PN Junction Diode Analysis

Six worked problems on the pn junction diode: barrier potential from the doping, the Shockley equation used in both directions, static versus dynamic resistance, a silicon-germanium diode network, load-line analysis compared across the ideal, constant-0.7 V and piecewise-linear models, and the design of a series resistor for a specified diode current.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

An abrupt pn junction is formed with \( N_A = 1\times10^{17}~\text{cm}^{-3} \) on the p-side and \( N_D = 1\times10^{16}~\text{cm}^{-3} \) on the n-side, at 300 K where \( V_T = 25.85~\text{mV} \).

1. Find the built-in (barrier) potential if the material is silicon, \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \).

2. Repeat for germanium with the same doping, \( n_i = 2.5\times10^{13}~\text{cm}^{-3} \).

3. By how much does \( V_{bi} \) of the silicon junction rise if \( N_D \) is raised to \( 1\times10^{18}~\text{cm}^{-3} \)?

• \( N_A = 1\times10^{17} \), \( N_D = 1\times10^{16}~\text{cm}^{-3} \), fully ionised

• \( V_T = 25.85~\text{mV} \) at 300 K

• Abrupt junction, non-degenerate doping, thermal equilibrium (no applied bias)

Relation

The built-in potential is the amount by which the bands must bend to line up the two Fermi levels:

Part (a) — Silicon

\( V_{bi} = 0.753~\text{V} \)

Part (b) — Germanium, identical doping

\( V_{bi} = 0.369~\text{V} \)

Part (c) — Heavier n-side doping in silicon

\( V_{bi} \) rises to \( 0.872~\text{V} \), an increase of \( 119~\text{mV} \)

Every decade of doping on either side adds only \( V_T\ln 10 = 59.5~\text{mV} \), so \( V_{bi} \) is stubbornly close to \( 0.7~\text{V} \) for practical silicon junctions. The whole gap between the silicon and germanium answers, \( 0.384~\text{V} \), comes from \( n_i \) alone — the same reason a germanium diode turns on near \( 0.3~\text{V} \) and a silicon one near \( 0.7~\text{V} \).

A silicon diode has a reverse saturation current \( I_S = 1\times10^{-13}~\text{A} \) and an ideality factor \( \eta = 1 \). Take \( V_T = 25.85~\text{mV} \) at 300 K.

1. Find the current at a forward bias of \( 0.6~\text{V} \).

2. Find the forward voltage that gives \( 10~\text{mA} \), and the voltage at \( 100~\text{mA} \).

3. Find the current at a reverse bias of \( 0.5~\text{V} \).

4. How much extra forward voltage multiplies the current by ten?

• \( I = I_S\!\left(e^{V/\eta V_T}-1\right) \), \( I_S = 1\times10^{-13}~\text{A} \), \( \eta = 1 \)

• \( V_T = 25.85~\text{mV} \); temperature constant throughout (no self-heating)

Part (a) — Current at 0.6 V

\( I = 1.203~\text{mA} \)

Part (b) — Voltage at a given current

With \( I \gg I_S \) the \( -1 \) is irrelevant, so invert the exponential:

\( V = 0.6547~\text{V} \) at 10 mA and \( 0.7143~\text{V} \) at 100 mA

Part (c) — Reverse bias

\( I \approx -0.1~\text{pA} \) — the exponential is dead within a few \( V_T \) of reverse bias, so the current simply saturates at \( -I_S \)

Part (d) — Decade of current

\( \Delta V = 59.5~\text{mV} \) per decade of current

That is the practical meaning of the "0.7 V" rule: between 1 mA and 100 mA the drop moves by only two decades \( \times\ 59.5~\text{mV} \approx 119~\text{mV} \). Over the whole useful range the diode voltage barely stirs, which is why treating it as a constant is usually good enough.

The same silicon diode (\( I_S = 1\times10^{-13}~\text{A} \), \( \eta = 1 \), \( V_T = 25.85~\text{mV} \)) is biased in the forward direction.

1. Find the DC (static) resistance and the AC (dynamic) resistance at a bias of \( 5~\text{mA} \).

2. Tabulate both at 0.1, 1, 5 and 20 mA and comment on how differently they behave.

3. A small signal of \( 5~\text{mV} \) peak is superposed on the 5 mA bias. Find the peak signal current, and check that the small-signal assumption is reasonable.

