Amplifiers principles,
frequency effects
&
some other parameters
Jerzy S. Witkowski
Response of an AC amp
AV [V/V]
voltage amplification - gain
AVmid
0.707AVmid
f  [Hz]
f1 10f1 0.1f2 f2
frequency
2
Low/Mid/High frequency
AV [V/V]
AVmid
0.707AVmid
Midband
f  [Hz]
f1 10f1 0.1f2 f2
frequency
Below Midband
Above Midband
AV H" AVmid H" const
1
Low/Mid/High frequency
AV [V/V]
AVmid
0.707AVmid
f  [Hz]
frequency
f1 10f1 0.1f2 f2
Below Midband
AVmid Midband
Above Midband
AV H"
2
f1
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
f
íÅ‚ Å‚Å‚
Low/Mid/High frequency
AV [V/V]
AVmid
0.707AVmid
f  [Hz]
frequency
f1 10f1 0.1f2 f2 Above Midband
AVmid
Below Midband Midband
AV H"
2
f
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
f2
íÅ‚ Å‚Å‚
Low/Mid/High frequency
AVmid
AVmid
AV H"
Ô!
2 2
ëÅ‚1+ 1 öÅ‚
f1 f
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
(1+ R2C2s)
1+ 1+
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
f f2 ìÅ‚ R1C1s ÷Å‚
íÅ‚ Å‚Å‚
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
Assumption:
One dominant capacitor is producing lower cutoff frequency
One dominant capacitor is producing high cutoff frequency
- first order poles
2
DC amplifier
AV [V/V]
AVmid
0.707AVmid
f  [Hz]
frequency
0.1f2 f2
Bandwith High frequecies
DC amplifier
AVmid
AV H"
AVmid
2
Ô!
f
ëÅ‚ öÅ‚
(1+ R2C2s)
1+
ìÅ‚ ÷Å‚
f2
íÅ‚ Å‚Å‚
Assumption:
One dominant capacitor is producing high cutoff frequency
Decibel power gain
review of logarithms
x = 10y Ô! y = log10(x) = log(x)
log(1) = 0
log(1) = log(100 ) = 0
log(0.1) = -1
log(10) = 1
log(0.01) = -2
log(100) = 2
log(0.001) = -3
log(1000) = 3
log(ab) = log(a) + log(b)
a
log( ) = log(a) - log(b)
b
9
3
Decibel power gain - Ap(dB)
power gain:
Pout
AP =
Pin
power gain in dB:
ëÅ‚ öÅ‚
Pout
ìÅ‚ ÷Å‚
AP(dB) = 10logìÅ‚ ÷Å‚ = 10log(AP )
Pin
íÅ‚ Å‚Å‚
Decibel power gain
 important numbers
AP (factor) AP(dB)
=10*10 100 20
=10*2*2 40 +16
=10*2 20 +13
10 10 +10
=2*2 4 +6
2 2 +3
1/2 0.5 -3
=(1/2) * (1/2) 0.25 -6
=(1/10) 0.1 -10
=(1/2)*(1/10) 0.05 -13
=(1/2)*(1/2)*(1/10) 0.025 -16
=(1/10)*(1/10) 0.01 -20
Decibel power gain
 important numbers
AP (factor) AP(dB)
10 10 +10
amplification
2 2 +3
1/2 0.5 -3
attenuation
1/10 0.1 -10
4
Decibel voltage gain
voltage gain:
Uout
AV =
Uin
voltage gain in dB:
ëÅ‚ öÅ‚
ìÅ‚Uout ÷Å‚
AP(dB) = 20logìÅ‚ ÷Å‚ = 20log(AV )
Uin
íÅ‚ Å‚Å‚
Decibel voltage gain
 important numbers
AV (factor) AV(dB)
=10*10 100 40
10 10 +20
=2*2 4 +12
2 2 +6
=1.41 "2 +3
=0.707 1/"2 -3
1/2 0.5 -6
=(1/2) * (1/2) 0.25 -6
=(1/10) 0.1 -20
=(1/10)*(1/10) 0.01 -40
Decibel voltage gain
 important numbers
AV (factor) AV(dB)
10 10 +20
2 2 +6
=1.41 "2 +3
=0.707 1/"2 -3
1/2 0.5 -6
1/10 0.1 -20
5
Response of an AC amp
AP [W/W] AV [V/V]
Avmid APmid
APmid AVmid
½ APmid 0.707AVmid
f1 10f1 0.1f2 f2 f  [Hz]
frequency
Response of an AC amp
AP [dB] AV [dB]
Avmid APmid
Apmid AVmid
Apmid-3dB AVmid-3dB
f1 10f1 0.1f2 f2 f  [Hz]
frequency
TIP :!!!!!!!
