By the end of this chapter you should be able to:
- Explain why a circuit containing a diode cannot be solved by the linear methods of network analysis.
- Construct a load line from the supply and the Thévenin resistance, and locate the Q-point as its intersection with the device characteristic.
- Solve a diode circuit iteratively, and explain why the iteration converges in two or three passes.
- Compare model-based answers with the iterative answer numerically, and predict when a model's error will become unacceptable.
- Analyse series, parallel and series-parallel diode circuits, including the problem of current sharing between paralleled diodes.
- Separate the DC and AC analysis of a diode carrying a signal on top of a bias, and use \(r_d\) correctly.
- Predict the movement of the Q-point with temperature and explain why a stiff supply protects the bias current.
Every circuit analysed so far in this course has been linear, and linear circuits are easy in a specific technical sense: superposition applies, Thévenin equivalents exist, and the simultaneous equations Kirchhoff produces can be solved by elimination. Put one diode in the loop and all three of those properties vanish. The equation that describes a supply, a resistor and a diode in series is
and no rearrangement will give \(I_D\) in terms of the other symbols. It is transcendental: the unknown appears both inside and outside a logarithm. This is not a failure of technique but a permanent feature of circuits containing devices, and every method in this chapter is a way of living with it.
There are three such ways, and a working engineer uses all of them. Graphical analysis draws the device characteristic and the circuit constraint on the same axes and reads off the intersection; it is slow and imprecise, but it is the only method that shows you at a glance what is going on, and it generalises directly to the transistor bias problem of Chapter 18. Iterative analysis guesses, computes, and repeats until the guess stops changing; it is what a simulator does, and it is far quicker by hand than its reputation suggests. Model-based analysis replaces the diode by the linear approximations of Chapter 6 and solves the resulting linear circuit exactly; it is what you will do ninety-five times out of a hundred, and the useful skill is knowing how much error it costs.
1 A Circuit with One Non-Linear Element
Take the simplest possible case: a supply \(V_S\), a resistor \(R\) and a diode in series. Two facts are available. Kirchhoff's voltage law gives a relation between the diode's voltage and its current that comes entirely from the rest of the circuit,
A straight line in the \((V_D, I_D)\) plane, with intercepts \(V_S\) on the voltage axis and \(V_S/R\) on the current axis. It contains no information about the diode at all — it would be identical if the diode were replaced by a thermistor or a lamp.
The device gives a second relation, the Shockley equation of Chapter 6, which contains no information about the circuit. The operating point must satisfy both simultaneously, and since one is linear and the other exponential, the pair has no algebraic solution.
It is worth pausing on why this is not a contrived difficulty. Every active device is non-linear; that is what makes it active. A resistor obeys \(I = V/R\) and can only attenuate. Amplification, rectification, switching, oscillation and logic all require a device whose characteristic bends, and the price of that bend is that the circuit equations stop being solvable in closed form. The methods of this chapter are therefore the standard methods of all device electronics, and every one of them recurs in Chapter 18 when the same problem appears with a transistor and two more terminals.
Notice also what the linear relation is. Looking out of the diode's two terminals into the rest of the circuit, whatever that rest contains — sources, resistors, ladders — it can be replaced by its Thévenin equivalent \(V_{Th}\) in series with \(R_{Th}\), because that part of the circuit is linear. So the general form of the constraint is
Which is why a single technique handles every circuit with one diode in it, however complicated the surrounding network: reduce the network to two numbers, then solve.
The three worked cases of this chapter all use \(V_S = 10\) V and \(R = 1\) k\(\Omega\) with a diode of \(I_S = 10\) fA and \(n = 1\) at 300 K, so that the answers can be compared directly. The exact solution, which we will obtain by iteration in a moment, is \(V_D = 0.7124\) V and \(I_D = 9.288\) mA. Keep those two numbers in view: every method below is judged against them.
2 The Load Line and the Q-Point
Plot the diode's measured or computed characteristic on axes of \(I_D\) against \(V_D\). Then plot the circuit constraint on the same axes. Because it is a straight line, two points suffice, and the two easiest are the axis intercepts:
- Set \(I_D = 0\): the diode is open, no current flows through \(R\), no voltage is dropped across it, and \(V_D = V_S = 10\) V. That is the voltage-axis intercept.
- Set \(V_D = 0\): the diode is a short, and the whole supply appears across \(R\), giving \(I_D = V_S/R = 10\) mA. That is the current-axis intercept.
