Lecture 17: BJT/FET Mixers/Mixer Noise

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1 EECS 142 Lecture 17: BJT/FET Mixers/Mixer Noise Prof. Ali M. Niknejad University of California, Berkeley Copyright c 2005 by Ali M. Niknejad A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 1/35

2 A BJT Mixer C 3 L 3 LO C 1 L 1 IF RF C C C 2 L 2 The transformer is used to sum the LO and RF signals at the input. The winding inductance is used to form resonant tanks at the LO and RF frequencies. The output tank is tuned to the IF frequency. Large capacitors are used to form AC grounds. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 2/35

3 AC Eq. Circuit IF LO L 1 C 3 L 3 C 1 RF C 2 L 2 The AC equivalent circuit is shown above. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 3/35

4 BJT Mixer Analysis When we apply the LO alone, the collector current of the mixer is given by ( I C = I Q 1 + 2I 1(b) I 0 (b) cos ωt + 2I ) 2(b) cos 2ωt + I 0 (b) We can therefore define a time-varying g m (t) by g m (t) = I C(t) V t = qi C(t) kt The output current when the RF is also applied is therefore given by i C (t) = g m (t)v s i C = qi Q kt ( 1 + 2I 1(b) I 0 (b) cos ωt + 2I 2(b) I 0 (b) cos 2ωt + ) ˆV s cos ω s t A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 4/35

5 BJT Mixer Analysis (cont) The output at the IF is therefore given by i C ωif = ˆV s qi Q kt }{{} g mq I 1 (b) I 0 (b) cos(ω 0 ω }{{} s )t ω IF The conversion gain is given by g conv = g mq I 1 (b) I 0 (b) g conv g mq b = ˆV i A. M. Niknejad University of California, Berkeley V EECS 142 Lecture 17 p. 5/35

6 LO Signal Drive For now, let s ignore the small-signal input and determine the impedance seen by the LO drive. If we examine the collector current ( I C = I Q 1 + 2I 1(b) I 0 (b) cos ωt + 2I ) 2(b) cos 2ωt + I 0 (b) The base current is simply I C /β, and so the input impedance seen by the LO is given by Z i ω0 = ˆV o i B,ω0 = β ˆV o i C,ω0 = β ˆV o I Q 2I 1 (b) I 0 (b) = βbv t I Q 2I 1 (b) I 0 (b) = b 2 β I 0 (b) g mq I 1 (b) = β G m A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 6/35

7 RF Signal Drive The impedance seen by the RF singal source is also the base current at the ω s components. Typically, we have a high-q circuit at the input that resonates at RF. i B (t) = i C(t) β = 1 β qi Q kt ( ˆV s cos ω s t + 2I ) 1(b) I 0 (b) cos(ω 0 ± ω s )t + The input impedance is thus the same as an amplifier R in = ˆV s component in i B at ω s = β kt qi Q = β g mq A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 7/35

8 Mixer Analysis: General Approach If we go back to our original equations, our major assumption was that the mixer is a linear time-varying function relative to the RF input. Let s see how that comes about I C = I S e v BE/V t where or I C = I S e V A/V t v BE = v in + v o + V A e b cos ω 0t e ˆ Vs V t cos ω s t If we assume that the RF signal is weak, then we can approximate e x 1 + x A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 8/35

9 General Approach (cont) Now the output current can be expanded into ( I C = I Q 1 + 2I 1(b) I 0 (b) cos ωt + 2I 2(b) cos 2ωt + I 0 (b) ( 1 + ˆV s V t cos ω s t ) ) In other words, the output can be written as = BIAS + LO + Conversion Products In general we would filter the output of the mixer and so the LO terms can be minimized. Likewise, the RF terms are undesired and filtered from the output. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 9/35

10 Distortion in Mixers Using the same formulation, we can now insert a signal with two tones v in = ˆ V s1 cos ω s1 t + ˆ V s2 cos ω s2 I C = I S e V A/V t e b cos ω 0t e ˆ V s1 V cos ω V s1 t+ ˆ s2 cos ω t V s2 t t The final term can be expanded into a Taylor series I C = I S e V A/V t e b cos ω 0t ( 1 + Vs1 cos ω s 1t + V s2 cos ω s 2t + ( ) 2 + ( ) 3 + ) The square and cubic terms produce IM products as before, but now these products are frequency translated to the IF frequency A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 10/35

11 Another BJT Mixer LO + v 0x C 0 C 3 L 3 IF RF L s C s C The signal from the LO driver is capacitively coupled to the BJT mixer A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 11/35

12 LO Capacitive Divider C 0 C 0 + v 0x L s C s + v 0x C s Assume that ω LO > ω RF, or a high side injection Note beyond resonance, the input impedance of the tank appears capacitive. Thus C s is the effective capacitance of the tank. The equivalent circuit for the LO drive is therefore a capacitive divider v o = C o C o + C s v ox A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 12/35

