AC Analysis / Input Impedance

Thread Starter

Dalaran

Joined Dec 3, 2009
168
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!
 

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#12

Joined Nov 30, 2010
18,224
If you pretend you are inside the chip and you are dc bias current, looking out at 75 ohms to ground and 75 ohms to input places the 2 resistors in parallel.

The parallel combination of those 2 resistors is 37.5 ohms.
37.5 is very similar to 37.4 ohms

This accomplishes having the chip see equal DC impedances on its inputs.

I don't know it this does you any good.

edit: with those capactiors, it doesn't do any good for DC. Maybe it helps with the AC considerations.
 
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Thread Starter

Dalaran

Joined Dec 3, 2009
168
Thanks, I guess that is where I am getting confused... why does 75R to ground and 75R to input = parallel combination of 75R//75R? Does input = ground in this case then?
 

Ron H

Joined Apr 14, 2005
7,063
The problem with this circuit is that we don't know what the load is on the right side of the schematic. I would be willing to bet that the 75Ω||6.2nH is either compensation for the load (to extend the bandwidth), or part of a low pass filter. If it is the latter, the load would constitute the remainder of the filter.
 
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