Join them. The result is the load line, and its slope is \(-1/R\). The intersection with the device characteristic is the quiescent point or Q-point, the one pair \((V_D, I_D)\) consistent with both the device and the circuit.
Panel (a) of Figure 7.1 shows the honest full-scale picture and it is instructive precisely because it looks so unhelpful. On a 10 V axis the diode characteristic is indistinguishable from a vertical line at about 0.7 V; on a 10 mA axis the load line is a gentle diagonal. They cross almost exactly at the current-axis intercept. This is a geometrical statement of why the models of Chapter 6 work so well: when \(V_S \gg V_D\), the Q-point current is set almost entirely by \(V_S\) and \(R\), and the diode contributes only a small correction.
Panel (b) magnifies a window 30 mV wide and 1.5 mA tall around the crossing. At this scale the roles reverse: the diode characteristic is steep and the load line is nearly flat. The lesson is worth stating as a slogan — the diode sets the voltage, the resistor sets the current — and it explains why a diode makes a passable voltage reference and a hopeless current source.
Reading a Q-point off a plot is accurate to perhaps two significant figures, which is often enough. The reason to draw the picture is not precision but the behaviour it makes visible:
- Increasing \(V_S\) slides the load line right, parallel to itself, and moves Q up the curve.
- Increasing \(R\) pivots the line about the fixed voltage intercept, lowering the current intercept and moving Q down.
- Heating the diode slides the curve left, which is the subject of the last section of this chapter.
- If the source is an AC waveform, the line sweeps back and forth, and the locus of Q traces the output waveform — the construction that Chapters 8 and 11 use to explain rectification and clipping.
The same construction handles a reverse-biased diode, a Zener in breakdown (Chapter 12 draws the load line in the third quadrant), and a transistor's output characteristics in Chapter 18, where a family of curves is crossed by one line and the Q-point choice becomes a design decision rather than a calculation.
3 Iteration, and Why It Converges So Fast
The transcendental equation can be solved to any accuracy you like by successive substitution, and by hand it takes about a minute. Rearrange it so that each of the two relations is used to improve the other:
Start from any sensible guess — \(V_D^{(0)} = 0.7\) V is the natural one — and alternate. The first expression uses the circuit; the second uses the device.
\(V_S = 10\) V, \(R = 1\) k\(\Omega\), \(I_S = 10^{-14}\) A, \(n = 1\), \(V_T = 25.852\) mV. Start at \(V_D = 0.700\) V.
| Pass | \(I_D = (10 - V_D)/1000\) | \(V_D = 0.025852\ln(I_D/10^{-14})\) | Change in \(V_D\) |
|---|---|---|---|
| 1 | 9.300000 mA | 0.712441 V | — |
| 2 | 9.287559 mA | 0.712406 V | −35 µV |
| 3 | 9.287594 mA | 0.712407 V | +1 µV |
| 4 | 9.287593 mA | 0.712407 V | < 1 µV |
Two passes give six correct figures. The answer is \(V_D = 0.7124\) V, \(I_D = 9.2876\) mA.
The speed is not luck, and understanding it tells you when iteration will misbehave. Each pass multiplies the error by a factor \(-r_d/R\), where \(r_d = nV_T/I_D\) is the dynamic resistance of Chapter 6. Here \(r_d = 25.852/9.2876 = 2.78\ \Omega\) against \(R = 1000\ \Omega\), so the error shrinks by a factor of 360 per pass; a 12 mV starting error becomes 33 µV, then 0.1 µV. The alternating sign of the correction in the table is the minus sign in that factor.
The condition for convergence is therefore \(r_d < R\), which is another way of saying the load line must be flatter than the device curve at the crossing. That is true in every ordinary biasing circuit, because a circuit in which \(r_d > R\) is one where the diode dominates and the current is wildly unpredictable — a circuit you would not design. If you ever meet the divergent case, iterate the other way round: guess \(I_D\), compute \(V_D\) from the device, then compute a new \(I_D\) from the circuit.
Two practical points. First, use a starting guess from the constant-drop model rather than zero; starting at \(V_D = 0\) gives \(I_D = 10\) mA on the first pass, which is still fine, but starting at \(V_D = 10\) V gives a negative current and a domain error in the logarithm. Second, iteration is the only method that stays accurate when the supply is comparable with the diode drop. With \(V_S = 2\) V and the same resistor, iteration gives \(V_D = 0.6623\) V and \(I_D = 1.3377\) mA in five passes, and there is no model in Chapter 6 that gets within 2 % of that.
