The Designer's Guide Community
Forum
Welcome, Guest. Please Login or Register. Please follow the Forum guidelines.
Nov 28th, 2020, 3:25am
Pages: 1
Send Topic Print
RF Mixer working mechanism (Read 6105 times)
wccheng
Junior Member
**
Offline



Posts: 31

RF Mixer working mechanism
May 18th, 2004, 6:07am
 
Dear all,

      I have read some paper discussing about the RF mixer. I have found that some of them working in the multiplication method. Some of them working as chopping method.

      Actually, what is the different between two mechanisms? Could you introduce some books talking about this in detail?

Best Regards,

wccheng
Back to top
 
 
View Profile   IP Logged
City
Junior Member
**
Offline



Posts: 19

Re: RF Mixer working mechanism
Reply #1 - May 24th, 2004, 10:19am
 
Hi,

I believe you are talking about multipliers (of course multiplication) and mixers (chopping).

Lots of multipliers have been described in the past but I think that you don't want to use that for a downconversion. Most basic of the multiplier is based on translinear loops that you can do with BJTs or MOSTs in weak inversion (at rather low freq.) Those kind of circuit toplogies are very noisy because they operate in region where almost all of the transistors are operating at the same time.

An integrated mixer usually requires a rather low/average noise figure depeding on how much gain preceeds in the LNA. In a practical circuit I don't think that you should choose linear multiplication.

Mixers is a very similar to multipliers in the sense that it downconverts or upconverts the desired signal with a multiplication effect. Current steering is a very known topology (also called tree mixers) in which you simply choose to steer the current in different paths at the LO rate. A high LO drive usually reduces the switching pair noise. A large-signal analysis demonstrates this.

You can refer to B. Razahvi's book on RF microelectronics which contains references to other mixers papers.

City
Back to top
 
 
View Profile   IP Logged
nano_RF
Community Member
***
Offline



Posts: 50
madison
Re: RF Mixer working mechanism
Reply #2 - May 24th, 2004, 11:07am
 
hello city,

I did'nt understand the chopping thing.Did u mean Switching as chopping.

As for as i know in Mixer u multiply RF signal with ur chosen LO signal .This multiplication terms generate one difference(LO- RF) and one (LO + RF) signal,and than u can filter out the undesired band.

In this LO overdrive has to be large for the reason of better switching.Basically the current driven by RF signal is switched off at the rate of LO frequency.

One suggestion- To check ur result try to get the Fourier transform of the output waveform,which will clearly tell the downcoverted or upconverted frquency components.

--Vikas
Back to top
 
 
View Profile   IP Logged
City
Junior Member
**
Offline



Posts: 19

Re: RF Mixer working mechanism
Reply #3 - May 25th, 2004, 4:20am
 
Vikas,

yes switching / chopping for me is similar. If you look at the current without filtering, it is chopped at the LO rate, which is a process similar to sampling. The reason why you want high LO drive is that it reduces noise coming from the switching pair and you also have better conversion gain. I suppose that is what you mean by better switching. I just wanted to comment that multipliers and mixers are fundamentaly differents.

City
Back to top
 
 
View Profile   IP Logged
wccheng
Junior Member
**
Offline



Posts: 31

Re: RF Mixer working mechanism
Reply #4 - May 25th, 2004, 9:17am
 
Dear all,

     I have asked this question in another forum. Someone answer me that:

======================
     In multiplier mode, the LO signals are sinusoids with amplitudes such that it can switch the transistors from sat to linear region. In this case only multiplication of the RF signal frequency and the LO frequency occurs.
     In the chopping mode, the LO signals are square waves with amplitude close to supply voltage which can switch the transistors from sat to cutoff as a result the input RF frequency is multiplied by the fourier series of the square wave LO signal.
=======================

     Actually, is this explanation correct? I am so confused between chopping and multiplication mode of the mixer.  ??? PLEASE HELP  :-/

Best Regards,

wccheng
Back to top
 
 
View Profile   IP Logged
City
Junior Member
**
Offline



Posts: 19

Re: RF Mixer working mechanism
Reply #5 - May 26th, 2004, 12:23am
 
wcheng,

This is what I (actually badly) tried to explain. In the multiplication mode the input can be sinewave but with a low amplitude so that you stay in a linear region (which has nothing to do with the operating region of the transistor, I'll argue later on ...) where the system will not produce any additionnal harmonics. So the only multiplication terms come from sin(wlo*t)*sin(wRf*t).
If you use high amplitude signals that forces your system to go into nonlinear regions, that will chop your RF signal depending on the drive again.

Let me take an example: Consider a Tree mixer with a tail current source, an input differential pair, 4 transistors in a cross-coupled configuration and resistive loads with a filtering capacitor (I suppose we are downconverting) and for simplicity say we work with BJTs or MOSTs in weak inversion. The RF is fed at the differential pair input and the LO is fed at the 4 four cross-coupled transistors. Now, I you drive the LO amplitude << than 4*Ut, the transistor stays in a linear mode because to a deltaLO corresponds a deltaIout which can be described with deltaI=deltaLO*G right ?
If the RF is also small enough (again << 4Ut) then the output voltage in the bandwidth of the filter can be written as Vout(t)=VRf*VLo*Rl*G*sin(wLO*t)*sin(wRF*t) which is the case that was described to you in the other forum.

If you increases the LO amplitude >> 4Ut (RF still small), the cross-coupled transistors will successively turn ON and OFF (i.e switching mode) and if you probe the collector current, you'll see a the input RF input Vin*sin(wRF*t)*Gm,in chopped at the LO rate which gives you a current in the form Vin*Gm,in*sin(wRF*t)*Fourier(square_wave(wLO*t)). This is the chopping mode.

So this does not depend on square-wave drive or not, it depends if you forces your system into linear or non-linear regions. Of course, a square-wave in this case is the most efficient but it can be prove that a sine-wave with high amplitude can also do a good job ... but at high voltage supply of course.

So what is the difference between both? Well you can see that even if the circuit is identical in multiplication and chopping region, gain will be different for both of them. Also Noise will be VERY different and very bad for mulitplication mode as described earlier. This is why those cells are used in the chopping mode and are called MIXERS rather than MULTIPLIERS, even if we can use them in both configuration.

Hope it was more clear.

City
Back to top
 
 
View Profile   IP Logged
wccheng
Junior Member
**
Offline



Posts: 31

Re: RF Mixer working mechanism
Reply #6 - May 26th, 2004, 5:32am
 
Dear City,

     Your explanation is very very clear and in detail. I am very thankful your reply.

     Actually, you have use the notation "Ut". What is it stand for?

Best Regards,

wccheng
Back to top
 
 
View Profile   IP Logged
City
Junior Member
**
Offline



Posts: 19

Re: RF Mixer working mechanism
Reply #7 - May 27th, 2004, 12:23am
 
Ut stands for k*T/q k is the Boltzmann constant T temperature and q is the electron charge.
It is the thermodynamic potential.
If you trace a Bipolar differential pair differential output current it is given by:

Iout,diff/Iq = tanh(Vin,diff/(2*Ut)). if you approximate
the tanh curve with a a piecewise linear function, you'll see that your equivalent linear input dynamic range is about 4*Ut.

regards,

City
Back to top
 
 
View Profile   IP Logged
Pages: 1
Send Topic Print
Copyright 2002-2020 Designer’s Guide Consulting, Inc. Designer’s Guide® is a registered trademark of Designer’s Guide Consulting, Inc. All rights reserved. Send comments or questions to editor@designers-guide.org. Consider submitting a paper or model.