Chapter 3: Thyristors — SCR Characteristics and the Thyristor Family

Where this sits

Part 1 · Power Semiconductor Devices
Chapter 3 of 30

You should already know

The diode's structure and reverse recovery from Chapter 2, and basic bipolar transistor action.

By the end you can

Explain latching from the two-transistor model, distinguish \(I_L\) from \(I_H\), and size \(L_s\) and the RC snubber.

Time

≈ 60 min reading · ≈ 50 min problems

i What you'll learn

Why the SCR is called semi-controlled: the gate commands turn-on but has no authority over turn-off.
The static v–i characteristic — forward blocking, break-over, forward conduction, reverse blocking — and the meaning of latching and holding current.
The two-transistor model, the regeneration condition \( \alpha_1 + \alpha_2 \to 1 \), and why it explains every turn-on mechanism at once.
The five ways an SCR can turn on — only one of which you actually want.
Turn-off time \( t_q \), and the difference between converter-grade and inverter-grade devices.
How to size the series inductor for \( di/dt \) and the RC snubber for \( dv/dt \) — the two ratings that quietly destroy thyristors.
The wider family: TRIAC, DIAC, GTO, IGCT, MCT, LASCR and where each is still used.

Section 3-1

The Semi-Controlled Switch

A thyristor, or silicon controlled rectifier (SCR), is a four-layer \(p\!-\!n\!-\!p\!-\!n\) device with three terminals: anode (A), cathode (K), and gate (G). The four layers form three junctions, and their behaviour explains the whole device. With the anode positive, junctions \(J_1\) and \(J_3\) are forward biased while the middle junction \(J_2\) is reverse biased — so the device blocks. Inject a small current into the gate and \(J_2\) collapses; the SCR latches into conduction and stays there. Reverse the anode voltage and \(J_1\) and \(J_3\) block instead.

The crucial word is latches. Once conduction is established, removing the gate current changes nothing — the device holds itself on through internal regeneration (Section 3-3). To turn it off, the external circuit must reduce the anode current below a threshold, which in an AC circuit happens naturally at every zero crossing and in a DC circuit requires a deliberate commutation circuit. This is the definition of semi-controlled: turn-on commanded, turn-off not.

◇Intuition · a starter pistol, not a throttle

This is the single mental model that makes the whole chapter click. Compare three devices by asking what does the control terminal actually do?

Device	What the control terminal is	Turn-off
Diode	There isn't one — the circuit decides everything	Circuit
SCR	A starter pistol. It fires the race, then has no further say	Circuit only
MOSFET / IGBT	A throttle, held down for as long as you want conduction	Gate command

Think of a mousetrap. A light touch releases it; after that, pressing the trigger again — or letting go — changes nothing. The trap stays sprung until someone resets it from outside.

Every awkward thing about thyristor circuits, and the entire subject of commutation in Chapter 4, follows from this one property. When you meet a confusing thyristor problem, ask: am I trying to make the gate do a job only the circuit can do?

Four layers, three junctions, three terminals

Three stable regions, one unstable one, and two current thresholds

!Common mistake

"To switch the SCR off, remove the gate signal."

Wrong The gate turned it on, so the gate must turn it off — like a transistor.

Right Once latched, the SCR sustains itself through internal regeneration. Removing gate current — or even driving it negative — does nothing. Only the anode current falling below \(I_H\) turns it off.

This is the most consequential misconception in the chapter, because it hides an entire design problem. In an AC circuit the supply reverses every half cycle and turns the device off for free. In a DC circuit nothing reverses, so you must build a circuit whose only job is to force the anode current to zero — which is why Chapter 4 devotes half its length to commutation, and why bulky commutating capacitors were the price of DC thyristor drives for thirty years.

Section recap. Four layers, three junctions, three terminals. With the anode positive, \(J_2\) blocks until the gate fires the device; then it latches and the gate loses all authority. Turn-on is commanded, turn-off is not — that is what "semi-controlled" means.

Section 3-2

Static Characteristics: Latching and Holding

Follow the characteristic from left to right. In the third quadrant the SCR blocks in reverse exactly like a diode, passing only leakage until avalanche breakdown at \(V_{BR}\). In the first quadrant with no gate current it blocks in the forward direction too — this is the state a diode does not have, and it is what makes controlled rectification possible. Raise the anode voltage far enough and the device breaks over at the break-over voltage \(V_{BO}\), snapping through a region of negative resistance into forward conduction, where it behaves like a diode with a drop of one to two volts.

Applying gate current lowers the break-over voltage — the dashed curves — and with adequate gate drive the device turns on at almost any forward voltage. This is the normal mode of operation: the gate does not "control" the anode current in any proportional sense, it simply chooses the instant at which conduction starts. That instant, expressed as an angle on the supply waveform, is the firing angle \(\alpha\) that governs every phase-controlled converter in Part 2.

Two current thresholds define the boundary between the OFF and ON states, and students confuse them constantly. The latching current \(I_L\) is the minimum anode current required to establish the latch during turn-on; until the anode current exceeds it, removing the gate signal lets the device fall back to blocking. The holding current \(I_H\) is the minimum anode current required to maintain conduction once established; fall below it and the SCR turns off. Because regeneration is already fully developed by then, \(I_H\) is smaller — typically \(I_L \approx 2\!-\!3\,I_H\).

🔑

The two thresholds, and the design rule that follows

\[ I_L \approx 2\ \text{to}\ 3\,I_H \qquad\Longrightarrow\qquad \text{gate pulse must last until } i_A \gt I_L \]

In a resistive circuit the anode current rises instantly and a short trigger pulse suffices. In an inductive circuit it rises at \(di/dt = V/L\), so a narrow pulse can end before the latch takes hold and the device simply fails to fire. The cure is a wider pulse, or a train of pulses — the reason gate-drive circuits are a topic in their own right (Chapter 4).

