Workshop (workshop@pcm.co.za) wrote: > Hi there > This is to all those who know anything about impedance matching! > I have a circuit using a ne602 ic.The input impedance is is quoted > as bieng 1.5k.I have a Phillips application note (AN1993) > that uses this ic.The input to the mixer of the ic is from a 50 ohm source. > The 50 ohm source has been matched to the ic by the following circuit. > To pin 1 of > | mixer > | > |----------- > | | > | | > === 47p | > === () > | ()0.28uH > To 50 ohm ____| () To pin 2 of mixer > source === () | > === 220p | | > | | | > | | | > ------------ | > |_______________| > | > ===100nF > === > | > GND > Can anyone out there tell me how these values are calculated.The frequency of interest > is 45Mhz I believe the mixer has a differential input. The 100nF cap just shorts pin 2 of the mixer to ground so that the input is single-ended. You can take that cap out of the picture. There are formulas for calculating the values of the rest components, based on the assumption that the bandwidth of the matching network B < fo/10, where fo = center freq of the circuit. |------------------- Rt (>R2) | | | | === C1 | === () L | () R2 ____| () === () === C2 | | | | | ------------ | --- - 1. Calculate: Qt = fo/B C = 1/(2*pi*B*Rt) L = 1/((2*pi*fo)^2*C) N^2 = Rt/R2 2. Calculate: Qp = sqrt(Qt^2/N^2 - 1) 3. C1 and C2 are given by: C2 = Qp/(2*pi*fo*R2) Cse = C2*(Qp^2+1)/Qp^2 C1 = Cse*C/(Cse - C) Regards -T L
Date: 17 Mar 1996 21:00:25 GMT
Original Subject: Re: Impedance matching