫ﹰ‬ ‫ﺩﺭ ﺻﻔﺤﺔ ﺁﻟﻮﻣﻴﻨﻴﻮﻣﻲ ﻣﻲ ﻛﻨﻨﺪ . ﺑﺎ ﻓﺮﺽ ﮐﺎﻣﻼ ﻣﻘﺎﻭﻣﺘﻲ ﺑﻮﺩﻥ ﻣﺴﻴﺮ ﺟﺮﻳﺎﻥ ﻭ‬ ‫ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﻣﺴﻴﺮ :‬ ‫1‪dϕ‬‬ ‫1‪e‬‬ ‫ﮐﻪ 1‪ R‬ﻣﻘﺎﻭﻣﺖ ﻣﺴﻴﺮ ﺍﺳﺖ.‬ ‫75‬ ‫1‪R‬‬ ‫= ‪if‬‬ ‫1‬ ‫‪dt‬‬ ‫− = 1‪e‬‬
  • 58.
    ‫) ‪ϕ1 =Φ1 sin( ωt‬‬ ‫ﺍﮔﺮ:‬ ‫1 ‪ωΦ‬‬ ‫1 ‪ωΦ‬‬ ‫− = 1‪⇒ i f‬‬ ‫= ‪cos ωt‬‬ ‫) 09 − ‪sin(ωt − 90 ) = I f1 sin( ωt‬‬ ‫1‪R‬‬ ‫1‪R‬‬ ‫~‬ ‫1‪Φ‬‬ ‫~‬ ‫‪If‬‬ ‫ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ~ ، 09 ﺩﺭﺟﻪ ﭘﺲ ﻓﺎﺯ ﺍﺯ ~‬ ‫‪If‬‬ ‫2‪ ϕ‬ﺍﺳﺖ. ﻣﻼﺣﻈﻪ ﻣﻲ ﮐﻨﻴﺪ ﮐﻪ ﺑﺎ‬ ‫‪~ I‬‬ ‫ﺗﻮﺟﻪ ﺑﻪ ﺯﺍﻭﻳﻪ 09 ﺩﺭﺟﻪ ﺍﻱ ﺑﻴﻦ ‪ ~f‬ﻭ 2‪ ϕ‬ﮔﺸﺘﺎﻭﺭ ﺣﺎﺻﻞ ﺍﺯ ﺁﻧﻬﺎ ﺻﻔﺮ ﺍﺳﺖ )ﺩﺭ‬ ‫1‬ ‫2‬ ‫2‬ ‫ﻭﺍﻗﻊ ﻣﺴﻴﺮﻫﺎﻱ ﻓﻮﮐﻮ ﮐﺎﻣﻼ ﻣﻘﺎﻭﻣﺘﻲ ﻧﻴﺴﺘﻨﺪ ﻭ ﺍﻳﻨﻬﺎ ﻗﺪﺭﻱ ﺍﻳﺠﺎﺩ ﺗﺮﻣﺰ ﻣﻲ ﮐﻨﻨﺪ(.‬ ‫~‬ ‫~ ~‬ ‫~‬ ‫ﺍﻣﺎ ﺍﮔﺮ 1‪ ϕ‬ﻭ 2‪ ϕ‬ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﺍﺛﺮ ﻣﺘﻘﺎﺑﻞ ‪ ~f‬ﻭ 1‪ ϕ‬ﻭ ﻧﻴﺰ ‪ ~f‬ﻭ 2‪ϕ‬‬ ‫‪I‬‬ ‫‪I‬‬ ‫ﺍﻳﺠﺎﺩ ﮔﺸﺘﺎﻭﺭ ﻣﺤﺮﮎ ﻣﻲ ﮐﻨﺪ .‬ ‫2‬ ‫85‬ ‫1‬
  • 59.
    ϕ1 ϕ2 ϕ1 > 0,i1 < 0 ,ϕ2 < 0 ,i2 < 0 ⇒ F ‫ ﻫﻢ ﺟﻬﺖ‬T12 ,T21 ϕ1 ϕ2 i2 i1 F 59
  • 60.