-3dB= power decreases twice and voltage "2 times
Cutoff frequencies
Cutoff frequencies
Cutoff frequencies
TIP :!!!!!!!
-3dB= power decreases twice
and voltage decreases"2H"0.707 times
6
[dB] vs.[V/V]
Vin AV1 AV2 Vout
Uout
AV = = AV1 Å" AV 2
Uin
AV (dB) = AV1(dB) + AV 2(dB)
[dB] vs.[V/V]
an example
Vout
Vin
26dB 40dB
AV (dB) = AV 1(dB) + AV 2(dB) = 26 + 40 = 66dB
66dB = 6dB + 20dB + 40dB
66dB ×2×10×100 = 2000[V /V ]
Decibels above a REFERENCE
dBm, dBµ, dBV& ..
P[W ]
P[dBm] = 10logëÅ‚ öÅ‚
ìÅ‚ ÷Å‚
1mW
íÅ‚ Å‚Å‚
ëÅ‚U[V ] öÅ‚
U[dBµ] = 20logìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚
1µV
íÅ‚ Å‚Å‚
U[V ]
U[dBV ] = 20logëÅ‚ öÅ‚
ìÅ‚ ÷Å‚
1V
íÅ‚ Å‚Å‚
7
dBm an exampl
10dBm •" 23dBm =
= 10mW + 200mW = 210mW = 23.2dBm
13dB
(x20W/W)
(x4.47V/V)
Pin = 10dBm
Pout= 23dBm = 10dBm+13dB
(10 mW)
(200mW)
Summary
" cutoff frequecies ?
" dB for voltage and power gains
" dB vs. factors (3,6,10,20 dB)
" dBm, dBu, dBV ?
Bode plots
8
Bode plots
-logarithmic scales
AV [V/V]
AV [dB]
voltage amplification - gain
AVmid
0.707AVmid
f  [Hz]
f1 10f1 0.1f2 f2
frequency
0
0.1 10 100
1
decades
Bode plot
AV [dB]
AVmid
AVmid
AVmid-3dB
f  [Hz]
f1 10f1 0.1f2 f2
frequency
-20dB/decade
20dB/decade
Bode plot  low f
AV [dB]
AVmid
AVmid
AVmid-3dB
f  [Hz]
20dB/decade f1 10f1 0.1f2 f2
frequency
fb fa
-20dB/decade
9
Bode plot
 20dB/decade ???
AVmid
AV H"
2
f1
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
f
íÅ‚ Å‚Å‚
2
ëÅ‚ f1 öÅ‚
1+
ìÅ‚ ÷Å‚
fa fa2 + f12 fb 2
AV ( fa ) íÅ‚ Å‚Å‚
AV (dB) / decade = 20log = 20log = 20log
2
2
AV ( fb)
f1 fb + f12 fa2
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
fb
íÅ‚ Å‚Å‚
fb
1decade = = 10
fa
fa , fb << f1
fa 2 + f12 fb 2 fb
AV (dB) / decade = 20log H" 20log = 20log(0.1) = +20[dB / dec]
fb2 + f12 fa 2 fa
Bode plot  high f
AV [dB]
AVmid
AVmid
AVmid-3dB
f  [Hz]
20dB/dec f1 10f1 0.1f2 f2
frequency
fa fb
-20dB/dec
Bode plot
 20dB/decade ???
