CXS Select the graph that corresponds to the imaginary stroke l. l k Imaginary stroke l = (Stroke) + k + l k: Distance between the center and end of the plate 6 (Example) 1. When using CXSJM6-10 and l = 15 mm: Imaginary stroke l = 10 + 2.75 + 15 = 27.75 Therefore, the graph used for your model selection should be the one for CXSJM6-30 (6) ). 2.75 mm 10 4 mm 15 5 mm 20 6 mm 25 2.
L k Imaginary stroke L' = (Stroke) + k + L k: Distance between the center and end of the plate 6 (Example) q When using CXSJM6-10 and L = 15 mm: Imaginary stroke L' = 10 + 2.75 + 15 = 27.75 Therefore, the graph used for your model selection should be the one for CXSJM6-30 6 ). 2.75 mm 10 4 mm 15 5 mm 20 6 mm 25 w When using CXSJL25-50 and L = 10 mm: Imaginary stroke L' = 50 + 6 + 15 = 71
L k Imaginary stroke L' = (Stroke) + k + L k: Distance between the center and end of the plate 6 (Example) q When using CXSJM6-10 and L = 15 mm: Imaginary stroke L' = 10 + 2.75 + 15 = 27.75 Therefore, the graph used for your model selection should be the one for CXSJM6-30 6 ). 2.75 mm 10 4 mm 15 5 mm 20 6 mm 25 w When using CXSJL25-50 and L = 10 mm: Imaginary stroke L' = 50 + 6 + 15 = 71
)-B/AR25(K)-B t e q IN OUT t Modular F.R.L. e q r IN OUT w r w AC-A AC-B AR30(K)-B/AR40(K)-B q t AR50(K)-B/AR60(K)-B e IN OUT q t r w e IN OUT r w Component Parts No.
V V K* K* K* -XA46 -XA51 Change of long shaft length (Without keyway) V V V V T, J, K* K* T, K* J* K* MSQ MSZ CRQ2X MSQX -XA52 Change of short shaft length (Without keyway) V V K* K* K* -XA53 Change of double shaft length (Both without keyway) V V V K* -XA54 Change of long shaft length (With four chamfers) V V V X, Z* X, Z* -XA55 Change of short shaft length (With four chamfers)
K V 1 10 XLA S P1P2 XL XL Conductances combined When each of the separate conductances are given as C1, C2 and Cn, the composite conductance C is expressed as: C=1/(1/C1+1/C2++1/Cn) when in series, and C=C1+C2++Cn, when in parallel.
K V 1 10 XLA S P1P2 XL XL Conductances combined When each of the separate conductances are given as C1, C2 and Cn, the composite conductance C is expressed as: C=1/(1/C1+1/C2++1/Cn) when in series, and C=C1+C2++Cn, when in parallel.
(If not specifying dimension C2, indicate instead.) Applicable shaft type: K Equal dimensions are indicated by the same marker. Minimum machining diameter for d1 is 0.1 mm. Applicable shaft types: K, T Minimum machining diameter for d1 is 0.1 mm.
(If not specifying dimension C2, indicate instead.) Applicable shaft type: K Equal dimensions are indicated by the same marker. Minimum machining diameter for d1 is 0.1 mm. Applicable shaft types: K, T Minimum machining diameter for d1 is 0.1 mm.
For further details, refer to mounting positions under Replacement heaters/Part Nos. on page 46. 24 High Vacuum Angle Valve Series XLD, XLDV Dimensions (mm) 1in=25.4mm XLDV-25/With solenoid valve 11 C Fn (KF flange) Fb (K flange) C Port R (exhaust port) K K 2.5 41.5 Initial exhaust solenoid valve Main exhaust solenoid valve G A B 8 10.5 Port P (pressure port) A Max 10.5 (except connector
For further details, refer to mounting positions under Replacement heaters/Part Nos. on page 46. 24 High Vacuum Angle Valve Series XLD, XLDV Dimensions (mm) 1in=25.4mm XLDV-25/With solenoid valve 11 C Fn (KF flange) Fb (K flange) C Port R (exhaust port) K K 2.5 41.5 Initial exhaust solenoid valve Main exhaust solenoid valve G A B 8 10.5 Port P (pressure port) A Max 10.5 (except connector
1 k Load OUT1 (Black) OUT2 (White) 12 to 24 VDC 12 to 24 VDC Load Load DC () (Blue) DC () (Blue) ZSE40(F) ISE40--30(L)-(M) With auto shift input ZSE40(F) ISE40--70(L)-(M) With auto shift input DC (+) (Brown) Auto shift input (Gray) OUT1 (Black) OUT2 (White) DC (+) (Brown) Auto shift input (Gray) 1.2 k 1.2 k Load 6.8 k 6.8 k Main circuit Main circuit Load OUT1 (Black) OUT2 (White) 12 to 24
Conductances combined When each of the separate conductances are given as C1, C2 and Cn, the composite conductance C is expressed as: C=1/(1/C1+1/C2+...1/Cn) when in series, and C=C1+C2+...Cn, when in parallel.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.
. = 16 V Load voltage at OFF = Leakage current x 2 pcs. x Load impedance = 1 mA x 2 pcs. x 3 k = 6 V Example: Power supply is 24 VDC Internal voltage drop in switch is 4 V. Example: Load impedance is 3 k.