EE05 Fall 205 Microelectronic Devices and Circuits Frequency Response Prof. Ming C. Wu wu@eecs.berkeley.edu 5 Sutardja Dai Hall (SDH)
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
Lower Cutoff Frequency (ω L ) Approximation: Short-Circuit 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 cut-off frequency can be approximated by n ω L R is C i i= Note: this is an approximation. The real low cut-off is slightly lower
Lower Cutoff Frequency (ω L ) Using SCTC Method for CS Amplifier SCTC Method: f L n 2π i= R is C i For the Common-Source Amplifier: f L # % 2π $ R S C + + R 2S C 2 R 3S C 3 & ( '
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
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 short-circuit 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
Coupling and Bypass Capacitor Design Common-Source Amplifiers For the C-S 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
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.
Capacitively Coupled CS Amplifier Find the pole frequency associated with each coupling/bypass capacitor:
Low Frequency Response Figure 0.7 Sketch of the low-frequency magnitude response of a CS amplifier for which the three pole frequencies are sufficiently separated for their effects to appear distinct.
High Frequency Response
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
(Simplified) High-Frequency Equivalent-Circuit Model for MOSFET Capacitances between source/body, C sb, and between drain/body, C db, are neglected
Intrinsic Response of FET: Unity-Gain Frequency, f T f T : defined as frequency at which short-circuit current gain = f T : a figure-of-merit 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 (short-circuit 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
Common-Source Voltage Amplifier Small-signal 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 2-port 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
CS Voltage Amp Small-Signal Model Two Nodes! Easy For now we will ignore C db to simplify the math
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 Low-frequency 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
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!
Method: The Miller Effect
The Miller Effect
Using The Miller Effect Effective input capacitance: C in = jωc Miller = A v,cgd jωc gd = jω ( A vcgd )C gd
CS Voltage Amp Small-Signal Model Modified Small-Signal 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
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 gain-bandwidth tradeoff
Common Drain Amplifier Calculate Bandwidth of the Common Drain (Source- Follower) Procedure:. Replace current source with MOSFET-based current mirror 2. Draw small-signal model with capacitors (for simplicity, we will focus on C gd and C gs ) 3. Find the DC small-signal gain 4. Use the Miller effect to calculate the input capacitance 5. Calculate the dominant pole
Small-Signal 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!
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
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 ))
Bandwidth of Source Follower Input low-pass 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
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!