• \( R_{DC} = V_D/I_D \) — the ratio of the total voltage to the total current at the operating point

• \( r_d = dV/dI = \eta V_T/I_D \) — the slope of the characteristic at that point

• Bulk and contact resistance neglected; \( \eta = 1 \), \( V_T = 25.85~\text{mV} \)

Part (a) — At \( I_D = 5~\text{mA} \)

First find the operating voltage from the Shockley equation:

\( R_{DC} = 127.4~\Omega \), \( r_d = 5.170~\Omega \)

Part (b) — Across the bias range

\( R_{DC} \) falls by a factor of roughly 159 over this range while \( r_d \) falls by exactly 200, in strict inverse proportion to current. The two are never equal and never should be compared: \( R_{DC} \) is a bias-point bookkeeping number, \( r_d \) is what the signal sees.

Part (c) — Small-signal response

\( i_d = 0.967~\text{mA} \) peak, riding on the 5 mA bias

Linearity requires \( v_d \ll V_T \). Here \( 5~\text{mV} \) is about 19 per cent of \( 25.85~\text{mV} \), so the second-order term in the expansion of \( e^{v/V_T} \) contributes only a few per cent — acceptable, but a 25 mV signal would not be.

Note that \( r_d \) does not depend on the diode's construction at all, only on the bias current: \( r_d = 25.85~\Omega \) at 1 mA, \( 5.170~\Omega \) at 5 mA. This is the same \( r_e = V_T/I_E \) that reappears in every small-signal transistor model.

Use the constant-voltage-drop model throughout: a conducting silicon diode holds \( 0.7~\text{V} \), a conducting germanium diode \( 0.3~\text{V} \), and a non-conducting diode is an open circuit.

1. A \( 12~\text{V} \) source drives a silicon diode, a germanium diode and a \( 5.6~\text{k}\Omega \) resistor, all in series and all forward-oriented. Find the current and the resistor voltage.

2. A \( 20~\text{V} \) source feeds a \( 2.2~\text{k}\Omega \) resistor, whose lower end drives a silicon diode and a germanium diode connected in parallel, both pointing to ground. Find the output voltage at that node, the resistor current, and the current in each diode.

3. Two identical silicon diodes are placed in parallel, in series with a \( 1~\text{k}\Omega \) resistor across a \( 10~\text{V} \) source. Find the current in each.

• Silicon: \( V_D = 0.7~\text{V} \) when on. Germanium: \( V_D = 0.3~\text{V} \) when on.

• Constant-voltage-drop (second approximation) model; bulk resistance neglected.

Part (a) — Series string

One loop, so one current. KVL round the loop:

\( I = 1.964~\text{mA} \), \( V_R = 11.0~\text{V} \)

Part (b) — Parallel pair: the germanium diode wins

Both diodes cannot conduct at once — they share the same node voltage. The germanium diode turns on at \( 0.3~\text{V} \) and clamps the node there; the silicon diode then sees only \( 0.3~\text{V} \), which is below its \( 0.7~\text{V} \) threshold, so it stays off.

\( V_o = 0.3~\text{V} \), \( I_{Ge} = 8.955~\text{mA} \), \( I_{Si} = 0 \)

The germanium diode alone dissipates \( (0.3)(8.955~\text{mA}) = 2.686~\text{mW} \).

Part (c) — Two identical silicon diodes in parallel

\( 4.65~\text{mA} \) in each diode, \( 9.30~\text{mA} \) total

Part (b) is the lesson worth keeping: paralleled diodes do not share — the one with the lowest turn-on takes everything. Part (c)'s tidy 50:50 split holds only for perfectly matched devices; a mismatch of \( 59.5~\text{mV} \) in \( V_D \) would send ten times as much current through one of them, which is why real paralleled diodes get individual ballast resistors.

A silicon diode is in series with \( R = 1~\text{k}\Omega \) across a \( 10~\text{V} \) DC supply, forward-biased. The diode obeys \( I = I_S(e^{V/V_T}-1) \) with \( I_S = 1\times10^{-13}~\text{A} \), \( \eta = 1 \), \( V_T = 25.85~\text{mV} \).

1. Write the load line and give its two intercepts.

2. Find the operating point using (i) the ideal-diode model, (ii) the constant \( 0.7~\text{V} \) model, and (iii) the piecewise-linear model with \( V_{T0}=0.7~\text{V} \) and bulk resistance \( r_B = 20~\Omega \).

