EE105 Fall 2015 Microelectronic Devices and Circuits Frequency Response. Prof. Ming C. Wu 511 Sutardja Dai Hall (SDH)


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1 EE05 Fall 205 Microelectronic Devices and Circuits Frequency Response Prof. Ming C. Wu 5 Sutardja Dai Hall (SDH)
2 Amplifier Frequency Response: Lower and Upper Cutoff Frequency Midband gain A mid and upper and lower cutoff frequencies ω H and ω L that define bandwidth of an amplifier are often of more interest than the complete transfer function Coupling and bypass capacitors (~ μμf) determine ω L Transistor (and stray) capacitances (~ pf) determine ω H
3 Lower Cutoff Frequency (ω L ) Approximation: ShortCircuit Time Constant (SCTC) Method. Identify all coupling and bypass capacitors 2. Pick one capacitor (CC ii ) at a time, replace all others with short circuits 3. Replace independent voltage source with short, and independent current source with open 4. Calculate the resistance (RR iiss ) in parallel with CC ii 5. Calculate the time constant, RR iiii CC ii 6. Repeat this for each of the n capacitor 7. The low cutoff frequency can be approximated by n ω L R is C i i= Note: this is an approximation. The real low cutoff is slightly lower
4 Lower Cutoff Frequency (ω L ) Using SCTC Method for CS Amplifier SCTC Method: f L n 2π i= R is C i For the CommonSource Amplifier: f L # % 2π $ R S C + + R 2S C 2 R 3S C 3 & ( '
5 Lower Cutoff Frequency (ω L ) Using SCTC Method for CS Amplifier Using the SCTC method: f L " $ 2π # R S C + + R 2S C 2 R 3S C 3 % ' & For C 2 : R 3S = R 3 + R D ( R id ) = R 3 + R D r o ( ) For C : R S = R I + R G ( R ig ) = R I + R G For C 3 : R 2S = R S R is = R S g m
6 Design: How Do We Choose the Coupling and Bypass Capacitor Values? Since the impedance of a capacitor increases with decreasing frequency, coupling/bypass capacitors reduce amplifier gain at low frequencies. To choose capacitor values, the shortcircuit time constant (SCTC) method is used: each capacitor is considered separately with all other capacitors replaced by short circuits. To be able to neglect a capacitor at a given frequency, the magnitude of the capacitor s impedance must be much smaller than the equivalent resistance appearing at its terminals at that frequency
7 Coupling and Bypass Capacitor Design CommonSource Amplifiers For the CS Amplifier: R in = R G R in CS For coupling capacitor C : C >> ( ) ω R I + R in For coupling capacitor C 3 : C 2 >> ( ) ω R 3 + R out R out = R D CS R out
8 Lower Cutoff Frequency f L Dominant Pole Design Instead of having the lower cutoff frequency set by the interaction of several poles, it can be set by the pole associated with just one of the capacitors. The other capacitors are then chosen to have their pole frequencies much below f L. The capacitor associated with the emitter or source part of the circuit tends to be the largest due to low resistance presented by emitter or source terminal of transistor and is commonly used to set f L. Values of other capacitors are increased by a factor of 0 to push their corresponding poles to much lower frequencies.
9 Capacitively Coupled CS Amplifier Find the pole frequency associated with each coupling/bypass capacitor:
10 Low Frequency Response Figure 0.7 Sketch of the lowfrequency magnitude response of a CS amplifier for which the three pole frequencies are sufficiently separated for their effects to appear distinct.
