Chapter 2 of 30
- Why the diode is the uncontrolled switch, and the three jobs it does in converters: rectifying, freewheeling, and blocking.
- The static v–i characteristic, and how to reduce a datasheet curve to the piecewise-linear model \( v_D = V_{T0} + r_D i_D \).
- Reverse recovery — \( t_{rr} \), \( I_{RM} \), the recovered charge \( Q_{rr} \), and the softness factor — and why a diode's turn-off costs energy in the transistor beside it.
- The four device families: general-purpose, fast-recovery, Schottky, and silicon-carbide diodes, and where each belongs.
- How to read the ratings that matter — \( V_{RRM} \), \( I_{F(AV)} \), \( I_{FSM} \), \( I^2t \), \( R_{th} \) — and how to size a diode with a sensible margin.
- Why diodes in series need voltage-sharing resistors and diodes in parallel need current-sharing resistances, and how to compute both.
The Uncontrolled Switch
Chapter 1 introduced the ideal switch. The power diode is the crudest realisation of it: a single PN junction with two terminals — anode (A) and cathode (K) — and no third terminal to command it. When the anode is positive with respect to the cathode the junction is forward biased and the diode conducts; when the polarity reverses it blocks. Conduction is decided entirely by the external circuit, which is why the diode is called an uncontrolled device.
Structurally a power diode is not a scaled-up signal diode. It carries a lightly doped \(n^{-}\) drift region between the \(p^{+}\) anode and the \(n^{+}\) cathode. That region is what holds off hundreds or thousands of volts when the device is reverse biased — and it is also what stores charge when the device conducts, which is the root cause of the reverse-recovery behaviour in Section 2-3. Blocking voltage and switching speed are therefore in direct conflict: a thicker drift region blocks more and switches slower. Almost every diode datasheet is a statement about where the manufacturer settled that conflict.
Three jobs a diode does in a converter
Whenever you see a diode in a schematic, ask which of these it is doing — the answer tells you which rating matters most.
| Job | What it does | Rating that decides the choice | Where |
|---|---|---|---|
| Rectifier | Converts AC to DC by conducting on one polarity only | \(V_{RRM}\) and \(I_{F(AV)}\) — recovery rarely matters at 50 Hz | Part 2 |
| Freewheeling | Gives inductive load current a path when the main switch opens. Without it, the collapsing \(L\,di/dt\) destroys the switch | \(Q_{rr}\) — because the recovery current lands on the transistor | Part 3 |
| Blocking / clamping | Prevents reverse current in battery, PV-string and snubber circuits | Leakage and \(V_{RRM}\) | Ch. 24, 28 |
Static Characteristics and the Piecewise-Linear Model
The theoretical junction obeys the Shockley equation, in which current rises exponentially with voltage:
This is exact and useless. At the currents power diodes handle — tens to thousands of amperes — the exponential has long since flattened into a nearly straight line, and the ohmic drop across the drift region dominates. So for power work we discard the exponential and fit a straight line to the datasheet curve at the operating current. The result is the piecewise-linear model: an ideal diode in series with a threshold voltage \(V_{T0}\) and a dynamic resistance \(r_D\).
Pick any two points on the forward curve near your operating current; the slope gives \(r_D\) and back-substitution gives \(V_{T0}\). Typical silicon values are \(V_{T0} \approx 0.7\!-\!1.0\) V with \(r_D\) of a few milliohms. Section 2-6 shows why this two-parameter fit — not the exponential — is what determines the heatsink.
In the reverse direction the diode passes only a small leakage current \(I_R\), typically microamperes to a few milliamperes, which roughly doubles for every 10 °C rise in junction temperature. Leakage is negligible for loss purposes but decisive for series operation (Section 2-7), where the diode with the lowest leakage ends up holding the highest voltage. Push the reverse voltage past the breakdown voltage \(V_{BR}\) and avalanche current rises without limit; unless the device is specifically avalanche-rated, this destroys it. The rated \(V_{RRM}\) always sits safely below \(V_{BR}\).
Reverse Recovery — Why a Diode Cannot Turn Off Instantly
While a power diode conducts, the drift region is flooded with excess minority carriers — this is what makes its forward drop so low despite the region being lightly doped. When the external circuit tries to reverse the current, that stored charge must first be swept out. Until it is, the junction cannot support reverse voltage, and the diode conducts backwards. The interval is called reverse recovery, and it is the diode's single most important dynamic property.
