RRB JE Junior Engineer · Technical Abilities (CBT-2)
Electronics Engineering (CBT-2)
Analog devices and op-amps, digital logic, communication, 8085, electronic measurements, and power electronics at diploma depth for RRB JE CBT-2 Technical Abilities (Electronics).
Six concepts for RRB JE CBT-2 Electronics at diploma depth. The same wayside hut keeps coming back: a sensor amplifier, interlock logic, a VHF radio, an 8085-class controller, a bench CRO, and an SCR lamp driver with a current loop. Each card is one independently testable idea.
- RRB JE Junior Engineer
- Medium level
- 6 concepts
- 44 practice questions
1Analog devices and the ideal op-amp
Analog electronics deals with voltages that can sit anywhere in a range, not just two levels. In the wayside hut the track-circuit sensor is a small continuous voltage. A silicon diode lets that signal through in one direction after about 0.7\text{ V}; a BJT then copies a small base current into a larger collector current with current gain \beta=I_C/I_B, but only while the transistor stays in the active region. Cut-off is off; saturation is fully on — neither is the linear amplifier region.
An ideal op-amp has huge open-loop gain and zero input current. With negative feedback closed, the only consistent state is a virtual short: the two input pins sit at the same voltage. For the inverting amplifier that means the inverting pin is held at 0\text{ V}, so the input current V_{in}/R_{in} is forced through R_f and the closed-loop gain is -R_f/R_{in}. The gain is set by the two resistors, not by the chip's open-loop number.
Figure. Topology only, not a schematic glyph. Rin feeds the inverting node; Rf returns from the output; the non-inverting pin is grounded. Feedback holds v− at 0 V, so gain is −Rf/Rin.
How it works
- One-way firstA diode conducts after its forward drop; reverse bias blocks (until breakdown). That is rectification, not amplification.
- BJT copies currentIn the active region I_C=\beta I_B. Bias the collector-base junction reverse, or the device leaves the linear region.
- Feedback sets gainNegative feedback forces v_-\approx v_+. The inverting gain is then -R_f/R_{in}; the non-inverting gain is 1+R_f/R_g.
Inverting gain of the sensor amp
The hut's sensor amplifier is an ideal inverting op-amp with R_{in}=10\text{ k}\Omega, R_f=47\text{ k}\Omega and V_{in}=0.80\text{ V}. Find the closed-loop gain and V_{out}.
- A_v=-R_f/R_{in}-47/10=-4.7
- V_{out}=A_v V_{in}-4.7\times 0.80=-3.76\text{ V}
Pro tip. The minus sign is a phase inversion, not a fault. If the output pins at a rail, the op-amp has saturated and the resistor formula no longer holds.
An ideal op-amp inverting amplifier uses R_{in}=8\text{ k}\Omega and R_f=24\text{ k}\Omega. The closed-loop voltage gain is
- -3
- +3
- the open-loop gain of the chip
Virtual short plus zero input current gives A_v=-R_f/R_{in}=-24/8=-3. The chip's open-loop gain is assumed huge; feedback, not the chip, sets the number.
2Digital logic and sequential memory
Digital electronics uses two named levels, usually called 0 and 1. Combinational logic — gates, multiplexers, adders — computes an output from the present inputs only. The hut's point-detection lamps are combinational: the lamp map is a Boolean function of the present switch contacts. NAND and NOR are universal: any other gate can be built from copies of one of them.
Sequential logic also depends on the past, because a flip-flop stores one bit. n flip-flops give 2^n distinct states, so a 4-bit binary counter is a MOD-16 machine. Clocked at f_{clk}, the most-significant bit toggles at f_{clk}/2^n. A T flip-flop toggles; a JK flip-flop with J=K=1 also toggles and is the usual exam 'universal' flip-flop.
Figure. State count is 2^n, not 2n and not n^2. The 4-bit hut counter is the tall bar: 16 states.
How it works
- Name the classNo memory → combinational. Memory of previous clocks → sequential.
- Count the statesn bits store 2^n patterns. A decade (MOD-10) counter is a 4-bit machine that is reset after state 1001.
- Read the clockEach flip-flop stage halves the frequency. The MSB of an n-bit binary ripple counter is f_{clk}/2^n.
| Inputs | AND | OR | NAND | XOR |
|---|---|---|---|---|
| 0, 0 | 0 | 0 | 1 | 0 |
| 0, 1 | 0 | 1 | 1 | 1 |
| 1, 0 | 0 | 1 | 1 | 1 |
| 1, 1 | 1 | 1 | 0 | 0 |
MSB of the interlock counter
A 4-bit binary ripple counter in the hut is clocked at 12.8\text{ kHz}. How many states does it have, and what is the frequency at the MSB?
