3. 1)     Calculate the depletion layer thickness of a GaAs p-n junction formed from an n-type

layer doped at 1016 cm-3 and a p-type layer doped at 1017 cm-3 .  (Assume that the built in voltage is 1 volt.)  If carriers cross the depletion layer with a saturation drift velocity 107 cm/s, what is the transit time?  If the thickness of the n-layer is 10m and the mobility is 1000 cm2/Vs, calculate the RC time constant for the structure.  How     will these results change if the junction is reverse biased with a voltage supply of 8     volts?

wdepl  = √           (      +      ) (Vbi  − V)

wdepl

= √ ( + ) (1 V − 0)

Remember 1F = 1 C/V …

wdepl  = √ 1.60x10−9 F V C−1  = 4.0x10−5cm  = 0.40 um

wdepl           4.0x10−5cm

Ctransit  =   vsat     = 1x107cm s −1  = 4.0 ps

Cdepl          ese0           13.2(8.85x10−14 F cm−1)

A         wdepl                          4.0x10−5cm

Since Nd < Na, I will neglect the resistance of the p-type layer.

RA = pL =  =  =  = 6.2x10−4Ω cm2 RC = RCdepl  = RA () = 6.2x10−4 Ω cm2 (29 nF cm−2) = 18 ps

In this case (V = 0), the speed of the diode is limited by the RC time constant.

For V = -8 V:

Under reverse bias, the depletion width increases. This decreases the capacitance and RC time constant, but increases the carrier transit time across the pn junction. In this case, the diode     response is now transit-time limited.

Cdepl

A

wdepl  = 1.20 um

Ctransit  = 12.0 ps

CRC  = 6. 1 ps

3.2)     Show that the total number of minority electrons injected into a diode with a long p-

base is given by np(0)Ln, where np(0) is the injected electron carrier density at the edge of the depletion region and Ln is the minority electron diffusion length.

anp (x)      1 aJn (x)     np (x) − np0

at          q    ax                   Tn

Jn (x) = qunn(x)E(x) + qDn                       ≅ qDn

where we have used the fact that E(x)  0 in the quasi-neutral p-base region. Substituting the 2nd equation into the first equation gives:

a2np (x)     np (x) − np0

ax2                           Tn

Let np(´)(x) ≡ np (x) − np0          gives:

a2np(´)(x)     np(´)(x)

ax2                  Tn

The solution to this equation is: Let np(´)(x) = A exp {} + B exp {} where Ln  ≡ √Dn Tn

If we assume that the n-base is on the right-hand side, then the carrier density must    decrease as the carriers diffuse towards the right and recombine. This means that B=0 for this case. Thus, we obtain:

np(´)(x) = np(´)(0)exp {}

n´p (x)

0            Lp                                                                                                          x

To obtain the total number of minority electrons in the p-base, we simply integrate their density across the width of the p-base:

∞

Total number of minority electrons = ∫ np(´)(0)exp{} dx

0

Total number of minority electrons == −np(´)(0)Lnexp{      }|    = −np(´)(0)Ln[0 − 1]

Total number of minority electrons = np(´)(0)Ln

3.3)     Consider a silicon pn diode with a junction area is 100m x 100m. The n- and p-type

regions have doping concentrations of Nd = 1x1016 cm-3 and Na = 1x1018 cm-3 with      corresponding layer thicknesses of Wn = 500m and Wp = 4m. The electron and hole mobilities are n = 1250 cm V2- 1s and p = 100 cm V2- 1s and the minority carrier           lifetimes are Tn = Tp = 10s. Assuming that a current of 10mA is flowing, do the          following:

[For this question, answers are given, but not the worked solutions. (You are expected to do these yourself…)]

a) Calculate the minority carrier diffusion lengths, Ln and Lp,

Answers:        Ln = 180 m    and    Lp = 50.9 m

b) Explain whether the n- and p-regions correspond to long or short bases.

Answers:        Long n-base and short p-base

c) Calculate the current density across the junction and the voltage that needs to be applied for this current to flow;

Answers:        J = 100 A cm-2         and    V = 0.806V

d) Calculate the excess minority carrier distributions in the n- and p-regions.

Answers:

Assuming that the p-side of the junction is on the left (x<0):

np(´)(x) = 4.36x1015 [1 + 2500x] cm −3

pn(´)(x) = 4.36x1017 exp(−196x)

where x is in cm …

e) Calculate the diffusion capacitances associated with the minority electrons in the p- base and the minority holes in the n-base.

Answers: Cdiff_n = 583 pF     and     Cdiff_p = 1.48 F

(n.b. Use the correct equations for short- and long- bases and the bias from Q3.3c …)

f) Calculate the depletion width, Wdepl .

Answer:  Wdepl = 50 nm

g) Calculate the junction capacitance, Cdepl .

Answer:  Cdepl = 2.08 nF

3.4)     Draw the small signal equivalent circuit of a discrete pn diode, being sure to include

parasitic circuit elements. Explain the physical origin of each term.

Lw              = bond wire inductance

Rs               = series resistance of the quasi-neutral

bases and the contact resistances

rd                = differential resistance, rd = nkT/qI

Cj               = depletion capacitance ofjunction E-field

Cdiff          = diffusion capacitance of injected

minority carriers

3.5)     You are asked to design a simple pn diode for an application requiring a breakdown

voltage of 1000V. For low doping levels, the avalanche breakdown field of silicon is EBR = 2x105 V/cm. If the p-type layer (top layer) has a doping density of Na = 1x1018 cm-3, what is the maximum n-type doping concentration your pn diode can have? (Be sure to leave a safety margin (e.g. 20%) in the breakdown voltage.) Plot the electric   field distribution across the depletion region.

Increasing the breakdown voltage by 20% give a design value of VBR=1200 V.

 + ) (Vbi  + 1200V)

xn  = (                ) wdepl  = √

but

Emax  =  = √ () (Vbi  + 1200V)

so

es e0 Em(2)ax                     Na Nd                1       1   −1

2q(Vbi  + 1200V)      Na  + Nd            Na        Nd

2q(Vbi  + 1200V)      1   −1

es e0 Em(2)ax                  Na

where

kT          Na Nd

q            ni(2)

If we guess Nd = 1015 cm-3, then

Vbi  = 0.0259ln () = 0.765 V

In turn, this gives:

Nd  = [                                                                                     −                         ]

Nd  = 1.086x1014 cm −3