AVmid
AV H"
2
ëÅ‚ f öÅ‚
1+
ìÅ‚ ÷Å‚
f2
íÅ‚ Å‚Å‚
2
fa
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
f2 fa2 + f22
AV ( fa ) íÅ‚ Å‚Å‚
AV (dB) / decade = 20log = 20log = 20log
2
AV ( fb )
fb fb2 + f22
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
f2
íÅ‚ Å‚Å‚
fa
1decade = = 0.1
fb
fa , fb >> f2
fa 2 + f22 fa
AV (dB) / decade = 20 log H" 20 log = 20 log(0.1) = -20[dB / dec]
fb2 + f22 fb
10
Bode plot
+ simplified frequency response
AV [dB]
AVmid
AVmid
AVmid-3dB
f  [Hz]
20dB/decade
frequency
f1 f2
-20dB/decade
20dB/dec = 6dB/oct
fa 2 + f22 fa
AV (dB) / decade = 20log H" 20log = 20log(0.1) = -20[dB / dec]
fb2 + f22 fb
fa 2 + f22 fa 1
AV (dB ) / oct = 20 log H" 20 log = 20 log( ) H" -6[dB / oct]
fb 2 + f22 fb 2
1
ëÅ‚1+ 1 öÅ‚
f1=100Hz ; f2=100kHz
(1+
ìÅ‚
R1C1s÷Å‚ R2C2s)
íÅ‚ Å‚Å‚
0
-10
-20
-30
-40
-50
-60
0,1 1 10 100 1e3 1e4 1e5 1e6 1e7 1e8
90
45
0
-45
-90
0,1 1 10 1e2 1e3 1e4 1e5 1e6 1e7 1e8
11
Bode plot
op amp example
20dB
decade
20dB/octave slope derivation
AVmid
AV H"
2
f1
ëÅ‚ öÅ‚
1+
ìÅ‚ ÷Å‚
f
íÅ‚ Å‚Å‚
Uout R 1 1 1 1
AV = = = = = = =
Uin 1 1 1 1+ É1 1+ f1
R + 1+
jÉC jÉCR RC
1+ jÉ jf
jÉ
ëÅ‚ f1 öÅ‚
actg ìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚
1 1 ëÅ‚ f1 öÅ‚
f
íÅ‚ Å‚Å‚
= e = ;Õ = artgìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚
2 2
f
f1 f1 íÅ‚ Å‚Å‚
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
1+ 1+
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
f f
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
20dB/octave slope derivation
AVmid
AV H"
1
2
f2 =
ëÅ‚ f öÅ‚
1+
ìÅ‚ ÷Å‚ 2Ä„RC
f2
íÅ‚ Å‚Å‚
1
Uout jÉC 1 1 1 1
AV = = = = = = =
1 jÉ f
Uin R + 1 1+ jÉCR
1+ 1+ j
1+ jÉ
jÉC
É2 f2
1
RC
ëÅ‚ f öÅ‚
-actg ìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚
1 1 ëÅ‚ öÅ‚
f
f2
íÅ‚ Å‚Å‚
= e = ;Õ = -artgìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚
2 2
f2
íÅ‚ Å‚Å‚
ëÅ‚ f öÅ‚ ëÅ‚ f öÅ‚
1+ 1+
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
f2 f2
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
12
Bode plots  phase
siplified plot
AVmid
AV [dB]
AVmid
AVmid-3dB
20dB/decade
f  [Hz]
frequency
f1 f2
00 +900
-450 +450
-900
-1350 -450
-1800 -900
Bode plot  phase
- real plot
+900
+450
-450
-900
f1 f2
0.1f1 10f1 0.1f2 10f2
Higher order Bode plots
simplified plot
AV [dB]
AVmid
AVmid-3dB
20dB/dec
40dB/dec
f11 f1
f2 f21 f  [Hz]
+1800
frequency
+900
+450
-450
-900
-1800
13
Slope vs. phase
frequency
Slope
+60 +40 [phone]
[dB/dec]
Phase
+270 +180 [phone]
[deg]
Transfer
~s3 ~s2 ~s const ~1/s ~1/s2 ~1/s3
function