- states =2^n=2^416 (MOD-16)
- f_{\mathrm{MSB}}=f_{clk}/1612800/16=800\text{ Hz}
Pro tip. Q2, three stages down, is f_{clk}/8=1.6\text{ kHz}. Dividing by n instead of 2^n is the usual miss.
An 8-to-1 multiplexer needs how many select lines?
- 3
- 8
- 1
2^s inputs require s select bits, so 8=2^3 needs three select lines. Eight select lines would address 256 inputs.
3Modulation, bandwidth and sampling
A low-frequency message cannot travel far as baseband. Modulation writes that message onto a high-frequency carrier so an antenna of practical size can radiate it. In amplitude modulation the carrier amplitude follows the message. The modulation index is \mu=V_m/V_c; over-modulation (\mu>1) folds the envelope and the receiver cannot recover the shape. AM always occupies 2fm of bandwidth, one sideband on each side of fc, regardless of μ.
Frequency modulation writes the message onto the carrier frequency. The deviation \Delta f and the message f_m set Carson's bandwidth 2(\Delta f+f_m). Sampling a band-limited message needs f_s>2f_{max} (Nyquist); an n-bit PCM stream then runs at n f_s bits per second. The hut's VHF telemetry is FM; the voice order-wire, if digitised, is an 8\text{ kHz}\times 8-bit PCM stream at 64\text{ kbit/s}.
Figure. Same 15\text{ kHz} message. AM needs 30\text{ kHz}. Narrowband FM (\beta=0.5) needs 45\text{ kHz}. Broadcast-like WBFM with \Delta f=75\text{ kHz} needs 180\text{ kHz}. Width, not 'FM is always better', is the point.
How it works
- AM envelope\mu=V_m/V_c. Total transmitted power is P_c(1+\mu^2/2); the two sidebands share P_c\mu^2/2.
- FM widthCarson's rule BW\approx 2(\Delta f+f_m) is the diploma estimate. \beta=\Delta f/f_m is the FM index, a different number from AM's \mu.
- Sample then encodeNyquist first (f_s>2f_{max}), then n bits per sample. Bit rate =n f_s, not n+f_s.
Sideband power on the hut radio
An AM transmitter in the hut runs P_c=400\text{ W} at \mu=0.5. Find the total radiated power and the power in both sidebands together.
- P_t=P_c(1+\mu^2/2)400(1+0.25/2)=400\times 1.125=450\text{ W}
- P_{sb}=P_c\mu^2/2400\times 0.125=50\text{ W}
Pro tip. Each sideband carries 25\text{ W}. \mu does not change the 2f_m bandwidth — it only changes how the power splits.
A 5\text{ kHz} message on AM occupies how much RF bandwidth?
- 5\text{ kHz}
- 10\text{ kHz}
- 2.5\text{ kHz}
USB and LSB sit at f_c\pm f_m, so BW=2f_m=10\text{ kHz}. \mu does not appear in the bandwidth.
4The 8085 as a bus machine
The 8085 is an 8-bit CPU. It names memory and I/O with a 16-bit address, so the address space is 2^{16}=64\text{ KB}. The low eight address bits share pins with the data bus (AD_0–AD_7); ALE strobes those bits into a latch at the start of every machine cycle so the same pins can then carry data. The hut controller is this machine: program in EPROM, status in RAM, lamps and contacts on I/O ports.
Five hardware interrupts, in falling priority, are TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR. TRAP is non-maskable. RST 7.5 is edge-triggered; RST 6.5 and 5.5 are level-triggered. Their restart addresses are 0024\text{H}, 003C\text{H}, 0034\text{H} and 002C\text{H}. A crystal on the 8085 is divided by two, so a 6\text{ MHz} crystal clocks the CPU at 3\text{ MHz}.
Figure. The CPU is the only master in this diploma picture. Memory and I/O are slaves on the buses. ALE and the multiplexed AD bus live in the prose; this figure is the three-box topology.
How it works
- Address then dataALE captures AD_0–AD_7. A_8–A_{15} stay on their own pins. \overline{RD} or \overline{WR} then moves the byte.
- Size the memoryN locations need \log_2 N address lines. 4\text{ KB}=4096=2^{12} needs twelve lines.
