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Berkeley Single Transistor Mixers and Large Signal Distortion Prof. Ali M. Niknejad U.C. Berkeley Copyright c 2015 by Ali M. Niknejad 1 / 35 BJT with Large Sine Drive IC vi VA vi = V̂i cos !t IC = IS e vBE Vt vBE = VA + V̂i cos !t Consider a bipolar device driven with a large sine signal. This occurs in many types of non-linear circuits, such as oscillators, frequency multipliers, mixers and class C amplifiers. 2 / 35 BJT Collector Current The collector current can be factored into a DC bias term and a periodic signal Vˆi I C = I S e V A V t e Vt cos !t IC = IS e a e b cos !t Where the normalized bias is a = Va /Vt and the normalized drive signal is b = V̂i /Vt . Since IC is a periodic function, we can expand it into a Fourier Series. Note that the Fourier coefficients of e b cos !t are modified Bessel functions In (b) e b cos !t = I0 (b) + 2I1 (b) cos !t + 2I2 (b) cos 2!t + · · · 3 / 35 BJT DC Current Assume that the bias current of the amplifier is stabilized. Then ✓ ◆ 2I1 (b) 2I2 (b) a IC = IS e I0 (b) 1 + cos !t + cos 2!t + · · · | {z } I0 (b) I0 (b) IQ 6 5 IC = IS e a e b cos !t IQ b cos !t = e I0 (b) 4 IC IQ 3 2 1 1 2 3 4 5 4 / 35 Collector Current Waveform 8 b = 10 7 6 IC IQ 5 4 3 b=4 2 b = .5 1 -0.4 -0.2 0 0.2 0.4 With increasing input drive, the current waveform becomes “peaky”. The peak value can exceed the DC bias by a large factor. 5 / 35 Harmonic Current Amplitudes 1 n=1 0.8 In (b) I0 (b) n=2 n=3 0.6 n=4 0.4 n=5 n=6 0.2 n=7 2.5 5 7.5 10 12.5 15 17.5 20 b The BJT output spectrum is rich in harmonics. 6 / 35 Small-Signal Region 0.01 0.008 n=1 In (b) I0 (b) 0.006 0.004 n=2 0.002 n=3 0.05 0.1 b 0.15 0.2 If we zoom in on the curves to small b values, we enter the small-signal regime, and the weakly non-linear behavior is predicted by our power series analysis. 7 / 35 BJT with Stable Bias Neglecting base current ( given by 1), the voltage at the base is R1 VA + vi R2 VA0 = RE CE R2 VCC R1 + R2 VE = VA0 VBE V 0 VBE VE = A RE RE We see that the bias is fixed since VBE does not vary too much. Typically VA0 is a few volts. IQ = 8 / 35 Di↵erential Pair with Sine Drive The large signal equation for IC 1 is given by IC 1 + IC 2 = IEE VBE 1 VBE 2 = Vi = Vt ln ✓ IC 1 IC 2 IC 1 = + Vi 2 Vi 2 + IEE ◆ IEE 1 + e vi /Vt vi = V̂i cos !t IC 1 = IEE 1 + e b cos !t 9 / 35 Di↵ Pair Waveforms 1 0.8 IC1 IEE 0.6 0.4 b =1 0.2 b =3 b =10 1 2 3 4 5 6 For large b, the waveform approaches a square wave ✓ ◆ IC 1 2 1 1 = + cos !t cos 3!t + cos 5!t + · · · IEE 2 ⇡ 3 5 10 / 35 Di↵ Pair Harmonic Currents Normalized Harmonic Currents 2 ⇡ n=1 0.6 0.5 0.4 0.3 n=3 0.2 (negative) n=5 0.1 n=7 2.5 5 7.5 10 12.5 (negative) 15 17.5 20 b As expected, the ideal di↵erential pair does not produce any even harmonics. 