產(chǎn)品詳情
型號
:甘南2098-DSD-020-DN羅克韋爾伺服驅(qū)動器多少錢
Allen Bradley 1398-DDM-009-DN
Allen Bradley 1398-DDM-005X-NV
Allen Bradley 1398-DDM-005X-DN
Allen Bradley 1398-DDM-009X
Allen Bradley 1398-DDM-019-DN
Allen Bradley 1398-DDM-009X-DN
Allen Bradley 1398-DDM-005X
Allen Bradley 1398-DDM-005-DN
Allen Bradley 1398-DDM-005
Allen Bradley 1398-DDM-075X
Allen Bradley 1398-DDM-075
Allen Bradley 1398-DDM-010X
Allen Bradley 1398-DDM-010
Allen Bradley 1398-DDM-150
Allen Bradley 1398-DDM-020X
Allen Bradley 1398-DDM-020
Allen Bradley 1398-DDM-030X
Allen Bradley 1398-DDM-030
Allen Bradley 1398-DDM-019X-DN
Allen Bradley 1398-DDM-019X
Allen Bradley 1398-DDM-150X
Allen Bradley 1398-DDM-019
Allen Bradley 1398-DDM-009
:甘南2098-DSD-020-DN羅克韋爾伺服驅(qū)動器多少錢上式為磁鐵激磁的步進電機產(chǎn)生的電磁轉(zhuǎn)矩,因此有下面的公式:E0=NdΦ/dtθ=ωtω=Nrωm式中,Φ為交鏈磁通,θ為轉(zhuǎn)子轉(zhuǎn)動角,ω為電氣角速度,N為相線圈匝數(shù)。E0=NdΦ/dt由法拉第定律得來。θ=ωt為機械角與電氣角的關系式,把上式代入到T=E0I/ωm可得:T=E0I/ωm=N(dΦ/dt)I/ωm=N(dΦ/dθ)(dθ/dt)I/ωm=N(ω/ωm)(dΦ/dθ)I(dΦ/dθ)=NNrI(dΦ/dθ)步進電機的轉(zhuǎn)矩由永磁體產(chǎn)生的交鏈磁通變化率與流過線圈電流之積產(chǎn)生為感應電動勢,圖表示如下:將此E0代入T=E0I/ωm,單相轉(zhuǎn)矩變?yōu)橄率剑篢1=2NIBLr依據(jù)圖,磁鐵激磁的步進電機轉(zhuǎn)矩公式為(T1=2NIBLr),當Nr=1時,轉(zhuǎn)矩公式與直流電機的轉(zhuǎn)矩公式(T=2NIBLr)相同,直流電機的氣隙磁通B,相當于步進電機的交鏈磁通的有效當量部分總和。