6.01 Course Notes

3) Circuits 2

Table of Contents

• 3) Circuits 2
• 3.1) Modularity and Abstraction in Circuits
• 3.1.1) Example: Voltage Divider
• 3.2) Equivalents
• 3.2.1) Linear Constraints
• 3.2.2) Thévenin
• 3.2.2.1) Finding Thévenin Equivalents
• 3.2.2.2) Alternative Methods
• 3.2.3) Example
• 3.3) Operational Amplifiers
• 3.3.1) "Ideal" Op-Amp Model
• 3.3.2) Buffer
• 3.3.2.1) Why Do We Care?
• 3.3.3) What Else Are Op-Amps Good For?
• 3.3.3.1) Non-inverting Amplifier
• 3.3.3.2) Inverting Amplifier
• 3.3.3.3) Others
• 3.3.4) Voltage-Controlled Voltage Source Model
• 3.3.5) Where Does the Current Come From?
• 3.4) Current Sources
• 3.4.1) What's In A Name?
• 3.5) Summary

3.1) Modularity and Abstraction in Circuits

We will find that thinking about modularity and abstraction in circuits is different from LTI or programming in a very fundamental way. In LTI, when we came up with a system functional to describe a system, that system would always behave that way, regardless of what other systems we connected it to. In circuits, however, we will find that connecting new components to a pre-existing circuit could potentially change all the currents and voltages in that circuit.

3.1.1) Example: Voltage Divider

In previous sections, we noted that resistors connected in series formed a "voltage divider", which can split up an "input" voltage so that some portion of the voltage dropped across the "top" resistor, and some dropped across the "bottom" resistor. We were able to solve and see that the voltage drop across each of these resistors was proportional to its resistance:

 

In the case when these two resistors are the same, we have a voltage divider that cuts an input voltage in half:

 

It is tempting to think of this like a gain block in LTI, which also takes an input and cuts it in half.

 

In LTI, if we cascaded two of these systems together, we would get a new system whose output was 1/4 its input:

Remember that our definition of a voltage divider was two resistors connected in series. In the first circuit, we had this, but when we added the two additional resistors to make the second resistance, we broke that abstraction (the original two resistors are no longer in series!). Note that we can still think about circuit pieces as modules, but we have to be careful, when making changes to the circuit, to make sure that those abstractions still hold.

In the next few sections, we will look at some examples of tools we can use to manage complexity when designing and analyzing circuits, and we will introduce two new kinds of components.

3.2) Equivalents

We just saw that pieces of circuits cannot be abstracted as input/output elements; the actual voltages and currents in them will depend on how they are connected to the rest of a larger circuit. However, we can still abstract them as sets of constraints on the values involved.

In fact, when a circuit includes only resistors and voltage sources¹, we can derive a much simpler circuit that induces the same constraints on currents and voltages as the original one. This is a kind of abstraction that's similar to the abstraction that we saw in linear systems: we can take a complex circuit and treat it as if it were a much simpler circuit.

We have already seen some applications of this idea. Two specific examples we have seen are series and parallel combinations, particularly of resistors. We saw, for example, that the following two one-ports are "equivalent", in the sense that any circuit we hook up to them will see the same relationship between the current I and the voltage V, regardless of which of the two we connect.

This is very interesting, and this abstraction has already proved very useful with some of the circuits we have analyzed so far. In this section, we will explore a more general type of circuit equivalent, and we will see how the series and parallel resistor combinations we have already discussed fit into this new framework.

3.2.1) Linear Constraints

In a circuit that contains only linear components², all of the current-voltage relationships will be linear. This means that, if we pick two arbitrary spots in a circuit, we can know that the relationship between the current I flowing between those two points, and the voltage V across those two points, will be a line!

That might not seem terribly exciting on its own, but what it means is that if we come up with a simpler circuit that has the same linear I-V constraint as an arbitrary subcircuit, then we can swap that simpler circuit in, and use that to do our analysis.

