Circuits
Competency 4.0 is the circuit itself: how series and parallel R, L and C behave on DC and AC, how to classify an impedance and write it in polar or rectangular form, resonance and Q, filters, oscillators and multivibrators, RC differentiators and integrators, clippers and clampers, and the phase-locked loop. This page follows ETA's list. Section 12 adds the thyristors (SCR, triac and diac), which ETA lists under Competency 3.6 but which are the switches in AC phase-control circuits, and section 13 is job practice. Build the series and parallel circuits in the Circuits Lab, and see impedance and resonance drawn in the Visual Learning Lab.
On the exam from this page
- the voltage drops around any closed loop add up to the applied voltage (KVL), and the current into any node equals the current out of it (KCL)
- a circuit is inductive when X_L is larger than X_C, capacitive when X_C is larger, and resistive when they cancel
- rectangular form, R ± jX, gives the resistance and the reactance; polar form, Z at θ, gives the size of the impedance and its phase angle
- an astable multivibrator has no stable state and free-runs; a monostable has one stable state and gives one pulse per trigger; a bistable has two and changes state only when triggered
- an RC integrator takes its output across the capacitor and needs a long time constant; an RC differentiator takes its output across the resistor and needs a short one
- a PLL compares the phase of its divided-down VCO output with a reference, filters the error into a control voltage, and so holds the VCO at N times the reference frequency
Series and parallel circuits on DC and AC
Competency 4.1 asks you to identify and describe common DC circuits, and 4.2 common AC circuits. In a series circuit there is one path, so the same current flows through every part and the voltage divides among them. In a parallel circuit every branch sits across the same two points, so each branch has the same voltage and the currents add. Most real circuits mix the two.
On DC, once the circuit has settled for about five time constants (the sheet's τ = R × C and τ = L / R), a capacitor is fully charged and passes no current, and an inductor drops only what its winding resistance drops, because the sheet's V = L(di/dt) is zero once the current stops changing. On AC the source keeps reversing, so capacitors and inductors keep reacting: their reactances, XL = 2π × f × L and XC = 1 / (2π × f × C) on the sheet, change with frequency, and they shift the current's phase. Kuphaldt's rule: every DC law still holds on AC, as long as you work in impedances with their phase angles (sections 3 and 4). Power is the exception.
Competency 4.3 and 4.4 ask how series and parallel R, L and C are used in equipment. ETA's sheet gives every combination:
| Connection | Resistors and inductors | Capacitors | Where you meet it |
|---|---|---|---|
| Series | Add: R = R1 + R2 + …, L = L1 + L2 + … | Reciprocals add: 1/C = 1/C1 + 1/C2 + …, so the total is less than the smallest | Dropping and current-limiting resistors, voltage dividers that set a bias point, RC and RL timing networks, and a series LC circuit that passes one frequency |
| Parallel | Reciprocals add: 1/R = 1/R1 + 1/R2 + …, so the total is less than the smallest | Add: C = C1 + C2 + … | Loads across a supply rail, each getting the full rail voltage and working on its own; and a parallel LC (tank) circuit, which has its highest impedance at resonance |
Worked example
A series-parallel network (Competency 4.17): 24 V feeds R1 = 2.2 kΩ in series with R2 = 4.7 kΩ and R3 = 10 kΩ in parallel. Reduce the parallel pair first: 4.7 × 10 / 14.7 ≈ 3.197 kΩ. The total is 2.2 + 3.197 = 5.397 kΩ, so the source current is 24 / 5.397 ≈ 4.447 mA. R1 drops 4.447 × 2.2 ≈ 9.78 V and the pair 4.447 × 3.197 ≈ 14.22 V. The pair's branch currents are 14.22 / 4.7 ≈ 3.03 mA and 14.22 / 10 ≈ 1.42 mA. Check: 9.78 + 14.22 = 24 V, and 3.03 + 1.42 = 4.45 mA, the source current to rounding.More practice, step by step, is in the Series Circuits and Parallel Circuits lessons in the Study Hub; the Circuits Lab simulates DC series and parallel circuits and analyzes RC, RL and RLC circuits on AC.
Kirchhoff's laws
Competency 4.10 asks you to explain Kirchhoff's laws and why they matter to a technician. Kuphaldt states them as:
- Kirchhoff's Voltage Law (KVL): the algebraic sum of all voltages in a loop must equal zero. Going around a series loop, the drops add up to the applied voltage.