!Common mistake

Treating latching current \(I_L\) and holding current \(I_H\) as the same thing.

They describe different moments, and a one-line test tells them apart:

\(I_L\) — getting in The minimum anode current needed to establish the latch. It matters during turn-on, while the gate pulse is still present. Fall short and the device never fires at all.

\(I_H\) — staying in The minimum anode current needed to maintain conduction. It matters during running, long after the gate pulse ended. Fall below it and the device drops out.

Why \(I_L > I_H\): establishing regeneration is harder than sustaining it. Once the feedback loop is fully developed, far less current keeps it going — hence \(I_L \approx 2\!-\!3\,I_H\).

A memory hook: it takes more effort to start a heavy flywheel spinning than to keep it spinning.

Why the gate loses authority. Once the internal regeneration of Section 3-3 has taken over, the gate junction is swamped by carriers arriving from the anode side; nothing you do at the gate terminal can starve the loop. A negative gate current will not turn a conventional SCR off — that ability had to be engineered in deliberately, and the result is the GTO of Section 3-7. Everything difficult about thyristor circuits traces back to this one fact.

⚙Where \(I_H\) bites in practice

Two real failures that look mysterious until you remember the holding current:

A lamp dimmer that flickers at low settings. Near the bottom of its range the load current dips below \(I_H\) each half cycle before the next firing, so the TRIAC drops out and re-fires erratically. Fitting a small resistor in parallel with the lamp — or using a dimmer rated for the load — keeps the current above \(I_H\).
An LED retrofit that will not dim. A 7 W LED lamp draws a fraction of the current of the 60 W incandescent the dimmer was designed for, so the TRIAC cannot hold. This is precisely why "dimmable LED" lamps carry a premium: they include a bleeder circuit that draws extra current purely to keep the thyristor latched.

A specification that looks like an obscure datasheet footnote turns out to explain a consumer complaint.

Section recap. Forward blocking is the state the diode does not have, and it is what makes controlled rectification possible. The gate chooses the instant of turn-on, not the magnitude of current. \(I_L\) gets you in, \(I_H\) keeps you in, and \(I_L \approx 2\!-\!3\,I_H\).

Section 3-3

The Two-Transistor Model

◇Intuition · two people holding each other up

Two exhausted runners lean against one another. Neither could stand alone — but each is held up by the other, so the pair is stable. Push one and you push both; remove the support from either and both collapse.

That is the thyristor latch:

\(Q_1\)'s collector current is \(Q_2\)'s base current.
\(Q_2\)'s collector current is \(Q_1\)'s base current.
Each therefore keeps the other in conduction, with no help from outside.

The gate pulse is just the initial shove. Once the two are leaning on each other, the shove is irrelevant — which is exactly why the gate loses authority. And the only way to separate them is to take away the current that both depend on: drop \(i_A\) below \(I_H\), and both transistors starve at once.

Split the four layers along a diagonal and the structure becomes two interleaved transistors: a \(pnp\) device \(Q_1\) (layers \(p\!-\!n\!-\!p\)) and an \(npn\) device \(Q_2\) (layers \(n\!-\!p\!-\!n\)), sharing the two middle layers. The collector of each feeds the base of the other. That is a positive-feedback loop, and it explains the entire device.

The latch: two transistors that hold each other on

Write the collector current of each transistor as \(\alpha\) times its emitter current plus leakage, note that \(I_{A} = I_{K}\) when no gate current flows, and solve for the anode current. The algebra is short and the result is the most quoted equation in thyristor theory:

🔑

Regeneration — one denominator explains every turn-on mechanism

\[ I_A = \frac{\alpha_2 I_G + I_{CBO1} + I_{CBO2}}{1 - (\alpha_1 + \alpha_2)} \]

At low current the transistor gains \(\alpha_1, \alpha_2\) are small, the denominator is close to unity, and \(I_A\) is just leakage — the device blocks. Anything that pushes \(\alpha_1 + \alpha_2\) toward 1 makes the denominator vanish and the anode current rise without internal limit; the SCR turns on and the external circuit alone decides the current. Since \(\alpha\) rises with emitter current, temperature, and applied voltage, this single expression predicts all five turn-on mechanisms of the next section.

→Why one equation covers everything

Read the expression as a fraction and ask what each part can be attacked by:

Numerator, \(\alpha_2 I_G\) — inject gate current. That is deliberate triggering.
Numerator, \(I_{CBO}\) — raise leakage. Heat does this (thermal triggering); so does light (LASCR); so does displacement current through \(J_2\)'s capacitance (\(dv/dt\) triggering).
Denominator, \(1-(\alpha_1+\alpha_2)\) — raise the gains. High anode voltage widens the depletion region and does exactly this (break-over triggering); so does temperature.

Five mechanisms, one equation. You do not have to memorise the list in Section 3-4 — you can derive it by asking what physical effect could touch the numerator or the denominator. That is what a good model buys you.

Section recap. Two cross-coupled transistors, each feeding the other's base, give \(I_A = (\alpha_2 I_G + I_{CBO})/[1-(\alpha_1+\alpha_2)]\). Turn-on is the denominator collapsing toward zero. Everything the device does, wanted or not, is a route to that collapse.

Section 3-4

Turn-On Mechanisms — One You Want, Four You Don't

Anything that raises \(\alpha_1 + \alpha_2\) to unity will fire the device. Only the first entry below is a design choice; the rest are failure modes that a converter must be engineered to avoid.

