EXAMPLE 5.1 OBJECTIVE Vbi = 0.695 V Calculate the built-in potential barrier of a pn junction. Consider a silicon pn junction at T = 300 K with doping concentrations of Na = 2  1016 cm-3 and Nd = 5  1015 cm-3. Solution The built-in potential barrier is determined from or Vbi = 0.695 V Comment The built-in potential barrier changes only slightly as the doping concentrations change by orders of magnitude because of the logarithmic dependence.

The space charge width extending into the p region is found to be EXAMPLE 5.2 OBJECTIVE Calculate the space charge widths and peak electric field in a pn junction. Consider a silicon pn junction at T = 300 K with uniform doping concentrations of Na = 2  1016 cm-3 and Nd = 5  1015 cm-3. Determine xn , xp , W , and max . Solution In Example 5.1, we determined the built-in potential barrier, for these same doping concentrations, to be Vbi = 0.695 V. or xn = 0.379  10-4 cm = 0.379 m The space charge width extending into the p region is found to be xp = 0.0948  10-4 cm = 0.0948 m

EXAMPLE 5.2 The total space charge width, using Equation (5.31), is or W = 0.474  10-4 cm = 0.474 m We can note that the total space charge width can also be found from W = xn + xp = 0.379 + 0.0948 = 0.474 m The maximum or peak electric field can be determined from, for example, max = 2.93  104 V/cm Comment We can note from the space charge width calculations that the depletion region extends farther into the lower-doped region. Also, a space charge with on the order of a micrometer is very typical of depletion region widths. The peak electric field in the space charge region is fairly large. However, to a good first approximation, there are no mobile carriers in this region so there is no drift current. (We will modify this statement slightly in Chapter 9.)

EXAMPLE 5.3 OBJECTIVE Solution Calculate width of the space charge region in a pn junction when a reverse-bias voltage is applied. Again, consider the silicon pn junction at T = 300 K with uniform doping concentrations of Na = 2  1016 cm-3 and Nd = 5  1015 cm-3. Assume a reverse-bias voltage of VR = 5 V is applied. Solution From Example 5.1, the built-in potential was found to be Vbi = 0.695 V. The total space charge width is determined to be or W = 1.36  10-4 cm = 1.36 m Comment The space charge width has increased from 0.474 m to 1.36 m at a reverse bias voltage of 5 V.

EXAMPLE 5.4 OBJECTIVE Comment Nd = 3.02  1015 cm-3 Design a pn junction to meet a maximum electric field specification at particular reverse-bias voltage. Consider a silicon pn junction at T = 300 K with a p-type doping concentration of Na = 1018 cm-3. Determine the n-type doping concentration such that the maximum electric field in the space charge region is max = 105 V/cm at a reverse bias voltage of VR = 10 V . The maximum electric field is given by Since Vbi is also a function of Na through the log term, this equation is transcendental in nature and cannot be solved analytically. However, as an approximation, we will assume that Vbi  0.75 V. We can then write which yields Nd = 3.02  1015 cm-3 We can note that the built-in potential for this value of Nd is Which is very close to the assumed value used in the calculation. So the calculated value of Nd is a very good approximation. Comment A smaller value of Nd than calculated results in a smaller value of max for a given reverse-bias voltage. The value of Nd determined in this example, then, is the maximum value that will meet the specifications.

EXAMPLE 5.5 OBJECTIVE Solution Comment Calculate the junction capacitance of a pn junction. Consider the same pn junction as described in Example 5.3. Calculate the junction capacitance at VR = 5 V assuming the cross-sectional area of the pn junction is A = 10-4 cm2. Solution The built-in potential was found to be Vbi = 0.695 V. The junction capacitance per unit area is found to be or C = 7.63  10-9 F/cm2 The total junction is found as C = AC = (10-4) (7.63  10-9) C = 7.63  10-2 F = 0.763 pF Comment The value of the junction capacitance for a pn junction is usually in the pF range, or even smaller.

EXAMPLE 5.6 OBJECTIVE Solution Comment Determine the impurity concentrations in a p+n junction given the parameters from Figure 5.12. Consider a silicon p+n junction at T = 300 K. Assume the intercept of the curve on the voltage axis in Figure 5.12 gives Vbi = 0.742 V and that the slope is 3.92  1015 (F/cm2)-2/V. Solution The slope of the curve in Figure 5.12 is given by 2/esNd , so we canwrite or Nd = 3.08  1015 cm-3 The built-in potential is given by Solving for Na , we find Na = 2.02  1017 cm-3 Comment The results of this example show that Na >> Nd ; therefore the assumption of a one-sided junction was valid.

EXAMPLE 5.7 OBJECTIVE Solution Comment Determine the diode current in a silicon pn junction dilde. Consider a silicon pn junction diode at T = 300 K. The reverse-saturation current is IS = 10-14 A. Determine the forward-bias diode current at VD = 0.5 V, 0.6 V, and 0.7 V. Solution The diode current is found from so for VD = 0.5 V, ID = 2.42 m and for VD = 0.6 V, ID = 0.115 m and for VD = 0.7 V, ID = 5.47 m Comment Because of the exponential function, reasonable diode currents can be achieved even though the reverse-saturation current is a small value.

EXAMPLE 5.8 OBJECTIVE Solution Comment Calculate the forward-bias voltage required to generate a forward-bias current density of 10A/cm2 in a Schottky diode and a pn junction diode. Consider diodes with parameters JsT = 6  10-5 A/cm2 and JS = 3.5  10-11 A/cm2. Solution For the Schottky diode, we hare Neglecting the (1) term, we can solve for the forward-bias voltage. We find For the pn junction diode, we have Comment A comparison of the two forward-bias voltages shows that the schottky diode has an effective turn-on voltage that, in this case, is approximately 0.37 V smaller than the turn-on voltage of the pn junction diode.

EXAMPLE 5.9 OBJECTIVE xn = 1.1  10-6 cm = 110 Ǻ Calculate the space charge width for a Schottky barrier on a heavily doped semiconductor. Consider silicon at T = 300 K doped at Nd = 7  1018 cm-3. Assume a Schottky barrier with B0 = 0.67 V. For this case, we can assume that Vbi  B0. Solution For a one-sided junction, we have for zero applied bias or xn = 1.1  10-6 cm = 110 Ǻ Comment In a heavily doped semiconductor, the depletion width is on the order of angstroms, so that tunneling is now a distinct possibility. For these types of barrier widths, tunneling may become the dominant current mechanism.