75-068614-06P Amphenol, 75-068614-06P Datasheet - Page 285

• Manufacturer Part Number: 75-068614-06P
• Description: Circular MIL / Spec Connectors 6P 97 Series MIL-DTL-5015
• Manufacturer: Amphenol
• Series: 97 Seriesr

Specifications of 75-068614-06P

• Mil Type: MIL-DTL-5015
• Product Type: Connectors
• Contact Style: Pin (Male)
• Shell Style: Plug
• Shell Size: 14S
• Number Of Contacts: 6
• Insert Arrangement: 14S-6
• Mounting Style: Panel
• Termination Style: Crimp
• Contact Type: Pin
• Shell Plating: Chromate over Cadmium, Olive Drab
• Contact Plating: Silver
• Voltage Rating: 250 V
• Body Material: Aluminum
• Mounting Angle: Straight
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

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Contact Amphenol Aerospace for more information at 800-678-0141 • www.amphenol-aerospace.com
The following formula and example are offered in order to
determine the expected filter performance in an impedance
system other than 50 ohms.
With the attenuation expressed in 50 ohms and the transfer
impedance curve shown in Figure 1 below, a designer can relate
the expressed attenuation to the input and output impedance of
his circuit.
Example:
Formula (Taken from Figure 1):
Amphenol
(1)
(2)
(3)
1.4 x 10 ohm = transfer impedance
for 65 dB in a 50 ohm system
140
120
100
80
60
40
20
10
0
Noise is 40dB above specification level at 100 MHz
Input and output impedance are 10 and 100 ohms
respectively
Amphenol
attenuation at 100 MHz and +25°C
-4
Aerospace
®
VHF 7000 pf filter has a 65 dB minimum
10
Attenuation vs Transfer Impedance in 50 Ohm System
-3
Impedance Matching Formula
(Your System to a 50 Ohm System)
Transfer Impedance - Z
10
-2
Figure 1
Atten (dB) = 20 log
Atten = filter performance in a system other than 50 ohms
Atten (dB) = 20 log
Attenuation = 56.3dB
In this case, the 7000 pf VHF filter will give 56.3 dB which is
16.3dB below the desired reduction in noise (40dB) as stated
in the above problem.
Z
Z
Z
S
L
12
= source impedance
= load impedance
10
= transfer impedance
12
-1
Ohms
10
10
1 +
1 +
10
1.4 x 10
Z
-0
12
(Z
Z
S
S
Z
10(100)
+ Z
L
–2
(10 + 100)
L
)
10
1