6.012 Recitation 23: Frequency Response of Common Collector & Common-Base Amplifier

Recitation 23 Frequency Response of C-C & C-B Amps 6.012 Spring 2009 Recitation 23: Frequency Response of Common Collector & Common-Base Amplifier Yesterday, we used OCT technique for the frequency response of Common-Drain and Common-Gate amplifiers. Today we will look at C-C, C-B frequency response. Common-Collector Amplifier One way to study the frequency response is to • First find the small signal equivalent model for the circuit • Do KCL, KVL nodal analysis, to find CO3dB • Or use OCT + Miller Approximation to find w3dB However, the small signal model of this circuit is quite complicated (as the C-D Amp. we talked about yesterday). What we can do is directly use the two-port model for the circuit, and add in the capacitances. So the methodology is as outlined below. Methodology 1. Start with low frequency two port model, obtain Av, Ai, Gm at low frequency 2. Identify the nodes (S/D/G/B for MOS; B/E/C for BJT) and add in capacitance in active device 3. Use Miller Approximation in conjunction with OCT to estimate bandwidth (w3dB ). Advantage: can directly use the “Rin ”,“Rout ” from two-port model, only need Av, Ai orGm much easier. 1 Recitation 23 Frequency Response of C-C & C-B Amps 6.012 Spring 2009 So taking the C-C Amplifier as an example, the two port model is: Rin = γπ + βo (γo ||γoc ||RL ) 1 Rs Rout = + βo gm Avo = 1 Rin RL Vout Av,LF = = · (1) · Vs Rs + Rin RL + Rout Large gm , βo will give desired resistances for voltage buffer. High Rin , low Rout = γπ + βo (γoc ||γo ||RL ) RL (1) RL + γπ + βo (γoc ||γo ||RL ) RL + g1 + m Rs Identify the B/E/C and add in capacitances Note: the other end of Cπ is to the right of Rout ! That is where “E” node is! Cπ is in the input/output feedback position. 2 Recitation 23 Frequency Response of C-C & C-B Amps 6.012 Spring 2009 Use Miller Approximation: where CM = Cπ (1−AvCπ ). AvCπ is the voltage gain across Cπ (not across overall amplifier). Vout Vout instead of What is the voltage gain across Cπ ? Vin Vs Vout Or, it is when Rs = 0. Vs Vout Vin ∴ AvCπ = = typically γo ||γoc RL   γπ + βo Vout  RL RL = (1)  Rs + γπ + βo RL Vs Rs =0 s RL + g1m + R  βo RL RL + 1/gm CM = Cπ (1 − AvCπ ) = Cπ   1/gm RL + 1/gm   = Cπ 1 1 + gm RL  1 (Rs ||Rin ) · (Cμ + CM ) In contrast to C-S or C-E amplifier, the Miller effect reduces the capacitance in this case, which will give better frequency response: (or another way to look at it, effect of Cπ is very small, since voltage gain across Cπ is ≈ 1. We do not need a lot of charges to go in/out the capacitor. And typically the movement of charges is the source to slow down the frequency response). If 1 gm  RL , then AvCπ −→ 1 =⇒ CM −→ 0 w3dB = • Therefore like C-D, Miller effect reduces capacitor value, =⇒ expect good frequency response. • Use of C-C: for multistage amplifiers, can enable high Rin , low Rout , won’t degrade frequency response 3 Recitation 23 Frequency Response of C-C & C-B Amps Common-Base Amplifier Current buffer: 1. Two port model (for current amplifier) Low frequency current gain iout vs Rin Rout Rs Rout (−1) Rs + Rin Rout + RL 1 = gm = γoc || [γo (1 + gm (γπ ||Rs ))] = 4 6.012 Spring 2009 Recitation 23 Frequency Response of C-C & C-B Amps 6.012 Spring 2009 2. Label B/E/C, add in capacitances No capacitor in the feedback position =⇒ Do not need Miller Approximation. Use OCT 1 gm = Rout ||RL = RL ||(γoc ||(γo + gm γo (γπ ||Rs ))) • Cπ : RTHCπ = Rs ||Rin = Rs || • Cμ : RTHCμ Let us try to make some simplifications (if conditions are met) for a on w3dB : If Rs not so small, since 1 gm is small (∼ 100Ω), 1 gm And if Rs  1 Cπ =⇒ TCπ = gm gm γπ (∼ 10 kΩ) Rout = γoc ||(γo + gm γo (γπ ||Rs )) gm γo (γπ ||Rs )) = gm γπ γo = βo γo Rs ||  Rout −→ γoc ||βo γo can be quite large =⇒ RTHCμ w3dB ≈  RL ||(γoc ||βo γo )  RL Cπ 1 can be approaching wT = + Cμ RL 5  gm Cπ + Cμ  - a good current buffer