Read the waveform in two intervals. During \(t_a\) the current falls linearly at the rate \(di/dt\) imposed by the external circuit, passes through zero and grows negative until the stored charge near the junction is exhausted, reaching the peak reverse recovery current \(I_{RM}\). During \(t_b\) the remaining charge recombines and the current decays back to the leakage level. The total reverse recovery time is \(t_{rr} = t_a + t_b\), and the area under the negative excursion is the recovered charge \(Q_{rr}\).
Approximating that area as a triangle gives three relations that let you find any two quantities from the third and the circuit's \(di/dt\):
Both \(I_{RM}\) and \(t_{rr}\) scale with the square root of the applied \(di/dt\) — so switching a circuit faster does not eliminate recovery, it converts recovery time into recovery current. The softness factor \(S = t_b/t_a\) describes the shape: a soft diode (\(S \ge 1\)) decays gently, while a snappy one (\(S \ll 1\)) collapses abruptly and rings against circuit inductance, producing voltage spikes and EMI.
A designer replaces a fast-recovery silicon diode with a SiC Schottky and finds the transistor runs 15 °C cooler, while the diode's own temperature barely changes. Explain.
Because the recovery energy was never being dissipated in the diode in the first place.
During recovery the diode is essentially a short circuit, so the reverse-recovery current \(I_{RM}\) flows through the transistor that is turning on, and it does so while that transistor still has nearly the full DC-link voltage across it. The product \(V \times I\) is therefore large in the transistor and small in the diode.
Removing \(Q_{rr}\) removes the transistor's extra turn-on loss. The diode was only ever passing charge, not dissipating much of it.
The practical lesson: when a converter's transistor is running hot, one of the first things to check is the freewheeling diode's \(Q_{rr}\) — the culprit is often not the component that is overheating.
Why is a "snappy" diode (softness factor \(S \ll 1\)) a problem, given that it recovers faster than a soft one?
Because the circuit is full of stray inductance, and \(v = L\,di/dt\) does not care whether you wanted that \(di/dt\).
A snappy diode collapses its reverse current abruptly at the end of recovery. That sudden \(di/dt\) acting on the loop inductance generates a large voltage spike across the device — which can exceed \(V_{RRM}\) and destroy it — and rings at the resonant frequency of the stray \(L\) with the junction capacitance, radiating EMI.
A soft diode lets the current decay gently, so \(di/dt\) at the end is small and so is the spike. This is why "faster" is not automatically "better" for a diode: you want the recovery to be short and to end gracefully, and manufacturers charge for both.
Types of Power Diode
Manufacturers resolve the blocking-versus-speed conflict differently for different jobs, giving four families you will meet constantly.
Wide drift region, \(t_{rr}\) of tens of microseconds. Perfectly adequate for 50/60 Hz line rectification where the supply reverses slowly. Available to several kilovolts and thousands of amperes.
Lifetime-controlled by gold or platinum doping or electron irradiation to give \(t_{rr}\) under a microsecond, at the price of a higher forward drop. The default freewheeling diode in choppers and inverters.
A metal–semiconductor junction with no minority carriers, so there is no stored charge and essentially no reverse recovery. Very low drop (0.3–0.5 V) but limited to roughly 100–200 V and higher leakage.
A wide-bandgap Schottky device that keeps zero recovery charge up to 650–1700 V. Costlier, but it removes recovery loss from high-frequency converters entirely — the enabling device in modern EV and solar hardware.
| Family | Typical \(V_{RRM}\) | Typical \(t_{rr}\) | Forward drop | Where it belongs |
|---|---|---|---|---|
| General purpose | 50 V – 5 kV | 25 µs and above | 0.7 – 1.2 V | Line-frequency rectifier bridges |
| Fast recovery | 50 V – 3 kV | 0.1 – 5 µs | 1.0 – 2.0 V | Freewheeling in choppers, inverters, SMPS |
| Schottky | 20 – 200 V | Effectively zero | 0.3 – 0.5 V | Low-voltage, high-current output rectifiers |
| SiC Schottky | 650 V – 1.7 kV | Effectively zero | 1.3 – 1.8 V | High-frequency PFC, solar and EV converters |
Ratings and Selection
A datasheet is a list of limits, each one a different failure mode. These are the ones that decide a design.
