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kelly

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passive mixer input impedance
Feb 28^th, 2011, 1:23pm

Hi All,

I think this is an old topic, but I still have some questions and would like to underatnd it more.

For a doubler balance passive mixer, if the output is high imppedance (the diff. load capacitance is less than 1p for example), you can set the input impedance by two 25 ohms pull up resistors to a common mode voltage (say you need to set the common mode voltage anyway). This way, the diff Z11 is 50ohms since the load is close to open circuit, right?

Now, if you have a load capacitor that's bigger, say 4 or 5p differrentially, now the Z11 is smaller because you have 50ohms in parallel with the impedance at the load. In order to get the 50ohms back, you will need to increase the 25ohms pull up resistors at the input. Am I thinking this correctly?

Thanks.

Kelly

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aaron_do

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Re: passive mixer input impedance
Reply #1 - Feb 28^th, 2011, 5:22pm

Hi Kelly,

If you are using cross coupled outputs (like in a Gilbert cell), then the input impedance is clearly not open. It depends on your biasing and switch sizes. Basically each switch will have an average conductance which will (for the most part) determine your input impedance.

If you are not using cross-coupled outputs, then you have something like a switched capacitor. So again you will not have an open circuit.

That said, 25 ohms may be small compared to your switch size and biasing conditions such that your passive mixer may look like an open anyway. But in that case, your 25 ohm resistors are likely to be loading down your source.

I believe the effect of the load capacitors will be to give you a bandpass response for your input impedance. i.e. at the center frequency you will still see approximately the same input resistance, but away from the center frequency it will get smaller. The reason is because your input signal gets downconverted and low-pass filtered by the average series resistance and load capacitance. If you assume the mixer is made of ideal switches with a fixed series resistor, then it is easy to visualize how that translates to a bandpass response for the input impedance. There was a paper by Cook. et. Al in JSSC I think dec. 2006 about a 400 mW receiver at 2.4 GHz which discussed this topic briefly. Sorry I can't remember all the details.

cheers,
Aaron

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there is no energy in matter other than that received from the environment - Nikola Tesla

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kelly

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Re: passive mixer input impedance
Reply #2 - Mar 1^st, 2011, 1:30pm

Hi Aaron,

Thanks for your post, you ansered one of my questions regarding the bandpass nature of the input impedance.

I guess I have two things I want to ask for your opnions.

1. Yes, you are right, the mixer output is cross coupled. So if you looking in from the input, you see the average conductance + load capacitance? If you do, for a small capacitance such that the average conductance + load capacitance is say around 200ohms, you can put a shunt 50 ohms at the input so now the input impedance is dominated by 50 ohms. As the load capacitances increases, you will need to increase the 50ohms to get closer to 50 ohms up to the point where the average conductance + load capacitance is real small so the shunt resistor doesn't do much. Am I think this right?

2. Back to the input impedance. So I get a bandpass responce as you said. Now I have a CLC filter at the output of the mixer. When I look at the input impedance, I still get the bandpass around the input frquenct, now it's much narrower. In addtion, I also get one more valley and hill on each side. (so the shape goes up and down 3 rimes around the input frequency and tapers out to a lower values on both side) I know for sure it's due to the inductor. if I disconnect the inductor, then I just get the normal bandpass. Do you think this is due to how the CLC filter process the input signal? But the CLC is just a low pass filter though......Don't undestand why if would change the shape of the input impedance like that.

Any comments are welcomed. Thanks.

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aaron_do

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Re: passive mixer input impedance
Reply #3 - Mar 1^st, 2011, 9:48pm

Hi Kelly,

I'll try and answer your questions to the best of my knowledge.

1. I think your theory is correct, but in the implementation, it would not work that way. The exact details will really depend on your particular design. If you are using a direct-conversion receiver, then obviously the bandpass response would be centered at your center frequency, and therefore the output capacitance would not affect the input impedance at the center frequency. If you are using a low-IF receiver, then you would want the output bandwidth to be more than the IF frequency (plus half the signal bandwidth). Therefore, for the input impedance, the 3 dB bandwidth of the input impedance would also cover your desired RF frequency. i.e. your capacitor needs to be small enough to cover your bandwidth.

2. So you have a CLC pi network at the output of your mixer. This network is a low-pass filter from the input to the output. However, the impedance looking into the CLC is not low-pass. It depends on your values of C and L, but it is probably low-pass with a notch. That's why the shape of your input impedance is changing.

cheers,
Aaron

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there is no energy in matter other than that received from the environment - Nikola Tesla

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kelly

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Re: passive mixer input impedance
Reply #4 - Mar 2^nd, 2011, 11:52am

Hi Aaron,

That makes total sense, thanks much.

The more I think about it, the more I think I need to put a buffer between the mixer and CLC filter if I can still meet the noise and IIP3. Anything I do to the mixer(to improve linearity) affects the impedance the filter see, anything I do to the filter affect the input impedance of the mixer. Now I can totally see why it's good to isolate the I/O.

Anyway, thanks again for all your help.

Kelly

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