Amplifiers
Amplifier circuits
Transistors used as switches
Electronics – The application of bipolar transistors
Prof. Márta Rencz, Gergely Nagy
BME DED
October 1, 2012
Amplifiers
Amplifier circuits
Transistors used as switches
Ideal voltage amplifier
On the previous lesson the theoretical methods of
amplification using bipolar transistor were overviewed.
Before we can start to analyse amplifiers, we need to
understand a few basic concepts.
An ideal voltage amplifier:
has no input current,
has an output voltage that is
proportional to its input
voltage:
vo = A · vi ,
can be modelled with a
voltage controlled voltage
source.
Amplifiers
Amplifier circuits
Transistors used as switches
Non-ideal voltage amplifiers I.
A non-ideal voltage amplifier:
has an input current that is proportional to its input voltage,
has an output voltage that is proportional to its load.
This non-idealities can be modelled with resistors:
Input resistor: Ri ,
Output resistor: Ro .
The gain (A) is called the nominal gain.
Amplifiers
Amplifier circuits
Transistors used as switches
Non-ideal voltage amplifiers II.
Amplifiers connected in series can be calculated using the
extensions seen on the previous slide.
Amplifier are usually built up of several stages.
In the circuit below:
the amplifier is driven by a voltage source vs that has an
internal resistance (RS ), that can be seen as the output
resistance of the previous stage.
The amplifier has an input (Ri ) and an output resistance (Ro )
and is loaded with RL .
Amplifiers
Amplifier circuits
Transistors used as switches
Calculating the gain
The voltage of source is divided by RS and Ri .
The input voltage of the amplfier: vi =
Ri
RS +Ri
· vs
The amplified voltage is also divided by two resistors
L
L
(Ro and RL ): vo = RoR+R
· A · vi = A · RoR+R
· RSRi
+Ri · vs
L
L
Thus the real gain is always smaller than the nominal value.
If Ri RS and Ro RL then the real gain is very close to
nominal gain.
Amplifiers
Amplifier circuits
Transistors used as switches
The Bode plot I.
The Bode plot is widely used in control theory.
It can be used to depict the transfer function of systems.
The transfer function: A(ω) = Vout /Vin .
The Bode plot consists of two plots: the absolute value
and the phase shift of the transfer function.
If the input signal is: Vin = V1 · cos(ωt) and the output is
Vout = V2 · cos(ωt + ϕ) then the transfer function’s
absolute value is: V2 /V1 ,
phase shift is: ϕ.
The amplitude and the phase shift are depicted as a function
of the frequency:
the amplitude is usually plotted on a log-log scale,
the phase shift is usually plotted on a log-lin scale,
the frequency is on a logarithmic scale in both cases.
Amplifiers
Amplifier circuits
Transistors used as switches
Bode plot II.
The amplitude’s unit is dB (decibel).
If the quantity in decibels is voltage or current, then
A|dB = 20 · lg(A),
if it is power, then
A|dB = 10 · lg(A).
The reason for 20 as a coefficient is that in linear systems:
P ∼ V 2 or P ∼ I 2 .
Amplifiers
Amplifier circuits
Transistors used as switches
Bode plot III.
The bode plot of an operational amplifier:
The nominal gain (AN ):
100 dB=100000
The cutoff frequency (fC ):
≈ 30 Hz
GBW: ≈ 3 MHz
Important points of the Bode plot:
Cutoff or break frequency (fC ): where√the gain goes below
the nominal gain minus 3 dB ( A|3 dB = 2 · AN ≈ 0.7 · AN ).
Transit frequency (fT or GBW ): the point where the
absolute value of the gain becomes 1.
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier I.
VCC = 12 V
RB1 = 47 k, RB2 = 5.1 k
RE = 470, RC = 4.7 k
CB = 22µ, CE = 470 µ, CC = 22 µ
The calculation:
1
2
first the DC operating point has to be found,
in order to determine the gain and the input and output
resistances, the small-signal model of the circuit is used.
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier II. – The operating point
The operating point is found using an approximation.
The input characteristic equation is approximated with a
DC voltage source (the same as with diodes). The voltage is:
VBE ≈ 0.7 V
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier III. – The operating point
The output characteristic equation is approximated as
follows:
in the normal active region (VCE > VCES ):
IC = B · IB
in saturation: VCE = VCES
where VCES is the saturation voltage – VCES = 0.1 − 0.3 V.
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier IV. – The OP calculation I.
The device is assumed to be operating in the normal active
region.
The base-emitter junction is substituted with a supply voltage
0.7 V.
2 One of the transistor’s currents is determined (the others can
be expressed using B and the KCL).
3 VCE is calculated.
4 If VCE > VCES the assumption about the operating mode was
correct, which means that the OP calculation is done.
1
If the device is in saturation:
1
2
The value of IB is correct but VCE = VCES .
In saturation IC 6= B · IB , IC has to be calculated using
VCE = VCES .
In amplifiers the transistor operates in the normal active
region, in switching application it operates in saturation when
switched on.
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier V. – The OP calculation II.
In DC capacitors are susbstitued
with an open circuit.
