Workshop wrote:
> 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
As long as the loaded Q is reasonable (say at least 10) then the circuit
behaves as a transformer at its resonance. The 100nF to ground is merely
an AC short to ground so that the DC at pin 1 isn't shorted to ground
through the inductor. It does not play a role in the transformer action.
Again, the following is only true at or near resonace. First, the
transformer voltage ratio is approximately
n = ( 47 / (47 + 220) ) ==> 0.176
Thus the impedance ratio is n**2 ==> 0.031.
So, the 1.5K at pin 1 will be transformed down to 1500*0.031 = 46ohms.
That is close enough to 50 ohms for all practical purposes.
Recall that I said earlier that the circuit must be resonant. The
inductor is chosen to resonate the series capacitor network. 47pF in
series with 220pF is 38.7pF. The resonant frequency of 38.7pF and 280nH
is about 49MHz -- close enough again for practical purposes. There is
also usually a few pF of stray capacitance an pin 1 which will lower the
resonance some. Normally it would be included in the calculation.
As a check, the loaded Q is calculated to be about 17 so the initial
assumption of Q>10 holds and the approximations are valid.
That's all there is to it!

**Date:** Wed, 13 Mar 1996 21:03:54 -0600

**Original Subject:** Re: Impedance matching

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