| Symbol | Name | What it limits |
|---|---|---|
| \(V_{RRM}\) | Peak repetitive reverse voltage | Blocking capability; must exceed the circuit PIV with margin |
| \(V_F\) | Forward voltage drop | Conduction loss at rated current |
| \(I_{F(AV)}\) | Average forward current | Continuous thermal rating; matched to the mounting case |
| \(I_{F(RMS)}\) | RMS forward current | Heating in the resistive part of the drop |
| \(I_{FSM}\) | Surge (non-repetitive) current | One-cycle fault survival, e.g. capacitor inrush |
| \(I^2t\) | Let-through energy | Fuse coordination — the fuse must clear below this |
| \(t_{rr},\ Q_{rr}\) | Recovery time and charge | Switching loss and EMI at the chosen \(f_s\) |
| \(R_{th(j-c)}\) | Junction-to-case thermal resistance | Heatsink sizing and \(T_j\) |
Selecting a diode, in order
The order matters. Each step eliminates candidates, so doing them in the wrong sequence wastes time comparing devices that were never viable.
- Find the peak inverse voltage the circuit actually applies For a single-phase bridge on a 230 V supply this is \(\sqrt{2}\times 230 = 325\) V. Read it off the circuit, not off the nameplate.
- Choose \(V_{RRM}\) at roughly twice the PIV The factor of two covers line transients, commutation spikes and lightning-induced surges. \(650\) V minimum here, so specify an 800 V part.
- Compute the average and RMS currents from the conduction waveform Not from the load current — from the fraction of the cycle this diode conducts. Apply a margin of about 1.5.
- Check the surge rating \(I_{FSM}\) against the worst inrush Usually the DC-link capacitor charging at switch-on. This is frequently the binding constraint, and it is the step most often skipped.
- Only now compare recovery figures Among the candidates that survived steps 1–4, pick on \(Q_{rr}\) and softness — and only if the switching frequency makes them matter.
Losses and Junction Temperature
Substituting the piecewise-linear model into the definition of average power gives the standard conduction-loss expression. Note that it needs both the average and the RMS current: the threshold voltage is a constant drop, so it multiplies the average current, while the dynamic resistance dissipates \(i^2 r_D\), which averages to the mean square.
The thermal equation is Ohm's law for heat: power is the current, thermal resistance the resistance, temperature rise the voltage. The chain \(R_{th(j-a)} = R_{th(j-c)} + R_{th(c-s)} + R_{th(s-a)}\) runs junction → case → sink → ambient, and the computed \(T_j\) must stay below the rated limit (typically 150 °C for silicon, 175 °C for SiC) with margin.
At switching frequencies above a few kilohertz a second term joins it. Each turn-off sweeps out \(Q_{rr}\) against the reverse voltage \(V_R\), so the energy lost per event is approximately \(Q_{rr}V_R\) and the average power is
with the total shared between the diode and the transistor that forced the recovery — the split depends on the circuit, and a common engineering practice is to assign roughly half to each. Unlike conduction loss, this term is proportional to frequency, which is precisely the trade-off named in Chapter 1. It is also why replacing a silicon fast-recovery diode with a SiC Schottky in a 100 kHz converter can lift efficiency by a full percentage point: the term disappears rather than shrinking.
Series and Parallel Operation
When a single device cannot meet the requirement, diodes are connected in series for higher voltage or in parallel for higher current. Neither works without help, because no two devices are identical.
In series, the same current flows through every device, so the reverse voltage divides in proportion to the reverse resistance — that is, inversely with leakage. The lowest-leakage diode takes the largest share and may exceed its \(V_{RRM}\) while its partner sits half-idle. Connecting a resistor \(R\) across each device swamps the difference: as long as the resistor current is much larger than the leakage mismatch, sharing is nearly equal. Writing the branch currents \(I = I_{s1} + V_{D1}/R = I_{s2} + V_{D2}/R\) gives the design relation
Smaller \(R\) equalises better but wastes more standing power — the usual compromise. Series strings also need a capacitor (often with a series resistor) across each device to equalise the dynamic voltage during reverse recovery, since diodes with different \(Q_{rr}\) stop conducting at different instants.
In parallel, every device sees the same voltage, so current divides according to the forward characteristics. The device with the lower \(V_F\) hogs the current, heats up, and — because \(V_F\) falls as temperature rises in the low-current region — takes even more. That positive feedback is called thermal runaway. A small series resistance in each branch forces sharing, since the branch carrying more current develops more drop:
In practice the same effect is achieved more cheaply by mounting matched devices on a common heatsink, so that any device tending to hog current warms its neighbours and equalises the characteristics, and by keeping the layout symmetric so stray inductances match. The quality of sharing is measured by the string efficiency, the ratio of the actual string rating to the arithmetic sum of the individual ratings — always less than one, and typically designed for around 0.8 to 0.9.
