Autotransformer and instrument transformers

Why it matters

Autotransformers appear wherever a modest voltage change is needed cheaply: variacs on every lab bench, induction-motor starters and interconnecting 400 kV and 220 kV grids. Instrument transformers are how every meter, relay and data-acquisition system in a substation or plant sees thousands of amperes and kilovolts safely. An instrumentation engineer must know their ratios, errors and the safety rules that go with them.

Key ideas

Autotransformer

• One winding per phase with a tapping. The part shared by input and output is the common winding; the remaining part is the series winding. Input and output are electrically connected — there is no isolation.
• With transformation ratio k = V_LV/V_HV (< 1), the common winding carries only the difference of the input and output currents. Part of the power is passed conductively (directly through the copper) and only the remainder inductively (through the core).
• For a step-down autotransformer with output V2 and input V1: conductively transferred power = S·k and inductively transferred power = S·(1 − k), where k = V2/V1.
• Copper needed (for the same output and current density) is (1 − k) times that of a two-winding transformer, so the copper saving is k times. The saving is large only when k is close to 1, i.e. when the two voltages are close. That is why autotransformers are used for ratios up to about 2–3.
• Advantages: less copper and core, smaller losses, higher efficiency, better regulation, smaller size and cost.
• Disadvantages: no isolation (a break in the common winding puts full input voltage on the load); lower series impedance so higher fault current; and a fault or surge on the HV side reaches the LV side directly.
• A two-winding transformer reconnected as an autotransformer (windings in series, additive polarity) handles a much larger kVA with the same winding currents and the same losses.

Instrument transformers. Small, precise transformers that reduce line current or voltage to standard values (5 A or 1 A for CTs, 110 V line-to-line for VTs/PTs) and isolate the instruments from the high-voltage circuit.

• Current transformer (CT): primary in series with the line, often a single turn (bar primary); secondary feeds an ammeter or relay coil — a near short circuit. Primary current is set by the line, not by the CT burden. Nominal ratio is stated as currents, e.g. 200/5 A, so turns ratio N2/N1 ≈ 40.
• Never open-circuit a CT secondary while the primary carries current. With no secondary ampere-turns to oppose it, the whole primary MMF magnetises the core: flux rises to saturation, a dangerously high peak voltage appears across the open terminals, and core heating and residual magnetism ruin accuracy. Short the secondary before removing an instrument.
• Voltage / potential transformer (VT/PT): primary across the line, secondary feeds voltmeters and potential coils — a high impedance, so it operates nearly at no load, like a small power transformer. The secondary must not be short-circuited; it is fused.
• Errors. Because of magnetising and core-loss current (CT) or winding drops (PT), the actual ratio differs from the nominal ratio (ratio error) and the secondary phasor is not exactly 180° from the primary (phase-angle error). Phase-angle error matters for wattmeters and energy meters. CTs are specified by accuracy class (e.g. 0.5, 1 for metering; 5P, 10P for protection) and by rated burden in VA — take the class limits from the relevant standard.
• Protection CTs must not saturate at fault currents many times rated; metering CTs are designed to saturate early so the meter is protected.

Formulas

• k = V2 / V1 — autotransformer ratio (step-down: output/input, < 1).
• I_common = I2 − I1 — current in the common winding (A), step-down; (I1 − I2) for step-up with the roles swapped.
• S_conductive = S·k ; S_inductive = S·(1 − k) — power transferred conductively and inductively (VA).
• Cu_auto / Cu_two-winding = 1 − k — copper weight ratio; copper saving = k × (two-winding copper).
• S_auto = S_2w·(V_HV/V_series) — kVA rating when a two-winding transformer is reconnected as an autotransformer; V_series is the voltage of the winding used in series.
• Kn = I_p,rated / I_s,rated — CT nominal ratio; ratio error = (Kn·Is − Ip)/Ip × 100 %.
• Kn = V_p,rated / V_s,rated — PT nominal ratio; ratio error = (Kn·Vs − Vp)/Vp × 100 %.

Worked examples

Example 1 (standard): step-down autotransformer. Given: an autotransformer supplies a 9 kVA load at 180 V from a 230 V supply. Neglect losses and magnetising current.

1. Ratio: k = 180/230 = 0.783.
2. Currents: I1 = 9000/230 = 39.1 A (input, series winding); I2 = 9000/180 = 50.0 A (load).
3. Common winding: I_common = 50.0 − 39.1 = 10.9 A — only a small current in most of the winding.
4. Power transferred inductively = 9000 × (1 − 0.783) = 1.96 kVA; conductively = 9000 × 0.783 = 7.04 kVA.
5. Copper saving = k = 78 % of the copper of an equivalent two-winding transformer.

Example 2 (GATE level): two-winding transformer reconnected as an autotransformer, and a CT reading. Given: a 10 kVA, 2000/200 V transformer with core loss 100 W and full-load copper loss 150 W is connected as a 2000/2200 V step-up autotransformer (windings in series, additive).

1. Rated winding currents: HV winding 10000/2000 = 5 A; LV winding 10000/200 = 50 A.
2. The 200 V winding is now the series winding and carries the output current, so output current = 50 A at 2200 V.
3. Autotransformer rating: S = 2200 × 50 = 110 kVA (11 times the two-winding rating). Input current = 110000/2000 = 55 A; the common 2000 V winding carries 55 − 50 = 5 A — its rated value.
4. Losses are unchanged because winding currents and core flux are unchanged. Full-load efficiency at unity pf: two-winding 10000/10250 = 97.56 %; autotransformer 110000/110250 = 99.77 %.
5. CT part: a 200/5 A CT feeds an ammeter that reads 3.6 A. Line current = 3.6 × (200/5) = 144 A (ignoring ratio error).

Common mistakes

• Opening a CT secondary to change an ammeter with the line live — always short it first.
• Writing a CT "turns ratio 100:5" when it is the current ratio 100/5 A; turns are in the inverse ratio of currents.
• Assuming an autotransformer isolates the load; it does not, so it is never used for safety isolation or where the LV side must be earthed separately.
• Taking copper saving as (1 − k) instead of k.
• Forgetting the polarity when reconnecting a two-winding transformer — subtractive connection gives 1800 V, not 2200 V.
• Treating a PT like a CT: a PT secondary must never be shorted; a CT secondary must never be opened.

For GATE IN

Expect numericals on autotransformer currents, kVA rating after reconnection, copper saving and efficiency, plus CT/PT ratio and ratio-error calculations and conceptual questions on why a CT secondary must not be opened. Instrument transformers also appear in the Measurements part of the paper (extension of meter range, phase-angle error in power measurement), so practise both together.

Quick check

1. An autotransformer steps 240 V down to 200 V. What fraction of copper is saved?
2. A 400/5 A CT's ammeter reads 4 A. What is the line current?
3. What is the danger of opening a CT secondary on load?
4. Why can an autotransformer not be used to step 11 kV down to 230 V for a domestic supply?

Answers: 1. k = 200/240 = 0.833, i.e. 83 %. 2. 320 A. 3. Core saturation, very high secondary voltage, overheating and loss of accuracy. 4. The large ratio gives little saving, and with no isolation the consumer could see 11 kV if the common winding fails.