To Draw The I-V Characteristics Curve Of P-N Junction In Forward Bias - Practice Questions & MCQ

Aim-  To draw the I-V characteristic curve of a p-n junction in forward bias 
Apparatus : A p-n junction (semi-conductor) diode, a 3 volt battery, a 50 volt battery, a high resistance rheostat, one 0.3 volt voltmeter, one 0.50 volt voltmeter, one 0-100 mA ammeter, one 0.700   ammeter, one way key, connecting wires and pieces of sand paper. 

Theory : Forward bias characteristics. When the p-section of the diode is connected to positive terminal of a battery and n-section is connected to negative terminal of the battery then junction is said to be forward biased. With increase in bias voltage, the forward current increases slowly in the beginning and then rapidly. At about 0.7 V for Si diode (0.2 V for Ge), the current increases suddenly. The value of forward bias voltage, at which the forward current increases rapidly, is called cut in voltage or threshold voltage.

Diagram 

Procedure
1. Make circuit diagram as shown in diagram.
2. Make all connections neat, clean and tight.
3. Note least count and zero error of voltmeter (V) and milli-ammeter (mA).
4. Bring moving contact of potential divider (rheostat) near negative end and insert the key K. Voltmeter
Vand milli-ammeter mA will give zero reading.

5. Move the contact a little towards positive end to apply a forward-bias voltage $\left(V_{\mathrm{F}}\right)$ of 0.1 V current remains zero.
6. Increase the forward-bias voltage upto 0.3 V for Ge diode. Current remains zero, (it is due to junction potential barrier of 0.3 V ).
7. Increase VF to 0.4 V . Milli-ammeter records a small current.
8. Increase $\mathrm{V}_{\mathrm{F}}$ in steps of 0.2 V and note the corresponding current. Current increases first slowly and then rapidly, till $\mathrm{V}_{\mathrm{F}}$ becomes 0.7 V .
9. Make $V_F=0.72 \mathrm{~V}$. The current increases suddenly. This represents "forward break-down" stage.
10. If the $V_F$ increases beyond forward breakdown' stage, the forward current does not change much. Now take out the key at once.

Calculations
For forward-bias
Plot a graph between forward-bias voltage $V_F$ (column 2) and forward current If (column 3 ) taking $V_F$ along xaxis and If along $Y$-axis.

This graph is called forward-bias characteristic curve a junction diode.

From graph, For change from point A to B

$$
\Delta V_F=2.4-2.0=0.4 \mathrm{~V}, \Delta I_F=30-20=10 \mathrm{~mA}
$$

Hence junction resistance for forward-bias,

$$
r=\frac{\Delta V_F}{\Delta I_F}=\frac{0.4 \mathrm{~V}}{10 \mathrm{~mA}}=40
$$
 

Result 
Junction resistance for forward-bias is 40 ohms.