EECS 105 Spring 2017, Module 4 Frequency Response Prof. Ali M. Niknejad Department of EECS
Announcements l HW9 due on Friday 2
Review: CD with Current Mirror 3
Review: CD with Current Mirror 4
Review: CD with Current Mirror 5
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 6
Common-Source Voltage Amplifier Small-signal model: C sb is connected to gnd 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 (1-2 nodes) 7 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 8
Frequency Response KCL at input and output nodes; analysis is made complicated V out = g m r o R L V in 1+ jω /ω p1 [ ]( 1 jω /ω z ) ( )( 1+ jω /ω p2 ) Zero Low-frequency gain: ( ) ( )( 1+ j0) V out = g r R m o L 1 j0 V in 1+ j0 Two Poles g m r o R L ü Zero: ω z = g m C gs + C gd 9
Calculating the Poles ω p1 1 R s C gs + ( 1+ g m R out )C gd { } + R out C gd ω p2 R out / R S R S C gs + ( 1+ g m R out )C gd { } + R out C gd Usually >> 1 Results of complete analysis: not exact and little insight 10 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 11
The Miller Effect 12
Using The Miller Effect Effective input capacitance: C in = 1 jωc Miller = 1 1 A v,cgd 1 jωc gd = 1 jω ( 1 A vcgd )C gd 13
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 14
Comparison with Exact Analysis Miller result (calculate RC time constant of input pole): Exact result: ω p1 1 = R S C gs + ( 1+ g m R out ʹ )C gd ω p1 1 = R S C gs + ( 1+ 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 15
Common Drain Amplifier Calculate Bandwidth of the Common Drain (Source- Follower) 16 Procedure: 1. 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
Two-Port CC Model with Capacitors R S 17 -Find DC Gain -Find Miller capacitor for C gs -- note that the gate-source 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 ) 1 v out + g m r o r = g v m in oc v out v in = r o g m 1 + g r m oc = g m (r o r oc ) 1+ g m (r o r oc ) = A vcgs 18
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 + (1 A vcgs )C gs Miller Cap C in = C gd + (1 g m (r o r oc ) 1+ g m (r o r oc ) )C gs 1 C in = C gd + ( 1+ g m (r o r oc ) )C gs 19 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 1 p = R S C gd + gs 1+ g m (r o r oc ) Substitute favorable values of R S, r o : R 1/ S g m r o >>1/ g m ω p 1 1/ g m ( ) C gd + C gs C gd / g m 1+ BIG Very high frequency! Model not valid at these high frequencies ω p g m / C gd 20
Some Examples Common source amplifier: A vcgd = Negative, large number (-100) C Miller = (1 A V,Cgd )C gd 100C gd Miller Multiplied Cap has detrimental impact on bandwidth Common drain amplifier: A vcgs = Slightly less than 1 C Miller = (1 A V,Cgs )C gs! 0 Bootstrapped cap has negligible impact on bandwidth! 21