    ~ ~ :‫ ﺩﺭﺟﻪﺍﺧﺘﻼﻑ ﻓﺎﺯ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ، ﺩﺍﺭﻳﻢ‬β ‫ ﺑﺎﻧﺪﺍﺯﻩ‬ϕ2 ‫ ﻭ‬ϕ1 ‫ﺍﮔﺮ‬ β β ~ If2 ~ I f1 ~ ϕ1 ~ ϕ2 T12 ∝ Φ1I f cos( 90 + β ) ∝ Φ1Φ 2 cos( 90 + β ) 2 T21 ∝ Φ 2 I f cos( 90 − β ) ∝ Φ1Φ 2 cos( 90 − β ) 1 60
  • 61.
    ‫ﺑﺎ ﻧﻮﺷﺘﻦ ﺭﻭﺍﺑﻂﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ :‬ ‫) ‪T ∝ T21 − T12 ∝ Φ1Φ 2 sin( β‬‬ ‫* ﺍﮔﺮ 0 = ‪ β‬ﺑﺎﺷﺪ 0=‪ T‬ﺍﺳﺖ. ﺑﻴﺸﺘﺮﻳﻦ ﮔﺸﺘﺎﻭﺭ ﻭﻗﺘﻲ ﺍﺳﺖ ﻛﻪ °09 = ‪. β‬‬ ‫16‬
  • 62.
    ‫ﺍﺻﻮﻝ ﮐﺎﺭ ﮐﻨﺘﻮﺭ:‬ ‫ﺩﺭ ﮐﻨﺘﻮﺭ ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ ﻳﮏ ﺳﻴﻢ ﭘﻴﭻ 1‪ ϕ‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺟﺮﻳﺎﻥ ﻭ 2‪ ϕ‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ‬ ‫ﻭﻟﺘﺎﮊ ﻭﻟﻲ ﺑﺎ ﺍﺧﺘﻼﻑ ﻓﺎﺯ 09 ﺩﺭﺟﻪ ﺍﺳﺖ )ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﺍﻧﺪﻭﮐﺘﺎﻧﺲ ﺯﻳﺎﺩﻱ‬ ‫ﺩﺍﺭﺩ(. ﻣﻲ ﺧﻮﺍﻫﻴﻢ ‪ T‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺑﺎﺷﺪ ﭘﺲ ﻛﺎﻓﻴﺴﺖ 1‪ Φ‬ﻣﺘﻨﺎﺳﺐ‬ ‫ﺑﺎ ﺟﺮﻳﺎﻥ ﻭ 2 ‪ Φ‬ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﻭﻟﺘﺎﮊ ﺑﺎﺭ ﺑﺎﺷﺪ ﻭﻟﻲ ﺑﺎ 09 ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﻧﺴﺒﺖ ﺑﻪ‬ ‫ﻭﻟﺘﺎﮊ ﺑﺎﺭ ﺍﺳﺖ. ﻟﺬﺍ :‬ ‫‪T ∝ IV sin( 90 − ϕ ) = IV cos( ϕ ) = P‬‬ ‫~‬ ‫‪V‬‬ ‫‪ϕ‬‬ ‫~‬ ‫1‪ϕ‬‬ ‫26‬ ‫~‬ ‫‪I‬‬ ‫‪β‬‬ ‫~‬ ‫2‪ϕ‬‬
  • 63.
    ‫ﺩﺭ ﻭﺍﻗﻊ ﺩﺭﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ ﺍﺧﺘﻼﻑ ﻓﺎﺯ ﻭﻟﺘﺎﮊ ﻭ ﺟﺮﻳﺎﻥ ﻗﺪﺭﻱ ﮐﻤﺘﺮ ﺍﺯ‬ ‫09 ﺩﺭﺟﻪ ﻣﻲ ﺑﺎﺷﺪ ﮐﻪ ﺑﺮﺍﻱ ﺟﺒﺮﺍﻥ ﺁﻥ ﺗﻤﻬﻴﺪﺍﺗﻲ ﺷﺪﻩ ﺍﺳﺖ.)ﺳﻴﻢ‬ ‫ﭘﻴﭻ ﺳﺎﻳﻪ ﺍﻧﺪﺍﺯ ﺩﺭ ﮐﻨﺎﺭ ﺳﻴﻢ ﭘﻴﭻ ﭘﺘﺎﻧﺴﻴﻞ(.‬ ‫ﺑﺮﺍﻱ ﺍﻳﺠﺎﺩ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺍﺯ ﺁﻫﻨﺮﺑﺎﻱ ﺩﺍﺋﻤﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ .‬ ‫36‬
  • 64.
