I have attached the schematic of the circuit in question. The 1uF capacitors each lead to one leg of a differential input. I am expecting to see 75ohms differential across the inputs but can not figure out how to derive this.
I understand that for AC the 1uF caps are considered shorts and the inductor an open circuit for simplicity. It looks to me like the differential impedance seen across the inputs is just 75R + 37R4 = 102R4, since the far left resistor is open?
Should I be calculating by 75R//75R + 37R4 = ~75R? This is what I am expecting by why would the far left resistor be included in the calculation?
edit: the circle is a BNC connector with centre pin connected to 75R/6n2 and case connected to ground.
Thanks!
I understand that for AC the 1uF caps are considered shorts and the inductor an open circuit for simplicity. It looks to me like the differential impedance seen across the inputs is just 75R + 37R4 = 102R4, since the far left resistor is open?
Should I be calculating by 75R//75R + 37R4 = ~75R? This is what I am expecting by why would the far left resistor be included in the calculation?
edit: the circle is a BNC connector with centre pin connected to 75R/6n2 and case connected to ground.
Thanks!
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