3/4/2008 Lecture 11 Ideal Current Source Circuit Symbol I-V Characteristic Equivalent Circuit OUTLINE I I • Cascode Stage + I R V N N I1 V Reading: Chapter 9.1 - −I1 • An ideal current source has infinite output impedance. How can we increase the output impedance of a BJT that is used as a current source? EE105 Spring 2008 Lecture 11, Slide 1Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 2Prof. Wu, UC Berkeley Boosting the Output Impedance Cascode Stage • Recall that emitter degeneration boosts the impedance • In order to relax the trade‐off between output seen looking into the collector. impedance and voltage headroom, we can use a – This improves the gain of the CE or CB amplifier. However, transistor instead of a degeneration resistor: headroom is reduced. Rout = [1+ gm (rO2 || rπ1 )]rO1 + rO2 || rπ1 Rout ≈ gm1rO1 ()rO2 || rπ1 VIRE = EE IC 2 = I E1 ≅ IC 2 if β1 >>1 Rout = []1+ g m (RE || rπ ) rO + RE || rπ • VCE for Q2 can be as low as ~ 0.4V (“soft saturation”) EE105 Spring 2008 Lecture 11, Slide 3Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 4Prof. Wu, UC Berkeley Maximum Bipolar Cascode Output Impedance PNP Cascode Stage Rgrrout,max≈ m 1 01π 1 Rrout,max ≈ β101 • The maximum output impedance of a bipolar cascode Rgrrrrr= [1++ ( || )] || is bounded by the ever‐present r between emitter out m 02π 1 01 02π 1 π Rgrrr≈ || and ground of Q1. out m101() 02π 1 EE105 Spring 2008 Lecture 11, Slide 5Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 6Prof. Wu, UC Berkeley EE105 Fall 2007 1 3/4/2008 False Cascodes Short‐Circuit Transconductance • The short‐circuit transconductance of a circuit is a measure of its strength in converting an input voltage signal into an output current signal. ⎡⎤⎛⎞11 Rg=+1 |||| rrrrr + |||| out⎢⎥ m121121⎜⎟ Oπ O O π i ⎣⎦⎢⎥⎝⎠ggmm22 out Gm ≡ ⎛⎞g 1 m1 vin Rrrout≈+⎜⎟12 O11 + ≈ O vout =0 ⎝⎠ggmm22 • When the emitter of Q1 is connected to the emitter of Q2, it’s no longer a cascode since Q2 becomes a diode‐ connected device instead of a current source. EE105 Spring 2008 Lecture 11, Slide 7Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 8Prof. Wu, UC Berkeley Voltage Gain of a Linear Circuit Example: Voltage Gain • By representing a linear circuit with its Norton equivalent, the relationship between Vout and Vin can be expressed by the product of Gm and Rout. Norton Equivalent Circuit iout vx Gm ≡ = gm1 Rout ≡ = ro1 Computation of v i vout = −iout Rout = −Gmvin Rout in v =0 x short-circuit out output current: vout vin = −Gm Rout Avm=−gr101 EE105 Spring 2008 Lecture 11, Slide 9Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 10 Prof. Wu, UC Berkeley Comparison of CE and Cascode Stages Voltage Gain of Cascode Amplifier • Since the output impedance of the cascode is higher • Since rO is much larger than 1/gm, most of IC,Q1 flows than that of a CE stage, its voltage gain is also higher. into diode‐connected Q2. Using Rout as before, AV is easily calculated. iout= gv m1 in vout = −gm1vinrO1 ⇒=Ggmm1 Av = −Gm Rout = −g ()1+ g ()r || r r + r || r VA m1[]m2 O1 π 2 O 2 O1 π 2 Av = −gm1rO1 = − Av ≈ −gm1rO2 gm2 (rO1 rπ 2 ) VT ≈ −g m1[]g m2 ()rO1 || rπ 2 rO 2 EE105 Spring 2008 Lecture 11, Slide 11 Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 12 Prof. Wu, UC Berkeley EE105 Fall 2007 2 3/4/2008 Alternate View of Cascode Amplifier Practical Cascode Stage Rrgrrrout≈ O32212|| m O ( O ||π ) • A bipolar cascode amplifier is also a CE stage in series • Since no current source can be ideal, the output with a CB stage. impedance drops. EE105 Spring 2008 Lecture 11, Slide 13 Prof. Wu, UC Berkeley EE105 Spring 2008 Lecture 11, Slide 14 Prof. Wu, UC Berkeley Improved Cascode Stage Rout ≈ [][]gm3rO3 (rO4 || rπ 3 ) || gm2rO2 (rO1 || rπ 2 ) Av = −gm1Rout • In order to preserve the high output impedance, a cascode PNP current source is used. EE105 Spring 2008 Lecture 11, Slide 15 Prof. Wu, UC Berkeley EE105 Fall 2007 3.