11 / 35 Mixer Analysis As we have seen, a mixer has three ports, the LO, RF, and IF port. Assume that a circuit is “pumped” with a periodic large signal at the LO port with frequency !0 . From the RF port, though, assume we apply a small signal at frequency !s . Since the RF input is small, the circuit response should be linear (or weakly non-linear). But since the LO port changes the operating point of the circuit periodically, we expect the overall response to the RF port to be a linear time-varying response io (t) = g (t)vin 12 / 35 Mixer Assumptions gm (t) The transconductance varies periodically and can be expanded in a Fourier series g (t) = g0 + g1 cos !0 t + g2 cos 2!0 t + · · · Applying the input vin = Vˆ1 cos !s t io (t) = (g0 + g1 cos !0 t + g2 cos 2!0 t + · · · ) ⇥ Vˆ1 cos !s t 13 / 35 Mixer Output Signal Expanding the product, we have g1 g2 io (t) = g0 Vˆ1 cos !s t+ Vˆ1 cos(!0 ±!s )t+ Vˆ1 cos(2!0 ±!s )t+· · · 2 2 The first term is just the input amplified. The other terms are all due to the mixing action of the linear time-varying periodic circuit. Let’s say the desired output is the IF at !0 conversion gain is therefore defined as gconv = !s . The |IF output current| g1 = |RF input signal voltage— 2 14 / 35 A BJT Mixer C3 LO L3 IF L1 C1 C RF C C2 L2 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. 15 / 35 AC Eq. Circuit IF LO L1 C3 L3 C1 RF C2 L2 The AC equivalent circuit is shown above. 16 / 35 BJT Mixer Analysis When we apply the LO alone, the collector current of the mixer is given by ✓ ◆ 2I1 (b) 2I2 (b) IC = IQ 1 + cos !t + cos 2!t + · · · I0 (b) I0 (b) We can therefore define a time-varying gm (t) by gm (t) = IC (t) qIC (t) = Vt kT The output current when the RF is also applied is therefore given by iC (t) = gm (t)vs ✓ ◆ qIQ 2I1 (b) 2I2 (b) iC = 1+ cos !t + cos 2!t + · · · ⇥ Vˆs cos !s t kT I0 (b) I0 (b) 17 / 35 BJT Mixer Analysis (cont) The output at the IF is therefore given by qIQ I1 (b) iC |!IF = Vˆs cos(!0 !s )t | {z } kT I0 (b) |{z} !IF gmQ The conversion gain is given by 1 0.8 gconv gmQ gconv = gmQ I1 (b) I0 (b) 0.6 0.4 0.2 2 4 b= V̂i V 6 8 10 18 / 35 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 ✓ ◆ 2I1 (b) 2I2 (b) IC = IQ 1 + cos !t + cos 2!t + · · · I0 (b) I0 (b) The base current is simply IC / , and so the input impedance seen by the LO is given by Zi | ! 0 = Vˆo iB,!0 = = Vˆo iC ,!0 = Vˆo (b) IQ 2II01(b) = bVt 2I1 (b) IQ I0 (b) b I0 (b) = 2 gmQ I1 (b) Gm 19 / 35 RF Signal Drive The impedance seen by the RF singal source is also the base current at the !s components. Typically, we have a high-Q circuit at the input that resonates at RF. iB (t) = = 1 qIQ kT ✓ iC (t) 2I1 (b) Vˆs cos !s t + cos(!0 ± !s )t + · · · I0 (b) ◆ The input impedance is thus the same as an amplifier Rin = Vˆs = |component in iB at !s | kT = qIQ gmQ 20 / 35 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 IC = IS e vBE /Vt where vBE = vin + vo + VA or Vˆs IC = IS e VA /Vt ⇥ e b cos !0 t ⇥ e Vt cos !s t If we assume that the RF signal is weak, then we can approximate e x ⇡ 1 + x 21 / 35 General Approach (cont) Now the output current can be expanded into ✓ ◆ 2I1 (b) 2I2 (b) IC = IQ 1 + cos !t + cos 2!t + · · · I0 (b) I0 (b) ! Vˆs ⇥ 1+ cos !s t Vt 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. 