4 Model-Based Analysis and the Size of Its Error
Now do the same circuit three more times, each with a linear model, and compare against the iterative answer of 9.2876 mA.
Ideal model. The diode is a closed switch, so the whole supply appears across \(R\):
Constant-voltage-drop model. The diode is a 0.7 V battery:
Piecewise-linear model, with \(V_{D0} = 0.65\) V and \(r_B = 10\ \Omega\):
Three things in that comparison deserve comment. The constant-drop model wins, and it wins because 0.7 V happens to be within 12 mV of the true drop at this current. The piecewise-linear model, despite having twice as many parameters, is worse here: its chord was fitted over a wide current range and does not pass through this particular operating point, and it overstates the voltage by 30 mV. More parameters do not guarantee more accuracy — only a better fit at the current of interest does. And the ideal model's 7.67 % error is exactly \(0.7124/(10-0.7124)\), the neglected drop as a fraction of the drop across the resistor, which is the general formula for that model's error.
Ideal-model error \(\approx V_D/(V_S - V_D)\). Constant-drop error \(\approx (V_D - 0.7)/(V_S-V_D)\). With \(V_S = 10\) V those are 7.7 % and 0.13 %; with \(V_S = 2\) V and the same 1 k\(\Omega\) the true answer is 1.3377 mA, the ideal model gives 2.000 mA (+49.5 %) and the constant-drop model gives 1.300 mA (−2.82 %). Below about 3 V, stop modelling and iterate.
A related question is which value of \(V_D\) to use in the constant-drop model, and the answer follows from the 60 mV/decade rule. If your circuit's current is within a factor of three of a milliampere, use 0.65 V; within a factor of three of 10 mA, use 0.70 V; at an ampere in a power rectifier, use 0.9 to 1.0 V and add the bulk resistance. Quoting 0.7 V for a diode running at 20 µA overstates its drop by about 130 mV, and in a 3 V circuit that is a 5 % error in the current for no reason at all.
Finally, a caution about what "error" means here. All of these figures compare models against the Shockley equation with assumed parameters, not against a real diode. A real 1N4148 has \(n\) somewhere between 1.6 and 1.9, an \(I_S\) that varies by an order of magnitude between production lots, and a junction that self-heats. Against a real device, all four answers above are within the part-to-part spread. That is the practical justification for the constant-drop model: refining the arithmetic beyond about 1 % is refining a number the component tolerance has already destroyed.
5 Series, Parallel and Series-Parallel Circuits
Diodes in series. Two identical diodes in series carry the same current, so their voltages add, and the constant-drop model simply doubles: \(I_D = (10 - 1.4)/1000 = 8.600\) mA. Solving exactly — the iteration is unchanged except that \(V_D = 2nV_T\ln(I_D/I_S)\) — gives 8.579 mA with 0.7104 V across each, so the model is 0.24 % high. Series stacks of diodes are used deliberately as crude references (three in series make a 2.1 V rail) and as level shifters, and in Chapter 20 a pair of them biases a push-pull output stage against crossover distortion.
Diodes in parallel. Here the interesting behaviour appears. Parallel diodes share a voltage, not a current, and because current depends exponentially on voltage, the sharing is unstable in a way that catches people out. Two identical diodes across the same 10 V and 1 k\(\Omega\) settle at 0.6945 V and split the 9.306 mA evenly, 4.653 mA each. Now make one of them twice as leaky — \(I_{S2} = 2I_{S1}\), well within a normal production spread:
A 2:1 mismatch in \(I_S\) produces an exact 2:1 mismatch in current, because both diodes see the same voltage and the current is proportional to \(I_S\) at fixed \(V\). One device takes 67 % of the load.
In a signal circuit that hardly matters. In a power rectifier it is dangerous, because the diode taking more current runs hotter, and a hotter diode has a larger \(I_S\), so it takes still more current. That is a positive feedback loop with no natural limit, and it is why paralleled power diodes are given small series ballast resistors: a few tens of milliohms each converts the shared-voltage problem into a shared-current one. Twenty milliohms carrying 10 A drops 0.2 V, which is comparable with the mismatch it is correcting, and that is the design criterion.