1 Gate triggering

Inject \(I_G\) into the numerator directly. Fast, controllable, and repeatable — the only method used deliberately. Requires the gate signal to persist until \(i_A\) exceeds \(I_L\).

2 Forward voltage (break-over)

Raise \(v_{AK}\) toward \(V_{BO}\); the widening depletion region increases \(\alpha\) until regeneration starts. Uncontrolled, and it stresses the junction. Avoided by rating \(V_{DRM}\) well above the circuit peak.

3 dv/dt triggering

The reverse-biased \(J_2\) has capacitance \(C_j\), so a rapidly rising anode voltage drives a displacement current \(i = C_j\,dv/dt\) that acts exactly like gate current. Spurious, common, and the reason snubbers exist.

4 Thermal triggering

Leakage roughly doubles per 10 °C, and \(\alpha\) rises with it. A device running too hot can fire itself — and because conduction raises the temperature further, the failure is self-reinforcing.

5 Light triggering

Photons generate carriers in the junction. Undesirable in a normal package, but engineered deliberately in the LASCR used for HVDC valves, where optical firing gives free isolation at hundreds of kilovolts.

6 The turn-on interval

Turn-on is not instant: a delay time while the gate current builds carriers, a rise time as the anode current climbs, and a spread time while conduction expands across the wafer. Together, a few microseconds.

◇Intuition · the crowd through a narrow door

The last card matters more than it sounds, and a picture fixes it.

When the SCR fires, conduction does not begin across the whole wafer at once. It starts as a small patch next to the gate and spreads outward at roughly 0.1 mm per microsecond — like a crowd entering a stadium through one narrow door and then dispersing.

For the first few microseconds, the entire anode current is squeezed through that small patch.
Current density there is enormous — hundreds of times the average.
If the external circuit forces the current up too fast, the patch melts before the crowd has dispersed.

This is why a device with a perfectly adequate average current rating can still be destroyed on its first pulse. The average rating assumes the whole wafer is conducting; during the spread interval it is not.

Two consequences worth carrying forward: a series inductor limits how fast the circuit can push current in (Section 3-6), and a harder, faster-rising gate pulse turns on a larger initial area, which is why gate drive and \(di/dt\) capability are the same problem seen from two ends (Worked Example 5).

!Common mistake

"The \(dv/dt\) rating is about how fast the supply voltage changes."

It is about the rate of rise of voltage across the device itself, and the two are frequently very different.

A 50 Hz sine reaches at most \(\omega V_m = 2\pi(50)(325) \approx 0.1\) V/µs — trivially below any rating.
But when a neighbouring thyristor in the same bridge commutates, the voltage across this device can step by hundreds of volts in a microsecond.
A nearby contactor opening, a transformer energising, or a lightning surge does the same.

The displacement current \(i = C_j\,dv/dt\) through the reverse-biased \(J_2\) then acts exactly like gate current, and the device fires when nobody asked it to. In a bridge that means a shoot-through — two devices conducting that should never conduct together.

Hence the snubber. It is not there to protect against the supply; it is there to protect against the rest of the converter.

Section recap. Five routes push \(\alpha_1+\alpha_2\) to unity: gate current (wanted), and forward break-over, \(dv/dt\), heat and light (unwanted, except in the LASCR). Turn-on also takes finite time, and during the spread interval the whole current crowds into a small patch — the origin of the \(di/dt\) limit.

Section 3-5

Turn-Off, Turn-Off Time, and Device Grades

To turn an SCR off, the anode current must fall below \(I_H\) and stay there long enough for the stored charge in the four layers to recombine. Simply returning the anode voltage to zero is not enough; if forward voltage reappears too soon, the residual charge re-triggers the device. The interval required is the circuit-commutated turn-off time \(t_q\), measured from the instant the anode current crosses zero to the instant forward voltage may safely be reapplied.

Within \(t_q\) there are two stages. First the reverse recovery of Chapter 2 sweeps out the charge near the junctions — the same \(Q_{rr}\) mechanism, the same reverse current spike. Then a gate recovery stage allows the remaining charge in the inner layers to recombine, which no external current can hasten. The circuit must therefore hold the device reverse-biased for the whole of \(t_q\), with margin.

◇Intuition · letting the device forget

Getting the anode current to zero is not enough. The four layers are still full of stored charge, and if forward voltage returns before that charge has recombined, the device simply re-triggers itself — no gate pulse required.

So the circuit has to hold the thyristor reverse-biased for long enough that it forgets it was ever conducting. That interval is \(t_q\).

An analogy that survives scrutiny: a sprinter who has just crossed the line cannot start the next race immediately. Stopping their legs is instant; recovering their breath is not. Ask them to run again too soon and they collapse back into the previous state.

Note the two stages inside \(t_q\): reverse recovery (the same \(Q_{rr}\) sweep-out as Chapter 2, which reverse current can hasten) followed by gate recovery (recombination in the inner layers, which nothing external can hurry). You can pump the charge out of the doorway; you cannot pump it out of the middle of the room.

The commutation requirement

\[ t_c \ge t_q \quad\text{(design for } t_c \approx 1.5\,t_q\text{)} \]

where \(t_c\) is the circuit turn-off time actually provided. In an AC circuit the supply reverses on its own and gives milliseconds — this is natural or line commutation, and it is free. In a DC circuit nothing reverses by itself, so an auxiliary circuit must force it, usually by dumping a pre-charged capacitor across the device. That is forced commutation, and Chapter 4 develops its classes in full.