- Stack and callPUSH or CALL writes a 16-bit value and decrements SP by 2. RET pops it back. The stack grows downward.
| Interrupt | Maskable | Vector |
|---|---|---|
| TRAP | No | 0024H |
| RST 7.5 | Yes | 003CH |
| RST 6.5 | Yes | 0034H |
| RST 5.5 | Yes | 002CH |
| INTR | Yes | external |
Address lines for the hut EPROM
The hut maps a 4\text{ KB} EPROM starting at 0000\text{H}. How many address lines does that chip need, and what is its last address?
- 4\text{ KB}=4096=2^{12}12 address lines
- start 0000\text{H}, length 0FFF+1last address 0FFF\text{H}
Pro tip. The 8085 still has sixteen address pins. The unused high bits become chip-select logic; they are not extra memory on that EPROM.
A 6\text{ MHz} crystal on an 8085 produces an internal clock of
- 6\text{ MHz}
- 3\text{ MHz}
- 12\text{ MHz}
The 8085 divides the crystal by two. 6\text{ MHz} in gives 3\text{ MHz} out. Doubling is the 8086-era confusion.
5Measuring with a CRO and a bridge
A CRO draws voltage against time. Horizontal scale is time/division; vertical scale is volts/division. Period T is (divisions of one cycle) times (time/div); frequency is 1/T. Peak-to-peak voltage is (vertical divisions) times (volts/div). For a sine, V_{rms}=V_{pp}/(2\sqrt{2}). The hut's maintenance bench uses the CRO to see the sensor-amp output, not a multimeter average.
A bridge compares an unknown against a known ratio. Wheatstone balances resistance (P/Q=R/S). Maxwell's inductance bridge and Schering's capacitance bridge are the diploma names for L and C. Accuracy is closeness to the true value; precision is how tightly repeats cluster. An LVDT is a displacement transducer — a voltage that tracks position — not a voltmeter.
Figure. One 6\text{ V} peak-to-peak sine, four named voltages. V_m=3\text{ V}, V_{rms}=2.12\text{ V}, full-wave average 2V_m/\pi=1.91\text{ V}. The CRO shows V_{pp}; a rectifier meter leans toward the average.
How it works
- Read the CROT=(\text{div})\times(\text{time/div}), f=1/T, V_{pp}=(\text{div})\times(\text{V/div}). Do not swap the two knobs.
- Balance the bridgeAt null, the ratio of one arm pair equals the other. Write the two ratios before substituting numbers.
- Name the transducerLVDT → displacement. Thermocouple → temperature. Strain gauge → strain. The instrument after them is a different question.
CRO on the sensor sine
One cycle of the hut sensor occupies 5 horizontal divisions at 0.2\text{ ms/div} and 4 vertical divisions at 5\text{ V/div}. Find f and V_{rms} assuming a sine.
- T=5\times 0.2\text{ ms}1.0\text{ ms}
- f=1/T1000\text{ Hz}
- V_{pp}=4\times 520\text{ V}
- V_{rms}=V_{pp}/(2\sqrt{2})7.07\text{ V}
Pro tip. Using V_{pp}/\sqrt{2} forgets that V_m is already half of V_{pp}. That slip gives 14.1\text{ V} — twice the true rms.
A Wheatstone bridge at balance has P/Q=R/S with P=1\text{ k}\Omega, Q=3\text{ k}\Omega and R=2\text{ k}\Omega. S is
- 6\text{ k}\Omega
- 1.5\text{ k}\Omega
- 3\text{ k}\Omega
1/3=2/S so S=6\text{ k}\Omega. Crossing the product the other way (S=PQ/R) is the trap and gives 1.5\text{ k}\Omega.
6Thyristors, converters and closed-loop control
Power electronics switches large power with semiconductor latches. An SCR (thyristor) turns on when it is forward-biased and a gate pulse arrives; it then stays on by itself until the anode current falls below the holding current. A TRIAC is the bidirectional cousin, used on AC. A chopper is DC-to-DC (V_{avg}=\delta V_s); an inverter is DC-to-AC. For a single-phase half-wave controlled rectifier on a resistive load, V_{avg}=(V_m/2\pi)(1+\cos\alpha). At \alpha=0 this is the familiar V_m/\pi.
A control system decides the firing angle — or any other actuator — from a measurement. Open-loop never looks at the result; closed-loop subtracts the measured output from the reference and drives the error toward zero. A first-order plant with time constant \tau reaches about 63\% of its final value in one \tau and is treated as settled by 5\tau. In the hut, the SCR lamp driver is the plant; a current transformer plus a comparator is the closed loop that holds lamp current.