3. Solve the exact transcendental equation by iteration and compare all four answers.

4. Repeat the comparison with the supply reduced to \( 1~\text{V} \).

• \( V_S = 10~\text{V} \), \( R = 1~\text{k}\Omega \), \( I_S = 1\times10^{-13}~\text{A} \), \( \eta = 1 \)

• The diode and the resistor carry the same current — one loop, KVL applies.

Step 1 — The load line

Its intercepts on the diode's \( I\!-\!V \) plane are

• \( V_D = 0 \Rightarrow I = 10.0~\text{mA} \) (short-circuit current)

• \( I = 0 \Rightarrow V_D = 10~\text{V} \) (open-circuit voltage)

The operating point is where this straight line crosses the diode curve.

Step 2 — The three approximations

Step 3 — Exact solution by iteration

The exact statement, \( (10-V_D)/1000 = I_S e^{V_D/V_T} \), has no closed form. Guess \( V_D = 0.7~\text{V} \), get \( I \) from the load line, get a better \( V_D \) from \( V_D = V_T\ln(I/I_S) \), and repeat:

It converges in two passes, because the load line is nearly vertical where it meets the diode curve.

Exact: \( V_D = 0.6530~\text{V} \), \( I = 9.347~\text{mA} \)

Step 4 — Comparison at \( V_S = 10~\text{V} \)

Every model lands within a few per cent of the current, even the ideal one, because \( V_S \gg V_D \): the resistor, not the diode, is setting the current. (The piecewise-linear entry is the worst here only because \( r_B = 20~\Omega \) is a deliberately generous bulk resistance — the exact model above has none.)

Step 5 — The same circuit at \( V_S = 1~\text{V} \)

At \( 1~\text{V} \) the constant-drop model is in error by -29.7 per cent, against a fraction of a per cent at \( 10~\text{V} \)

The moral of the load line: the accuracy you need from the diode model depends entirely on how much of the supply is dropped across it. When the diode holds a large share of the loop voltage, solve the exponential; when the resistor dominates, \( 0.7~\text{V} \) is plenty.

Design problem. A silicon diode is to be biased at \( 20~\text{mA} \) from a \( 9~\text{V} \) supply through a single series resistor. The diode is described by the piecewise-linear model \( V_D = 0.7 + I_D r_B \) with \( r_B = 10~\Omega \), and its absolute maximum forward current is \( 25~\text{mA} \). The supply is specified to \( \pm10 \) per cent.

1. Compute the ideal resistor value and choose the nearest E24 standard value from 300, 330, 360, 390, 430, 470, 510 \( \Omega \).

2. Verify the current at nominal supply and at both supply extremes against the 25 mA rating.

3. Specify the power rating of the resistor and state the diode dissipation.

• \( V_S = 9~\text{V} \pm 10 \) per cent, target \( I_D = 20~\text{mA} \), \( I_{D,\max} = 25~\text{mA} \)

• Diode: \( V_D = 0.7 + I_D(10~\Omega) \) — piecewise-linear, valid over the milliampere range

• Resistor tolerance ignored; only the supply tolerance is budgeted.

Step 1 — Ideal resistance

Equivalently, \( R = (V_S - 0.7)/I_D - r_B = 415.0 - 10 = 405~\Omega \).

Step 2 — Choose a standard value

\( 405~\Omega \) is not a stock value; the E24 neighbours are 390 and 430 \( \Omega \). The nearest is \( 390~\Omega \), and the smaller resistor is the one that keeps the current closer to the 20 mA target:

Choose \( R = 390~\Omega \), giving \( I_D = 20.75~\text{mA} \)

Step 3 — Supply-tolerance check

Worst-case current \( 23.00~\text{mA} < 25~\text{mA} \) — the design passes

Had \( 430~\Omega \) been chosen instead, the worst case would be only \( 20.91~\text{mA} \) — safer, but the nominal current would sit 5.7 per cent below specification.

Step 4 — Power ratings

Use a \( 390~\Omega \), \( \tfrac{1}{2}~\text{W} \) resistor; the diode dissipates about \( 21.4~\text{mW} \)

A quarter-watt resistor would be running at 82.5 per cent of its rating at maximum supply, which is poor practice — the usual rule is to derate to about half, so the next size up is the right call. Note also how insensitive the current is to the diode model here: because \( V_S \gg V_D \), a \( \pm50~\text{mV} \) error in the assumed drop shifts \( I_D \) by only about 0.60 per cent.