The value of \(t_q\) splits the product range into two grades. Converter-grade (phase-control) thyristors have \(t_q\) of 50–100 µs, which is irrelevant when the supply gives 10 ms; they are optimised instead for the lowest possible on-state drop, and they dominate rectifiers, AC controllers and cycloconverters. Inverter-grade (fast-switching) thyristors are lifetime-controlled to reach \(t_q\) of 5–50 µs so they can be force-commutated at kilohertz rates, at the cost of a higher forward drop and a higher price. Specifying the wrong grade is one of the classic ways to build a converter that works on the bench and fails in service.

!Common mistake

"I'm building a converter, so I need a converter-grade thyristor."

The names describe the commutation method, not the application. Read them this way:

Grade	Really means	Turn-off time \(t_q\)	On-state drop	Use it when
Converter (phase-control)	"The AC line will commutate me"	50–100 µs	Lowest available	Rectifiers, AC controllers, cycloconverters, soft starters
Inverter (fast-switching)	"I must be force-commutated"	5–50 µs	Higher	DC choppers and inverters with commutation circuits

Ask instead: what turns this device off? If the answer is "the supply reverses on its own", any converter-grade device will do, and you should take the lower forward drop. If the answer is "a capacitor I have to build", you need inverter grade — and Worked Example 6 shows what a converter-grade \(t_q\) would cost you in capacitor size.

Fitting a converter-grade device into a force-commutated circuit produces a converter that works when cold and lightly loaded, then fails to commutate as it warms up. That intermittency is the classic symptom.

Section recap. Turn-off requires \(i_A < I_H\) and enough reverse-biased time for the stored charge to recombine: \(t_c \ge t_q\), designed at about \(1.5\,t_q\). AC supplies do this for free (natural commutation); DC circuits must force it. \(t_q\) splits the product range into converter and inverter grades.

Section 3-6

Ratings, and the Two That Actually Kill Devices

Symbol	Name	What it limits
\(V_{DRM}\)	Peak repetitive off-state forward voltage	Forward blocking without break-over
\(V_{RRM}\)	Peak repetitive reverse voltage	Reverse blocking capability
\(I_{T(AV)},\ I_{T(RMS)}\)	Average and RMS on-state current	Continuous thermal rating
\(I_{TSM},\ I^2t\)	Surge current, let-through energy	Fault survival and fuse coordination
\(I_L,\ I_H\)	Latching and holding current	Successful firing and continued conduction
\(t_q\)	Circuit-commutated turn-off time	Maximum switching frequency; commutation design
\((di/dt)_{max}\)	Critical rate of rise of current	Local heating during the spread interval
\((dv/dt)_{max}\)	Critical rate of rise of voltage	Spurious turn-on through junction capacitance
\(I_{GT},\ V_{GT},\ P_{G(AV)}\)	Gate trigger current, voltage, mean power	Gate-drive design

Voltage and current ratings get checked routinely. The two rate-of-change ratings are the ones inexperienced designers omit — and they destroy devices that were correctly sized in every other respect. Both have simple, one-component cures.

Where the two limits come from

\(di/dt\): the conduction area is small at first Turn-on starts beside the gate and spreads at about 0.1 mm/µs. Force current in faster than the patch can grow and the current density there melts the silicon — a local failure with a perfectly healthy average rating.
Cure: a series inductor An inductor opposes change of current. Across it the supply voltage produces a slope \(di/dt = V_s/L_s\), so choosing \(L_s = V_s/(di/dt)_{max}\) caps the rate the circuit can impose.
\(dv/dt\): the blocking junction is a capacitor Reverse-biased \(J_2\) has junction capacitance \(C_j\). A fast-rising anode voltage drives a displacement current \(i = C_j\,dv/dt\) straight into the gate region — indistinguishable from a real gate pulse. The device fires spuriously.
Cure: an RC snubber across the device The capacitor cannot change its voltage instantly, so it absorbs the step. Charging through \(R_s\) gives an initial slope \(V_s/(R_sC_s)\); set that below the rated \(dv/dt\) and choose \(C_s = V_s/[R_s(dv/dt)_{max}]\).

Two components, two failure modes averted

🔑

Sizing the two protective components

\[ L_s = \frac{V_s}{(di/dt)_{max}} \qquad\qquad C_s = \frac{V_s}{R_s\,(dv/dt)_{max}} \]

The inductor sees the full supply voltage at turn-on, so it fixes the initial slope of the anode current. The snubber capacitor charges through \(R_s\) when the device turns off, and the initial slope of that charging is \(V_s/(R_sC_s)\) — set it below the rated \(dv/dt\). Choose \(R_s\) first from the discharge current it must survive when the device fires: \(I_{discharge} = V_s/R_s\) adds directly to the anode current, so too small an \(R_s\) trades a \(dv/dt\) failure for a \(di/dt\) one.

!Common mistake

"A bigger snubber capacitor gives better \(dv/dt\) protection, so bigger is safer."

It protects better against \(dv/dt\) and makes the \(di/dt\) problem worse. The two cures fight each other:

While the device blocks, \(C_s\) charges up to the supply voltage.
When the device fires, \(C_s\) dumps that charge through the device, limited only by \(R_s\).
That discharge current \(V_s/R_s\) adds to the load current, arriving in the first microseconds — exactly when conduction has not yet spread.

So the snubber you fitted to prevent spurious turn-on can destroy the device on legitimate turn-on. In Worked Example 4 the snubber contributes a 20 A spike on top of the load current.

The design is a negotiation, not an optimisation: choose \(R_s\) large enough that \(V_s/R_s\) is a tolerable spike, then choose \(C_s\) to meet the \(dv/dt\) limit, then confirm the total \(di/dt\) still passes with \(L_s\) in place. If it does not, you need a different device — not a cleverer snubber.