3.2.2) Thévenin

Consider this circuit (and the associated I-V characteristic), which we will refer to as a Thévenin equivalent circuit³:

It's nothing too complicated, but this circuit will open some very powerful doors for us. Let's start by thinking about its I-V characteristic. We could think about this in terms of specific V_T and R_T values, but the interesting thing to note is the general relationship:

Note that this circuit can be made to produce any⁴ line in the I-V plane: adjusting the voltage V_T adjusts the "V intercept" (the point where the curve intersects the horizontal axis), and adjusting R_T adjusts the slope.

This gives us a convenient new avenue for simplifying circuits: if we can find an appropriate subcircuit and solve for the I-V relationship between its terminals, we can then construct a Thévenin that has that same I-V relationship, and substitute it in for the more complicated circuit.

3.2.2.1) Finding Thévenin Equivalents

In order to construct an appropriate Thévenin equivalent, we need first to characterize the line representing the I-V characteristic of the circuit we wish to replace. In a throwback to high-school geometry, we know that in order to characterize the line, we either need to know 2 points on the line, or one point and a slope.

Generally, the two intercepts will be (relatively) easy points to find. So one straightforward strategy is to find:

1. the voltage V when I=0, and
2. the current I when V=0.

Once we have those two points, we can use them to back-solve for the Thévenin resistance R_T.

So we will need to solve the circuit twice, under these difference conditions. So now the question becomes: how do we ensure that I=0 or V=0?

For I=0, we need to make sure that no current is flowing into the circuit we wish to replace. The easiest way to do that is just to leave the circuit disconnected from anything. So if we leave the circuit floating and solve for the voltage V, we will have found our first point.

Then, we will need to set V=0. V=0 implies that the two terminals are at the same potential. What's the easiest way to ensure this? We can simply connect them together with a wire! Once we have done that, solving for I will give us another point.

3.2.2.2) Alternative Methods

Of course, this is not the only way to solve for V_T and R_T. We could set V to two separate, arbitrary values (by hooking up two different voltage sources across the terminals) and solve for the currents in each of those cases, and this would still give us the two points we need.

Another method is to solve for the resistance R_T directly. In this method, we "turn off" all sources in the circuit (voltage sources become wires, and current sources become opens) and solve for the equivalent resistance between the two terminals.

Yet another method would be to solve for I in terms of V directly, rather than using two specific values of V.

All of these methods should get you to the same result, and so it might be good to practice with all of them. In some situations, a certain method may be easier than another, and vice versa.

3.2.3) Example

We can see the usefulness of Thévenin equivalents by looking back at the circuit from "Check Yourself 19", but with a specific input voltage:

We saw one way of solving this circuit before, which involved several resistor combinations. Here we will solve it again, but using a Thévenin equivalent instead of the series/parallel combinations from before.

Here, we will choose to replace the portion of the circuit in the dashed box below with its Thévenin equivalent:

Let's take a look at the boxed piece in isolation:

Now we can solve for the voltage when I=0. To do this, we simply leave the circuit disconnected, and solve for V. Since this is a voltage divider with two equal resistances, we know that that voltage (and, thus, V_T) must be 5V.

We can then solve for the current when V=0. We do this by intorducing a wire between the two terminals:

Here, the bottom 1k\Omega resistor is in parallel with a wire, so all of the current will flow down that path (the combination is equivalent to a wire). Thus, the total current is I=-10V/1k\Omega = -10mA.

We can use this to solve for the resistance. By Ohm's Law, R_T = -V_T / I_{sc}, where I_{sc} is the current we just solved for (the "short circuit" current). So in this case, we have R_T = 5V / 10mA = 500\Omega. So the following two circuits are equivalent when viewed from the port on the right (with I and V labeled):

 

Why did we go to this trouble? Well, now that we have our equivalent circuit, we can use it as a replacement for the boxed portion of the original circuit, which gives us an easier circuit to analyze!