- Kirchhoff's Current Law (KCL): the algebraic sum of all currents entering and leaving a node must equal zero. What flows into a junction flows out.
On the exam
Exam point: the voltage drops around any closed loop add up to the applied voltage (KVL), and the current into any node equals the current out of it (KCL). Together they let you find a voltage or current you didn't measure, as in the check at the end of section 1.Why a technician cares. The two laws turn a meter into a fault finder: if the readings around a loop don't add up to the supply, or the branch currents don't add up to the total, something between the points you measured is not what the schematic says. NEETS uses KVL to explain the RC wave-shaping circuits in section 9: at every instant the resistor's and the capacitor's voltages add up to the input.
Trap
Exam trap: KVL still holds when a series string has failed open. No current flows, so every good part drops 0 V and the whole source voltage appears across the open. The part that reads the full supply is the open one, not a short.Resistive, inductive or capacitive?
Competency 4.7 asks you to classify circuits as inductive, capacitive or resistive. In a series circuit with both kinds of reactance, XL and XC work against each other, and NEETS subtracts them: X = XL − XC, and the result carries the sign of the larger one. ETA's sheet gives the size of the series impedance, and of R and X in parallel:
| The circuit is | When | Current compared with voltage | Phase angle |
|---|---|---|---|
| Resistive | No net reactance: only R, or XL = XC (resonance) | In phase | 0° |
| Inductive | XL is larger than XC | Lags the voltage | Between 0° and +90° |
| Capacitive | XC is larger than XL | Leads the voltage | Between 0° and −90° |
The phase angle is the shift between the voltage and the current. A perfect resistor's is 0°, a perfect inductor's +90° (its voltage leads its current) and a perfect capacitor's −90° (its voltage lags). The classification depends on frequency: NEETS shows that a series LC circuit looks capacitive below its resonant frequency and inductive above it, and a parallel (tank) circuit the other way round.
Worked example
One circuit, two frequencies: R = 80 Ω, L = 10 mH and C = 1 µF in series. At 1 kHz, XL = 6.28 × 1000 × 0.01 = 62.8 Ω and XC = 1 / (6.28 × 1000 × 0.000001) ≈ 159.2 Ω, so X = 62.8 − 159.2 = −96.4 Ω: capacitive, and Z = √(80² + 96.4²) ≈ 125.3 Ω. At 2 kHz, XL = 125.6 Ω and XC ≈ 79.6 Ω, so X ≈ +46.0 Ω: inductive, and Z ≈ 92.3 Ω. The circuit resonates in between, at 1 / (6.28 × √(0.01 × 0.000001)) ≈ 1592 Hz, where it is resistive and Z = 80 Ω.On the exam
Exam point: a circuit is inductive when X_L is larger than X_C, capacitive when X_C is larger, and resistive when they cancel. In an inductive circuit the current lags the voltage; in a capacitive one it leads.Polar and rectangular notation
Competency 4.9 asks you to describe the polar and rectangular presentations of L, R, C circuits. An impedance has a size and a direction, so it is written as a complex number, in either of two forms (Kuphaldt):
- Rectangular form, R + jX: the real part is the resistance and the imaginary part, marked j, is the reactance. Inductive reactance is +j, capacitive reactance is −j. A 100 Ω resistor is 100 + j0 Ω; an inductor with XL = 50 Ω is 0 + j50 Ω; a capacitor with XC = 50 Ω is 0 − j50 Ω.
- Polar form, Z at θ: the magnitude and the phase angle (textbooks print an angle sign between the two; this page writes “at”). The same three parts are 100 Ω at 0°, 50 Ω at 90° and 50 Ω at −90°.
Both describe the same right triangle: R is the adjacent side, X the opposite side and Z the hypotenuse. Converting is trigonometry. Only the magnitude formula is on ETA's sheet (section 3); the angle and the back-conversion are not:
Why two forms? Kuphaldt: rectangular form suits addition and subtraction (add the real parts, add the j parts), which is how series impedances combine; polar form suits multiplication and division (multiply the magnitudes and add the angles; divide the magnitudes and subtract the angles), which is how Ohm's law, I = V / Z, works on AC.
