7.01  bandwidth
The bandwidth of an amplifier is the range of frequencies within the power gain that does not fall below half of its maximum value, since;
the bandwidth is also the range of frequencies within which the voltage gain does not fall below 1/âˆš2 (0.7) of its maximum value.
The gain of a capacitor coupled with an amplifier decreases at the lower frequencies due to the increasing resistance of the capacitor, and at the upper frequencies due to stray capacitance in the circuit.
A typical voltage gain  frequency curve is shown below, with a LOG scale used on the x axis; to accommodate its large range.
100 _____________
 ..' '.
 ./ `\
80 ./ `\ x = frequency / Hz
70 +./`\ y = voltage gain, power
 ./ \
60 /  ^  \
/    \
  BANDWIDTH   \
40    \
  <>  \
    \
20    
    
  V  
0++++++++
10 10^2 10^3 10^4 10^5 10^6
To estimate the bandwidth of an amplifier, either voltage of power, do:
V/Pmax / (1/âˆš2) ~= 0.7 V/Pmax
Then draw a line on the Y axis parallel to the X axis. Where the 0.7 line intersects with the bandwidth curve, draw two X lines, parallel to the Y axis down, as shown in the above diagram. The frequency between the two X lines that now intersect the X axii origin, is the bandwidth.
7.02  inverting amplifier
Below is a diagram of an inverting amplifier. The power supply connections are omitted, but the circuit it assumed to be operating from a dualrail supply.
Rf
+####+
 
R1 P '\ 
####+\ 
Vin  >+Vout
++/
 ./
OV  0V
+
Since the openloop gain is very large, if the output voltage is less than
the power supply, then the voltage difference between the two input
terminals is very small.
The positive (noninverting) terminal is connected to the 0V rail, and
so the voltage at the negative (inverting) input is virtually 0V, this is
known as a Virtual Earth Point, P.
The input voltage, V_{in}, appears across the resistor, R_{1},
so a current of V_{out}/R_{1} passes through R_{1}.
This therefore means that the input resistance of the circuit is equal to
R_{1}.
It is assumed that because the input impedance (internal resistance) is so
large that no current passes through the input terminals, therefore the only
path for current passing through R_{1} must go through R_{f}
, because the negative input terminal is also a virtual earth point; the
output voltage, V_{out}, also appears across R_{f}. Therefore
a current of V_{out}/R_{f} to pass through R_{f}.
This means that the voltage gain is determined only by the two resistors,
R_{1} and R_{f}, the negative sign on the above equation,
indicates that the amplifier is inverting.
The voltage gain of the opamp has not been reduced, but the input
signal has been reduced by cancelling part of it out, thus reducing the
overall circuit voltage gain (closed loopgain).
At high frequencies however the voltage gain will decrease in line with
the frequency of the opamp. The bandwidth of the amplifier will depend
on the product of the closed loop gain of the circuit and the bandwidth.
The product of the voltage gain and bandwidth is assumed to be 10^6.
Therefore an amplifier with a closedloop gain of 100, the bandwidth
will be 10kHz.
The below waveform shows the input and output waveforms, with the
output voltage being amplified and with a phase shift of 180 degrees, notice
the peaks and troughs are the same, whilst the wavelength is still the same.
 x = time
 y = voltage _.._
\ /` `\
 `\ /` `\ input
 `.___.` \

 _.._
 / \
 / \
/ \

 \ /
 \ / output
 \ /
 `''`
7.03  summing amplifiers
This variation of an amplifier has several resistors, instead of one. The circuit diagram is shown below:
V3####+
R3  Rf
+####+
V2####+ 
R2  '\ 
V1####+\ 
R1  >+Vout
++/
 ./

0V+0V
The negative terminal of the opamp is Virtual Earth Point, and so the current at this point, is the sum of all the currents passing through the individual input resistors.
Therefore this current is equal to the current passing through R_{f} as a result of V_{out}
If R_{1} = R_{2} = R_{3} = R; then
if V_{1} = V_{2} V_{3} = V; then
V_{out} = R_{f} * V * [1/R_{1} + 1/R_{2} + 1/R_{3}]
This circuit can be used to make a simple Digital to Analogue converter (DAC) as is shown in the diagram below:
D3####+
R3  R3 = Rf,
 R2 = 2Rf
D2####+ R1 = 4Rf,
R2  Rf R0 = 8Rf
+####+
 
D1####+ 
R1  
 '\ 
D0####+\ 
R0  >+Vout
++/
 ./

0V+0V
The logic inputs D_{3}  D_{0} all have the same input voltage, with D_{3} being the most significant bit of data, since it has the lowest series resistor. The output voltage will therefore be directly related to the digital number applied to the inputs, but it will be a negative voltage, since the inputs go to the inverting input.
7.04  noninverting amplifier
The input goes to the noninverting input, the circuit uses negative feedback to reduce the overall voltage gain of the circuit by cancelling out part of the input signal. The advantage of sending the input to the noninverting input is that it provides an input impedance for the circuit equal to the input impedance of the opamp itself, which can be as high as 10^12 ohms.
'\
Vin+\ Vout
 >+
+/ 
 ./ #
 # Rf
 #
++

#
# R1
#

0V+0V
7.05  difference amplifier
Rf
+###+ R1 = R2
  R3 = Rf
R1  '\ 
###+\  Vout
 >+
###++/
R2  ./
#
#
# R3

0V++0V