To Study The Characteristics Of A Common Emitter Non (or PNP) Transistor And To Find Out The Values Of Current And Voltage Gains. - Practice Questions & MCQ

Aim:-
To study the characteristics of a common emitter npn (or pnp) transistor and to find out the values of
current and voltage gains.

Apparatus:-
An n-p-n transistor, a three volt battery, a 30 volt battery, two high resistance rheostats, one 0-3 volt voltmeter, one 0-30 volt voltmeter, one 0-50  micro-ammmeter, one 0 - 50 mA milli-ammeter, two one way keys, connecting wires.

Theory:-
In common-emitter circuit of a transistor, emitter-base make input section and emitter-collector make
output section. As usual, base junction (input junction) is forward biased and collector junction (output
junction) is reverse biased.

Resistance offered by base junction, is called input resistance (R) It has a very small resistance due to forward biasing.
Resistance offered by collector junction, is called output resistance (R₀) . \text { It has a high value due to reverse } biasing.

Due to high output resistance (resistance in output section), a high resistance can be used as load resistance ( $R_{\mathrm{L}}$ ). Generally $R_L=R_0$.
The ratio $\frac{R_L}{R_I}=\frac{R_0}{R_I}$ measures resistance gain of the common emitter transistor. It is of the order of one thousand. The current equation, $I_e=I_b+I_c$
Also emitter current ( le ) divides itself into base current $\left(I_b\right)$ and collector current $\left(I_c\right)$. In n-p-n transistor, $I_c$ is about $98 \%$ of $I_e$, base current $I_b$ remains only $2 \%$ of $l_{\mathrm{e}}$. A little change in $l_{\mathrm{b}}$ causes a large change (about 49 times) in $l_{\mathrm{c}}$. The ratio of change in collector current to the corresponding change in base current, measures current gain in common emitter transistor. It is represented by symbol $\beta$.

$
\beta=\frac{\Delta I_c}{\Delta I_b}
$

Formula used:-
Input resistance,

$
R_I=\frac{\Delta V_b}{\Delta I_b}
$

Output resistance,

$
R_o=\frac{\Delta V_c}{\Delta I_c}
$

Resistance gain $=\frac{R_0}{R_I}$

Current gain,

$
\beta=\frac{\Delta I_c}{\Delta I_b}
$
 

Voltage gain $=$ Current gain $\times$ Resistance gain

$
A_V=\beta \cdot \frac{R_0}{R_I}
$
 

Diagram:-

Procedure:-

1. Make circuit diagram as shown in figure.
2. Make all connections neat, clean and tight.
3. Note least count and zero errors of voltmeters and ammeters.
4. Make voltmeter readings zero in V₁ and V₂ and insert the keys.

For input characteristics
5. Apply forward bias voltage on base junction. Read base voltage V_b and base current I_b.

6. Go on increasing $\mathrm{V}_{\mathrm{b} \text { till }} \mathrm{l}_{\mathrm{b}}$ rises suddenly. Note corresponding values of $\mathrm{l}_{\mathrm{b}}$ for each value of $\mathrm{V}_{\mathrm{b}}$.
7. Make collector voltage $\mathrm{V}_{\mathrm{C}}=10 \mathrm{~V}$ and repeat steps 5 and 6 .
8. Repeat step 7 with $\mathrm{V}_{\mathrm{C}}=20 \mathrm{~V}$ and 30 V .
9. Make all readings zero.

For output characteristics
10. Keep collector voltage $\left(\mathrm{V}_{\mathrm{c}}\right)$ zero. Adjust base voltage $\mathrm{V}_{\mathrm{b}}$ to make base current $\mathrm{l}_{\mathrm{b}}=10 \mu \mathrm{~A}$. Though collector voltage $\mathrm{V}_{\mathrm{c}}$ is zero; but there is collector current I ${ }_c$ \{Note it. \}
11. Make collector voltage $10 \mathrm{~V}, 20 \mathrm{~V}$ and 30 V and note the corresponding collector currents.
12. Repeat steps 10 and 11 with $\mathrm{l}_{\mathrm{b}}=20 \mu \mathrm{~A}, 30 \mu \mathrm{~A}$, and $40 \mu \mathrm{~A}$.
13. Record your observations as given below:

Observations:-
Least count of voltmeter, $\quad \mathrm{V}_1=\ldots . . \mathrm{V}$
Zero error of voltmeter, $\quad V_1=\ldots . . \mathrm{V}$
Least count of voltmeter, $\quad \mathrm{V}_2=\ldots . . \mathrm{V}$
Zero error of voltmeter, $\quad \mathrm{V}_2=\ldots . \mathrm{V}$
Least count of milli-ammeter $=\ldots . . \mathrm{mA}$
Zero error of milli-ammeter $\quad=\ldots \mathrm{mA}$
Least count of micro-ammeter $=\ldots \ldots \mu \mathrm{A}$
Zero error of micro-ammeter $\quad=\ldots \ldots . \mu \mathrm{A}$

Calculations:-

1. Calculation for input resistance (R_I)
Plot a graph between base voltage V_b and base current  l_b  for  zero collector voltage V_c, taking V_b  along X-axis and l_b along Y-axis. Plot graphs for different values of V_c. The graphs come as shown. These graphs are called 'input characteristics' of the transistor.

The slope of graphs becomes large at the ends. The slope gives value of $\frac{\Delta I_b}{\Delta V_b}$. Its reciprocal $\frac{\Delta V_b}{\Delta I_b}$ gives input resistance $\mathrm{R}_{\mathrm{I}}$. As graphs run parallel near the ends, all give same value of $R_1$.

2. Calculation for output resistance $\left(\mathrm{R}_0\right)$

Plot a graph between collector voltage $\mathrm{V}_{\mathrm{c}}$ and collector current $\mathrm{I}_{\mathrm{c}}$ for $10 \mu \mathrm{~A}$ base current $\mathrm{l}_{\mathrm{b}}$, taking $\mathrm{V}_{\mathrm{c}}$ along X -axis and $I_{\mathrm{c}}$ along Y -axis. Plot graphs for different values of $\mathrm{l}_{\mathrm{b}}$.

These graphs are called 'output characteristics' of the transistor.

near the ends, all give same values of $R_0$.
3. Calculation for current gain $(\beta)$

Plot a graph between base current $\mathrm{I}_{\mathrm{b}}$ ) and corresponding collector current $\mathrm{I}_{\mathrm{c}}$ ( for 30 volts collector voltage ) $\mathrm{V}_{\mathrm{c}}, \mathrm{t}$ along Y -axis. The graph comes to be a straight line. The graph is called current gain characteristic of the common

The slope of the straight line gives value of $\frac{\Delta I_c}{\Delta I_b}$ which is the value of current gain $\beta$ of the common emitter transistor

4. Calculation for voltage gain $A_v$

From relation, Voltage gain $=$ Current gain $\times$ Resistance gain

$
A_v=\beta \times \frac{R_0}{R_I}
$

Result
For the given common emitter transistor,
Current gain, $\beta=\ldots$.
Voltage gain, $\quad A_v=\ldots \ldots$