Worked example
Section 3's circuit at 1 kHz: in rectangular form Z = 80 − j96.4 Ω. In polar form the magnitude is √(80² + 96.4²) ≈ 125.3 Ω and the angle arctan(−96.4 / 80) ≈ −50.3°, so Z ≈ 125.3 Ω at −50.3°. Back again: 125.3 × cos(−50.3°) ≈ 80.0 and 125.3 × sin(−50.3°) ≈ −96.4. Now put 10 V at 0° across it. I = V / Z in polar form: divide the magnitudes, 10 / 125.3 ≈ 79.8 mA, and subtract the angles, 0° − (−50.3°) = +50.3°. So I ≈ 79.8 mA at +50.3°: the current leads the voltage, as a capacitive circuit's does. Going the other way, 25 Ω at 60° is 25 × cos 60° = 12.5 Ω of resistance and 25 × sin 60° ≈ 21.65 Ω of inductive reactance: 12.5 + j21.65 Ω.On the exam
Exam point: rectangular form, R ± jX, gives the resistance and the reactance; polar form, Z at θ, gives the size of the impedance and its phase angle. A positive j term or angle means inductive; a negative one means capacitive.Trap
Exam trap: the angle is arctan(X / R), not arctan(R / X): with 30 Ω of resistance and 40 Ω of inductive reactance the impedance is 50 Ω at 53.1°, not at 36.9°. And the magnitude is never R + X (70 Ω here), because the two sides are at right angles.The impedance triangle and phasor views in the Visual Learning Lab draw the same triangle as you change R, XL and XC.
Resonance, Q and bandwidth
Competency 4.8 asks you to explain resonance and calculate the resonant frequency. At one frequency XL equals XC: a series LC circuit then has its minimum impedance, and a parallel (tank) circuit its maximum. ETA's sheet gives two forms; the second uses kilohertz, microhenries and microfarads:
Competency 4.15 asks how bandwidth and Q relate. Q measures how sharp the resonance is. Kuphaldt gives Q = X / R for the coil's reactance and the circuit's resistance, and the bandwidth, measured between the half-power (70.7%) points, is the resonant frequency divided by Q. Neither formula is on ETA's sheet:
So a high Q means a narrow bandwidth and a selective circuit, and series resistance lowers Q (for a parallel tank, heavier loading, that is a lower shunt resistance, does the same): NEETS notes that less resistance gives a narrower bandwidth, which is better selectivity. In a series resonant circuit the voltage across L or C is Q times the applied voltage (Kuphaldt).
Worked example
Both sheet forms, then Q: L = 100 µH and C = 0.001 µF (1000 pF). The kHz form gives 159.2 / √(100 × 0.001) = 159.2 / 0.3162 ≈ 503.4 kHz. The hertz form, with 2π = 6.28, gives 1 / (6.28 × √(0.0001 × 0.000000001)) ≈ 503.5 kHz; the two differ only by the rounding of their constants. With 10 Ω of coil resistance, XL = 6.28 × 503,400 × 0.0001 ≈ 316.1 Ω, so Q = 316.1 / 10 ≈ 31.6 and BW = 503.4 / 31.6 ≈ 15.9 kHz.The Resonance lesson in the Study Hub has more on series and parallel resonance, and the resonant frequency calculator is in the Calculator Suite.
Filter circuits
Competency 4.13 asks you to describe filter circuits, why and how they are used. Kuphaldt: a filter separates some frequencies from others in a mixed signal, as in audio equalizers and loudspeaker crossover networks. It works because reactance changes with frequency.
| Filter | Passes | Simple RC or LC form (Kuphaldt) |
|---|---|---|
| Low-pass | Low frequencies; blocks high | An inductor in series with the load, or a series resistor with a capacitor across the load |
| High-pass | High frequencies; blocks low | A capacitor in series with the load, or a series resistor with an inductor across the load |
| Band-pass | One band; blocks frequencies too low or too high | A low-pass and a high-pass section in cascade, or a series LC circuit in the signal path, which passes its resonant frequency |
| Band-stop (notch) | Everything except one band | A low-pass and a high-pass section in parallel (the twin-T), or a series LC circuit across the signal path, which shorts out its resonant frequency |
The cutoff frequency is where the output has fallen to 70.7% of the input. For a single RC section it is where XC equals R, which gives a formula that is not on ETA's sheet but comes straight from its XC formula:
Worked example
An RC low-pass filter: R = 4.7 kΩ in series, C = 0.01 µF across the output. fc = 1 / (6.28 × 4700 × 0.00000001) ≈ 3388 Hz, about 3.39 kHz; at that frequency XC = 1 / (6.28 × 3388 × 0.00000001) ≈ 4.7 kΩ, equal to R. Swap the two parts (C in series, R across the output) and the same values make a high-pass filter with the same cutoff.Power supply filters (L, T and pi) are in the Power Supplies guide, and the Filter Calculator works RC, RL and LC sections for you.
