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.

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

V_BB = I_BR_B + V_BE … (1) &

V_CE = V_CC – I_CR_C … (2)

Now, we will treat V_BB as the input voltage (V_i) and V_CE as the dc output voltage (V_o). Therefore, we get,

V_i = I_BR_B + V_BE

V_o = V_CC – I_CR_C

Next, let’s observe the change in V_o as V_i increases from zero

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

Also, V_o 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 V_i further, V_o decreases but never becomes zero. Here is a plot of V_o vs. V_i (also called the transfer characteristics of a base-biased 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 V_i is low and unable to forward-bias the transistor, then V_o is high (=V_CC).
2. If V_i is high enough to move the transistor into saturation, then V_o 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 V_o vs. V_i 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,

V_o = V_CC-I_CR_CAND NOT I_CR_C.

Now, if we consider ΔV_o and ΔV_i as small changes in the output and input voltages, then the small signal gain of the amplifier (A_V) is

A_v = ΔV_o/ ΔV_i

If the V_BB 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 ΔV_o/ ΔV_i.

The voltage gain is expressed as follows:

We know that V_o = V_CC – I_CR_C; hence, ΔV_o = 0 – R_CΔI_C

Also, V_i = I_BR_B + V_BE; hence ΔV_i = R_BΔI_B + ΔV_BE

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

A_v = – R_CΔI_C/R_BΔI_B = – β_ac(R_C/R_B) … since β_ac = ΔI_C/ΔI_B

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 (I_B) and collector current (I_C) are constant. Also, the dc voltage (V_CC) remains constant. And, the operating values of V_CE and I_B determine the operating point of the amplifier.

Imagine a sinusoidal voltage (having amplitude v_s) 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 I_B. Hence, the collector current will also have sinusoidal variations superimposed on I_C.

Eventually, these will lead to a change in the value of V_o. 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 (v_i) on the dc bias (V_BB) as shown below.

The output is measured between the collector and ground.

Let’s look at how the amplifier works.

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

V_CC = V_CE + I_CR_L

And, for the input loop, we get,

V_BB = V_BE + I_BR_B

If v_i is not zero, then

V_BE + v_i = V_BE + I_BR_B + ΔI_B (R_B + r_i)

Now, a change in V_BE can be related to the input resistance r_i and a change in I_B. Hence,

Vi = ΔI_B (R_B + r_i) = rΔI_B

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

β_ac = ΔI_C/ΔI_B = I_C/I_B

also known as ac current gain A_i. In the linear region of the output characteristics, β_ac is nearly equal to β_dc. Also, a change in I_C (due to a change in I_B) affects a change in V_CE. This causes the voltage to drop across resistor R_L since V_CC is constant. So,

ΔV_CC = ΔV_CE + R_LΔI_C = 0 OR

ΔV_CE = –R_LΔI_C

We also know that the change in V_CE is the output voltage V_o. Hence, we get

V_o = ΔV_CE = – β_acR_LΔI_B

And the voltage gain of the amplifier is

A_v = V_o/V_i = ΔV_CE/rΔI_B = – β_acR_L/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 A_p as follows:

A_p = β_ac × A_v

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.