Worked Examples
Problem. A diode's forward curve passes through (1.00 V, 20 A) and (1.40 V, 100 A). Find \(r_D\) and \(V_{T0}\), then predict the drop at 60 A.
Solution. The slope of the line through the two points is the dynamic resistance:
Two datasheet points, and the device is now a component you can put into a loss calculation.
Problem. The diode of Example 1 freewheels in a chopper with \(D = 0.7\) and a well-filtered load current of 50 A. With \(R_{th(j-a)} = 1.2\) °C/W and \(T_a = 45\) °C, find the conduction loss and \(T_j\).
Solution. The diode conducts for the fraction \((1-D) = 0.3\) of each cycle at 50 A:
Comfortably inside a 150 °C limit. Note that using 15 A in both terms would have under-predicted the loss by about 2.6 W — the RMS current is what heats the resistive part.
Problem. A fast-recovery diode has \(Q_{rr} = 5\ \mu\text{C}\) at \(di/dt = 100\ \text{A}/\mu\text{s}\). Find \(t_{rr}\) and \(I_{RM}\). If it blocks 400 V and switches at 10 kHz, estimate the recovery loss.
Solution. Convert the slew rate to SI: \(100\ \text{A}/\mu\text{s} = 10^{8}\) A/s.
A 31.6 A spike rides on top of the load current every cycle, and 20 W is dissipated — most of it in the transistor. At 100 kHz it would be 200 W, which is why SiC exists.
Problem. Two diodes in series block 5 kV. Their reverse leakage currents are 30 mA and 35 mA. Each carries a 100 kΩ sharing resistor. Find the voltage across each, and the value of \(V_{RRM}\) you would specify.
Solution. From the sharing relation with \(I_{s1} = 30\) mA and \(I_{s2} = 35\) mA:
The lower-leakage device holds the higher voltage, so specify against 2750 V, not the 2500 V an equal split would suggest — a 3.3 kV part with margin. Halving \(R\) to 50 kΩ would narrow the gap to 250 V but double the standing loss in the resistors.
Problem. Two diodes share a 200 A load. At that current their forward drops would be 1.00 V and 1.10 V. Each is fitted with a 10 mΩ series resistance. Find \(I_1\) and \(I_2\).
Solution. Equating the two branch voltages:
A 5% imbalance instead of the runaway that would occur with no sharing resistance — at a cost of \(2 \times I^2 R \approx 200\) W in the resistors, which is why heatsink-based matching is preferred wherever it is feasible.
Problem. A single-phase bridge on a 230 V, 50 Hz supply feeds a highly inductive load drawing a constant 40 A. Specify the diodes.
Solution. Each diode blocks the peak of the supply and conducts for one half cycle:
With a 1.5 margin on the average current, specify a 800 V, 30 A general-purpose diode — fast recovery is unnecessary here, because the supply reverses at 50 Hz and \(t_{rr}\) of tens of microseconds is invisible on a 10 ms half cycle. Finally, check \(I_{FSM}\) against the inrush of the DC-link capacitor at switch-on, which is usually the binding constraint.
Summary & Formula Sheet
The diode has no gate — the circuit decides conduction. Its jobs are rectifying, freewheeling, and blocking.
\(v_D = V_{T0} + r_D i_D\), fitted from two datasheet points. Forget the exponential at power currents.
Stored charge must be swept out: \(Q_{rr} \approx \frac12 I_{RM}t_{rr}\), with both \(I_{RM}\) and \(t_{rr}\) scaling as \(\sqrt{di/dt}\).
General purpose for line frequency, fast recovery for freewheeling, Schottky for low voltage, SiC for high frequency.
\(P_{cond} = V_{T0}I_{F(AV)} + r_D I_{F(RMS)}^2\); add \(Q_{rr}V_Rf_s\) above a few kHz; then \(T_j = T_a + PR_{th}\).
Series needs voltage-sharing resistors (lowest leakage takes most volts); parallel needs current sharing (lowest \(V_F\) hogs current).
Key terms
- Drift region
- The lightly doped \(n^{-}\) layer that holds off reverse voltage. Thicker means higher blocking and slower recovery.
- Piecewise-linear model
- An ideal diode in series with \(V_{T0}\) and \(r_D\). The two-parameter fit every power loss calculation uses.
- Reverse recovery
- The interval after the current reverses during which stored charge is swept out and the diode conducts backwards.