Oscillators, crystals and the piezoelectric effect
Competency 4.5 asks you to explain the purpose of oscillators and the use of crystals. NEETS: an oscillator is an amplifier that provides its own input through feedback, and its purpose is to generate a waveform at a constant amplitude and a specific frequency. It needs three things:
- Amplification, to make up the circuit's losses.
- Regenerative (positive) feedback, enough to sustain the oscillation.
- A frequency-determining device: an LC tank, an RC network or a crystal. Without one the stage oscillates at a random frequency.
LC oscillators (NEETS's Armstrong, Hartley and Colpitts) use a tuned tank; RC oscillators use a phase-shifting RC network. A load that changes pulls the frequency and amplitude, so a stable low-power oscillator is often followed by a buffer amplifier that isolates it from the load.
Competency 4.16 asks you to explain the piezoelectric effect. NEETS: it is the property of a crystal by which mechanical forces produce electrical charges and, conversely, electrical charges produce mechanical forces. Quartz shows it less strongly than Rochelle salt, but it is used for frequency control because it is mechanically stronger. A quartz crystal is a thin wafer ground to a thickness that sets its natural resonant frequency: for a given cut, the thinner the crystal, the higher the frequency.
| Crystal fact (NEETS) | What it means in a circuit |
|---|---|
| It behaves like an R, L and C in series, with the holder's capacitance across them | It can act as a series-tuned circuit (lowest impedance at series resonance) or, slightly higher in frequency, a parallel-tuned circuit (highest impedance) |
| Below series resonance it acts capacitively; between series and parallel resonance, inductively | In NEETS's series-mode crystal oscillator, the impedance rising away from resonance cuts the feedback, so the circuit won't oscillate at any other frequency |
| Its Q is very high: 5,000 to 30,000 for commercial crystals | Frequency stability, which is why crystal oscillators keep radio transmitters within their assigned limits |
| Temperature and drive change it | The most stable transmitters keep the crystal in a temperature-controlled oven; too much drive can crack it or age it early. NEETS (an older Navy text) gives 50 to 200 mA as the safe current for the crystals of its day; a modern crystal's datasheet gives a maximum drive level in microwatts, and Renesas's example crystal runs at about 1.74 mA (57 µW) |
Multivibrators
Competency 4.6 asks you to tell oscillators from multivibrators. The oscillators in section 7 make sine waves, set by a tank, an RC network or a crystal. A multivibrator is a relaxation oscillator: NEETS describes it as two amplifiers with regenerative feedback, one conducting while the other is cut off, switching quickly between saturation and cutoff. Its output is a square or rectangular wave, and in the astable and monostable types an RC time constant, not a resonant circuit, sets its timing (a bistable is not timed: it waits for a trigger). NEETS classifies multivibrators by their number of stable states:
| Type | Stable states | What it does | Typical use |
|---|---|---|---|
| Astable (free-running) | None | Switches back and forth by itself at a rate set by its RC time constants; it needs no input | Clock and pulse generators |
| Monostable (one-shot) | One | Rests in its stable state; a trigger flips it, and after a time set by its RC components it returns by itself | Uniform pulses from irregular triggers, and pulse stretching |
| Bistable (flip-flop) | Two | Stays in either state until a trigger flips it to the other | Memory: the flip-flops and latches in digital circuits |
On the exam
Exam point: an astable multivibrator has no stable state and free-runs; a monostable has one stable state and gives one pulse per trigger; a bistable has two and changes state only when triggered. A multivibrator makes square or rectangular waves; a sine-wave oscillator's frequency is set by a tank or a crystal.The 555 timer is the usual IC multivibrator. TI's NE555 datasheet gives its timing, which is not on ETA's sheet: a monostable pulse of tw = 1.1 × RA × C, and an astable frequency of 1.44 / ((RA + 2RB) × C), with the output high for 0.693 × (RA + RB) × C and low for 0.693 × RB × C.
