Frequency Response
Impact of Coupling Capacitors (1) v R = i i + + ω v R R 1 /( j C ) sig i sig c 1 v R 1 = × i i + − ω + v R R 1 j / [( R R ) C ] sig i sig i sig c 1 v R 1 = × i i + − ω ω v R R 1 j / sig i sig p 1 1 ω = + p 1 R R C ( ) i sig c 1 High Pass filter with pole at ω p 1 F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (31/42)
Impact of Coupling Capacitors (2) v R = o L + + ω A v R R 1 /( j C ) v 0 i L o c 2 v R 1 = × × o L A + − ω ω v 0 v R R 1 j / i L o p 2 1 ω = + p 2 ( R R ) C L o c 2 High Pass filter with pole at ω p 2 F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (32/42)
Impact of Coupling Capacitors (3) Mid-band gain (caps are short): v R R = = × × o i L A A + + v v 0 v R R R R sig i sig L o Low-frequency Response (included): v R 1 = × i i + − ω ω v R R j 1 / sig i sig p 1 v R 1 = × × o L A + − ω ω v 0 v R R 1 j / i L o p 2 v R R 1 1 = × × × × o i L A + + − ω ω − ω ω v 0 v R R R R 1 j / 1 j / sig i sig L o p p 1 2 v 1 1 Each coupling capacitor = × × o A − ω ω − ω ω v v 1 j / 1 j / introduced a pole sig p 1 p 2 F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (33/42)
Each Capacitor introduces a pole Each capacitor introduces a pole: 1) Coupling capacitor at the input: 1 = π f + p 1 2 ( R R ) C i sig c 1 2) Coupling capacitor at the output: 1 = π f + p 2 2 ( R R ) C o L c 2 3) By-pass capacitor (CE wit R E , CS with R S , CB)* 1 = f π p 3 2 R C − − by pass by pass Exact calculation of the lower cut-off frequency can only be done numerically. A surprisingly good approximation is ≈ + + f f f ... p p 1 p 2 * Formulas for R by-pass are given in the Amp. summary sheets. F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (34/42)
Multi-Stage Amplifiers
Gain of a multi-stage amplifier Example: A 3-stage amplifier R = R o o , 3 v R v = = = v = = = i , 1 i , 1 i R R v , v v v v + i i , 1 o o , 3 o , 1 i , 2 o , 2 i , 3 v v R R sig sig i , 1 sig v v v v v = = × × o , 3 o , 1 o , 2 o , 3 o v R v v v v v = × × × × o i i i i o o , 1 , 1 , 1 , 1 , 2 A A A ... + v 1 v 2 v 3 v R R sig i sig = = v v v v o 1 i 2 o 2 i 3 But we need to know R L, 1 , R L, 2 , v v v v = × × = × × o , 1 o , 2 o , 3 o R L, 3 , … in order to find A M s. A A A v 1 v 2 v 3 v v v v i i i i , 1 , 1 , 2 , 3 F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (36/42)
What are R L and R sig for each stage? Amp Model for Stage j-1 Amp Model for Stage j+1 = R R − sig , j o , j 1 = R R + L , j i , j 1 R L for each stage is the input resistance of the following stage. R sig for each stage is the output resistance of the previous stage. F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (37/42)
Procedure for Solving multi-stage Amplifiers Gain & R i : 1.Start from the load side (n th stage), o Find the gain A v,n = ( v o /v i ) n and R i,n . 2.For (n-1) th stage, set R L,n -1 = R i,n o Find the gain A v,n- 1 = ( v o /v i ) n- 1 and R i ,n- 1 . 3. Repeat until reaching to the first stage. Then, v R = = × × × × o i R R A A A ... + i i , 1 v 1 v 2 v 3 v R R sig i sig R o : 1. Start from the source side (1 st stage). Find R o, 1 . 2. Go to the second stage. Set R sig, 2 = R o, 1 . Find R o, 2 3.Continue to the last stage (n th stage). Then, R o = R o,n F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (38/42)
What is “Buffer” Mid-band gain: v R R = × × o i L A + + v 0 v R R R R sig i sig L o Sever loss in gain if R L is small compared to R o o Main gain cells (CE or CS amps) have an R o of several to 10s of k Ω . This can be resolves by using a “buffer”: o Ideal Buffer: A vo = 1 , R i = ∞ and R o = 0. o Practical Buffer: A vo ≈ 1 , large R i (several to 10s of M Ω ) and small R o (10s or 100s of Ω ). o CD amplifier (source follower) and CC amplifiers (emitter follower) are examples of practical buffers * There is also loss in gain if R i is small compared to R sig . Buffer circuits can also solve this issues F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (39/42)
Buffer circuit removes the impact of R L = A 1 vo , 2 = R 0 o , 2 = ∞ R i , 2 v R R v = × = × o L , 1 , 1 L , 2 o A A + + vo , 1 vo , 2 v R R v R R i , 1 L , 1 o , 1 i L o , 2 , 2 , 2 v R v R = × = = × = o , 1 i , 2 o L A A 1 1 + + vo , 1 vo , 1 v R v R R 0 i , 2 L i , 1 i , 2 o , 1 v v v v = × × o , 1 o i o v v v v sig sig i , 1 i , 2 v R = × × o i With a buffer : A 1 + vo , 1 Buffer circuit has removed v R R sig i sig the dependence on R o & R L . v R R = × × Without a buffer : o i L A + + v 0 v R R R R sig i sig L o F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (40/42)
Buffer circuit can also be used to remove the impact of low R i = A 1 vo , 1 = R 0 o , 1 = ∞ R i , 1 v R R v = × = × o L , 1 , 1 L , 2 o A A + + v 0 , 1 vo , 2 v R R R v R R v = = i , 1 i , 1 L , 1 o , 1 i L o , 2 , 2 i 1 + v R R v R v R = × = = × sig i , 1 sig o , 1 L , 1 o L A 1 1 + + vo , 2 v R R v R 0 i , 2 L o i , 1 L , 1 v v v v = × × o , 1 o i o v v v v sig sig i , 1 i , 2 v R = × o L With a buffer : A + vo , 2 Buffer circuit has removed the v R R sig L o dependence on R i & R sig . v R R = × × Without a buffer : o i L A + + vo v R R R R sig i sig L o F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (41/42)
Cut-off frequency of a multi-stage amplifier Similar to a single-stage amplifier, each capacitor introduces a pole: 1) Coupling capacitor at the input: 1 = π f + p 1 2 ( R R ) C i sig c 1 2) Coupling capacitor at the output: 1 = π f + po 2 ( R R ) C o L co The pole due to the C cj , coupling 3) Coupling capacitor between stages j-1 and j capacitor between stages j-1 and j, can be found by considering as the 1 = π f input capacitor to stage j (using + pj 2 ( R R ) C − i j o j cj , , 1 formula from part 1) noting that R sig, j = R o, j-1 . 4) By-pass capacitors for stage j (if exists) 1 = f π p , bypass 2 R C − − by pass by pass ≈ + + 5) Then: f f f ... p p 1 p 2 F. Najmabadi, ECE65, Winter 2013, Discrete Amplifiers (42/42)
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