Now we can look at this as a voltage divider! 5V drops across the series combination of those 3 resistors, and so the voltage drop we have labeled as V_o must be:

Arguably, this solve (even including the time it takes to find the Thevenin equivalent) is faster than our original method of combining only resistors!

3.3) Operational Amplifiers

So far, we have considered circuits with resistors, voltage sources, and current sources. Now we are going introduce a new component, called an operational amplifier (or "op-amp," for short).

Op-amps are a crucial tool for decoupling some aspects of complex circuits, so that they can be designed independently and then connected. They are also important for amplifying small voltage differences into larger ones, and can be used to scale, invert, and add voltages.

In our circuit diagrams, we will depict an op-amp like this:

Note that, unlike other components we have seen before, the op-amp has three terminals: a non-inverting input (denoted with a +), an inverting input (-), and an output. Despite this difference, we will find that they fall quite nicely into our framework for solving circuits, which we developed in the last section of the readings.

3.3.1) "Ideal" Op-Amp Model

Op-amps are actually complex assemblages of transistors and other circuit components, but, as we have done before in 6.01, we will make some simplifying assumptions. We'll need a simpler model than that of what op-amps do, so that we can analyze circuits that contain them and design new circuits that use them.

The first model we will introduce is likely to be the most-commonly used model in 6.01, which we will call the "ideal" op-amp model.

Op-amps are typically connected in a feedback configuration like the one shown below, where one of the inputs is tied to the output in some way.

When the op-amp is properly connected in negative feedback (i.e., the feedback loop goes back to the inverting (-) input rather than the non-inverting input), then the op-amp approximately obeys the following two rules:

1. No current flows into or out of the input terminals of the op-amp, and
2. The potentials at the two inputs are equal.

3.3.2) Buffer

The configuration we have just seen (reproduced below) is often referred to as a "buffer" or a "voltage follower". Let's go ahead and analyze how this circuit behaves. Note that we can use our same tools from before (in particular, our framework for solving circuits), even despite the new type of component.

Let's take a look at the buffer, and assume an input voltage V_i and an output voltage V_o.

3.3.2.1) Why Do We Care?

That's a nice little result, but, on the face of it, it might not seem too interesting or relevant. The importance of this circuit, though, is that it allows us to "isolate" pieces of a circuit from one another.

Consider again our attempt at cascading two voltage dividers together:

The problem with adding those new components was that they broke our original abstraction of the first two resistors being in series (because, after the addition, there was a path for current to flow out from in between them). We can use a buffer, though, to prevent the flow of current out from in between those two resistors.

Before op-amps, we necessarily had a problem that connecting new components to a circuit would change other voltages and currents throughout the circuit. Now, op-amps allow use to modularize our circuit in a way, keeping certain pieces separate from others and minimizing the effect of adding new components.

It is important to note that, while no current flows into or out of the input terminals of the op-amp, current can (and almost certainly will) flow into or out of the output terminal of an op-amp. In this way, the output of the op-amp behaves like a voltage constraint; the current can be positive, negative, or zero (whatever current is required in order to maintain its constraint on the potential at its output).

3.3.3) What Else Are Op-Amps Good For?

So far, we have seen one good use of op-amps (the buffer configuration). What else is an op-amp good for? As we mentioned in the very beginning of this section, op-amps can also be made to amplify voltages, to invert them, or to add or subtract them. Here, we will look at several difference op-amp configurations and see how they behave.

3.3.3.1) Non-inverting Amplifier

3.3.3.2) Inverting Amplifier

3.3.3.3) Others

These are not all of the amplifiers we will look at in 6.01, but we will explore the others through labs and exercises, rather than through the readings.

3.3.4) Voltage-Controlled Voltage Source Model

Throughout this section, we have been using a particular model of an op-amp called the "ideal" op-amp model. In some cases, we may find it helpful to go a bit beyond this model. A slightly-more-complicated model we would use is to model the op-amp as a voltage-controlled voltage source.

In this model, we can view the output of the op-amp as a voltage source whose voltage is K(V_+ - V_-), where K is some very large gain, V_+ is the potential at the + input terminal, and V_- is the potential at the - input terminal.