13 Harmonic Mixer RF Second Harmonic Mixer IF LO We can use a harmonic of the LO to build a mixer. Example, let LO = 500MHz, RF = 900MHz, and IF = 100MHz. Note that IF = 2LO RF = = 100 A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 13/35

14 Harmonic Mixer Analysis The nth harmonic conversion tranconductance is given by IF current out g conv,n = input signal voltage = g n 2 For a BJT, we have g conv,n = g mq I n (b) I 0 (b) The advantage of a harmonic mixer is the use of a lower frequency LO and the separation between LO and RF. The disadvantage is the lower conversion gain and higher noise. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 14/35

15 FET Large Signal Drive I D v o V Q Consider the output current of a FET driven by a large LO signal I D = µc ox 2 W L (V GS V T ) 2 (1 + λv DS ) where V GS = V A + v LO = V A + V o cos ω 0 t. Here we implicitly assume that V o is small enough such that it does not take the device into cutoff. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 15/35

16 FET Large Signal Drive (cont) That means that V A + V 0 cos ω 0 t > V T, or V A V 0 > V T, or equivalently V 0 < V A V T. Under such a case we expand the current I D ( (V A V T ) 2 + V 0 cos 2 ω 0 t + 2(V A V T )V 0 cos ω 0 t ) A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 16/35

17 FET Current Components The cos 2 term can be further expanded into a DC and second harmonic term. Identifying the quiescent operating point I D = I DQ + µc ox W L I Q = µc ox W 2 L (V A V T ) 2 (1 + λv DS ) V cos2 ω 0 t }{{} 1 4 V 0 2 }{{} bias point shift LO 2nd harmonic + (V A V T )V o cos ω 0 t + }{{} LO modulation (1 + λv DS) A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 17/35

18 FET Time-Varying Transconductance The transconductance of a FET is given by (assuming strong inversion operation) g(t) = I D V GS = µc ox W L (V GS V T )(1 + λv DS ) V GS (t) = V A + V 0 cos ω 0 t W g(t) = µc ox L (V A V T + V 0 cos ω 0 t)(1 + λv DS ) ) V 0 g(t) = g mq (1 + cos ω 0 t (1 + λv DS ) V A V T This is an almost ideal mixer in that there is no harmonic components in the transconductance. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 18/35

19 MOS Mixer v s v o V Q IF Filter We see that we can build a mixer by simply injecting an LO + RF signal at the gate of the FET (ignore output resistance) i 0 = g(t)v s = g mq (1 + ) V 0 cos ω 0 t V s cos ω s t V A V T i 0 IF = g mq 2 V 0 V A V T cos(ω 0 ± ω s )tv s g c = i 0 IF V s = g mq 2 V 0 V A V T A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 19/35

20 MOS Mixer Summary But g mq = µc ox W L (V A V T ) g c = µc ox 2 W L V 0 which means that g c is independent of bias V A. The gain is controlled by the LO amplitude V 0 and by the device aspect ratio. Keep in mind, though, that the transistor must remain in forward active region in the entire cycle for the above assumptions to hold. In practice, a real FET is not square law and the above analysis should be verified with extensive simulation. Sub-threshold conduction and output conductance complicate the picture. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 20/35

21 Dual Gate Mixer + v o i o M2 G2 D2 shared junction (no contact) V A2 + v s M1 G1 S1 V A1 The dual gate mixer, or more commonly a cascode amplifier, can be turned into a mixer by applying the LO at the gate of M2 and the RF signal at the gate of M1. Using two transistors in place of one transistor results in area savings since the signals do not need to be combined with a transformer or capacitively. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 21/35

22 Dual Gate Mixer Operation saturation triode LO Without the LO signal, this is simply a cascode amplifier. But the LO signal is large enough to push M1 into triode during part of the operating cycle. The transconductance of M 1 is therefore modulated periodically g m sat = µc ox W L (V GS V T ) g m triode = µc ox W L V DS A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 22/35

23 Dual Gate Waveforms v LO,1 v LO,2 Active Region Triode Region g m (t) g m,max V GS2 is roughly constant since M1 acts like a current source. g(t) = { V D1 = v LO V GS2 = V A2 + V 0 cos ω 0 t V GS2 µc ox W L (V GS1 V T ) V D1 > V GS V T µc ox (V A2 V GS2 V 0 cosω 0 t ) V D1 < V GS V T A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 23/35

24 Realistic Waveforms A more sophisticated analysis would take sub-threshold operation into account and the resulting g(t) curve would be smoother. A Fourier decomposition of the waveform would yield the conversion gain coefficient as the first harmonic amplitude. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 24/35

25 Mixer Analysis: Time Domain A generic mixer operates with a periodic transfer function h(t + T) = h(t), where T = 1/ω 0, or T is the LO period. We can thus expand h(t) into a Fourier series y(t) = h(t)x(t) = c n e jω 0nt x(t) For a sinusoidal input, x(t) = A(t) cosω 1 t, we have y(t) = c n 2 A(t) ( e j(ω 1+ω 0 n)t + e j( ω 1+ω 0 n)t ) Since h(t) is a real function, the coefficients c k = c k are even. That means that we can pair positive and negative frequency components. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 25/35

