4.4 Autotransformers, Instrument Transformers (CTs & VTs) & Special Transformers
Key Takeaways
- An autotransformer utilizes a single continuous tapped winding per phase, where power transfers through both direct electrical conduction (S_cond) and magnetic core induction (S_trans), drastically increasing the effective apparent power rating.
- The power boost factor of an autotransformer relative to its two-winding physical core rating is S_auto / S_two_winding = 1 / (1 - 1/a) = V_H / (V_H - V_L), yielding massive savings in cost, physical size, weight, and core/copper losses when the transformation ratio a is close to 1.0.
- The per-unit impedance of an autotransformer is significantly reduced by the ratio (1 - 1/a), which lowers internal voltage drop but causes dramatically higher short-circuit fault currents (I_sc = I_rated / Z_pu,auto) and poses safety hazards due to lack of galvanic isolation.
- Current Transformers (CTs) step down high AC currents to standard secondary levels (5 A or 1 A); opening a live CT secondary circuit eliminates the demagnetizing secondary MMF, creating extreme core flux saturation and lethal kilovolt-level peak voltages across the open terminals.
- ANSI/IEEE C-class CT ratings (e.g., C400 or C800) specify the maximum secondary terminal voltage the CT can deliver at 20 times rated secondary current (100 A for a 5A CT) without exceeding 10% ratio error: V_terminal = 20 * I_nom * (R_CT + R_burden).
4.4 Autotransformers, Instrument Transformers (CTs & VTs) & Special Transformers
Executive Overview: Specialized transformer designs fulfill critical roles in high-voltage interconnects, power system protection, and grounding stabilization. Autotransformers exploit direct electrical conduction alongside magnetic induction to achieve dramatic reductions in physical footprint, cost, and losses for near-unity voltage transformation. Instrument Transformers—Current Transformers (CTs) and Voltage Transformers (VTs/PTs)—scale extreme primary currents and voltages to standard low-energy instrumentation signals ($5\text{ A}, 120\text{ V}$) for protective relays and revenue metering. Grounding Transformers establish stable system neutral points on ungrounded delta networks. On the NCEES PE Power examination, candidates must expertly compute autotransformer power boosts, verify CT saturation compliance under ANSI C-class limits, calculate secondary loop burdens, and identify the severe hazards of open-circuited CT secondaries.
1. Autotransformer Theory & Power Transfer Mechanics
Unlike a conventional two-winding transformer which relies solely on mutual electromagnetic induction between physically isolated windings, an autotransformer utilizes a single continuous tapped copper winding per phase shared between the high-voltage and low-voltage circuits.
AUTOTRANSFORMER STEP-DOWN CONFIGURATION SCHEMATIC:
High-Voltage Terminal (H1)
o
|
| I_H ->
( ( (
Series ( ( (
Winding (N_se)( ( (
Voltage V_se ( ( (
|
Common Junction +-------------------------o Low-Voltage Terminal (X1)
| -> I_L = I_H + I_c
( ( (
Common ( ( (
Winding (N_c) ( ( (
Voltage V_L ( ( (
( ( (
|
+-------------------------o Common Neutral (H2 / X2)
o
Mathematical Formulation: Conduction vs. Transformation Power
Let the high-voltage terminal be $V_H$ with current $I_H$, and the low-voltage terminal be $V_L$ with current $I_L$. Defining the autotransformer turns ratio:
The voltage across the series winding is $V_{se} = V_H - V_L$, and the current in the common winding is $I_c = I_L - I_H$.
