Correction:
In my haste (yikes!), I erred and grabbed the first 2 values for X+jY...
...so in hindsight, from the load side a series inductor or transmission
line and then shunt capacitor works easily.
But... unless you need to be better than 1.25:1, why not just use just 1
component - a shunt capacitor?
apologies,
Dave N3AC
=========================================
On Sat, Feb 2, 2019, 16:37 Dave Haring <dcharing@gmail.com wrote:
> Any Smith Chart program will help you visualize it. From load side, shunt
> 360 pF then series 1100 pF.
>
> 73, Dave N3AC
>
> On Sat, Feb 2, 2019, 10:30 Richard Solomon <dickw1ksz@gmail.com wrote:
>
>> I did a quick Google on this,
>> but didn't see exactly what I
>> needed.
>>
>> I want to match a load with
>> the following characteristics:
>> Rs=25.67
>> Rp=39.26
>> Xs=18.68
>> Xp=53.96
>> To a 50 ohm source at a fixed
>> frequency of 3.725 Mhz.
>>
>> Can someone point me to a
>> calculator that will give L & C
>> values I will need ?
>>
>> Tnx, Dick, W1KSZ
>> _______________________________________________
>> Amps mailing list
>> Amps@contesting.com
>> http://lists.contesting.com/mailman/listinfo/amps
>>
>
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|