22 / 35 Distortion in Mixers Using the same formulation, we can now insert a signal with two tones vin = Vˆs1 cos !s1 t + Vˆs2 cos !s2 IC = IS e VA /Vt ⇥ e b cos !0 t ⇥ e Vˆs1 Vt cos !s1 t+ Vˆs2 Vt cos !s2 t The final term can be expanded into a Taylor series IC = IS e VA /Vt ⇥ e b cos !0 t ⇥ 1 + Vs1 cos !s 1t + Vs2 cos !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 23 / 35 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 = 1000 900 = 100 24 / 35 Harmonic Mixer Analysis The nth harmonic conversion tranconductance is given by gconv ,n = |IF current out| gn = |input signal voltage| 2 For a BJT, we have gconv ,n = gmQ In (b) I0 (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. 25 / 35 FET Large Signal Drive ID vo VQ Consider the output current of a FET driven by a large LO signal µCox W ID = (VGS VT )2 (1 + VDS ) 2 L where VGS = VA + vLO = VA + Vo cos !0 t. Here we implicitly assume that Vo is small enough such that it does not take the device into cuto↵. 26 / 35 FET Large Signal Drive (cont) 0.5 1 1.5 2 2.5 3 That means that VA + V0 cos !0 t > VT , or VA V0 > VT , or equivalently V0 < VA VT . Under such a case we expand the current ID / (VA VT )2 + V02 cos2 !0 t + 2(VA VT )V0 cos !0 t 27 / 35 FET Current Components The cos2 term can be further expanded into a DC and second harmonic term. Identifying the quiescent operating point IQ = ID = IDQ + µCox µCox W (VA 2 L 0 W B B L @ VT )2 (1 + VDS ) 1 2 V 4 0 |{z} bias point shift 1 + (VA | VT )Vo cos !0 t + {z } LO modulation C V02 cos(2!0 t)C A (1 + VDS ) | 4 {z } LO 2nd harmonic 28 / 35 FET Time-Varying Transconductance The transconductance of a FET is given by (assuming strong inversion operation) g (t) = @ID W = µCox (VGS @VGS L VT )(1 + VDS ) VGS (t) = VA + V0 cos !0 t W (VA L ✓ g (t) = gmQ 1 + g (t) = µCox VT + V0 cos !0 t)(1 + VDS ) ◆ V0 cos !0 t (1 + VDS ) VA VT This is an almost ideal mixer in that there is no harmonic components in the transconductance. 29 / 35 MOS Mixer vs We see that we can build a mixer by simply injecting an LO + RF signal at the gate of the FET (ignore output resistance) IF Filter vo VQ i0 = g (t)vs = gmQ i0 |IF = ✓ 1+ V0 VA VT ◆ cos !0 t Vs cos !s t gmQ V0 cos(!0 ± !s )tVs 2 VA VT gc = i0 |IF gmQ V0 = Vs 2 VA VT 30 / 35 MOS Mixer Summary But gmQ = µCox W L (VA VT ) gc = µCox W V0 2 L which means that gc is independent of bias VA . The gain is controlled by the LO amplitude V0 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. 31 / 35 “Dual Gate” Mixer io M2 + vo VA G2 + vs M1 G1 D2 shared junction (no contact) S1 VA 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 . 32 / 35 Dual Gate Mixer Operation tri o de saturation 0.5 LO 1 1.5 2 2.5 3 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 M1 is therefore modulated periodically gm |sat = µCox W (VGS L gm |triode = µCox VT ) W VDS L 33 / 35 Dual Gate Waveforms vLO,1 vLO,2 Active Region Triode Region gm (t) gm,max VGS2 is roughly constant since M1 acts like a current source. VD1 = vLO g (t) = ⇢ VGS2 = VA2 + V0 cos !0 t VGS2 µCox W VT ) VD1 > VGS L (VGS1 µCox (VA2 VGS2 |V0 cos !0 t|) VD1 < VGS VT VT 34 / 35 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. 35 / 35