Series-parallel circuits. When the network around the diode is more elaborate, reduce it. Figure 7.2 shows a 12 V supply, a 1 k\(\Omega\) series resistor to a node A, and two branches to ground from A: a diode in series with 2.2 k\(\Omega\), and 3.3 k\(\Omega\) alone.
Assume the diode conducts and model it as 0.7 V. Kirchhoff's current law at node A:
Multiplying through by 6600 gives \(79.2 - 6.6V_A = 3V_A - 2.1 + 2V_A\), so \(11.6V_A = 81.3\) and \(V_A = 7.0086\) V. Then
- \(I_{R1} = (12 - 7.0086)/1000 = 4.9914\) mA
- \(I_D = (7.0086 - 0.7)/2200 = 2.8676\) mA
- \(I_{R3} = 7.0086/3300 = 2.1238\) mA, and the two branch currents sum to 4.9914 mA as they must.
Solving the same circuit with the full Shockley equation gives \(V_A = 7.0040\) V, \(V_D = 0.6821\) V and \(I_D = 2.8736\) mA. The constant-drop model is 0.21 % low in the diode current and 0.07 % high in the node voltage. A piecewise-linear model with \(V_{D0}=0.65\) V and \(r_B = 10\ \Omega\) gives 2.8747 mA, high by 0.04 %.
Two habits make this reliable. Always state the assumption — here, that the diode conducts — and always check it: the answer gives \(I_D = +2.87\) mA, positive, so the diode is indeed forward biased and the assumption stands. If the current had come out negative, the assumption was wrong; replace the diode by an open circuit and solve again. With \(m\) diodes there are \(2^m\) possible conduction states, and the analysis is a search over them, though in practice inspection settles all but one or two.
6 Diodes Carrying AC on Top of DC
Most diodes in signal circuits carry a steady bias with a small alternating component riding on it, and the standard treatment splits the problem in two. First find the Q-point using only the DC sources, with all capacitors open and all AC sources set to zero. Then find the response to the signal using only the AC sources, with the DC sources set to zero and the diode replaced by its dynamic resistance \(r_d = nV_T/I_D\) evaluated at that Q-point. Superposition is being applied to a non-linear circuit, which is illegal — and it works only because the second step has been linearised about the first.
A 1 k\(\Omega\) resistor and a diode (\(I_S = 10\) fA, \(n=1\)) are fed from a source \(v_S = V_S + v_s\) with \(V_S\) chosen to bias the diode at exactly 1 mA. From the Shockley equation \(V_D = 0.025852\ln(10^{-3}/10^{-14}) = 0.6548\) V, so \(V_S = 0.6548 + (0.001)(1000) = 1.6548\) V, and \(r_d = 25.852/1 = 25.85\ \Omega\).
The small-signal circuit is a plain divider, so for a peak signal \(v_s\):
Comparing the linear prediction with an exact solution of the full non-linear circuit:
| Signal amplitude | \(i_d\) predicted | \(i_d\) exact, +ve peak | \(i_d\) exact, −ve peak | \(v_d\) predicted | \(v_d\) exact, ± |
|---|---|---|---|---|---|
| 10 mV | 9.748 µA | 9.749 µA | 9.747 µA | 0.252 mV | +0.251 / −0.253 mV |
| 100 mV | 97.48 µA | 97.59 µA | 97.35 µA | 2.520 mV | +2.407 / −2.648 mV |
| 500 mV | 487.4 µA | 489.7 µA | 482.9 µA | 12.60 mV | +10.30 / −17.05 mV |
At 10 mV the linear model is right to four figures. At 500 mV the current is still right to about 0.5 %, but the diode voltage swing is asymmetric by two to one — +10.3 mV on one peak and −17.1 mV on the other — which is severe distortion. The current is well behaved only because the 1 k\(\Omega\) resistor, not the diode, is setting it. Drive the same diode from a current source and the voltage distortion is all you would see.
The rule of thumb that follows is worth memorising: the small-signal treatment is good for signal amplitudes across the diode below about 10 mV, which is \(0.4V_T\). Above that, the exponential's curvature produces harmonics, and above \(V_T\) the notion of a single \(r_d\) has no meaning at all. This is the same criterion that will define "small signal" for a transistor in Chapter 19, for the same reason: it is the same junction.
