If you are not familiar with op-amps it might be worth reading the first article in this series: Operational Amplifiers - Basic Concepts.
Closing the loop - Negative feedback
The first article in this series only looked at the open loop configuration of an op-amp, while this has its uses, most applications require some form of feedback.
In a closed loop system some part of the system's output is fed back to the input. Figure 1 below shows a closed loop op-amp circuit. In this image beta represents the transfer function applied to the feedback path.
Figure 1 can be re-drawn so that it is easier to visualise, as shown in Figure 2.
In this diagram A(OL) is the open loop gain of the op-amp, which is typically assumed to be greater than 100,000.
Using Figure 2 the gain transfer function for the system can be derived:
It can be seen from equation 6 that the closed loop gain is only dependant on the feedback path from the amplifier, and not the open loop gain. This is very useful since a precise value for the open loop gain is not something that is normally known.
Example - Voltage follower
So how does it all work? Let's take the simplest example of an op-amp configured as a voltage follower (also called a buffer amplifier). This is where the closed loop gain is set to 1 by directly connecting the output to the inverting input (V-). This is shown in figure 3 below.
If a 1V step increase is applied to V(IN) the output voltage will rise as a result of the amplifier gain. This rise in output voltage causes a rise in the feedback voltage, which acts to decrease the differential voltage between the V+ and V- inputs. The system quickly stabilises at point were V+ is roughly equal to V-.
In most circuit application it is fine to say that V+ is so close to V- that they are essentially equal. Ideal op-amp equations make the assumption that the gain of the amplifier is infinite, if this was true then the difference between V+ and V- would actually be 0V.
In the interest of fully understanding how the system operates it is useful to know how it works in the real world. As previously mentioned the gain of the amplifier is not infinite, however it can be assumed to be very large, say 100,000. Also if the differential voltage between V+ and V- was really 0V the output of the amplifier would also be 0V. As a result the difference between V+ and V- must be large enough that when multiplied by the amplifier gain it produces an output voltage roughly equal to V(IN).
For example if V(IN) is set to 1V and the open loop gain of the amplifier A(OL) = 100,000 we can use equation 5 to calculate the actual gain.
Closed loop gain A(CL) = A(OL)/( 1+Beta*A(OL) ) = 100,000/(1+1*100,000) = 0.999990000099999
From this the expected output voltage can be calculated:
V(out) = Vin * A(CL) = 1*0.999990000099999... = 0.999990000099999...V
In a voltage follower V- = V(out), therefore the differential voltage between V+ and V- =
1 - 0.999990000099999... = 9.99990000099999...uV (NB. I have used ... to indicate that the sequence continues)
It can be see that this tiny differential voltage is enough for the amplifier to generate the desired 1V output, or at least very close to it!
As mentioned previously this example was more to help with understanding the concept, it is rare that it is necessary to do these calculations for a real situation, especially since many op-amps have much larger gains than 100,000.
This article has covered the use of negative feedback with op-amps. It is also possible to use positive feedback (I.e the feedback loop connected to the non-inverting input) however this will be covered in a future article.
The next article covers ideal op-amp circuits using negative feedback, including the inverting and non-inverting amplifier circuits.
Click here to proceed to the next article in the series: Operational Amplifiers - Ideal Circuits