- Recovered charge, \(Q_{rr}\)
- The charge removed during recovery — the area under the negative current excursion. Sets the recovery energy \(Q_{rr}V_R\).
- Softness factor, \(S\)
- \(t_b/t_a\). Soft (\(S \ge 1\)) decays gently; snappy (\(S \ll 1\)) collapses abruptly, causing overshoot and EMI.
- Freewheeling diode
- The diode that carries inductive load current when the main switch opens. Chosen mainly on \(Q_{rr}\).
- Peak inverse voltage (PIV)
- The largest reverse voltage the circuit applies. Choose \(V_{RRM} \ge 2\times\) PIV.
- Thermal resistance, \(R_{th}\)
- °C of temperature rise per watt. Junction-to-case, case-to-sink and sink-to-ambient add in series.
- String efficiency
- Actual rating of a series or parallel string divided by the arithmetic sum of individual ratings. Always < 1; design for 0.8–0.9.
Test Yourself
Answer each before revealing it. These test whether you can reason about a diode, which is what the numerical problems then let you practise.
Why can a Schottky diode not be made with a 1200 V rating in silicon, when a PN diode easily can?
A Schottky junction is metal-to-semiconductor, and its blocking ability comes from the barrier height at that interface plus the depletion region in a lightly doped, thin layer. To block more voltage you need a thicker, more lightly doped layer — but in a Schottky device that layer is a pure majority-carrier resistor with no conductivity modulation to help it. Its resistance rises steeply, and the forward drop becomes unacceptable well before 1200 V. Reverse leakage also climbs sharply with barrier lowering.
Silicon carbide breaks the deadlock because its critical electric field is roughly ten times that of silicon, so the same blocking voltage needs about a tenth the drift thickness. That is why SiC Schottky diodes exist at 650–1700 V while silicon ones stop around 200 V.
A freewheeling diode in a chopper runs at \(D = 0.9\) with a load current of 100 A. Another runs at \(D = 0.1\) with the same load current. Which diode needs the bigger heatsink?
The one at \(D = 0.1\), by a wide margin — the freewheeling diode conducts for \((1-D)\).
- At \(D = 0.9\): \(I_{F(AV)} = 10\) A, \(I_{F(RMS)} = 100\sqrt{0.1} = 31.6\) A.
- At \(D = 0.1\): \(I_{F(AV)} = 90\) A, \(I_{F(RMS)} = 100\sqrt{0.9} = 94.9\) A.
With \(V_{T0} = 1\) V and \(r_D = 5\) mΩ that is about 15 W against 135 W — a factor of nine.
The design lesson: a converter that spends most of its life at low duty ratio stresses its diode, not its transistor. Always size the diode at the worst-case duty ratio the application actually reaches, not at the nominal one.
Two diodes in series each rated 2 kV are asked to block 3.5 kV. Their leakages are 10 mA and 12 mA and each has a 200 kΩ sharing resistor. Do they survive?
\(V_{D1} - V_{D2} = R(I_{s2}-I_{s1}) = (2\times10^{5})(0.002) = 400\) V.
With \(V_{D1}+V_{D2} = 3500\): \(V_{D1} = 1950\) V and \(V_{D2} = 1550\) V.
Marginally yes — 1950 V against a 2 kV rating is only 2.5% of margin, which is not enough. Line transients or an unequal transient split during recovery would take it over.
Two fixes: halve the sharing resistors to 100 kΩ (which halves the imbalance to 200 V, giving \(V_{D1} = 1850\) V, at the cost of doubling the standing loss), or specify 2.5 kV devices. In practice, do both — and add the RC networks that handle the dynamic imbalance during reverse recovery, which this static calculation says nothing about.
You measure a diode's case temperature at 70 °C and the datasheet says \(T_{j,max} = 150\) °C. Is the device safe?
You cannot tell from case temperature alone — you have measured the wrong node.
The junction sits above the case by \(P \times R_{th(j-c)}\). If the device dissipates 40 W and \(R_{th(j-c)} = 0.9\) °C/W, then \(T_j = 70 + 36 = 106\) °C — safe. But if it dissipates 100 W through the same path, \(T_j = 70 + 90 = 160\) °C, and it is already past the limit while the case still reads a comfortable 70 °C.
This is why thermal design is always done from the loss calculation upward, not from a thermocouple on the case. The case is simply the last point you can reach with an instrument.
In a 50 Hz bridge rectifier, does reverse-recovery loss matter?
Essentially not, and the reason is worth stating precisely.