As in our previous model, no current flows into or out of the input terminals.

Here, we can see the op-amp as amplifying the voltage difference V_+ - V_-. This might feel weird at first, but consider wiring the op-amp in a buffer configuration as before:

Here, we have:

As K\to\infty, that gain \frac{K}{1+K}\to 1.

When we use the ideal op-amp model (or "make the ideal op-amp assumption"), we are assuming that K is infinitely large, which has the effect of minimizing the potential difference at the input terminals.

3.3.5) Where Does the Current Come From?

In all of the models we have seen so far, we've left out an important point. As you know from lab, op-amps, in fact, need two more connections, which is to V_{\rm CC}, the main positive voltage supply, and to V_{\rm EE/GND}, which is the negative power supply (which can be ground). Why? If no current flows "through" the op-amp, from V_+ or V_- to the other side, then how can it maintain the necessary voltage difference on the output side, and supply current to, e.g., drive the motor? The answer is that it uses current from the power supplies to regulate the output voltage and to supply the necessary output current.

One metaphorical way of thinking about this is that V_{\rm CC} is a big main water-supply pipe and that V_+ - V_- is controlling the handle of a faucet. As we change the size of the difference V_+ - V_-, we open or close the faucet; but the water is being supplied from somewhere else entirely.

This metaphor is appropriate in another way: the total voltage on the output of the op-amp is limited by V_{\rm CC}. So, if the V_{\rm CC} = 10, then the output of the op-amp is limited to be between 0 and 10. If we connect the ground connection of the op-amp to a negative supply voltage, then the output of the op-amp can be negative as well as positive; this has many applications.

3.4) Current Sources

Finally, we will introduce one final component: the current source. A current source is a component whose current is a constant (i.e., I = I_0), independent of the voltage drop across it. We will denote a current source by a circle enclosing an arrow. A number beside the current source will indicate the constant current maintained by the source, as indicated below:

This current source will make sure that the current flowing through it is always I_0. Importantly, the voltage across a current source is unconstrained by the source itself; rather, it is determined by the other components to which the source is connected. The current source will develop whatever voltage drop it must (positive, negative, or 0) to maintain its constraint on the current.

Below, we plot this device's I-V Characteristic; this is a graph that shows the relation between the variables v and i. Note that, for this particular component, the voltage can take any value; the current will always be I_0. The plot below shows the relation for a 1.5 Amp current source (I_0 = 1.5V).

3.4.1) What's In A Name?

We call this idealized device a current source, but it is important to keep in mind that its job is to enforce a constraint on the current, not to add some amount of current to whatever is already there. In many ways, it would make more sense to call this device a current constraint, as it always constrains the current through it to be constant.

3.5) Summary

In this chapter, we introduced the idea of circuits, with a particular focus on electrical circuits. We introduced the concepts of nodes and components, and we saw four different examples of specific common components: voltage sources, resistors, op-amps, and current sources. We noticed that when multiple components are connected together, all of their constraints must still be satisfied.

We introduced a general process for solving circuits and looked at a couple of specific examples of using that method to solve somewhat-complicated circuits.

We also introduced the one port abstraction, whereby components are represented by their I-V Characteristic curve. We saw how multiple components could be replaced with equivalent one-ports, in particular by looking at series and parallel combinations. We then expanded on this idea to look at more general equivalent circuits by looking at Thévenin equivalents, and we described a process for finding equivalent circuits.

 

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Footnotes

¹And current sources, which are discussed in later sections (click to return to text)

²I.e., components whose I-V characteristic curve is linear. We will only deal with linear components in 6.01. (click to return to text)

³This is one of those interesting cases where the person whose name is associated with an idea is not the person who originally came up with the idea. The ideas in this section were originally derived by Hermann von Helmholtz several decades before Thévenin independently derived them. (click to return to text)

⁴Well, almost any. A straight horizontal line cannot be represented by a Thévenin. (click to return to text)