Where the AC swing is large the diode is no longer a small-signal element but a switch, and the right tool is the sweeping load line of Section 2 rather than \(r_d\). That is the regime of the rectifiers of Chapters 8 and 9, in which the signal is tens of volts and the diode is either fully on or fully off for whole half-cycles at a time. The two treatments are the two extremes of the same picture: a Q-point that barely moves, or a Q-point that traverses the entire characteristic every cycle.
7 Temperature and the Q-Point
The load line is fixed by resistors and supplies, both of which are nearly temperature independent. The device characteristic is not: Chapter 6 established that the forward voltage falls by roughly 2 mV for every kelvin at constant current, because \(I_S\) rises much faster than \(V_T\) does. The whole curve therefore translates to the left as the diode warms, and the Q-point slides along the fixed load line to meet it.
Figure 7.3 shows the result for the standard 10 V, 1 k\(\Omega\) circuit, computed with \(I_S \propto T^3e^{-E_g/kT}\):
| Temperature | \(V_D\) | \(I_D\) | Change in \(I_D\) |
|---|---|---|---|
| 300 K (27 °C) | 0.7124 V | 9.2876 mA | — |
| 325 K (52 °C) | 0.6718 V | 9.3282 mA | +0.44 % |
| 350 K (77 °C) | 0.6308 V | 9.3692 mA | +0.88 % |
Over a 50 K rise the diode voltage falls by 81.6 mV — an average of \(-1.63\) mV/K, in line with the rule of thumb — but the current changes by less than one per cent. The geometry explains why. The load line here is nearly horizontal in the region of the crossing, so translating the curve sideways moves the intersection sideways too, and the vertical coordinate barely responds. The extra current is simply the 81.6 mV released from the diode reappearing across the 1 k\(\Omega\) resistor: \(0.0816/1000 = 81.6\ \mu\text{A}\), which is the 0.88 % observed.
Repeat the exercise with a supply close to the diode drop and the conclusion reverses. With \(V_S = 2\) V and the same 1 k\(\Omega\), the current rises from 1.3377 mA at 300 K to 1.4260 mA at 350 K, an increase of 6.60 %. The absolute change in current is identical — the same 81.6 mV across the same resistor — but it is now 6.6 % of a much smaller current. This is the general principle of bias stability, and it is the whole content of Chapter 18 in transistor form:
Fractional current drift \(\approx \Delta V_D/(V_S - V_D)\). With \(V_S = 10\) V that is 81.6 mV over 9.29 V, or 0.88 %; with \(V_S = 2\) V it is 81.6 mV over 1.34 V, or 6.1 %. Design a bias network so the device drop is a small part of the total, and temperature stops being a problem.
Two further temperature effects deserve a mention even though they rarely dominate at these currents. Reverse leakage doubles every 10 °C, so a reverse-biased diode used as a switch in a high-impedance node can lose its blocking action entirely at 125 °C — a 5 µA leak into a 1 M\(\Omega\) node is 5 V. And self-heating makes the temperature a function of the answer: a diode dissipating \(V_DI_D = 0.712\times9.29\ \text{mA} = 6.6\) mW with a thermal resistance of 300 K/W runs 2 K above ambient, which is negligible; the same diode at 1 A and 1 V dissipates 1 W and runs 300 K hotter, which is not, and must be solved as a coupled electrical and thermal problem.
With the analysis methods in place, the next three chapters apply them to the one circuit that made the diode a commercial product. Chapter 8 takes a single diode and a sinusoid and asks what comes out; Chapter 9 improves the topology; Chapter 10 makes the result usable.