The transistor is assumed to be
operating in the normal active
region.
If the base current is neglected, the
base potential is simply the supply
voltage divided by the RB1 and
RB2 :
VB = VCC ·
RB2
= 1.17 V
RB1 + RB2
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier VI. – The OP calculation III.
The emitter’s potential is equal to
that of the base minus the forward
voltage of the B-E junction (0.7 V),
thus the emitter current is:
IE =
VB − VBE
= 1 mA
RE
If the base current is neglected
then IC = IE = 1 mA.
VCE can be determined by writing
the KVL for the VCC -RC -BJT-RE
loop:
VCE = VCC −IC ·RC −IE ·RE = 6.8 V
VCE = 6.8 V > VCES = 0.1 V
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier VII. – AC calculation I.
At the input signal’s frequency the capacitors can be treated
as short circuits (their impedance is very small – the effective
1
, thus it is inversely proportional
resistance of a capacitor is ωC
to the frequency of the signal).
The small-signal model is created according to the following
rules:
1
2
3
The DC supply voltages are substituted with short circuits.
Capacitors are also substitued with short circuits.
Non-linear devices are substitued with their small-signal
models.
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier VIII. – AC calculation II.
The parameters of the small-signal model:
β = B = 150, re = VIET = 26
1 = 26 Ω (VT = 26 mV)
The input and output voltages should be expressed with ib :
vin = (β + 1) · re · ib
vout = −β · ib · RC where the negative sign is due to the
opposite direction of the current and the output voltage.
The gain can be expressed using the two equations above:
β
= β+1
· RreC = −180
A = vvout
in
Amplifiers
Amplifier circuits
Transistors used as switches
Common-emitter amplifier IX. – AC calculation III.
The input resistance: three resistors in parallel can be seen
from the input
rin = RB1 × RB2 × (β + 1) · re = 2.1 kΩ
The output resistance: only RC can be seen form the
output as the current source is an open circuit
rout = RC = 4.7 kΩ
Amplifiers
Amplifier circuits
Transistors used as switches
The emitter follower I.
The output resistance of the common-emitter amplifier is a
large value, while in an ideal amplifier it is a very small value
(so that most of the output voltage is dropped on the load
resistor).
The circuit below, which is also called analog buffer, copies
its input to its output (it gets shifted by a forward voltage)
but its output resistance is much smaller.
VB is the DC offset of the input signal.
VE = VB − VBE = VB − 0.7
IE = (VB − VBE ) /RE
VCE = VCC − IE · RE
Amplifiers
Amplifier circuits
Transistors used as switches
The emitter follower II.
The current of the RE resistor is iRE = ib + β · ib = (β + 1) · ib
The gain can be calculated by expressing vin and vout with ib :
(β+1)·RE
E
= (β+1)·(r
= reR
A = vvout
+RE ≈ 1
e +RE )
in
The input resistance similarly:
in
rin = viin
= (β+1)·(rieb+RE )·ib = (β + 1) · (re + RE )
The output resistance (calculation not given):
RG
rout = re + β+1
where RG is the resistance of the driving
stage.
Thus this circuit divides the input resistance by β.
Amplifiers
Amplifier circuits
Transistors used as switches
The emitter follower III. – An example
VCC = 12 V
RB1 = 47 k, RB2 = 5.1 k
RE1 = 470, RC = 4.7 k
CB = 22µ, CC = 22 µ
RE2 = 3.3 k, CE2 = 22 µ
A common-emitter amplifier is amended by an emitter
follower stage.
The OP of Q2’s base potential:
VB2 = VCC − IC1 · RC1 = 12 − 4.7 = 7.3 V
The emitter current can now be calculated:
BE
IE2 = VB2R−V
= 2 mA
E2
The output resistance:
u
C1
rout = igg = re2 + R
β+1 ≈ 44 Ω
Amplifiers
Amplifier circuits
Transistors used as switches
The switching operating mode
The transistor switches the current of the load represented by
RC and is controlled by its base potential.
If the input is 0 V, there is no base current, hence the current
of the load is zero as well.
If the input voltage is above VBE0 , then:
IB = (VCC − V BE) /RB .
If the resistances are appropriate, the transistor gets into
saturation: VCE = VCES .
The current of the load: IC = (VCC − VCES ) /RC
Amplifiers
Amplifier circuits
Transistors used as switches
The switching operating mode – Example I.
VCC = 5 V
VCES = 0.2 V
VBE = 0.7 V
If Vin = 0 then IB = 0.
If Vin = 5 then
BE
IB = VinR−V
= 5−0.7
10 = 4.3 mA
B
VCE has to be calculated to determine the operating mode of
the transistor:
VCE = VCC − RC · IC = −640 V < VCES thus the transistor
saturates.
Vout = VCE = VCES = 0.2 V
−VCES
IC = VCCR
= 5−0.2
= 4.8 mA
1
C
Amplifiers
Amplifier circuits
Transistors used as switches
The switching operating mode – Example II.
In this example an load of 1 kΩ is switched on and off.
It can also be seen as the input being inverted: this circuit is
also a logic inverter.