7 - Operational Amplifiers

2016-04-27 21:56:00 +0000, 2 years and 3 months ago

7.01 - 741 operational amplifier

A logic gate is able to decide if an analogue signal is 1 or 0, high or low, however there is a region of uncertainty (10uV~) which can be up to ~100mV. An Op amp overcomes this issue.

An Op amp is a voltage amplifier which amplifies the differences between two input voltages to decide if the output should be high or low depending on these inputs. Op amps usually require a dual rail, DC supply.

The symbol of an Op amp is shown below:

Where V+ = non-inverting input
V- = inverting input
+Vs = positive supply rail
-V- = negative supply rail

Vout is dictated by using the following calculation:

Where A = open loop gain (10^6) A theoretical op-amp is assumed to have the following behaviours:

7.02 - Op-amp as a voltage comparator

Vout = A(V+ - V-)
Where A = 10^6 (in reality it’s +Vs - 2V and -Vs + 2V)

V+ = non-inverting input
V- = inverting input

Since the open loop gain is very large (A), only a very small difference in voltage is required for the op-amp to saturate.
The theoretical & realistic outputs of an op-amp are shown below:

theoretical:                 realistic:   
               V+ > V-                             
     +Vs  |   /------            +Vs  |   V+ > V-
          |  /.|             +Vs -2V  |  /-------
          | /..|                      | /.|
          |/...| 10μV                 |/..| 10μV
----------+----------       ----------+----------    
10μV |.../|                   10μV|../|
     |../ |                       |./ |
     |./  |                 -------/  |  -Vs + 2V
------/   | -Vs             V+ < V-   |  -Vs
V+ < V-

Theoretical: Vout should be equal to +/-Vs

Realistic: Vout saturates -/+ 2V of +/-Vs, respectively.

The dotted regions of the graph, show the regions of uncertainty (10uV~), the characteristic graph shows how the Op-amp is able to compare two voltages at its input terminals (V-, V+).

If V+ is greater than V- then the Op-amp saturates to +Vs - 2V

If V+ is less than V- then the Op-amp saturates to -Vs + 2V

This behaviour can be used to compare two voltages, a ‘set (reference)’ voltage and a varying input voltage such as an LDR or thermistor.

7.03 - dual rail supply comparator

+Vs----+----+----->-------+
       |    |             |
      +++  +++            |
      | |  | |            |
   R1 | |  | | R2         V
      +++  +++            |
       |    |             | +Vs 
       |    |             |
       *----+-->----+-----+----+    
       |    |       |Comparator+-----> Vout ^
       |    *-->----+-----+----+            |
       |    |             |                 |
       |    |             ^                 |
     +-+-+ +++            |                 |
     |LDR| | |         -Vs                  |
     +-+-+ | | R3                           |
       |   +++                              V
       |    |     
0V ----+----+-------------------------------+

The Op-amp operates on the +Vs and -Vs supplies, but the rest of the circuit operates on +Vs and 0V.
The reason for doing this is to overcome the issue of when the Op-amp saturates into to the negative supply, it actually swings +2V of -Vs, this 2V is enough to perhaps power an LED. If the -Vs is -2V, then when the Op-amp saturates to -Vs, the true output voltage will be 0V, this is because the -2V cancels out the +2V swinging.

This issue can also be solved by adding a logic gate to the output of the Op-amp, since a logic gate switches high or low at 1/2 of Vs, as shown:

   +---+
   |+Vs|
   +-+-+
     |
+----+-----+   +------+   +----------+   +------------+
|Comparator+->-+(Vout)+->-+Logic gate+->-+Correct Vout|
+----+-----+   +------+   +----------+   +------------+
     |
   +-+-+
   |-Vs|
   +---+

7.04 - single power rail comparator

The single rail supply is essentially the same as the dual rail supply, except -Vs is now 0V, so if the Op-amp saturated to the negative, then Vout would be 2V, because a real Op-amp swings +2V of the negative supply.

+Vs----+----+----->-------+
       |    |             |
      +++  +++            |
      | |  | |            |
   R1 | |  | | R2         V
      +++  +++            |
       |    |             | +Vs    voltage drop across
       |    |             |        across LED is 1.8V
       *----+-->----+-----+----+   +---+
       |    |       |Comparator+->-+LED+-> Vout 
       |    *-->----+-----+----+   +---+        ^ 
       |    |             |                     |
       |    |             ^  Vout = 2V - 1.8V   |
     +-+-+ +++            |  Vout = 0.2V        |
     |LDR| | |            |                     |
     +-+-+ | | R3         | notice -Vs          |
       |   +++            | connected to 0V     V
       |    |             |
0V ----+----+-------------+----------<------

Another way of getting around the issue of +2V swinging high, would be to place a red LED in series with Vout, since the voltage drop a red LED is around 1.8V, it would negate the effect of this +2V swing.

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