8 Summary and Key Results
| Method or quantity | Expression | Result / error |
|---|---|---|
| Circuit constraint | \(I_D = (V_{Th}-V_D)/R_{Th}\) | Load line: intercepts 10 V and 10 mA, slope −1/R |
| Q-point (exact) | Intersection of load line and characteristic | \(V_D = 0.7124\) V, \(I_D = 9.2876\) mA |
| Iteration | \(I_D \leftarrow (V_S-V_D)/R\), \(V_D \leftarrow nV_T\ln(I_D/I_S)\) | 6 figures in 2 passes; error × \(r_d/R\) = 1/360 per pass |
| Ideal model | \(I_D = V_S/R\) | 10.000 mA, +7.67 % |
| Constant-drop model | \(I_D = (V_S-0.7)/R\) | 9.300 mA, +0.13 % |
| Piecewise-linear model | \(I_D = (V_S-V_{D0})/(R+r_B)\) | 9.257 mA, −0.33 % (\(V_{D0}\)=0.65 V, \(r_B\)=10 \(\Omega\)) |
| Low supply (\(V_S=2\) V) | same circuit, 1 k\(\Omega\) | exact 1.3377 mA; ideal +49.5 %, constant-drop −2.82 % |
| Two diodes in series | \(V_D = 2nV_T\ln(I_D/I_S)\) | 8.579 mA exact, 8.600 mA by model (+0.24 %) |
| Two diodes in parallel | Equal \(V\), current \(\propto I_S\) | Identical: 4.653 mA each; 2:1 \(I_S\) mismatch: 3.105 / 6.211 mA |
| Series-parallel (Fig. 7.2) | KCL at node A | \(V_A\) = 7.004 V, \(I_D\) = 2.874 mA exact; model 2.868 mA (−0.21 %) |
| Small-signal resistance | \(r_d = nV_T/I_D\) | 25.85 \(\Omega\) at 1 mA; valid for \(v_d < 10\) mV |
| Q-point drift | \(dV_D/dT \approx -1.6\) mV/K | 0.7124 V → 0.6308 V over 300–350 K |
| Current drift | \(\Delta I_D/I_D \approx \Delta V_D/(V_S-V_D)\) | +0.88 % at \(V_S\) = 10 V; +6.60 % at \(V_S\) = 2 V |
9 Common Mistakes
The load line contains nothing about the diode. Its intercepts are \(V_{Th}\) and \(V_{Th}/R_{Th}\) — the open-circuit voltage and short-circuit current of the network the diode is plugged into — and it would be drawn identically for a lamp or a thermistor. Students who try to "include the 0.7 V" in the intercept end up with a line that no longer represents Kirchhoff's law and a Q-point that satisfies neither relation.
Model-based analysis begins by assuming a conduction state, and the assumption must be verified from the answer. If you assumed conduction, the computed \(I_D\) must be positive; if you assumed cut-off, the computed voltage across the diode must be less than the turn-on value. In a circuit with two diodes and two supplies of opposite sign this is not a formality — the obvious assumption is wrong often enough that the check is the only thing standing between you and a confidently reported impossible answer.
\(r_d = nV_T/I_D\) is the slope at one point, valid for excursions of a few millivolts. At a 500 mV signal the diode's voltage swing is +10.3 mV on one peak and −17.1 mV on the other against a linear prediction of ±12.6 mV — the model has quietly stopped applying. Equally, using the static resistance \(V_D/I_D\) = 655 \(\Omega\) in a small-signal divider understates the signal by a factor of 25. Decide first which regime you are in, then choose the tool.
10 Chapter Review
1. A 5 V supply feeds a 470 \(\Omega\) resistor in series with a silicon diode having \(I_S = 10^{-14}\) A and \(n = 1\). Find the Q-point by iteration and compare it with the constant-drop and ideal models.
Start at \(V_D = 0.7\) V. Pass 1: \(I_D = (5-0.7)/470 = 9.149\) mA, \(V_D = 0.025852\ln(9.149\times10^{-3}/10^{-14}) = 0.71203\) V. Pass 2: \(I_D = (5-0.71203)/470 = 9.1233\) mA, \(V_D = 0.71196\) V. Pass 3: \(I_D = 9.1235\) mA, \(V_D = 0.71196\) V — converged. The Q-point is \(V_D = 0.7120\) V, \(I_D = 9.123\) mA. The constant-drop model gives \((5-0.7)/470 = 9.149\) mA, high by 0.28 %; the ideal model gives \(5/470 = 10.638\) mA, high by 16.6 %. The ideal error is larger than in the 10 V case because the neglected 0.71 V is a larger fraction of a 5 V supply — it is \(0.712/(5-0.712) = 16.6\) %, as the general formula predicts. Note also that the iteration converged in two passes again: \(r_d = 25.852/9.123 = 2.83\ \Omega\) against 470 \(\Omega\), so the error contracts by a factor of 166 per pass.
2. Sketch what happens to the load line and the Q-point when (a) the supply is doubled, (b) the series resistor is doubled, (c) the diode is heated by 50 K. In which case does the current change most?
