wideband RF impedance matching

First, that input impedance is for the signal, not the LO input. The LO input is designed for a fairly low power 50 ohm source.

If the input impedance is complex then its also frequency dependent. The value given sounds right for the high end of the frequency range but is quoted at 50MHz -- which contradicts the Smith chart in figure 9 of the data sheet.

I'm not sure what frequency range you want on the input, but you should probably just consider a simple balun and live with the inductive reactance.

--

Tim Wescott
Wescott Design Services
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Tim,

Yes, the impedance I gave was for the signal input. I'm trying to create a 0-500MHz frequency generator and I've got a 250-500MHz VCO, so I was going to mix the VCO output with a 250MHz LO signal from a MAX2620 fixed frequency oscillator.

Thanks to everyone for their replys, they've been really helpful!

Steve

I read in sci.electronics.design that steve wrote (in ) about 'wideband RF impedance matching', on Sun, 11 Sep 2005:

[snip]

Wind some turns at right-angles to the others.(;-)

Is it really true that the j term in your AD8343 input impedance is constant over your bandwidth?

If, instead, the input looks like R (= 2.7 ohms) and L in series, you can connect a CR series combination across the input. If the added resistor is 2.7 ohms, and LC = 2.7^2, the input will magically be converted to 2.7 + j0 at all frequencies. Now you can match your 50 ohm source to that with a transformer or equivalent, with turns ratio sqrt(50/2.7) = 4.3.

The added CR is called a Zobel network. More complicated networks can be used to convert more complicated impedances to pussy-cat resistive.

--
Regards, John Woodgate, OOO - Own Opinions Only.
If everything has been designed, a god designed evolution by natural selection.
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Hi,

The impedance you quote is for the signal input at 50MHz only and not for the local oscillator which is presumably non-critical since it is squared-off internally. You should only need to transform your VCO's output to a simple differential signal just as in the data sheet. If the VCO is choosey impedance-wise, then maybe a simple resistive isolating pad between it and the balun would help.

Cheers - Joe

steve wrote:

The signal inputs go straight to the emitters of the switching transistors, by the look of it. The impedance they give would be for small-signal operation e.g. in a receiver. Do you mind if there are harmonics in the input current? You could drive it hard, if you want lots of output power. The impedance would be different then but it doesn't matter, all that matters is the input current will come out the top of the mixing cell chopped up by the LO waveform. If you just use a centre-tapped mini-circuits transformer or something like that, with a resistor from the centre tap to ground to sink 20mA or so out of the centre tap, that ought to do. You wouldn't need much swing at the mixer inputs to fully switch the current from one side to the other, 100-200mV ought to do it. You will however need plenty of current so a transformer which steps down the voltage and steps up the current should help. The only problem is that if you drive it hard like this, then you will get other mixing products such as 3LO +/- 3RF, which would be hard to get rid of. If you don't want these then you could put resistors in between the transformer and the input pins so that the current is sinusoidal again. The 3LO, 5LO etc. frequencies will still be present in the LO however, because the limiting action of the LO buffer will create them. For that reason, with your VCO, I would try using a 500MHz fixed oscillator, so that 3LO+/- fVCO will be so high that you can filter it out effectively.

You might want to reconsider the oscillator frequencies, I don't know if it matters for you if there is some VCO leakage in the output. If you used a

750-1250MHz VCO and a 750MHz fixed oscillator, then it would be easier to get rid of any LO and VCO leakage.

Chris

Hello Steve,

Just a word of caution: If you do this and your output needs to be, say,

249.9MHz you will also have 250.1MHz. Unless you employ an I/Q scheme.

Regards, Joerg

Another hint : With the IQ, where the LO and the IF are required as 0 and 90 degrees, you'll get together with your 249.9 also 250.0 plus 250.1, but attenuated, say -30dB. They'll be there ! While the fixed frequency is doable with some effort as 0 and 90 degrees, the sweepable is rather hard to get as 0 and 90 degrees broadband.

Rene

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Hello Rene,

You can get better than -30dB but yes, they'll be there.

Unless the VCO itself runs at 4F. Which would make it more expensive.

I would consider DDS. The AD9858 may be able to serve a range of

0-400MHz but there should be higher clock speed chips somewhere. Of course, it'll cost.

Regards, Joerg