Section recap. \(L_s = V_s/(di/dt)_{max}\) in series protects the small initial conduction area; \(C_s = V_s/[R_s(dv/dt)_{max}]\) across the device stops spurious firing through \(J_2\)'s capacitance. The two interact through the snubber discharge current \(V_s/R_s\), so they must be designed together.

Section 3-7

The Thyristor Family

The SCR is the ancestor. Its descendants each relax one of its restrictions — unidirectional conduction, or the inability to command turn-off — and each pays for it somewhere else.

Device	Turn-on	Turn-off	Typical rating	Where it is used
SCR	Gate pulse	Circuit only	Up to 8 kV, 6 kA	Rectifiers, HVDC, soft starters
TRIAC	Gate, either polarity	Current zero	Up to 1 kV, 40 A	Fan and lamp dimmers, small AC controllers
DIAC	Break-over, either polarity	Current zero	Low power	Trigger device for TRIAC gates
GTO	Positive gate pulse	Large negative gate pulse	Up to 6 kV, 6 kA	Traction drives, large inverters (now largely displaced)
IGCT	Gate pulse	Hard gate drive, unity gain	Up to 6.5 kV, 4 kA	Medium-voltage drives, grid converters
MCT	MOS gate, negative pulse	MOS gate, positive pulse	Moderate	Niche; low drive power, low on-state drop
LASCR	Light	Circuit only	Very high voltage	HVDC valves — optical firing gives free isolation
RCT	Gate pulse	Circuit only	Moderate	SCR with integral antiparallel diode for inverter legs

Two entries deserve comment. The TRIAC is effectively two SCRs in inverse parallel on one chip, so it conducts in both directions and can be gated by either polarity — which makes it ideal for cheap AC power control and useless above a few hundred hertz, since it must recover in both directions each half cycle. The GTO is the device that finally gave the family a commanded turn-off, at the price of a brutal gate drive: turning off 1000 A may require a 200 A negative gate pulse, a turn-off gain of only about five. Its successor the IGCT integrates that drive into the package and pushes the gain to unity, and it remains the switch of choice in the multi-megawatt range where IGBTs still struggle.

⚙Thyristors you have already used

Light dimmer / fan regulator — a TRIAC fired by a DIAC through an RC phase-shift network. The knob sets the RC delay, which sets the firing angle, which sets the RMS voltage.
Motor soft starter — back-to-back SCRs in each phase, firing angle ramped from 180° to 0° over a few seconds, so a large induction motor accelerates without drawing six times full-load current.
HVDC converter station — stacks of LASCRs, fired by light down optical fibres. The fibre provides hundreds of kilovolts of isolation for free, which no copper gate lead could.
Crowbar protection — an SCR deliberately fired across a supply rail to short it out when an overvoltage is detected, blowing the fuse and saving the load. The latching that is normally a nuisance is here the whole point: once fired, it stays fired.
Battery-charger and electroplating rectifiers — thousands of amperes at a few volts, where the SCR's very low on-state drop is worth more than any switching capability.

Interactive · firing angle

The gate chooses when conduction starts. Here is what that buys you: a half-wave controlled rectifier feeding a resistive load. Move \(\alpha\) and watch the output collapse. This is a preview of Part 2 — for now, just note that a timing decision has become a voltage decision, with nothing dissipated to achieve it.

Firing angle \(\alpha\) 60°

Firing angle α60°

V_avg / V_m0.239

V_rms / V_m0.448

Conduction120°

At \(\alpha = 0\) the thyristor behaves exactly like a diode. As \(\alpha \to 180°\) the output falls to zero. Note again that average and RMS fall at different rates — the Chapter 1 distinction, now with a control knob on it.

Is the thyristor obsolete? No — it is specialised. The IGBT of Chapter 5 has taken almost everything below a few megawatts because it turns off on command. But the SCR still holds two records that matter: the lowest on-state drop of any controllable device, and the highest single-device voltage and current ratings ever built. That is why HVDC converter stations, static VAR compensators, large industrial soft starters and cycloconverter mill drives remain thyristor territory, and why Part 2 of this book is built almost entirely on the device you have just met.

Section recap. The family relaxes one SCR restriction each: the TRIAC conducts both ways, the GTO and IGCT add commanded turn-off, the LASCR fires optically. Each pays somewhere — the TRIAC in frequency, the GTO in gate-drive brutality. The SCR itself survives because nothing beats its on-state drop or its ratings.

Section 3-8

Worked Examples

1 Regeneration in the two-transistor model

Problem. An SCR has \(\alpha_1 = 0.28\), \(\alpha_2 = 0.65\) and a total leakage of 2 mA. Find the anode current with no gate drive, and with \(I_G = 20\) mA. Comment on what happens as \(\alpha_1 + \alpha_2 \to 1\).

Solution. The denominator is \(1 - (0.28 + 0.65) = 0.07\).

Working

\[ I_A\big|_{I_G=0} = \frac{2\ \text{mA}}{0.07} = 28.6\ \text{mA} \]

\[ I_A\big|_{I_G=20\,\text{mA}} = \frac{(0.65)(20) + 2}{0.07} = \frac{15}{0.07} = 214\ \text{mA} \]

The gate current has multiplied the anode current sevenfold. As the rising current pushes \(\alpha_1 + \alpha_2\) toward unity the denominator collapses and \(I_A\) is limited only by the external circuit — the latch has closed, and the gate is no longer needed.

2 Will a short gate pulse latch the device?

Problem. An SCR with \(I_L = 50\) mA switches a 100 V supply into a 0.5 H inductive load. A 100 µs gate pulse is applied. Does it latch? If not, find the minimum pulse width.