11 High Frequency Response
12 Capacitors in MOS Device C gs = (2 / 3)WLC ox + C ov C gd = C ov C sb = C jsb (area + perimeter) junction C db = C jdb (area + perimeter) junction
13 (Simplified) HighFrequency EquivalentCircuit Model for MOSFET Capacitances between source/body, C sb, and between drain/body, C db, are neglected
14 Intrinsic Response of FET: UnityGain Frequency, f T f T : defined as frequency at which shortcircuit current gain = f T : a figureofmerit for transistor speed V gs = KCL: I o + I i sc gs + sc gd V gs / sc gd = g m V gs I o = g m V gs sc gd V gs g m V gs = g m I i sc gs + sc gd Drain is grounded (shortcircuit load) As gate length reduces in advanced technology node, C gs reduces and f T increases A I = I o I i = s = jω I o I i = ω T = g m sc gs + sc gd g m ω(c gs + C gd ) g m C gs + C gd
15 CommonSource Voltage Amplifier Smallsignal model: C sb is connected to ground on both sides, therefore can be ignored R S Can solve problem directly by nodal analysis or using 2port models of transistor OK if circuit is small (2 nodes) We can find the complete transfer function of the circuit, but in many cases it s good enough to get an estimate of the 3dB bandwidth
16 CS Voltage Amp SmallSignal Model Two Nodes! Easy For now we will ignore C db to simplify the math
17 Frequency Response KCL at input and output nodes;; analysis is made complicated V out = g m r o R L V in + jω /ω p [ ]( jω /ω z ) ( )( + jω /ω ) p2 Zero Lowfrequency gain: ( ) ( )( + j0) V out = g r R m o L j0 V in + j0 g m Two Poles r o R L Zero: ω z = g m C gs + C gd
18 Calculating the Poles ω p ω p2 { } + R s C gs + ( + g m R out )C gd R out / R S R S C gs + ( + g m R out )C gd { } + R out C gd R out C gd Usually >> Results of complete analysis: not exact and little insight These poles are calculated after doing some algebraic manipulations on the circuit. It s hard to get any intuition from the above expressions. There must be an easier way!
19 Method: The Miller Effect
20 The Miller Effect
21 Using The Miller Effect Effective input capacitance: C in = jωc Miller = A v,cgd jωc gd = jω ( A vcgd )C gd
22 CS Voltage Amp SmallSignal Model Modified SmallSignal Model with Miller Effect: C gs +C Miller We can approximate the first pole by using Miller capacitance This gives us a good approximation of the 3dB bandwidth
23 Comparison with Exact Analysis Miller result (calculate RC time constant of input pole): Exact result: ω p ω p = R S C gs + ( + g m R out ʹ )C gd = R S C gs + ( + g m R out ʹ )C gd + R out ʹ C gd As a result of the Miller effect there is a fundamental gainbandwidth tradeoff
24 Common Drain Amplifier Calculate Bandwidth of the Common Drain (Source Follower) Procedure:. Replace current source with MOSFETbased current mirror 2. Draw smallsignal model with capacitors (for simplicity, we will focus on C gd and C gs ) 3. Find the DC smallsignal gain 4. Use the Miller effect to calculate the input capacitance 5. Calculate the dominant pole
25 SmallSignal Model with Capacitors R S Find DC Gain Find Miller capacitor for C gs  note that the gatesource capacitor is between the input and output!
26 Voltage Gain Across C gs Write KCL at output node: v out r o r oc = g m v gs = g m (v in v out ) v out + g m r o r = g v m in oc v out v in = r o g m + g r m oc = g m (r o r oc ) + g m (r o r oc ) = A vcgs
27 Compute Miller Effected Capacitance Now use the Miller Effect to compute C in : Remember that C gs is the capacitor from the input to the output R S C in = C gd + C M C in = C gd + ( A vcgs )C gs Miller Cap C in = C gd + ( g m (r o r oc ) + g m (r o r oc ) )C gs C in = C gd + ( + g m (r o r oc ) )C gs C in C gd (for large g m (r o //r oc ))
28 Bandwidth of Source Follower Input lowpass filter s 3 db frequency: C ω p = R S C gd + gs + g m (r o r oc ) Substitute favorable values of R S, r o : R» / S g m ω p / g m ( ) C gd + C gs r o >>/ g m C gd / g m + BIG ω p g m / C gd Very high frequency! Model not valid at these high frequencies
29 Common Source Amplifier: Some Examples A vcgd = Negative, large number (00) C Miller = ( A V,Cgd )C gd 00C gd Miller Multiplied Cap has detrimental impact on bandwidth Common Drain Amplifier: A vcgs = Slightly less than C Miller = ( A V,Cgs )C gs! 0 Bootstrapped cap has negligible impact on bandwidth!
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