Worked example
A 555 both ways: as a one-shot with RA = 47 kΩ and C = 10 µF, each trigger gives a pulse of 1.1 × 47,000 × 0.00001 ≈ 0.517 s. As an astable with RA = 1 kΩ, RB = 10 kΩ and C = 0.1 µF, f = 1.44 / (21,000 × 0.0000001) ≈ 686 Hz. The output is high for 0.693 × 11,000 × 0.0000001 ≈ 0.762 ms and low for 0.693 × 10,000 × 0.0000001 ≈ 0.693 ms, a duty cycle of about 52.4%; it can't reach 50% because the capacitor charges through RA + RB but discharges through RB alone.Try other values in the 555 Timer Calculator.
Differentiators and integrators
Competency 4.11 asks for the purposes and types of differentiator and integrator circuits. NEETS calls them shaping circuits: a series RC (or RL) network whose time constant is chosen against the length of the input pulse, so that it changes the shape of square, rectangular or sawtooth waves. They do not change the shape of a pure sine wave. The names come from calculus, although the RC circuits only approximate the operations.
| RC integrator | RC differentiator | |
|---|---|---|
| Output taken across | The capacitor | The resistor |
| Time constant | Long: NEETS uses at least 10 times the pulse duration | Short: NEETS's example is one-tenth of the pulse duration |
| Square wave in gives | A slowly rising and falling, sloped wave: the capacitor never has time to charge fully | A sharp positive spike at each rising edge and a negative spike at each falling edge |
| Works like | A low-pass filter: the capacitor drops less of each higher harmonic, which rounds the edges | A high-pass filter: only the fast edges get through |
| RL version | Output across the resistor, long time constant | Output across the inductor, short time constant |
Uses: the differentiator turns edges into the sharp spikes that timing and trigger circuits need, and the integrator turns a square wave into a sloped, ramp-like wave; NEETS notes the RC integrator is used as a wave-shaping network in communications, radar and computers. An op-amp with a capacitor in its feedback loop makes a true integrator, and one with a capacitor at its input a true differentiator; Kuphaldt: a differentiator gives a constant output for a steadily changing input, and an integrator a steadily changing output for a constant input.
Worked example
Short or long? A 1 kHz square wave, 10 V high, has 500 µs halves. With R = 4.7 kΩ and C = 0.01 µF, τ = 4700 × 0.00000001 = 47 µs, and 500 / 47 ≈ 10.6, so the time constant is short. Taken across the resistor, each rising edge gives a 10 V spike that decays to 36.8% of it (about 3.68 V) after one time constant, about 1.35 V after two and about 0.50 V after three, and is gone after five (235 µs), well before the next edge: a differentiator. To integrate the same wave you need τ of at least 10 × 500 µs = 5 ms; R = 100 kΩ and C = 0.1 µF give 10 ms, twenty times the half-period, with the output across the capacitor.On the exam
Exam point: an RC integrator takes its output across the capacitor and needs a long time constant; an RC differentiator takes its output across the resistor and needs a short one. Fed a square wave, the integrator gives a sloped wave and the differentiator gives spikes at the edges.Wave shaping: clippers and clampers
Competency 4.14 asks you to explain wave-shaping circuits and their purposes. Besides the RC shapers in section 9, the two diode circuits ETA expects are the clipper and the clamper.
Clippers (limiters) remove the part of a waveform beyond a set level. NEETS: a limiter keeps some part of a waveform from exceeding a specified value, and is used for wave shaping and circuit protection. In a series limiter the diode is in series with the output; in a parallel (shunt) limiter it is across the output, after a series resistor. Kuphaldt's examples: one silicon diode across the output clips one peak at about 0.7 V; two in anti-parallel clip both peaks at about ±0.7 V; a DC source in series with the diode moves the clipping level to that voltage plus the diode drop; and a zener can replace the diode and the source. Uses: keeping voice peaks from overdriving a radio transmitter, squaring up a sine wave, and protecting IC inputs with diodes to the supply rails.
Clampers (DC restorers) keep the waveform's shape but shift its DC level, so one peak sits at a fixed voltage. A series capacitor and a shunt diode do it: the diode charges the capacitor to the peak on one half-cycle, and the capacitor's charge then adds to the signal. The diode's direction decides which peak is clamped; a DC reference in series with the diode moves the clamp level. NEETS also calls it a base-line stabilizer; Kuphaldt's application is restoring the DC level of composite video in television transmitters and receivers.
