Chapter 7: pn Junction Diode: Small-Signal Admittance

Note on the source: the OCR of this chapter is noisy. Where a symbol or exponent was lost I restored it from physics and cross-checks, and I say so when it matters. For example, the area of the 1N5472A is $A = 3.72 \times 1 0^{- 3} c m^{2}$ , because that is the value that reproduces the book’s quoted $N_{A} = 8.84 \times 1 0^{15} c m^{- 3}$ .

7.1 Introduction

Picture a diode sitting quietly at some DC operating point. Now tap it very gently: add a tiny sinusoidal voltage $v_{a} = V_{0} sin ω t$ on top of the DC bias $V_{A}$ , and listen for the current that comes back, $i$ . This “tap and listen” experiment is the small-signal response. Because the tap is small (tens of millivolts or less), the diode answers linearly, and the whole answer is packaged in one complex number, the small-signal admittance

$Y = \frac{i}{v _{a}} .$

The junction region responds in two ways. It can store charge (a capacitor) and it can pass current (a conductor). So in general

$\boxed{Y=G+j\omega C}\tag{7.1}$

with $j = - 1$ and $ω$ the angular frequency in rad/s.

Equivalent circuit. The circuit has a conductance $G$ in parallel with a capacitance $C$ (the junction admittance), in series with the series resistance $R_{S}$ of the contacts and quasineutral regions (introduced in Chapter 6). $C$ and $G$ both depend on the DC bias, which is why the book draws arrows through them. $R_{S}$ is normally tiny compared with the junction impedance except at large forward bias, so from here on we assume $R_{S} \approx 0$ unless stated.

Plan of attack.

Reverse bias: the conductance is tiny, so $Y \approx j ω C$ . Only majority carriers wiggle.
Forward bias: conductance matters, and minority carriers join the dance, producing the diffusion admittance.

Analogy: a diode is like a drum. Under reverse bias you tap the drum skin (the depletion edges) and it flexes like a spring (capacitance). Under forward bias there is also a leaky pipe through it (conductance), and water sloshing in the pipe (minority carriers) lags behind your tapping when you tap fast.

7.2 Reverse-Bias Junction Capacitance

7.2.1 General Information

Under reverse bias the pn diode is functionally a capacitor. Many “capacitors” inside integrated circuits are really reverse-biased pn junctions. It differs from an ordinary capacitor in one striking way: its capacitance decreases monotonically as the reverse bias increases (the sample data in Fig. 7.3, from a 1N5472A abrupt-junction diode, fall from about 120 pF at 0 V to a few pF near $- 25$ V).

Thought experiment: what happens during one a.c. cycle? The total junction voltage is $V_{A} + v_{a}$ .

Positive half cycle ( $v_{a} > 0$ ): the reverse bias is slightly reduced, the depletion width shrinks a little, so some charge is uncovered then removed from each side. A few majority carriers rush back toward the junction and neutralize edge charge.
Negative half cycle ( $v_{a} < 0$ ): reverse bias increases, the depletion region widens, and more ionized dopants are uncovered on both sides.

So the depletion width $W$ breathes in and out around its DC value, and a charge density oscillation $Δ ρ$ appears. Because $v_{a}$ is tiny, the $Δ ρ$ is effectively a thin sheet of $+$ charge at one edge of the depletion region and a thin sheet of $-$ charge at the other, a distance $W$ apart. That is exactly a parallel-plate capacitor:

$\boxed{C_J=\frac{K_S\varepsilon_0A}{W}}\tag{7.2}$

The subscript $J$ means junction (or depletion-layer) capacitance. Under reverse bias $C = C_{J}$ since nothing else oscillates significantly. Since $W$ grows with reverse bias, $C_{J} \propto 1/ W$ falls, matching the experiment.

(Figure 7.4 shows (a) the depletion edge moving, (b) the step-junction charge density $\pm q N$ profile with edges shifting, (c) $Δ ρ$ for $v_{a} > 0$ , (d) $Δ ρ$ for $v_{a} < 0$ ; the $Δ ρ$ plots are the shifted profile minus the DC profile.)

