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Semiconductor Electronics: Materials, Devices and Simple Circuits

Junction Transistor as a Device

A junction transistor can be used as a device application like a switch, amplifier etc. depending on many parameters.

  • The configuration used – CB (Common Base), CC (Common Collector) or CE (Common Emitter)
  • Reverse/Forward biasing of the E-B and B-C junctions
  • Operation region – Cutoff, active or saturation

Our focus is the CE configuration and we will concentrate on the biasing and operation region to understand how the device works.

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Junction Transistor as a Switch

When a junction transistor is used in the cutoff or saturation state, it acts as a switch. Let’s look at a base-biased transistor, as shown in the figure below, in the CE configuration and analyze its behaviour to understand the operation of a transistor as a switch.

junction transistor

Now, let’s apply Kirchhoff’s voltage rule to the input and output sides of this circuit. It is easy to understand that

VBB = IBRB + VBE … (1) &

VCE = VCC – ICRC … (2)

Now, we will treat VBB as the input voltage (Vi) and VCE as the dc output voltage (Vo). Therefore, we get,

Vi = IBRB + VBE

Vo = VCC – ICRC

Next, let’s observe the change in Vo as Vi increases from zero

A Silicon junction transistor remains in a state of Cutoff as long as Vi is less than 0.6 V. Also, IC = 0. Hence, Vo = VCC. When Vi goes beyond 0.6 V, the transistor moves into an active state. Also, IC > 0 and Vo decreases (since ICRC increases). Initially, IC increases almost linearly with increasing Vi.

Also, Vo decreases linearly till its value falls below 1 V. Post this, the change becomes non-linear and the transistor moves into the saturation state. On increasing Vi further, Vo decreases but never becomes zero. Here is a plot of Vo vs. Vi (also called the transfer characteristics of a base-biased transistor):

junction transistor

In the above graph, we can see that the transition from Cutoff to Active state and Active to Saturation state is not sharply defined. You can see non-linear regions in these areas.

How to operate a junction transistor as a switch?

Two things to remember here:

  1. If Vi is low and unable to forward-bias the transistor, then Vo is high (=VCC).
  2. If Vi is high enough to move the transistor into saturation, then Vo is very low (~0).

Also, when a transistor is not conducting, it is switched off. On the other hand, when it is in the saturation state, it is switched on. Putting these elements together, imagine a transistor where the low and high states are defined below and above certain voltage levels.

These levels correspond to the cutoff and saturation of the transistor. In such a scenario, we can say that a low input switches the transistor off and a high input switches it on. These circuits are designed so that the transistor does not remain in an active state. This is how it can operate as a switch.

Junction Transistor as an Amplifier

To use a junction transistor as an amplifier we will focus on the active region of the Vo vs. Vi curve shown above. If you observe this region, the linear part of the curve is the change of the output with the input. However, the output decreases as input increases because,

Vo = VCC-ICRC AND NOT ICRC.

Now, if we consider ΔVo and ΔVi as small changes in the output and input voltages, then the small signal gain of the amplifier (AV) is

Av = ΔVo/ ΔVi

If the VBB voltage has a fixed value corresponding to the mid-point of the active region, then the circuit will behave like a CE amplifier having a voltage gain of ΔVo/ ΔVi.

The voltage gain is expressed as follows:

We know that Vo = VCC – ICRC; hence, ΔVo = 0 – RCΔIC

Also, Vi = IBRB + VBE; hence ΔVi = RBΔIB + ΔVBE

However, ΔVBE is negligibly small as compared to RBΔIB in this circuit. Hence, the voltage gain of this CE amplifier is

Av = – RCΔIC/RBΔIB = – βac(RC/RB) … since βac = ΔIC/ΔIB

The Configuration of a Junction Transistor as an Amplifier

It is important to fix the operating point of a junction transistor in the middle of its active region to enable it to function as an amplifier. This will ensure that the dc base current (IB) and collector current (IC) are constant. Also, the dc voltage (VCC) remains constant. And, the operating values of VCE and IB determine the operating point of the amplifier.

Imagine a sinusoidal voltage (having amplitude vs) superimposed on a dc base bias. If the sinusoidal source of the signal is connected in series with the VBB supply, the base current will have sinusoidal variations superimposed on IB. Hence, the collector current will also have sinusoidal variations superimposed on IC.

Eventually, these will lead to a change in the value of Vo. AC variations can be measured by blocking the dc voltages by using large capacitors. Up until now, we have not spoken about ac signals. And the fact is, amplifiers are used to amplify alternating signals. So, let’s superimpose an ac input signal (vi) on the dc bias (VBB) as shown below.

junction transistor

The output is measured between the collector and ground.

Let’s look at how the amplifier works.

Initially, let’s assume that vi = 0. Applying Kirchhoff’s law, for the output loop we get

VCC = VCE + ICRL

And, for the input loop, we get,

VBB = VBE + IBRB

If vi is not zero, then

VBE + vi = VBE + IBRB + ΔIB (RB + ri)

Now, a change in VBE can be related to the input resistance ri and a change in IB. Hence,

Vi = ΔIB (RB + ri) = rΔIB

A change in IB causes a change in IC. We define a parameter βac, which is similar to the βdc defined earlier, as

βac = ΔIC/ΔIB = IC/IB

also known as ac current gain Ai. In the linear region of the output characteristics, βac is nearly equal to βdc. Also, a change in IC (due to a change in IB) affects a change in VCE. This causes the voltage to drop across resistor RL since VCC is constant. So,

ΔVCC = ΔVCE + RLΔIC = 0 OR

ΔVCE = –RLΔIC

We also know that the change in VCE is the output voltage Vo. Hence, we get

Vo = ΔVCE = – βacRLΔIB

And the voltage gain of the amplifier is

Av = Vo/Vi = ΔVCE/rΔIB = – βacRL/r

Please note that the negative sign means that the output voltage has an opposite phase as of the input voltage. From the above calculations, we can find the power gain Ap as follows:

Ap = βac × Av

Remember, a transistor is not a power generating device. The energy for the higher ac power at the output is supplied by the battery.

Solved Examples for You

Question: What are the parameters based on which a junction transistor is used as a device?

Solution: A junction transistor is used as a device based on the configuration used, biasing of the junctions, and the operation region. Configurations are Common Base, Collector, and Emitter. Biasing can be forward or reverse. Operation region includes cutoff, active and saturation regions.

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