Worked example
Clip, then clamp: an 8 V peak sine wave feeds a parallel clipper whose silicon diode returns to a 4.3 V DC source: the positive peaks are flattened at 4.3 + 0.7 = 5.0 V, and the negative half-cycles, which reverse-bias the diode, pass unchanged down to −8 V. Now feed a 5 V peak sine into a clamper whose silicon diode conducts on the negative peaks. The capacitor charges to 5 − 0.7 = 4.3 V, and the output is the input plus 4.3 V: it swings from −5 + 4.3 = −0.7 V to 5 + 4.3 = +9.3 V. Its shape and its 10 V peak-to-peak are unchanged. Reverse the diode and it swings from −9.3 to +0.7 V instead.Trap
Exam trap: a clipper changes the waveform's shape by cutting off part of it; a clamper doesn't change the shape or the peak-to-peak value at all, only where the waveform sits relative to 0 V. If the output has flat tops, it was clipped; if it is the same wave moved up or down, it was clamped.An overdriven amplifier clips too, as distortion; see the Amplifiers guide.
The phase-locked loop
Competency 4.12 asks you to describe the sections of a PLL and its uses. A phase-locked loop is a negative-feedback loop that holds a voltage-controlled oscillator at an exact multiple of a stable reference. TI's application note on PLL synthesizers describes the integer-N loop, the classic form, in these blocks:
| Section | What it does |
|---|---|
| Reference | A stable crystal oscillator, often temperature-compensated (TCXO), usually divided down by R to the comparison frequency Fr. |
| Phase (frequency) detector | Compares the phase of Fr with the divided-down VCO signal and puts out an error proportional to the difference. |
| Loop filter | A low-pass filter, usually R and C, that smooths the detector's output into a steady control voltage and keeps the detector's switching noise off the VCO. A narrower loop filter is cleaner but slower to change channel. |
| VCO | An oscillator whose tank is tuned by the control voltage across a varactor; it is the loop's output. |
| Divider (÷N) | Divides the VCO output by N before it goes back to the phase detector. Changing N changes the output frequency. |
When the loop is locked, the divided VCO frequency equals the reference, so the output is N times Fr, and the channel spacing is Fr. This is TI's equation, not one on ETA's sheet:
Worked example
A VHF synthesizer: a 12.8 MHz TCXO divided by R = 1024 gives Fr = 12,800,000 / 1024 = 12.5 kHz. With N = 12,000 the VCO locks at 12,000 × 12.5 kHz = 150.000 MHz; with N = 12,001 it relocks one channel up, at 150.0125 MHz. TI notes that the locked VCO has the reference's accuracy: a 1 ppm reference holds 150 MHz to within about ±150 Hz.On the exam
Exam point: a PLL compares the phase of its divided-down VCO output with a reference, filters the error into a control voltage, and so holds the VCO at N times the reference frequency. Its best-known use is the frequency synthesizer, which TI calls the dominant frequency-synthesis method in wireless communications: it gives a radio many crystal-accurate channels from one crystal reference. Other PLL uses include FM demodulation, tone decoding, data (clock) synchronization and motor-speed control (TI CD74HC4046A).The RF & Communications guide shows where the synthesizer sits in a transmitter and a receiver.
Thyristors: SCR, triac and diac
ETA lists the thyristors under Competency 3.6.1 (identify diacs, triacs and SCRs and explain their operation) and 3.6.2 (compare SCRs with other semiconductors). They are taught here because their main job is switching AC power in phase-control circuits. Kuphaldt and Littelfuse: a thyristor is a four-layer PNPN device that behaves like a PNP and an NPN transistor coupled in positive feedback, so once it turns on it latches, staying on with no further drive, and once off it stays off. Bipolar transistors and FETs don't latch: their output follows the input signal moment by moment.