    ‫ﺷﺒﻴﻪ ﺭﺍﺑﻄﻪ ﺍﻱﮐﻪ ﺑﺮﺍﻱ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺣﺎﺻﻞ ﺍﺯ ﮔﺮﺩﺵ ﺳﻴﻢ ﭘﻴﭻ‬ ‫ﻣﺘﺤﺮﮎ ﮔﺎﻟﻮﺍﻧﻮﻣﺘﺮ ﺩﺭ ﻣﻴﺪﺍﻥ ﻣﻐﻨﺎﻃﻴﺴﻲ ﺩﺍﺷﺘﻴﻢ) ‪، ( ( NBA )2 dθ‬‬ ‫‪R‬‬ ‫‪dt‬‬ ‫ﺩﺭ ﺍﻳﻨﺠﺎ ﻧﻴﺰ :‬ ‫ﺳﺮﻋﺖ ﭼﺮﺧﺶ‬ ‫‪dθ‬‬ ‫= ‪ ∝ n‬ﮔﺸﺘﺎﻭﺭﻣﻘﺎﻭﻡ‬ ‫‪dt‬‬ ‫ﻟﺬﺍ :‬ ‫‪ ⇒ n ∝ P‬ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ=ﮔﺸﺘﺎﻭﺭ ﻣﺤﺮﮎ‬ ‫46‬
  • 65.
    ‫ﭼﻮﻥ ﺳﺮﻋﺖ ﭼﺮﺧﺶﺻﻔﺤﻪ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺍﺳﺖ ، ﺗﻌﺪﺍﺩ ﺩﻭﺭ ﺯﺩﻩ‬ ‫ﺷﺪﻩ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺍﻧﺮﮊﻱ ﻣﺼﺮﻓﻲ ﺧﻮﺍﻫﺪ ﺑﻮﺩ .‬ ‫ﺍﮔﺮ ﺍﺯ ﻓﻨﺮ ﺑﺠﺎﻱ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﮐﺮﺩﻳﻢ ، ﻭﺍﺗﻤﺘﺮ ﺍﻧﺪﻭﮐﺴﻴﻮﻧﻲ‬ ‫ﺩﺍﺷﺘﻴﻢ.‬ ‫ﺑﺎ ﺗﻐﻴﻴﺮ ﻣﺤﻞ ﺁﻫﻨﺮﺑﺎﻱ ﺩﺍﺋﻤﻲ ﻣﻲ ﺗﻮﺍﻥ ﻣﻴﺰﺍﻥ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺭﺍ ﺗﻐﻴﻴﺮ ﺩﺍﺩ.‬ ‫ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ‪ R‬ﻓﺎﺻﻠﻪ ﺁﻫﻨﺮﺑﺎ ﺍﺯ ﻣﺮﮐﺰ ﺍﺳﺖ ﻭ ﺍﮔﺮ ﻓﺎﺻﻠﻪ ﺍﺯ‬ ‫ﻣﺮﻛﺰ ﺯﻳﺎﺩ ﮔﺮﺩﺩ ، ‪T‬ﻣﻘﺎﻭﻡ ﺑﻴﺸﺘﺮ ﺷﺪﻩ ، ﻭ ﺻﻔﺤﻪ ﻛﻨﺪﺗﺮ ﻣﻲ ﭼﺮﺧﺪ . ﺑﻪ ﺍﻳﻦ‬ ‫ﺗﺮﺗﻴﺐ ﻣﻲ ﺗﻮﺍﻥ ﮐﻨﺘﻮﺭ ﺭﺍ ﮐﺎﻟﻴﺒﺮﻩ ﮐﺮﺩ )ﮐﺎﻟﻴﺒﺮﺍﺳﻴﻮﻥ ﺩﺭ ﺑﺎﺭ ﮐﺎﻣﻞ(‬ ‫)ﻫﻤﭽﻨﻴﻦ ﮐﺎﻟﻴﺒﺮﺍﺳﻴﻮﻥ ﺩﺭ %01 ﺑﺎﺭ ﮐﺎﻣﻞ ﺗﻮﺳﻂ ﺗﻨﻈﻴﻢ ﺳﻴﻢ ﭘﻴﭻ ﺳﺎﻳﻪ‬ ‫ﺍﻧﺪﺍﺯ ﺻﻮﺭﺕ ﻣﻲ ﮔﻴﺮﺩ( .‬ ‫56‬
  • 66.
