Transistor amplifier
npn
Common emitter, amplifier with negative feedback
Voltage divider
High pass
High pass
Increased amplification gives larger IC which gives inreased voltage drop over RE
i.e VBE is reduced, which counteracts the amplification increase, i.e. stabilizes the
Amplifier. This is called negative feedback
Fysn15 electronics
Working point
Fysn15 electronics
Amplifier with
transistor
UKE
Fysn15 electronics
Model of diode
Fysn15 electronics
FET model
Large signal model
Small signal model
Fysn15 electronics
npn model
Large signal model
Small signal model
Fysn15 electronics
Amplifiers, general
Signal
source
amplifier
mic
Amplification
A(ω)=vut(ω)/vin(ω)
load
Loud
speaker
Matching an amplifier to its peripherals
Optimal power transfer
Optimal voltage transfer
Optimal current transfer
Fysn15 electronics
Amplifier model
Controlled
Voltage source
input
output
Best power transfer if Zth=Z*in
Best voltage transfer if Zin is high
Best current transfer if Zin is low
Fysn15 electronics
Amplifyer types
Type
Input Output Zin
Zout
Voltage amp
U
U
large
Small
Transadmittance
U
I
large
large
Transimpedance
I
U
small
small
Current amp
I
I
small
large
Fysn15 electronics
Feedback and gain
Amplifier with
raw gain A
Feedback net
with fraction
-β
The negative feedback, damps the input. This stabilizes
the output to a certain value determined by the feedback
fraction –β
xe= xi - βy
y=AXe
}
So the gain Af = y/xi=A/(1+ βA)
A: open loop gain (very large)
Af: closed loop gain
T= βA: loop gain
F= 1+ βA : return difference
For A→ ∞
Af∞ =1/β
Fysn15 electronics
Golden rules of OPamp with feedback
In order to understand how an operational amplifier works one should remember
-The OP-amplifier produces an an output which, via the feedback
to the negative input tries to minimize the voltage difference
between the inputs.
-Very little current goes into the inputs. In reality one can assume
That no current flows into the inputs.
Fysn15 electronics
Voltage amplifier
voltage output
Can be the thevenin equivalent of a two-terminal circuit
The voltage amplier shall have very high
Input impedance (ideally infnite) – i.e very
little current (ideally 0) should be drawn
from the external circuit whose terminal
Voltage we should amplify. This gives
low voltage drop over Rs.
The output voltage should be the same
Irrepesctive of the load RL.
Low output resistance (ideally 0)
gives small voltage drop for any output current
Ve=V+ - V- .
If V+ > V- VL becomes positive so that the feedback makes V- increase which makes Ve smaller.
If V+ < V- VL becomes negative so that the feedback makes V- decrease which makes Ve smaller
The smaller fraction of VL that the voltage divider returns in the feedback network, the higher output
Voltage (higher gain) has to be produced to cancel the difference between V+ och V-
Feedback: voltage to voltage
Fysn15 electronics
Calculate input resistance.
Find the Theveninekvivalent of the OP-amp.
The input resistance is the resistance seen
between the terminal of the Thev. Equi.
So Rinp =Rth for the equivlent circuit
In the method from lecture 1 we should
Replace all independent voltage sources
with short circuit and Rth could be found
as the resultant resistance of all resistors
in the circuit.
But this circuit has a dependent circuit which can not be replaced with a short circuit.
An alternative method is to apply a test voltage to the input and measure (calculate)
Hat current runs into the circuit at the actual voltage. The ratio U/I gives the resistance that
the external circuit experiences. So this is an alternative way of calculating Rth which works if the
Circuit is lienear.
Fysn15 electronics
Calculate the output resistance
The same method is used. Apply a test
voltage between the output terminals.
The independent voltage source has
been replaced by a short circuit. In
practice this means that we have chosen
0V test voltage. If the circuit is linear,
this voltaage is as representative as any
other voltage.
Fysn15 electronics
Current – voltage amplifier
Low inp resistans, low output resistans, called transimpedanse amplifier
Since the OP itself has high input
resistans the only way to obtain low
input resistans is to have a low
resistor parallel with rin.
For the low output resistansen we
Need as before resistor parallel with
som tidigare ha resistor parallel rut.
VL
Function: with the input current
In the direction of the arrow
V- is higher than V+. Then
VL responds in the negative
(inverting) direction and the current in the
feedback is directed towards the output.
Thus, the current into the OP-amp is reduced and Ve is decreased. We have negative feedback.
The higher R, The higher output voltage needed to level out Op’n V- and V+.
Feedback: voltage to current
Fysn15 electronics
Voltage to current amplifier
High input resistans, high output resistans, transadmittanse amplifier
Function: VL is determined by the nonInverting amplification of Ve . If
V+ > V- the VL goes positive.
Then IL increaseses and it goes mainly
through RL and R to ground
(since the current into V- negligible) .
V1 increases which decraeses Ve, i.e
negative feedback. Small R means
High gain to level out V+ och V-
IL V L
V1
RL och R constitute a voltage divider. We can not do the voltage divider as in the voltage-voltage amplifier
case since then the output current had gone through this voltage divider instead and not through the load.
Feedback: current - voltage
Fysn15 electronics
Current to current
amplifier
Low input resistanse, high output resistanse, transadmittanse amplifier
Function: with the input current
In the direction of the arrow
V- is higher than V+. Then
VL responds in the negative
(inverting) direction and the current in the
feedback is directed towards the output.
How much is fed back is determined by the ratio between R1 and R2.
Thus, the current into the OP-amp is reduced and Ve is decreased. We have negative feedback.
The higher R, The higher output voltage needed to level out Op’n V- and V+.
Fysn15 electronics