Solution. The anode current rises at \(di/dt = V/L\):

Working

\[ \frac{di}{dt} = \frac{100}{0.5} = 200\ \text{A/s} \;\Rightarrow\; i_A(100\ \mu\text{s}) = (200)(100\times10^{-6}) = 20\ \text{mA} \]

\[ t_{min} = \frac{I_L\,L}{V} = \frac{(0.05)(0.5)}{100} = 250\ \mu\text{s} \]

At 20 mA the device is well short of the 50 mA latching current, so it fails to fire — and the fault is intermittent, since a slightly warmer device with lower \(I_L\) may latch. Specify a pulse of at least 250 µs, or better, a pulse train lasting the whole conduction period.

3 On-state loss in a phase-controlled SCR

Problem. A thyristor in a three-phase bridge conducts a flat-topped 100 A for one third of each cycle. With \(V_{T0} = 1.0\) V and \(r_T = 1\) mΩ, find the average and RMS currents and the conduction loss.

Solution. Using the piecewise-linear model of Chapter 2 with a conduction fraction of \(1/3\):

Working

\[ I_{T(AV)} = \tfrac{1}{3}(100) = 33.3\ \text{A}, \qquad I_{T(RMS)} = 100\sqrt{\tfrac{1}{3}} = 57.7\ \text{A} \]

\[ P = V_{T0}I_{T(AV)} + r_T I_{T(RMS)}^2 = (1.0)(33.3) + (0.001)(3333) = 36.6\ \text{W} \]

The threshold term dominates because \(r_T\) is tiny — this is exactly the trade the converter-grade device was optimised for, and it is why thyristors still win on efficiency at very high current.

4 Sizing the di/dt inductor and the dv/dt snubber

Problem. A 400 V circuit uses a thyristor rated \((di/dt)_{max} = 50\ \text{A}/\mu\text{s}\) and \((dv/dt)_{max} = 200\ \text{V}/\mu\text{s}\). Choose \(L_s\), and with \(R_s = 20\ \Omega\) choose \(C_s\). Check the snubber discharge current.

Solution.

Working

\[ L_s = \frac{V_s}{(di/dt)_{max}} = \frac{400}{50\times10^{6}} = 8\ \mu\text{H} \]

\[ C_s = \frac{V_s}{R_s(dv/dt)_{max}} = \frac{400}{(20)(200\times10^{6})} = 0.1\ \mu\text{F} \]

\[ I_{discharge} = \frac{V_s}{R_s} = \frac{400}{20} = 20\ \text{A} \]

The 20 A discharge pulse adds to the load current every time the device fires, and it must flow through the same \(L_s\) — so the two components are not independent. Reducing \(R_s\) to sharpen the \(dv/dt\) protection would raise this spike, which is the compromise every snubber design negotiates.

5 Designing the gate drive

Problem. A gate source of 15 V fires a thyristor requiring \(V_{GT} = 1.5\) V at \(I_{GT} = 100\) mA. Find the series resistor. If the pulse duty cycle is 0.3, check the mean gate power against a 0.5 W rating.

Solution.

Working

\[ R_G = \frac{V_{source} - V_{GT}}{I_{GT}} = \frac{15 - 1.5}{0.1} = 135\ \Omega \]

\[ P_{G(peak)} = V_{GT}I_{GT} = 0.15\ \text{W}, \qquad P_{G(AV)} = \delta\,P_{G(peak)} = (0.3)(0.15) = 45\ \text{mW} \]

Well inside the 0.5 W limit, so the drive could safely be made harder — and a harder, faster-rising gate pulse spreads conduction across the wafer more quickly, which directly improves the \(di/dt\) capability. Gate drive and \(di/dt\) rating are the same problem seen from two ends.

6 Commutation capacitor from the turn-off time

Problem. An inverter-grade thyristor has \(t_q = 40\ \mu\text{s}\) and carries 50 A in a 200 V DC circuit. A commutating capacitor discharges at constant load current to reverse-bias the device. Size \(C\) for a 1.5 margin.

Solution. Require \(t_c = 1.5\,t_q = 60\ \mu\text{s}\). A capacitor discharging at constant current \(I_o\) holds the device reverse-biased for \(t_c = CV_s/I_o\):

Working

\[ C = \frac{I_o\,t_c}{V_s} = \frac{(50)(60\times10^{-6})}{200} = 15\ \mu\text{F} \]

Note how the requirement scales: doubling the load current doubles the capacitor, and a converter-grade device with \(t_q = 100\ \mu\text{s}\) would need nearly 38 µF. Commutation components are bulky, lossy and expensive — which is the whole reason the industry moved to self-commutating devices in Chapter 5.

Review

Summary & Formula Sheet

✓ Semi-controlled

Four layers, three junctions, three terminals. The gate commands turn-on; only the external circuit can turn it off.

✓ Two thresholds

\(I_L\) establishes the latch, \(I_H\) maintains it, with \(I_L \approx 2\!-\!3\,I_H\). Inductive loads need wide gate pulses.

✓ Regeneration

\(I_A = (\alpha_2 I_G + I_{CBO})/(1 - \alpha_1 - \alpha_2)\); turn-on is the denominator vanishing.

✓ Five turn-on routes

Gate (wanted); forward break-over, \(dv/dt\), heat and light (unwanted, except in the LASCR).

✓ Turn-off time

\(t_c \ge t_q\). Converter grade: 50–100 µs, lowest drop. Inverter grade: 5–50 µs, force-commutable.

✓ Two protections

\(L_s = V_s/(di/dt)_{max}\) in series; \(C_s = V_s/[R_s(dv/dt)_{max}]\) across. Omitting either destroys correctly rated devices.