    ‫ﺟﻨﺲ ﺁﻫﻨﺮﺑﺎ ﺑﺎﻳﺪﺑﺴﻴﺎﺭ ﺧﻮﺏ ﺑﺎﺷﺪ ﺗﺎ ﺩﺭ ﺍﺛﺮ ﮔﺬﺷﺖ ﺯﻣﺎﻥ ﺿﻌﻴﻒ ﻧﺸﻮﺩ.‬ ‫ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﺩﺭﺟﻪ ﺣﺮﺍﺭﺕ ﺧﺎﺻﻴﺖ ﺁﻫﻨﺮﺑﺎﻳﻲ ﮐﻢ ﻭ ﻣﻘﺎﻭﻣﺖ ﻣﺴﻴﺮ‬ ‫ﺍﻓﺰﺍﻳﺶ ﻣﻲ ﻳﺎﺑﺪ ، ﻳﻌﻨﻲ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﮐﻢ ﻣﻴﺸﻮﺩ .‬ ‫* ﺩﺭﮐﻨﺘﻮﺭ ﺳﻪ ﻓﺎﺯ ﻫﺮ ﻓﺎﺯ ﺳﻴﻢ ﭘﻴﭽﻬﺎ ﻭ ﺻﻔﺤﻪ ﺁﻟﻮﻣﻴﻨﻴﻮﻣﻲ ﻣﺨﺼﻮﺹ ﺑﻪ‬ ‫ﺧﻮﺩ ﺭﺍ ﺩﺍﺭﺩ ﻭﻟﻲ ﻫﻤﺔ ﮔﺸﺘﺎﻭﺭﻫﺎﻱ ﺍﻳﻦ ﺳﻪ، ﺑﻪ ﻳﻚ ﺷﻔﺖ ﻣﺘﺼﻞ ﻣﻲ‬ ‫ﺷﻮﻧﺪ ﮐﻪ ﺩﺭ ﻧﻬﺎﻳﺖ ﺳﺮﻋﺖ ﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﮐﻞ ﺗﻮﺍﻥ ﻣﺼﺮﻓﻲ ﺳﻪ ﻓﺎﺯ‬ ‫ﺍﺳﺖ .‬ ‫66‬
  • 67.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥﻭ ﺍﻧﺮﮊﻱ‬ ‫ﻣﻄﺎﻟﺐ :‬ ‫‐ ﻣﻘﺪﻣﻪ‬ ‫ـ ﻭﺍﺗﻤﺘﺮ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺗﻮﺍﻥ ﺩﺭ ﺳﻴﺴﺘﻢ 3 ﻓﺎﺯ‬ ‫ـ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺍﻧﺮﮊﻱ )ﮐﻨﺘﻮﺭ(‬ ‫ـ‪ CosФ‬ﻣﺘﺮ‬ ‫76‬
  • 68.
    ‫ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺿﺮﻳﺐﺗﻮﺍﻥ‬ ‫ﺩﺳﺘﮕﺎﻩ ‪ Cosφ‬ﻣﺘﺮ، ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮﻱ ﺍﺳﺖ ﻛﻪ ﺑﺠﺎﻱ ﻳﻚ ﺳﻴﻢ ﭘﻴﺞ‬ ‫ﻣﺘﺤﺮﻙ، ﺩﻭ ﺳﻴﻢ ﭘﻴﭻ ﻣﺘﺤﺮﻙ ﻣﺘﻘﺎﻃﻊ )ﻋﻤﻮﺩ ﺑﺮﻫﻢ( ﺩﺍﺭﺩ. ﻣﻲ ﺩﺍﻧﻴﻢ‬ ‫ﺩﺭ ﺍﻟﻜﺘﺮﻭ ﺩﻳﻨﺎﻣﻮﻣﺘﺮ:‬ ‫‪dM‬‬ ‫2 ‪I1 I‬‬ ‫‪dθ‬‬ ‫86‬ ‫=‪T‬‬
  • 69.