Quasistatic following. For the oscillation to happen, majority carriers must flow in and out of the quasineutral regions in sync with $v_{a}$ , responding as if $v_{a}$ were a DC bias. This is called following the signal quasistatically. Majority carrier response times in Si are about $1 0^{- 12}$ s or less (the dielectric relaxation time), so the quasistatic assumption is valid up to very high frequencies.

7.2.2 C-V Relationships

To get the exact $C$ versus $V_{A}$ we substitute the appropriate $W (V_{A})$ into (7.2). From Chapter 5:

$W=\left[\frac{2K_S\varepsilon_0}{qN_B}(V_{bi}-V_A)\right]^{1/2}\quad\text{(asymmetrical step)}\tag{7.3}$

$W=\left[\frac{12K_S\varepsilon_0}{qa}(V_{bi}-V_A)\right]^{1/3}\quad\text{(linearly graded)}\tag{7.4}$

where $N_{B}$ is the doping on the lightly doped side and $a$ is the grading constant. Rather than treat each separately, we generalize to a one-sided power-law profile: one side ( $x < 0$ ) heavily doped and the light side

$N_B(x)=b\,x^{m},\qquad x>0\tag{7.5}$

with $b > 0$ and $m$ constants. Special cases: $m = 0$ (step, $b = N_{B}$ ), $m = 1$ (one-sided linearly graded), and $m < 0$ (hyperabrupt: doping falls as you leave the junction; made by ion implantation or epitaxy). Fig. 7.5 shows (a) $m = 1$ , a straight line rising from the origin, (b) $m = 0$ , a flat line, (c) $m = - 1$ , a hyperbola falling away.

For $m > - 2$ the depletion width is

$\boxed{W=\left[\frac{(m+2)K_S\varepsilon_0}{qb}(V_{bi}-V_A)\right]^{1/(m+2)}}\tag{7.6}$

(This derivation is Problem 7.2; I derive it fully there.) Check: $m = 0, b = N_{B}$ gives (7.3); $m = 1$ with $b = a /4$ gives (7.4) (the " $a /4$ " accounts for the two-sided linear profile in the book’s (7.4) versus the one-sided one here).

Substituting (7.6) into (7.2):

$\boxed{C_J=\frac{K_S\varepsilon_0A}{\left[\dfrac{(m+2)K_S\varepsilon_0}{qb}(V_{bi}-V_A)\right]^{1/(m+2)}}}\tag{7.7}$

Define the zero-bias capacitance

$C_{J0}=C_J\big|_{V_A=0}=\frac{K_S\varepsilon_0A}{\left[\dfrac{(m+2)K_S\varepsilon_0V_{bi}}{qb}\right]^{1/(m+2)}}\tag{7.8}$

Dividing (7.7) by (7.8) the constants cancel:

$\boxed{C_J=\frac{C_{J0}}{\left(1-\dfrac{V_A}{V_{bi}}\right)^{1/(m+2)}}}\tag{7.9}$

Varactors. A diode built to exploit (7.7)/(7.9) is a varactor (“variable reactor”; reactance $= 1/ j ω C$ ). Uses: parametric amplification, harmonic generation, mixing, detection, voltage-variable tuning (think of the tuning knob of a radio being replaced by a voltage). The figure of merit is the tuning ratio

$\mathrm{TR}=\frac{C_J(V_{A1})}{C_J(V_{A2})}=\left[\frac{V_{bi}-V_{A2}}{V_{bi}-V_{A1}}\right]^{1/(m+2)}\;\xrightarrow{\;|V_A|\gg V_{bi}\;}\;\left(\frac{V_{A2}}{V_{A1}}\right)^{1/(m+2)}\tag{7.10}$

The exponent $1/ (m + 2)$ grows as $m$ becomes smaller, so TR increases going from linear ( $m = 1$ , exponent $1/3$ ) to step ( $m = 0$ , $1/2$ ) to hyperabrupt ( $m = - 1$ , exponent $1$ ). That is why hyperabrupt varactors are commercially valuable: the capacitance changes most per volt.

Exercise 7.1: Normalized C-V curves

Problem. Compute and plot $C_{J} / C_{J 0}$ versus $V_{A} / V_{bi}$ for the linearly graded, step, and $m = - 1$ hyperabrupt diodes, for $- 25 \leq V_{A} / V_{bi} \leq 0$ , and comment.