rise time vs. bandwith
t
Uout ëÅ‚ - RC öÅ‚
ìÅ‚
= e
ìÅ‚1- ÷Å‚
÷Å‚
+V
íÅ‚ Å‚Å‚
rise time vs. bandwith
t2
ëÅ‚ - öÅ‚
t2 üÅ‚
RC
ìÅ‚ ÷Å‚
0.9 = e Ò! - = ln(0.1)ôÅ‚
ìÅ‚1- ÷Å‚
RC
íÅ‚ Å‚Å‚ ôÅ‚
żłT = t2 - t1 = RC ln(9) H" 2.2RC
R
t1
ëÅ‚ - öÅ‚
t1
RC
ìÅ‚1- ÷Å‚
0.9 = e Ò! - = ln(0.9)ôÅ‚
ìÅ‚ ÷Å‚ ôÅ‚
RC
íÅ‚ Å‚Å‚ þÅ‚
1
f2 =
2Ä„RC
0.35
TR = = TF
2Ä„
f2 0.35 =
ln(90%10%)
Assumption:
One dominant capacitor is producing high cutoff frequency
14
Simulink analysis - low pass filter
A(0Hz)=-6dB
f2=5.134Hz
Simulink analysis  l.p. result
-6dB
-9dB
-45deg
Fall & Rise time  example
0.35 0.35
TR = TF = = = 68µs
f2 5134
TR TF
15
Simulink analysis - high pass filter
A("Hz)=-6dB
f1=1.283Hz
Simulink analysis  h.p. result
-6dB
-9dB
45deg
A few examples
R1 C R2 f1
Av(f=")
5k1 100n 10k 105Hz -3.6dB
10k 100n 5k1 105Hz -9.5dB
10k 220n 5k1 50Hz -9.5dB
10k 100n 51k 26Hz -1.6dB
16
A few examples
R1 R2 C f2
Av(f=0)
10k 5k1 1n 105Hz -9.5dB
5k1 10k 1n 105Hz -3.6dB
5k1 10k 2n2 53Hz -3.6dB
k51 10k 1n 26Hz -0.4dB
Compensated divider
Uin
Uout
>
Uout
<
Uout =
Bode plot - conclusions
" the absolut value and the phase of transfer
function are corelated
" for circuits with one dominant capacitor (first
order t. function) the gain rise/fall)
(low/high frequency) 20dB/dec =6dB/oct
" for cutoff frequencies (-3dB below maximum
gain) the pase is +/- 45deg
" when the gain rise/fall 20dB/dec the phase is
+/-90deg
" compensated divider- ideal pulse response-
flat frequency response
17
Problems
(Amplifiers principles, Bode plots)
" What is dB ?
" dB (Ä…3,Ä…6,Ä…10,Ä…20) vs. [V/V] and [W/W]
" What is dBµ, dBm, dBV ?
" What is idea of Bode plots ?
" Draw Bode plot for low-/high-pass first
order filter.
" What is the idea of compensated divider ?
Power
amplification
Voltage amplfication
unloaded amp
AV = AV 0
18
Voltage amplfication
loaded amp
Uout Eout Uout RL
AV = = = AV
Uin Uin Eout 0 RL + Rout
TIP:
This AV usually announced in advertisements
Effective voltage amplification
-unmatched circuit
Uout Eout Uout RL
AV = = = AV 0
Uin Uin Eout RL + Rout
def
Uout Uin Eout Uout Rin RL
AVeff = = = AV = Å‚AV
EG EG Uin Eout 1 Rin 04L 44
RG24 R R3
4+ 3142+ out
Å‚ AV
Power amplification
2 2
Uin Uout
Pin = Pout =
Rin RL
2
ëÅ‚ öÅ‚
Pout ìÅ‚Uout ÷Å‚ Rin Rin
2
AP = = = (AV )
Pin ìÅ‚ Uin ÷Å‚ RL RL
íÅ‚ Å‚Å‚
19
Voltage vs. power amplification
in dB case Rout=0; Rin="
Rin=" Rout=0
2
ëÅ‚ öÅ‚
Pout Uout Rin Rin
2
AP = = ìÅ‚ ÷Å‚ = (AV ) = "
Pin ìÅ‚ Uin ÷Å‚ RL RL
íÅ‚ Å‚Å‚
TIP:
Operational amplifier has infinite power gain
-input power is zero !!!!