    ‫ﺍﮔﺮ ﻣﻴﺪﺍﻥ ﺑﺼﻮﺭﺕﺑﺎﻻ ﺑﺎﺷﺪ:‬ ‫‪dM‬‬ ‫→ ‪M ∝ cos θ‬‬ ‫‪∝ sin θ‬‬ ‫‪dθ‬‬ ‫ﺟﻬﺖ ﺳﻴﻢ ﭘﻴﭽﻬﺎﻱ ‪ A‬ﻭ ‪ B‬ﻃﻮﺭﻱ ﺑﺴﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﮔﺸﺘﺎﻭﺭﻫﺎﻱ ﺁﻧﻬﺎ‬ ‫ﺑﺮﺧﻼﻑ ﻫﻢ ﺑﺎﺷﺪ . ﺩﺭ ﻭﺍﻗﻊ ﮔﺸﺘﺎﻭﺭ ﻣﻘﺎﻭﻡ ﺑﻪ ﺍﻳﻦ ﺻﻮﺭﺕ ﺍﻳﺠﺎﺩ ﻣﻴﺸﻮﺩ ﻭ‬ ‫ﺩﺳﺘﮕﺎﻩ ﻓﻨﺮﻱ ﻧﺪﺍﺭﺩ ﺑﺨﺎﻃﺮ ﻫﻤﻴﻦ ﻫﻢ ﻫﺴﺖ ﮐﻪ ﭘﺲ ﺍﺯ ﻗﺮﺍﺋﺖ ، ﻋﻘﺮﺑﻪ ﺳﺮ‬ ‫ﺟﺎﻳﺶ ﻣﻴﻤﺎﻧﺪ .‬ ‫96‬
  • 70.
    TA = KVIcos( ϕ ) sin( θ ) TB = KVI cos( 90 − ϕ ) sin( 90 + θ )= KVI sin( ϕ ) cos( θ ) :‫ﺩﺭﺣﺎﻟﺖ ﺗﻌﺎﺩﻝ‬ TA = TB ⇒ cos θ sin ϕ = cos ϕ sin θ ⇒ tgθ = tgϕ ⇒ θ = ϕ 70
  • 71.
    ‫ﺻﻔﺤﺔ ﺩﺳﺘﮕﺎﻩ ﺭﺍﺑﺮﺣﺴﺐ ‪ Cosφ‬ﻣﺪﺭﺝ ﻣﻲ ﻛﻨﻨﺪ . ﺑﺮﺍﻱ ‪ 50Hz‬ﮐﺎﻟﻴﺒﺮﻩ‬ ‫ﺷﺪﻩ ﺍﺳﺖ ﻭ ﺩﺭ ﻓﺮﮐﺎﻧﺴﻬﺎﻱ ﺩﻳﮕﺮ ﺧﻄﺎ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ.‬ ‫ﺩﺭ ‪ Cosφ‬ﻣﺘﺮ ﺳﻪ ﻓﺎﺯ، ﻓﺎﺯ 1 ﺭﺍ ﺍﺯ ﺳﻴﻢ ﭘﻴﭻ ﺛﺎﺑﺖ ﻋﺒﻮﺭ ﻣﻴﺪﻫﻨﺪ ﻭﺩﻭ ﺳﻴﻢ‬ ‫ﭘﻴﭻ ﺩﻳﮕﺮ ﺭﺍ ﺑﺎ ﻣﻘﺎﻭﻣﺖ ﺑﻪ 21‪ V‬ﻭ 31‪ V‬ﻭﺻﻞ ﻣﻲ ﮐﻨﻨﺪ. ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ‬ ‫ﻣﻴﺘﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﮐﻪ ﺑﺎ ﻓﺮﺽ ﻣﺘﻌﺎﺩﻝ ﺑﻮﺩﻥ ﻓﺎﺯﻫﺎ ‪ϕ = θ‬ﺧﻮﺍﻫﺪ ﺑﻮﺩ .‬ ‫ﻣﺮﺍﺟﻌﻪ ﮐﻨﻴﺪ ﺑﻪ ﺳﺎﻭﻧﻲ ﺻﻔﺤﻪ 226.‬ ‫17‬