Figure. Half-wave controlled rectifier, V_m=325\text{ V}. V_{avg}=(V_m/2\pi)(1+\cos\alpha): 103.5, 77.6, 51.7, 25.9\text{ V}. Delayed firing shrinks the average; \alpha=90^\circ is not off.
How it works
- Latch on, starve offGate starts an SCR; holding-current starvation (or forced commutation) stops it. Removing the gate alone does not turn it off.
- Name the converterAC→DC is a rectifier (controlled if \alpha is used). DC→DC is a chopper. DC→AC is an inverter.
- Close the loopMeasure, subtract from the set-point, act. Integral action kills a leftover offset; derivative damps overshoot.
Firing angle on the lamp rectifier
The hut lamp is fed from a 230\text{ V} rms single-phase half-wave controlled rectifier (V_m=325\text{ V}) with resistive load and \alpha=60^\circ. Find V_{avg}.
- V_{avg}=(V_m/2\pi)(1+\cos 60^\circ)(325/2\pi)(1+0.5)
- 325/(2\pi)51.73
- V_{avg}=51.73\times 1.577.6\text{ V}
Pro tip. At \alpha=0^\circ the same formula is V_m/\pi=103.5\text{ V}. At \alpha=90^\circ it is 51.7\text{ V}, not zero — zero average waits until \alpha=180^\circ.
A step-down chopper has V_s=110\text{ V} and duty cycle 0.6. The average output is
- 66\text{ V}
- 110\text{ V}
- 183\text{ V}
V_{avg}=\delta V_s=0.6\times 110=66\text{ V}. V_s/\delta would be a step-up confusion.
Notes
- Analog devices and the ideal op-amp: A diode passes signal one way past its forward drop, and a BJT copies base current into collector current, I_{C} = \beta I_{B}, only in the active region - cut-off and saturation are switch states, not amplification. With negative feedback closed the op-amp’s two inputs sit at the same voltage, so an inverting stage has gain -R_{f}/R_{in}, set by the two resistors and not by the chip’s open-loop gain.
- Combinational has no memory; sequential does: Gates, multiplexers and adders compute from the present inputs, and NAND and NOR are each universal. A flip-flop stores one bit, so n of them give 2^{n} states and a 4-bit binary counter is a MOD-16 machine; each stage halves the clock, so the most significant bit toggles at f_{clk}/2^{n}.
- Modulation, bandwidth and sampling: In amplitude modulation the index is \mu = V_{m}/V_{c}, over-modulation folds the envelope, and the bandwidth is 2f_{m} whatever the index is. Frequency modulation takes Carson’s estimate 2(\Delta f + f_{m}). Sampling needs f_{s} > 2f_{max}, after which an n-bit PCM stream runs at n f_{s} bits per second.
- The 8085 is a bus machine: A 16-bit address gives 2^{16} = 64 KB of space, and ALE strobes the low eight address bits off the shared AD0-AD7 pins so the same pins can then carry data. Hardware interrupt priority falls TRAP, RST 7.5, RST 6.5, RST 5.5, INTR, with TRAP non-maskable and vectored at 0024H, and the crystal is divided by two to clock the CPU.
- Reading a CRO and balancing a bridge: Period is divisions times time per division and peak-to-peak is divisions times volts per division, so for a sine V_{rms} = V_{pp}/(2\sqrt{2}). A bridge compares an unknown against a known ratio - Wheatstone for resistance at P/Q = R/S, Maxwell for inductance and Schering for capacitance. Accuracy is closeness to the true value; precision is how tightly repeats cluster.
- Thyristors, converters and closed loops: An SCR latches on when a gate pulse arrives while it is forward-biased, and it turns off only when the anode current falls below the holding current - removing the gate does not stop it. A chopper is DC to DC at V_{avg} = \delta V_{s}, an inverter is DC to AC, and a controlled rectifier’s average follows the firing angle. Open loop never looks at the result, a closed loop drives the measured error toward zero, and a first-order plant reaches about 63% of its final value in one time constant.
Formulas
- BJT active region: I_{C} = \beta I_{B}. Cut-off and saturation are switch states, not amplification. With feedback closed an op-amp holds both inputs at the same voltage, so an inverting stage gives -R_{f}/R_{in}.