Formula sheet · Chapter 3

Regenerationturn-on is this denominator vanishing

\( I_A = \dfrac{\alpha_2 I_G + I_{CBO1} + I_{CBO2}}{1-(\alpha_1+\alpha_2)} \)

The two thresholdsgetting in vs staying in

\( I_L \approx 2\ \text{to}\ 3\,I_H \)

Minimum gate pulse widthinductive load, \(di/dt = V/L\)

\( t_{min} = \dfrac{I_L\,L}{V} \)

Commutation requirementdesign at \(t_c \approx 1.5\,t_q\)

\( t_c \ge t_q \)

Commutating capacitorconstant-current discharge

\( C = \dfrac{I_o\,t_c}{V_s} \)

\(di/dt\) limiting inductorprotects the spreading conduction area

\( L_s = \dfrac{V_s}{(di/dt)_{max}} \)

\(dv/dt\) snubber capacitorchoose \(R_s\) first

\( C_s = \dfrac{V_s}{R_s\,(dv/dt)_{max}} \)

Snubber discharge currentadds to anode current at firing

\( I_{discharge} = \dfrac{V_s}{R_s} \)

Gate series resistorand mean gate power at duty \(\delta\)

\( R_G = \dfrac{V_{source}-V_{GT}}{I_{GT}}, \qquad P_{G(AV)} = \delta\,V_{GT}I_{GT} \)

On-state losssame model as the diode

\( P = V_{T0}I_{T(AV)} + r_T I_{T(RMS)}^{2} \)

Key terms

Semi-controlled: Turn-on can be commanded; turn-off cannot. The defining property of the SCR and the origin of commutation.
Latching: Self-sustaining conduction through internal regeneration. Once latched, the gate has no further authority.
Latching current, \(I_L\): Minimum anode current to establish the latch during turn-on. Matters while the gate pulse is present.
Holding current, \(I_H\): Minimum anode current to maintain conduction. Matters continuously; falling below it turns the device off.
Break-over voltage, \(V_{BO}\): Forward voltage at which the device fires without a gate signal. A failure mode, not an operating mode.
Firing angle, \(\alpha\): The instant of gate triggering expressed as an angle on the supply waveform. The control variable of every phase-controlled converter.
Turn-off time, \(t_q\): Circuit-commutated turn-off time: from the anode current zero to the earliest safe reapplication of forward voltage.
Natural commutation: The AC supply reverses and turns the device off unaided. Free, and the reason line-frequency thyristor converters are so simple.
Forced commutation: An auxiliary circuit drives the anode current to zero in a DC circuit. Bulky, lossy, and the reason self-commutating devices took over.
Snubber: An RC network across a device that limits \(dv/dt\) and absorbs turn-off energy. Its discharge current is the price paid.

Self-assessment

Test Yourself

Six questions that test reasoning rather than recall. Try each aloud before revealing the answer.

Chapter 3 · six questions answers hidden until you ask

An SCR in a DC chopper is conducting. You disconnect the gate lead entirely. What happens?

Nothing. The device continues conducting exactly as before.

Once regeneration has taken hold, \(Q_1\) and \(Q_2\) supply each other's base current. The gate terminal is simply not part of that loop any more — you could cut the wire off with pliers and the load would not notice.

To turn it off you must reduce the anode current below \(I_H\), which in a DC circuit means building a commutation circuit that forces it there. This is why a DC thyristor chopper is a fundamentally harder circuit than an AC thyristor rectifier, and why the industry abandoned it as soon as GTOs and IGBTs became affordable.

The same SCR type fires reliably into a resistive load but intermittently into a large inductor, with the same gate pulse. Why?

Because latching depends on the anode current reaching \(I_L\) before the gate pulse ends — and the inductor controls how fast that happens.

Resistive load: \(i_A = v/R\) appears essentially instantly. Even a 10 µs pulse is ample.
Inductive load: \(i_A\) ramps at \(V/L\). With \(V = 100\) V and \(L = 0.5\) H that is 200 A/s, so reaching a 50 mA latching current takes 250 µs (Worked Example 2).

The intermittency is the diagnostic clue: \(I_L\) falls with temperature, so a warm device latches where a cold one does not. Any fault that appears on a cold start and clears after a few minutes should make you suspect marginal latching.

Fix: widen the pulse, or use a high-frequency pulse train that lasts the whole intended conduction period. The train is usually preferred — it costs no more, and it also re-fires the device if it drops out during a load transient.

A thyristor bridge works perfectly on the bench but fires spuriously when a large contactor elsewhere in the plant opens. Diagnose it.

Almost certainly \(dv/dt\) triggering, and the snubber is either absent or under-sized.

Opening a contactor interrupts current in an inductive circuit, generating a fast transient that couples onto the supply. The step appears across the blocking thyristor, and the displacement current \(i = C_j\,dv/dt\) through junction \(J_2\) acts exactly like gate current.

Two clues that confirm it: the fault correlates with other equipment rather than with load, and it does not appear on the bench because a bench supply has no such neighbours.

Remedies, in order: fit or resize the RC snubber across each device; add supply-side transient suppression (MOV or line reactor); and check for gate-lead loops picking up radiated noise — the gate circuit should be twisted-pair or screened, and a gate-cathode resistor of a few hundred ohms shunts induced currents away from the junction.

A converter-grade SCR has \(t_q = 100\) µs; an inverter-grade one has \(t_q = 25\) µs. Both are used in a force-commutated chopper carrying 50 A from a 200 V supply. Compare the commutating capacitors needed.