26 Time Domain Analysis (cont) Take c 1 and c 1 as an example = c 1 e j(ω 1+ω 0 )t + e j( ω 1+ω 0 )t 2 A(t)+c 1 e j(ω 1 ω 0 )t + e j( ω 1 ω 0 )t 2 = c 1 A(t) cos(ω 1 + ω 0 )t + c 1 A(t) cos(ω 1 ω 0 )t + Summing together all the components, we have A(t)+ y(t) = c n cos(ω 1 + nω 0 )t Unlike a perfect multiplier, we get an infinite number of frequency translations up and down by harmonics of ω 0. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 26/35

27 Frequency Domain Analysis Since multiplication in time, y(t) = h(t)x(t), is convolution in the frequency domain, we have Y (f) = H(f) X(f) The transfer function H(f) = c nδ(f nf 0 ) has a discrete spectrum. So the output is given by Y (f) = c n δ(σ nf 0 )X(f σ)dσ = c n δ(σ nf 0 )X(f σ)dσ A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 27/35

28 Frequency Domain (cont) noise folding from 3LO noise folding from -3LO noise folding from 2LO noise folding from -2LO 3fo 2fo fs fo fif2fo fo fs 3fo By the frequency sifting property of the δ(f σ) function, we have Y (f) = cnx(f nf0) Thus, the input spectrum is shifted by all harmonics of the LO up and down in frequency. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 28/35

29 Noise/Image Problem Previously we examined the image problem. Any signal energy a distance of IF from the LO gets downconverted in a perfect multiplier. But now we see that for a general mixer, any signal energy with an IF of any harmonic of the LO will be downconverted! These other images are easy to reject because they are distant from the desired signal and a image reject filter will be able to attenuate them significantly. The noise power, though, in all image bands will fold onto the IF frequency. Note that the noise is generated by the mixer source resistance itself and has a white spectrum. Even though the noise of the antenna is filtered, new noise is generated by the filter itself! A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 29/35

30 Current Commutating Mixers V CC +LO +RF Q 2 Q 3 Q 1 R IF IF LO +LO +RF IF LO A popular alternative mixer topology uses a differential pair LO drive and an RF current injection at the tail. In practice, the tail current source is implemented as a transconductor. The LO signal is large enough to completely steer the RF current either through Q1 or Q2. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 30/35

31 Current Commutating Mixer Model V CC R IF IF +LO +RF Q 1 LO If we model the circuit with ideal elements, we see that the current I C1 is either switched to the output or to supply at the rate of the LO signal. When the LO signal is positive, we have a cascode dumping its current into the supply. When the LO signal is negative, though, we have a cascode amplifier driving the output. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 31/35

32 Conversion Gain s(t) +1 0 T LO t We can now see that the output current is given by a periodic time varying transconductance i o = g m (t)v s = g mq s(t)v s where s(t) is a square pulse waveform (ideally) switching between 1 and 0 at the rate of the LO signal. A Fourier decomposition yields i o = g mq v s ( π cos ω 0t 2 π ) 1 3 cos 3ω 0t + A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 32/35

33 Conversion Gain (cont) So the RF signal v s is amplified (feed-thru) by the DC term and mixed by all the harmonics i o = g mq V s 2 ( 1 2 cos ω st + 2 π cos(ω 0 ± ω s )t 2 ) 3π cos(3ω 0 ± ω s )t + The primary conversion gain is g c = 1 π g mq. Since the role of Q1 (or M1) is to simply create an RF current, it can be degenerated to improve the linearity of the mixer. Inductance degeneration can be employed to also achieve an impedance match. MOS version acts in a similar way but the conversion gain is lower (lower g m ) and it requires a larger LO drive. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 33/35

34 Differential Output IF +LO +RF LO This block is commonly known as the Gilbert Cell If we take the output signal differentially, then the output current is given by i o = g m (t)v s = g mq s 2 (t)v s A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 34/35

35 Differential Output Gain s 2 (t) +1 t 1 T LO The pulse waveform s 2 (t) now switches between ±1, and thus has a zero DC value s 2 (t) = 4 π cos ω 0t π cos 3ω 0t + The lack of the DC term means that there is ideally no RF feedthrough to the IF port. The conversion gain is doubled since we take a differential output g c /g mq = 2 π A. M. Niknejad University of California, Berkeley EECS 142 Lecture 17 p. 35/35

Berkeley. Mixers: An Overview. Prof. Ali M. Niknejad. U.C. Berkeley Copyright c 2014 by Ali M. Niknejad

Berkeley. Mixers: An Overview. Prof. Ali M. Niknejad. U.C. Berkeley Copyright c 2014 by Ali M. Niknejad Berkeley Mixers: An Overview Prof. Ali M. U.C. Berkeley Copyright c 2014 by Ali M. Mixers Information PSD Mixer f c The Mixer is a critical component in communication circuits. It translates information