Dependent & controlled sources

Suppose a circuit voltage vx controls another source. The statement

means the source output current io is constrained to equal gm times the signed control voltage vx. It does not, by itself, say that charge flows from the control terminals into the output terminals, nor that the control port supplies the output power.

The control variable is simply another variable of the same circuit model. Its reference direction or polarity must be defined explicitly, just like every other circuit variable.

This abstraction is extremely useful because it represents amplification and coupling while keeping Kirchhoff equations linear. Later device models can be built from controlled-source relations without forcing device physics into every circuit calculation.

There are two choices for the controlled output—voltage or current—and two choices for the controlling variable—voltage or current. Combining them gives four canonical families:

The units are not decoration. They are a fast dimensional check. A coefficient multiplying volts to produce amperes must have units A/V; a coefficient multiplying amperes to produce volts must have units V/A.

In Figure 2.2, vx is positive when the upper control terminal is above the lower one. The output-current arrow defines positive io upward through the controlled current source. The relation io=gmvx must be interpreted using both references.

If gm>0 and vx>0, io follows its reference arrow. If vx<0, the calculated io is negative and the physical current flows opposite that arrow. Likewise a negative gain coefficient reverses the sign relation.

The dashed dependency in the conceptual figure is deliberately not a circuit wire. Unless a particular model explicitly connects the control and output terminals electrically, the control law alone does not create a conductive path between them.

Consider an abstract two-part linear model. An independent 10 V source drives a 5 kΩ / 5 kΩ divider. Let vx be the midpoint voltage relative to ground. Because the control input is idealized here as sensing voltage without drawing current,

A separate voltage-controlled current source has transconductance

and its output arrow points upward from ground into an output node. Therefore

If a 2 kΩ load connects from the output node to ground, KCL requires 2.5 mA downward through the load:

The control voltage is 5 V, the controlled current is 2.5 mA, and the load voltage becomes 5 V. The output was obtained from the controlled-source equation plus the load law; no device physics was needed.

The 2 kΩ output load absorbs

The controlled-current arrow points upward while the output voltage is positive at the upper node. Current therefore leaves the source's positive-voltage terminal. Under passive-sign bookkeeping, the controlled source absorbs

so it delivers 12.5 mW to the load.

On the control side, the 10 V source drives 1 mA through the two 5 kΩ resistors. The two resistors absorb 5 mW each and the independent source delivers 10 mW.

The complete signed power check is 5 mW+5 mW+12.5 mW−10 mW−12.5 mW=0. The ideal dependent-source model does not say that its 12.5 mW output power was drawn through the voltage-sensing control variable. In a physical amplifier, energy comes from power supplies and internal device physics. Those mechanisms are outside this abstract controlled-source element unless the larger model includes them explicitly.

Take one unknown node V connected to a known 10 V node through 1 kΩ, to ground through 2 kΩ, and to a downward controlled current source with

Count currents leaving V. Kirchhoff's current law gives

Collect conductance terms:

The resistor to ground carries 2.5 mA and the controlled source carries another 2.5 mA downward. The 1 kΩ branch supplies 5 mA into the node, so KCL closes exactly.

This example also shows why a dependent source must not be treated as a known independent current injection: its current contains an unknown circuit variable and belongs algebraically with the other unknown-dependent terms.

The 1 kΩ resistor has a 5 V drop and carries 5 mA:

The 2 kΩ resistor has 5 V and 2.5 mA:

The controlled current source carries 2.5 mA downward from the positive 5 V node, so current enters its positive-voltage terminal:

The 10 V independent source supplies 5 mA out of its positive terminal:

The signed total is 25+12.5+12.5−50=0 mW. A dependent source can therefore absorb power in one operating point and deliver power in another; “dependent” says how its output value is controlled, not the sign of its power.

A relation such as

with constant gm is linear in the control variable: scaling vx scales io by the same factor, and sums of control signals produce sums of outputs.

Likewise vo=μvx, vo=rmix and io=βix are linear when their coefficients are constants within the model.

A relation such as

contains an independent offset I0 and is affine rather than homogeneous. The full circuit can still be handled with linear-network techniques by treating the independent offset as a separate source contribution, but one must not silently call the entire input-output map homogeneous.

A genuinely nonlinear law such as io=k(vx)² lies outside ordinary linear superposition.

When applying superposition in a linear circuit, independent source values are set to zero one at a time. A dependent source is different: its value is tied to a circuit variable, so its defining relation must remain in the equations.

For a VCCS, keep

even after independent sources have been deactivated. The resulting solution may happen to produce vx=0 and hence io=0, but that is a result of the reduced circuit—not an instruction to replace the controlled source by an open circuit by hand.

The same principle explains why a test source is often required to find resistance seen at a port of a circuit containing dependent sources: the controlled sources remain active while the test excitation probes the terminal relation.

1. A VCVS obeys vo=−4vx. If vx=0.6 V, find vo and interpret the sign.
2. A VCCS has gm=2 mS and vx=−0.25 V. Find io for the stated output-current reference.
3. A CCVS produces 3 V when its control current is 1.5 mA. Find the transresistance.
4. A CCCS obeys io=20ix. If io=8 mA, find ix.
5. In the worked VCCS example, change the load from 2 kΩ to 4 kΩ while keeping the ideal control side unchanged. Find io, vo and load power.
6. For the self-controlled nodal example, change g from 0.5 mS to 1.5 mS. Find V and the controlled-source current.
7. An independent voltage source is set to zero during superposition. A dependent VCCS elsewhere is controlled by vx. Should the VCCS be opened? Explain.
8. Why can a dependent source be useful in an amplifier model even though the diamond is not a literal transistor symbol?

Chapter 03 will make the word linear precise and use it to derive the superposition principle rather than presenting superposition as a recipe for “turning sources off.”