- Sequential capacity: n flip-flops give 2^{n} states, so a 4-bit binary counter is MOD-16 and its most significant bit toggles at f_{clk}/2^{n}. NAND and NOR are each universal.
- AM: index \mu = V_{m}/V_{c} and bandwidth 2f_{m} whatever the index. FM by Carson: 2(\Delta f + f_{m}). Sampling needs f_{s} > 2f_{max}, and an n-bit PCM stream runs at n f_{s} bits per second.
- 8085: a 16-bit address bus gives 2^{16} = 64 KB. ALE strobes the low address byte off AD0-AD7. Interrupt priority falls TRAP, RST 7.5, RST 6.5, RST 5.5, INTR, with TRAP non-maskable at 0024H.
- CRO and bridges: period = divisions x time/div, peak-to-peak = divisions x volts/div, and for a sine V_{rms} = V_{pp}/(2\sqrt{2}). Wheatstone balances at P/Q = R/S; Maxwell reads inductance, Schering capacitance.
- Power and control: an SCR latches on and turns off only below holding current. Chopper V_{avg} = \delta V_{s}, inverter is DC to AC, and a first-order plant reaches about 63% of final value in one time constant.
Exam traps & shortcuts
- I_{C} = \beta I_{B} holds only in the active region; in saturation the collector current is set by the external circuit.
- AM bandwidth is 2f_{m} whatever the modulation index — the index changes the envelope, not the width.
- A 4-bit counter is MOD-16, and the highest bit toggles at f_{clk}/16, not at f_{clk}/4.
- ALE exists because AD0-AD7 carry address then data; a question about "why multiplexed pins" is asking about ALE.
- A CRO shows peak-to-peak, so divide by 2\sqrt{2} for the RMS value of a sine.
- Removing the gate pulse does not turn an SCR off — the anode current must fall below the holding current.
Reference tables
One row per concept. The hut column is the running example, not a second syllabus.
| Concept | Object of study | Hut picture |
|---|---|---|
| Analog / op-amp | Diode, BJT β, virtual short, −Rf/Rin | Sensor amplifier |
| Digital / sequential | Combinational vs memory, 2^n states | Interlock counter |
| Communication | μ, 2fm, Carson, Nyquist, n fs | VHF telemetry |
| 8085 buses | 16-bit address, ALE, interrupt vectors | Hut controller |
| Measurements | CRO scales, bridge ratio, LVDT | Bench CRO |
| Power + control | SCR latch, Vavg(α), closed loop | Lamp driver |
Each formula answers one question. Do not borrow a neighbour's number.
| Ask this | Use this | Not this |
|---|---|---|
| Inverting gain | −Rf/Rin | open-loop A |
| AM bandwidth | 2 fm | μ or Δf |
| AM total power | Pc(1+μ²/2) | Pc(1+μ) |
| FM width (Carson) | 2(Δf+fm) | 2 fm |
| PCM bit rate | n fs | 2 fm |
| Half-wave controlled Vavg | (Vm/2π)(1+cos α) | Vm/π at every α |
| Chopper Vavg | δ Vs | Vs/δ |
| CRO frequency | 1 / (div × time/div) | div × V/div |
Recap
Read this the night before CBT-2 Electronics.
- Analog
- Active BJT: Ic=β Ib. Ideal inverting op-amp: virtual short, Av=−Rf/Rin. The chip's open-loop gain is not the answer.
- Digital
- Combinational has no memory. n flip-flops → 2^n states. MSB of a binary counter is fclk/2^n. NAND/NOR are universal.
- Comms
- AM bandwidth is always 2fm. Power split is Pc(1+μ²/2). FM width is Carson 2(Δf+fm). PCM rate is n fs after Nyquist.
- 8085
- 16-bit address, 64 KB, ALE demultiplexes AD0–7. Priority TRAP > 7.5 > 6.5 > 5.5 > INTR. Crystal ÷ 2. TRAP vector 0024H.
- Measure
- CRO: T from time/div, Vpp from V/div, Vrms=Vpp/(2√2) for a sine. Wheatstone P/Q=R/S. LVDT is displacement.
- Power / loop
- SCR latches; gate-off does not turn it off. Vavg=(Vm/2π)(1+cos α). Chopper δVs. Closed-loop uses the measured output.
Practise Electronics Engineering (CBT-2)
Reading is free and needs no account. Practice, mocks and progress live in the app.
- 44 exam-style questions on this topic, with explanations
- A 5-question practice set that ends the chapter
- Timed mocks scored with the real marking scheme
- Readiness tracked per topic, kept on your device