Power amplification
2 2
Uin Uout
Pin = Pout =
Rin RL
2
ëÅ‚ öÅ‚
Pout ìÅ‚Uout ÷Å‚ Rin Rin
2
AP = = = (AV )
Pin ìÅ‚ Uin ÷Å‚ RL RL
íÅ‚ Å‚Å‚
Voltage vs. power amplification
in dB
2
ëÅ‚ öÅ‚
Pout Uout Rin Rin
2
AP = = ìÅ‚ ÷Å‚ = (AV )
Pin ìÅ‚ Uin ÷Å‚ RL RL
íÅ‚ Å‚Å‚
2
ëÅ‚ öÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
Pout ëÅ‚ëÅ‚Uout Rin öÅ‚ Uout Rin Rin
÷Å‚
AP(dB) = 10 logìÅ‚ ÷Å‚ = 10logìÅ‚ìÅ‚ ÷Å‚ = 20 logìÅ‚ ÷Å‚ +10logìÅ‚ ÷Å‚ = AV (dB) +10 logìÅ‚ ÷Å‚
ìÅ‚ ÷Å‚ ÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
Pin ìÅ‚ìÅ‚ Uin RL ÷Å‚ Uin RL RL
íÅ‚ Å‚Å‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
íÅ‚íÅ‚ Å‚Å‚
ëÅ‚ öÅ‚
Rin
ìÅ‚ ÷Å‚
AP (dB) = AV ( dB) +10 logìÅ‚ ÷Å‚
RL
íÅ‚ Å‚Å‚
20
Avaliable input/output power
--)(maximum) avilable power gain
2
Uout
2
Pout max =
EG
Pin max = 4Rout
4RG
Available power gain= (Available output power)/(Available input power)
2 2
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
Pout max Uout RG 2 RG Rin RL RG
AP-available = = ìÅ‚ ÷Å‚ = (AVeff ) = ìÅ‚ AV 0 ÷Å‚
Pin max ìÅ‚ EG ÷Å‚ Rout Rout ìÅ‚ RG + Rin Rout + RL ÷Å‚ Rout
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
---- for matched circuits
RG = Rin & Rout = RL
Available power gain =
Available output power
=
2 2 2 2
Available input power
EG U Eout U
in out
Pin max = = Pout max = =
4RG Rin 4Rout RL
2 2
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
Pout max Eout RG Uout Rin
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
AP-available = = =
Pin max ìÅ‚ EG ÷Å‚ Rout ìÅ‚ Uin ÷Å‚ RL
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
RG = Rin
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
Rin RG
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
AP -availabla = AP (dB ) = AV (dB ) +10 logìÅ‚ ÷Å‚ = AV (dB ) +10 logìÅ‚ ÷Å‚
RL RL Rout = RL
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
---- for matched circuits
- the most offten case RG=RL=Rin=Rout=R
2 2
Uout Eout
2 2
Pout = = Pout max =
Uin EG
Pin = = Pin max = R 4R
R 4R
Available power gain= Available output power)/(Available input power)
2 2 2
2
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
Pout max Eout Pout Uout 1 Uout 2
ëÅ‚
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
AP-available = = = = = AV 0 öÅ‚ = AV = 4ìÅ‚ ÷Å‚ = 4(AVeff )
ìÅ‚ ÷Å‚
Pin max ìÅ‚ EG ÷Å‚ Pin ìÅ‚ Uin ÷Å‚ íÅ‚ 2 EG Å‚Å‚
Å‚Å‚
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚
AP(dB ) = AP (dB )-available = AV (dB ) = AVeff (dB ) + 6dB
21
---- for matched circuits
another approach
- the most offten case RG=RL=Rin=Rout=R
EG
Uin =
2
2
EG
P1 =
4R
EG 1
ëÅ‚
Uout = AV 0 öÅ‚
ìÅ‚ ÷Å‚
2 2
íÅ‚ Å‚Å‚
2 2
EG AV 0
P2 =
16R
Uout AV 0 P2 ëÅ‚ AV 0 öÅ‚2
AV = = AP = =
ìÅ‚ ÷Å‚
Uin 2 P1 íÅ‚ 2
Å‚Å‚
AP[dB] = AV [dB]
50Ohm system
100 100
10 20
PdB = AVdB = 100dB; AP = 10 = 1010; AV = 10 = 100000
PdB = AVdB = 23dB PdB = AVdB = 36dB PdB = AVdB = 41dB
23 36 41
10 10 10
AP = 10 = 200 AP = 10 = 3981 AP = 10 = 12589
23 26 41
20 20 20
AV = 10 =14.1 = 200 AV = 10 = 63.1 AV = 10 = 112
dBm vs. dBµ
PW
PdBm = 10log
1mW
PdBm 2
U
10
PW = 1mW Å"10 =
R
PdBm
10
U = 1mW Å" R Å"10
PdBm
10 PdBm
1mW Å" R Å"10 1mW Å" R
10
UdBµ = 20log = 20log + 20log 10
1µV 1µV
UdBµ = 10 log(109 R)+ PdBm = 80 +10log(10R)+ PdBm
UdBµ = 107 + PdBm
for R=50&!
22
Problems (power amplification)
" voltage gain vs. effective voltage gain
" Voltage vs. power amplification
in dB
" available power gain ?
" available power gain for matched systems
?
23