Solution logic. Equation (7.9) in normalized form is $y = (1 - x)^{- s}$ with $x = V_{A} / V_{bi}$ and $s = 1/ (m + 2)$ , so for $m = 1, 0, - 1$ we have $s = 1/3, 1/2, 1$ . These are “universal” curves: any one-sided power-law diode collapses onto one of them once axes are normalized. The code builds a matrix with one row per $m$ and plots all rows.

% Exercise 7.1 -- Normalized C-V curves
clear
m = [1 0 -1];            % linear, step, hyperabrupt
s = 1./(m+2);            % exponents 1/(m+2)
x = linspace(-25,0);     % x = VA/Vbi
y = [];                  % y = CJ/CJ0
for i = 1:3
    y = [y; 1./(1-x).^s(i)];
end
close
plot(x,y,'-'); grid
axis([-25 0 0 1])
xlabel('V_A/V_{bi}'); ylabel('C_J/C_{J0}')
text(-20,.42,'linear (m=1)')
text(-20,.27,'step (m=0)')
text(-20,.10,'hyperabrupt (m=-1)')

(The OCR lost the “1” in the vector m and the dot operators; the version above is the corrected one.)

Results and comments. At $V_{A} / V_{bi} = - 25$ the curves reach about $2 6^{- 1/3} = 0.34$ , $2 6^{- 1/2} = 0.20$ and $2 6^{- 1} = 0.038$ . The step-junction curve matches the 1N5472A data of Fig. 7.3 well if $V_{bi} \approx 1$ V. The tuning ratio with $V_{A 1} = 0$ and any $V_{A 2}$ is largest for the hyperabrupt diode and smallest for the linearly graded diode, as predicted by (7.10).

7.2.3 Parameter Extraction and Profiling

C-V data are used routinely to extract the average doping or the whole doping profile on the lightly doped side. Automated systems exist. The author’s lab uses an MSI C-V meter with a 15 mV rms, 1 MHz probing signal, four capacitance ranges (2, 20, 200, 2000 pF), and an internal DC supply with ranges $\pm 9.999$ V (1 mV steps) and $\pm 99.99$ V (10 mV steps). A PC controls bias and data; a probe box with a heated chuck (832 T-controller) holds wafers; encapsulated devices go into a spring-clip adapter. The meter measures and removes stray capacitance of cables, probe box and adapter. (Fig. 7.6 is a schematic of this system.)

Step-junction analysis. For an abrupt junction set $m \to 0, b \to N_{B}$ in (7.7). Invert and square:

$\frac{1}{C_J^2}=\frac{2(V_{bi}-V_A)}{qN_BK_S\varepsilon_0A^2}\tag{7.11}$

Step 1. $1/ C_{J}^{2}$ is linear in $V_{A}$ . Step 2. The slope is $- \frac{2}{q N _{B} K _{S} ε _{0} A ^{2}}$ , so a known area gives $N_{B}$ . Step 3. The intercept on the $V_{A}$ axis ( $1/ C_{J}^{2} = 0$ ) gives $V_{bi}$ . A straight $1/ C^{2}$ line also confirms that the junction is abrupt.

Exercise 7.2: Analysis of 1N5472A data

Problem. The 1N5472A is an $n^{+}$ - $p$ abrupt diode, $A = 3.72 \times 1 0^{- 3} c m^{2}$ , with about 2 pF stray capacitance in parallel with the junction. Using the Fig. 7.3 data, apply the $1/ C^{2}$ plot to confirm abruptness, find the $p$ -side doping and quote $V_{bi}$ .

Solution.

Subtract the 2 pF stray capacitance from every measured value to get $C_{J}$ .
Plot $1/ C_{J}^{2}$ versus $V_{A}$ (Fig. E7.2): the points fall on an almost perfect line, so the junction is abrupt.
Least squares gives

$\frac{1}{C _{J}^{2}} = (6.89 \times 1 0^{19}) - (9.78 \times 1 0^{19}) V_{A} (C_{J} in F, V_{A} in V)$

Compare with (7.11): slope magnitude $= 2/ (q N_{A} K_{S} ε_{0} A^{2})$ , so

Vault

Explorer

part07