wideband RF impedance matching

[steve]

Hi,

I'm trying to impedance match a 250-450MHz VCO (designed for a 50 Ohm
load) to an AD8343 mixer which has a 2.7+6.8j input impedance.

I'm fairly new to impedance matching in the wild (have only got limited
pure-resistance based impedance matching from university), so I'm
getting a bit stuck here.

I've read a bit about LC matching networks but it seems that they're
only for narrowband applications, unsuitable for my setup. I've also
come across using transformers for wideband matching applications, but
I'm unsure how to apply this. From what I've read, the impedance
relationship of a transformers primary and secondary coils is:

Zs/Zl = (Ns/Nl)^2

So for my application I need a transform with a turns ratio which
satisfies the following:

50/(2.7+6.8j) = 2.52-j6.35 = (Ns/Nl)^2

But this gives me a complex transformer ratio, which doesn't seem
possible.

I'm a little stuck at the moment, can anyone explain to me where I
should go from here?

Steve

 

[John Woodgate]

I read in sci.electronics.design that steve <[email protected]>

I'm trying to impedance match a 250-450MHz VCO (designed for a 50 Ohm
load) to an AD8343 mixer which has a 2.7+6.8j input impedance.
[snip]

So for my application I need a transform with a turns ratio which
satisfies the following:

50/(2.7+6.8j) = 2.52-j6.35 = (Ns/Nl)^2

But this gives me a complex transformer ratio, which doesn't seem
possible.

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

I'm a little stuck at the moment, can anyone explain to me where I
should go from here?

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.

 

[Tim Wescott]

Hi,

I'm trying to impedance match a 250-450MHz VCO (designed for a 50 Ohm
load) to an AD8343 mixer which has a 2.7+6.8j input impedance.

I'm fairly new to impedance matching in the wild (have only got limited
pure-resistance based impedance matching from university), so I'm
getting a bit stuck here.

I've read a bit about LC matching networks but it seems that they're
only for narrowband applications, unsuitable for my setup. I've also
come across using transformers for wideband matching applications, but
I'm unsure how to apply this. From what I've read, the impedance
relationship of a transformers primary and secondary coils is:

Zs/Zl = (Ns/Nl)^2

So for my application I need a transform with a turns ratio which
satisfies the following:

50/(2.7+6.8j) = 2.52-j6.35 = (Ns/Nl)^2

But this gives me a complex transformer ratio, which doesn't seem
possible.

I'm a little stuck at the moment, can anyone explain to me where I
should go from here?

Steve

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.

 

[Joe McElvenney]

Hi,

I'm trying to impedance match a 250-450MHz VCO (designed for a 50 Ohm
load) to an AD8343 mixer which has a 2.7+6.8j input impedance.

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]

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

 

[Chris Jones]

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

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

 

[Joerg]

Hello Steve,

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.

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

 

[Rene Tschaggelar]

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.

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

 

[Joerg]

Hello Rene,

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 !

You can get better than -30dB but yes, 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.

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