| Device | Conducts | How it turns on | How it turns off | Typical use |
|---|---|---|---|---|
| SCR (silicon-controlled rectifier) | One direction, anode to cathode | A gate pulse, positive with respect to the cathode, while it is forward biased (it can also be fired by exceeding its breakover voltage or by too fast a rise in voltage) | Anode current falling below the holding current. Removing the gate signal does not turn it off. | Controlled rectifiers and motor drives; Kuphaldt notes that larger power-control circuits favour SCRs |
| Triac | Both directions | A gate signal of either polarity (Littelfuse: quadrants I and III are the most used) | Current falling below the holding current, which on AC happens near every zero crossing | Lamp dimmers and other simple, low-power AC controls |
| Diac | Both directions, with no gate | Voltage across it reaching its breakover voltage, in either direction | Current falling below its dropout level as the AC current falls toward zero | Triggering a triac: Littelfuse calls it an economical trigger for dimmers and motor speed controls |
Phase control. On AC, an SCR or triac turns itself off as the current falls through zero at the end of each half-cycle (natural commutation) and must be fired again in the next. Firing it later in each half-cycle chops more of the sine wave away and delivers less power; a lamp dimmer's RC network delays the firing point. Kuphaldt explains why the diac is there: triacs often don't fire at the same gate voltage in each polarity, and the diac's breakover voltage is much more symmetrical, so it makes the triac fire at the same point on both half-cycles and cuts the harmonics. A gate turn-off thyristor (GTO) is an SCR variant built to be turned off by a reverse gate pulse, which takes a fairly high current (Kuphaldt: typically 20% of the anode current).
Trap
Exam trap: the gate only turns an SCR on. On DC it stays on after the gate signal is gone, until the anode current is interrupted or falls below the holding current; on AC it drops out by itself at each zero crossing. A transistor switch, by contrast, turns off when its base or gate drive is removed.The Semiconductor Basics guide covers the diodes and transistors these devices are built from.
On the job: measuring circuits that misbehave
Job practice, not exam content. These checks apply the sections above at the bench. The service manual and your employer's procedures come first.
- Let Kirchhoff find the fault. Measure the drops around a loop and compare them with the supply. In a dead series string the part that reads the full supply voltage is open (section 2). If one part reads near 0 V while the others read higher than the expected values, that part is shorted.
- Remember the parallel paths. An ohmmeter across a part still in the circuit reads that part in parallel with everything else connected to the same two nodes, so it reads low. Lift one end before trusting a resistance reading; the Test Equipment guide covers in-circuit testing.
- Don't let the probe pull the oscillator. NEETS: load changes shift an oscillator's frequency and amplitude, which is why a buffer follows it. Measure at the buffer's output where you can, and use a 10× probe rather than a 1× probe on the oscillator itself (see probe loading in the Amplifiers guide).
- Off-frequency synthesizer? Check the reference. A locked PLL is only as accurate as its reference (section 11), so a radio that is off frequency on every channel points at the reference oscillator, not the VCO. Follow the radio's service manual for the reference adjustment.
- Test an SCR with a simple DC circuit. Kuphaldt's test: the SCR in series with a load and a DC supply, a normally open pushbutton from the anode to the gate to latch it, and a normally closed pushbutton in series to unlatch it. It should latch when the gate button is pressed and stay on after it is released, and drop out when the series button breaks the current. If it won't stay latched, the load current may be below the SCR's latching current, the minimum that keeps it on right after the gate signal is removed (Littelfuse); Kuphaldt's test simply calls this drawing less than the holding current.
Safety
Phase-control circuits are line-connected: the triac in a lamp dimmer or motor speed control sits in series with the load across the AC line (as in Kuphaldt's dimmer circuit), so the whole circuit, trigger network included, is at line potential. Treat every part of it as live. De-energize, lock out and verify before you touch it (see the Safety guide), and use an isolation transformer when you must probe it live with grounded test equipment.Check yourself
Answer in your head first, then reveal.
What do Kirchhoff's two laws say?
the voltage drops around any closed loop add up to the applied voltage (KVL), and the current into any node equals the current out of it (KCL)
When is a circuit inductive, capacitive or resistive?
a circuit is inductive when X_L is larger than X_C, capacitive when X_C is larger, and resistive when they cancel
What does each notation tell you about an impedance?
rectangular form, R ± jX, gives the resistance and the reactance; polar form, Z at θ, gives the size of the impedance and its phase angle
How many stable states does each kind of multivibrator have?
an astable multivibrator has no stable state and free-runs; a monostable has one stable state and gives one pulse per trigger; a bistable has two and changes state only when triggered
Where is the output taken, and how long is the time constant, in an RC integrator and an RC differentiator?
an RC integrator takes its output across the capacitor and needs a long time constant; an RC differentiator takes its output across the resistor and needs a short one
How does a PLL hold its VCO on frequency?
a PLL compares the phase of its divided-down VCO output with a reference, filters the error into a control voltage, and so holds the VCO at N times the reference frequency
Practice this topic: Circuits questions are in the Study Hub quiz bank, and the key terms are in Flashcards.