(a) Doubling \(V_S\) from 10 to 20 V slides the load line to the right without changing its slope: both intercepts double, to 20 V and 20 mA. The Q-point climbs the curve to about 19.29 mA, and the diode voltage rises by \(V_T\ln(19.29/9.29) = 18.9\) mV to 0.731 V. (b) Doubling \(R\) to 2 k\(\Omega\) pivots the line about the fixed 10 V intercept, halving the current intercept to 5 mA. The Q-point falls to roughly 4.65 mA and the diode voltage drops by about 18 mV to 0.694 V. (c) Heating by 50 K translates the curve 82 mV to the left; the load line does not move, and the current rises only to 9.369 mA, a change of 0.88 %. So the supply and the resistor each change the current by a factor of about two, while a 50 K temperature rise changes it by under one per cent — because a stiff supply converts the diode's drift into a negligible extra drop across a large resistance.
3. Two diodes with \(I_{S1} = 10^{-14}\) A and \(I_{S2} = 2\times10^{-14}\) A are connected in parallel and fed through 1 k\(\Omega\) from 10 V. Find the current in each. What series resistance would make the split within 10 %?
In parallel the two share one voltage, so the total current obeys \(I = (I_{S1}+I_{S2})(e^{V/V_T}-1) = 3\times10^{-14}(e^{V/V_T}-1)\). Iterating with \(I = (10-V)/1000\) gives \(V = 0.6841\) V and \(I = 9.316\) mA. Since each diode's current is proportional to its own \(I_S\) at the shared voltage, \(I_1 = 3.105\) mA and \(I_2 = 6.211\) mA — an exact 1:2 split, so the leakier diode takes 67 % of the current. To equalise, put a ballast resistor \(R_b\) in series with each. The mismatch appears as a voltage difference of \(V_T\ln 2 = 17.9\) mV; for the currents to differ by no more than 10 % about 4.66 mA, the ballast must convert that 17.9 mV into no more than a 0.47 mA difference, so \(R_b \ge 0.0179/0.00047 = 38\ \Omega\). A 39 \(\Omega\) resistor in each leg would do it, at the cost of 0.18 V of extra drop. In a power circuit the same calculation with 10 A per diode gives a ballast of a few milliohms.
4. Explain why the iterative method converges in two passes for a 10 V supply but should be treated with more care if the series resistance were only 20 \(\Omega\).
Each pass multiplies the error in \(V_D\) by \(-r_d/R\), where \(r_d = nV_T/I_D\). With \(R = 1\) k\(\Omega\) the current is 9.29 mA, so \(r_d = 2.78\ \Omega\) and the contraction factor is 1/360: a 12 mV initial error becomes 33 µV after one pass and 0.1 µV after two. With \(R = 20\ \Omega\) the current would be about 465 mA, giving \(r_d = 0.056\ \Omega\) and a contraction factor of 1/360 again — so in fact this particular case still converges well, because \(r_d\) falls as fast as \(R\) does. The method only fails when \(r_d\) approaches \(R\), which needs a very small current through a small resistance: at \(I_D = 26\ \mu\text{A}\), \(r_d = 1\) k\(\Omega\), and with \(R = 20\ \Omega\) the iteration would diverge violently. The remedy is to iterate in the other direction — guess \(I_D\), get \(V_D\) from the device equation, then get a new \(I_D\) from the circuit — which has contraction factor \(R/r_d\) and converges exactly when the first form does not.
5. A diode is biased at 2 mA and carries a 20 mV peak sinusoid. Find the AC current, then state what would change if the bias were 200 µA and the source resistance were 100 \(\Omega\).
At 2 mA, \(r_d = 25.852/2 = 12.93\ \Omega\). If the 20 mV source has a series resistance \(R\), the AC current is \(i_d = 0.020/(R + 12.93)\); for \(R = 1\) k\(\Omega\) that is 19.74 µA peak, and the AC voltage across the diode is \(0.020\times12.93/1012.9 = 0.255\) mV — small enough that the linear model is excellent. At 200 µA the dynamic resistance rises tenfold to 129.3 \(\Omega\), and with a 100 \(\Omega\) source the divider now favours the diode: \(v_d = 0.020\times129.3/229.3 = 11.3\) mV peak. That is above the roughly 10 mV limit for small-signal behaviour, so the current waveform will be visibly distorted — the positive half-cycle compressed less than the negative half is compressed more — and \(r_d\) itself swings by a factor of \(e^{\pm11.3/25.85}\), that is between 0.64 and 1.55 times its quiescent value over the cycle. Either reduce the drive or raise the bias current; raising the bias is usually the right answer, since \(r_d\) is inversely proportional to it.