Question Video: Determining the Current Division in a Transistor Circuit Physics

Video Transcript

The following circuit diagram shows a transistor used as a switch. By inspecting the diagram, determine whether the switch is off or on and find the current division, 𝛼 𝑒. (A) The switch is on and the current division is 0.96. (B) The switch is on and the current division is 0.04. (C) The switch is off and the current division is 0.96. Or (D) the switch is off and the current division is 0.04.

This is the circuit symbol for the transistor. It helps us discern certain properties of the transistor thanks to this little arrow. Because the arrow is pointing outward, we know that this is an NPN transistor rather than a PNP transistor, which has an arrow pointing inward. We can also recall that in any transistor symbol, the arrowhead is always drawn so that it’s connected to the emitter region. That means that in our diagram, the emitter is here. We can then tell that the collector must be here and the base is in between the emitter and the collector.

So, now that we’ve labeled the transistor, let’s also notice these current values labeled on the diagram. We can tell that 72 milliamps is the collector current, so we’ll label this 𝐼 C. And the three-milliamp current must be the base current, so we’ll label it 𝐼 B. We can see that the base current here is much smaller than the collector current. This is what we’d usually expect from a transistor.

Recall that a transistor can act as a switch. The existence of a base current enables a collector current to flow. In this case, we’d say the transistor was on. But if there’s no base current, the collector current can’t flow, so the transistor is off. Because there is a base current here, we can say that the switch formed by this transistor is effectively on. Now, options (C) and (D) both say that the switch is off, so we can rule these options out.

Now, we also need to find the current division, which is represented by 𝛼 𝑒. We can do this using the formula 𝐼 C equals 𝐼 E times 𝛼 𝑒, where 𝐼 C is the collector current, 𝐼 E is the emitter current, and 𝛼 𝑒 is the current division. This equation effectively defines the current division as being the constant of proportionality between the collector current and the emitter current.

We can make the current division the subject of this equation by dividing both sides by 𝐼 E. This gives us 𝐼 C over 𝐼 E equals 𝛼 𝑒. And we can swap the left- and right-hand sides of this expression around to give us 𝛼 𝑒 equals 𝐼 C over 𝐼 E. So we could also think of the current division as being the ratio between the collector current and the base current.

Now, we’ve been given 𝐼 C, but we don’t yet know 𝐼 E. Fortunately, this is easy to calculate. We just need to remember that the emitter current, that’s here on our diagram, is equal to the sum of the collector current and the base current. Written as an equation, we have 𝐼 E equals 𝐼 C plus 𝐼 B.

Now, using the values for 𝐼 C and 𝐼 B given in the diagram, we can calculate that the emitter current is 75 milliamps. We’re now ready to solve for the current division. The collector current is 72 milliamps, and the emitter current is 75 milliamps. Typing this into a calculator, we find that the current division is equal to 0.96.

Note that this is a dimensionless number since the units of current cancel each other out in the numerator and denominator. This means the correct answer to the question is option (A). The switch is on and the current division is 0.96.