\(P_{rr} \approx Q_{rr}V_Rf_s\), and here \(f_s = 50\) Hz. Even a slow general-purpose diode with \(Q_{rr} = 50\ \mu\)C blocking 325 V gives \(P_{rr} = (50\times10^{-6})(325)(50) \approx 0.8\) W — negligible beside tens of watts of conduction loss.
There is also a second reason: the supply reverses over milliseconds, so the imposed \(di/dt\) is tiny, and a \(t_{rr}\) of tens of microseconds is invisible on a 10 ms half-cycle.
Consequence: specifying an expensive fast-recovery diode for a line-frequency rectifier buys nothing and costs a higher forward drop — i.e. it makes the converter less efficient. Recovery only starts to matter above roughly a kilohertz.
Why does mounting parallel diodes on a common heatsink improve current sharing, when it adds no components to the circuit at all?
It converts a regenerative feedback loop into a degenerative one, using thermal coupling instead of electrical components.
On separate heatsinks: the greedy diode takes more current → heats up alone → its \(V_F\) falls further → it takes even more. Runaway.
On a common heatsink: the greedy diode's extra heat warms its neighbours too. Their \(V_F\) falls as well, so they start taking more current, which pulls the sharing back toward equality.
It is free, lossless (unlike sharing resistors, which burned 200 W in Worked Example 5), and it is why manufacturers sell multi-die modules on a single baseplate. The remaining requirement is a symmetric layout, so stray inductances match and the sharing holds during transients as well as in steady state.
Problems
Work these with a method:
- Identify which job the diode is doing — rectifying, freewheeling, or blocking.
- Find the conduction fraction of the cycle, and the current during it.
- Compute \(I_{F(AV)}\) and \(I_{F(RMS)}\) — you will need both.
- Apply the model, loss, thermal, or sharing relation.
Problems 1–5 are direct application; 6–10 need judgement; 11–12 are design decisions worth discussing in a tutorial.
- State which diode family you would choose for: (a) a 50 Hz bridge rectifier, (b) a 5 V/60 A SMPS output rectifier, (c) a 100 kHz PFC boost stage at 400 V, (d) the freewheeling path of a 2 kHz DC drive.
- Explain in your own words why a power diode has a lightly doped drift region, and what that region costs at turn-off.
- A diode's curve passes through (0.95 V, 10 A) and (1.35 V, 90 A). Find \(V_{T0}\) and \(r_D\), and the drop at 45 A.
- The diode of Problem 3 carries a 40 A rectangular pulse for 40% of each cycle. Find \(I_{F(AV)}\), \(I_{F(RMS)}\), and the conduction loss.
- For Problem 4, the total thermal resistance to ambient is 1.5 °C/W and ambient is 40 °C. Find \(T_j\). What is the highest ambient at which \(T_j\) stays below 125 °C?
- A diode has \(t_{rr} = 3\ \mu\text{s}\) and \(I_{RM} = 40\) A. Estimate \(Q_{rr}\) and the \(di/dt\) that produced it.
- A diode with \(Q_{rr} = 2\ \mu\text{C}\) blocks 600 V. Find the recovery loss at 5 kHz, 50 kHz and 150 kHz, and comment on the device choice at each.
- Two diodes with softness factors \(S = 0.3\) and \(S = 1.2\) are otherwise identical. Explain which produces more EMI and voltage overshoot, and why.
- Three diodes in series block 6 kV. Their leakage currents are 20, 24 and 28 mA. With 80 kΩ sharing resistors, find the voltage across each and the string efficiency if each device is rated 2.5 kV.
- Three diodes in parallel share 300 A. Their forward drops at 100 A are 1.05, 1.00 and 1.10 V. With 8 mΩ series resistances, find each branch current and the power lost in the resistors.
- A three-phase bridge on a 415 V line supplies 100 A of well-filtered DC. Find the PIV and the average and RMS current per diode, then specify \(V_{RRM}\) and \(I_{F(AV)}\) with sensible margins.
- A 50 kHz boost converter operates at 400 V with 20 A of load current at \(D = 0.4\). Compare a silicon fast-recovery diode (\(V_{T0} = 1.2\) V, \(r_D = 8\) mΩ, \(Q_{rr} = 4\ \mu\text{C}\)) with a SiC Schottky (\(V_{T0} = 1.5\) V, \(r_D = 10\) mΩ, \(Q_{rr} \approx 0\)). Compute the total loss for each, decide which you would specify, and state at what switching frequency the answer would flip.