Using \(C = I_o t_c/V_s\) with \(t_c = 1.5\,t_q\):

Converter grade: \(t_c = 150\) µs → \(C = (50)(150\times10^{-6})/200 = \) 37.5 µF
Inverter grade: \(t_c = 37.5\) µs → \(C = (50)(37.5\times10^{-6})/200 = \) 9.4 µF

A factor of four in capacitance — and the capacitor must be a pulse-rated type handling the full load current, so that is a factor of four in volume, cost and commutation loss.

It also caps the switching frequency: the capacitor must recharge between commutations, so the slower device limits the chopper to a lower \(f_s\), which in turn means bulkier filter magnetics. One datasheet parameter propagates into the size of three separate components. That cascade is why \(t_q\) dominates device selection for force-commutated circuits.

Why can a TRIAC handle a 50 Hz lamp dimmer but not a 20 kHz converter?

Because a TRIAC must recover in both directions, every half cycle, and it has no help doing so.

It is effectively two SCRs in inverse parallel on a single chip. Each time the current crosses zero, the half that was conducting has to clear its stored charge before the other half is asked to block the rising voltage of the opposite polarity. The available time is set by the supply frequency.

At 50 Hz there are milliseconds — vastly more than the tens of microseconds needed.
At 20 kHz there are 25 µs per half cycle, which is comparable to \(t_q\) itself.

The device then fails to regain blocking capability and conducts continuously — a phenomenon called commutation failure. The same limitation is why TRIACs struggle with highly inductive loads even at 50 Hz: the current zero and the voltage zero no longer coincide, so the device is asked to block a substantial voltage the instant it stops conducting. That is exactly what the \(dv/dt\) snubber across a TRIAC is for.

Given that IGBTs turn off on command and thyristors do not, why do HVDC converter stations still use thyristors?

Three reasons, all of which outrank commanded turn-off in that application:

Ratings. A single thyristor reaches roughly 8 kV and 6 kA. No IGBT comes close, so a thyristor valve needs far fewer series devices — and each series device is a reliability risk and a voltage-sharing problem.
On-state drop. At these currents conduction loss dominates everything. A thyristor's drop is the lowest of any controllable device, and in a 1000 MW station a few tenths of a volt per device is megawatts.
Commutation is free anyway. A line-commutated converter is connected to a strong AC grid that reverses 50 times a second. You are being handed turn-off at no cost — paying for a device that can do it itself is paying for something you already have.

The nuance: modern voltage-source HVDC (VSC-HVDC, used for offshore wind and for feeding weak grids) does use IGBTs — because it needs capabilities a line-commutated converter cannot offer, such as black-start and independent reactive-power control. So the honest answer is that the two technologies serve different jobs, and "newer" does not mean "replaces".

Practice

Problems

Before reaching for a formula, place the question in one of three regimes — they use entirely different relations:

Turning on — regeneration, \(I_G\), \(I_L\), gate pulse width, \(di/dt\).
Staying on — \(I_H\), conduction fraction, \(V_{T0}\) and \(r_T\), loss and temperature.
Turning off — \(t_q\), \(t_c\), commutating capacitor, \(dv/dt\) on reapplication.

Problems 1–6 are direct application; 7–10 need judgement; the remainder are design questions worth discussing in a tutorial.

State which junction blocks in each case: (a) anode positive, no gate signal, (b) anode negative, (c) device conducting.
Explain, using the two-transistor model, why a negative gate current cannot turn a conventional SCR off.
An SCR has \(\alpha_1 = 0.3\) and \(\alpha_2 = 0.6\) with 1.5 mA total leakage. Find \(I_A\) with no gate current, and the \(I_G\) needed to raise \(I_A\) to 500 mA.
A thyristor has \(I_H = 18\) mA. Estimate \(I_L\), and state which of the two matters when the load current briefly dips during a fault.
An SCR with \(I_L = 40\) mA switches 200 V into 0.8 H. Find the minimum gate pulse width for reliable firing.
Repeat Problem 5 with a resistive load of 100 Ω. Comment on why the answer changes so drastically.
A thyristor conducts a constant 80 A for 40% of each cycle. With \(V_{T0} = 1.1\) V and \(r_T = 1.5\) mΩ, find \(I_{T(AV)}\), \(I_{T(RMS)}\) and the conduction loss.
A 600 V circuit uses a device rated \((di/dt)_{max} = 100\ \text{A}/\mu\text{s}\) and \((dv/dt)_{max} = 500\ \text{V}/\mu\text{s}\). Find \(L_s\), and \(C_s\) for \(R_s = 30\ \Omega\). Find the snubber discharge current.
A gate source of 12 V drives a device needing \(V_{GT} = 1.2\) V at 200 mA, with a peak gate power limit of 5 W and a mean limit of 0.5 W. Find the series resistor and the maximum permissible duty cycle.
A device with \(t_q = 25\ \mu\text{s}\) carries 120 A in a 400 V circuit. Size the commutating capacitor for a 1.5 margin, and repeat for a converter-grade device with \(t_q = 90\ \mu\text{s}\).
Compare a TRIAC and a pair of inverse-parallel SCRs for a 5 kW AC voltage controller. Give two advantages of each, and state which you would specify at 50 kW and why.
A designer replaces a converter-grade thyristor with an inverter-grade device of the same voltage and current rating in a 50 Hz phase-controlled rectifier, expecting better performance. Explain what actually changes — consider on-state drop, losses, cost and \(t_q\) — and state whether the substitution is justified.

Tip: when a thyristor problem gives you an inductance, it is almost always a firing question — the anode current rises at \(V/L\) and you are being asked whether it reaches \(I_L\) in time. When it gives you a capacitance, it is a commutation question — the capacitor holds the device reverse-biased for \(CV/I\) and you are being asked whether that exceeds \(t_q\). Spotting which of the two component types appears in the problem statement usually identifies the required relation before you read any further.