The total apparent power rating delivered by the autotransformer is:
This total power splits into two distinct energy transfer mechanisms:
- Transformed Power ($S_{trans}$ / $S_{ind}$): Power transferred magnetically through core induction across the series and common windings (corresponding to the physical physical size and core copper rating of the two-winding core):
- Conducted Power ($S_{cond}$): Power transferred directly through electrical conductor connection without traversing the magnetic core:
The Power Advantage (Boost) Factor
The ratio of autotransformer output capacity to the physical two-winding transformer size is the Power Advantage Factor:
+---------------------------------------------------------------------------------------------------+
| AUTOTRANSFORMER POWER ADVANTAGE FACTOR ACROSS COMMON VOLTAGE RATIOS |
+----------------------------+-----------------------+-----------------------+----------------------+
| Voltage Transformation | Turns Ratio ($a$) | Boost Factor ($S_a/S_{2w}$)| Conducted Power % |
+----------------------------+-----------------------+-----------------------+----------------------+
| **$138\text{ kV} \to 115\text{ kV}$** | $a = 1.200$ | **$6.00\times$** | $83.3\%$ Conducted |
| **$230\text{ kV} \to 115\text{ kV}$** | $a = 2.000$ | **$2.00\times$** | $50.0\%$ Conducted |
| **$500\text{ kV} \to 230\text{ kV}$** | $a = 2.174$ | **$1.85\times$** | $46.0\%$ Conducted |
| **$480\text{ V} \to 120\text{ V}$** | $a = 4.000$ | **$1.33\times$** | $25.0\%$ Conducted |
+----------------------------+-----------------------+-----------------------+----------------------+
Design Takeaway: When $V_H$ and $V_L$ are close (e.g., $138\text{ kV} / 115\text{ kV}$ or $500\text{ kV} / 345\text{ kV}$ grid tie), an autotransformer delivers up to $6\times$ the power of a standard transformer of the same physical dimensions and weight, dramatically lowering capital costs and losses.
Per-Unit Impedance & Short-Circuit Fault Current Implications
The per-unit impedance of an autotransformer on its own autotransformer rating base ($Z_{pu,auto}$) is reduced by the identical ratio $(1 - 1/a)$ compared to its two-winding base impedance:
+---------------------------------------------------------------------------------------------------+
| CRITICAL ENGINEERING TRADE-OFFS OF AUTOTRANSFORMERS |
+-------------------------------------------------+-------------------------------------------------+
| Major Advantages | Serious Disadvantages & Operational Hazards |
+-------------------------------------------------+-------------------------------------------------+
| - Vastly higher MVA rating for same core size. | - **Zero Galvanic Isolation:** Direct copper |
| - Higher full-load efficiency (>99.5%). | path allows HV surges/transients onto LV bus. |
| - Substantially lower percent voltage drop. | - **Extreme Short-Circuit Currents:** Low Z_pu |
| - Smaller physical footprint, weight, and cost. | drives severe fault levels (I_sc = I_r / Z_pu)|
+-------------------------------------------------+-------------------------------------------------+
2. Current Transformers (CTs): Relaying & Metering Fundamentals
Current Transformers (CTs) step down high primary currents ($I_p$) to standardized secondary levels ($I_s = 5\text{ A}$ nominal in North America, $1\text{ A}$ nominal in IEC systems) for protective relays, digital fault recorders, and ammeters.
CURRENT TRANSFORMER RELAYING CIRCUIT & BURDEN MODEL:
Primary Line (Ip)
o===================+===================o
|
+---+---+ (Toroidal Magnetic Core)
| ( C ) |
+---+---+
| Secondary Winding (Ns Turns, R_ct)
+-------------[ R_ct ]---------[ R_lead ]---------+
| |
(~) E_s (Internal Induced EMF) [ Z_relay ] Relay Burden
| |
+------------------------------[ R_lead ]---------+
CT Polarity & The Dot Convention
- Terminals are designated $H_1 - H_2$ on the primary and $X_1 - X_2$ on the secondary.
- Polarity Rule: When primary current instantaneously enters the polarity terminal $H_1$, secondary current instantaneously leaves the polarity terminal $X_1$.
Total Secondary Burden Impedance ($Z_b$)
The burden is the total connected electrical impedance across the secondary terminals, expressed in ohms or volt-amperes ($VA = I_s^2 Z_b = 25 Z_b$ for a 5A CT):
Where the lead loop resistance for a two-wire run of distance $L$ (feet) with conductor resistance $r$ ($\Omega/1000\text{ ft}$) is:
3. ANSI/IEEE C-Class Rating & CT Saturation Analysis
Under high-current short-circuit faults, the massive primary current can drive the CT ferromagnetic core into magnetic saturation. Once saturated, the secondary current becomes severely distorted (chopped peaks), causing protective relays to misoperate or fail to trip.
ANSI/IEEE C-Class Definition (IEEE C57.13)
The ANSI/IEEE standard designates relaying CTs by a letter and a voltage rating, e.g., C100, C200, C400, C800:
- Letter "C" (Calculated): Signifies that the CT leakage flux is negligible (bushing/toroidal CT with uniformly distributed secondary winding), allowing performance to be calculated directly from excitation curves.
- Voltage Number (e.g., C400, C800): The maximum secondary terminal voltage that the CT can deliver at 20 times rated secondary current ($20 \times 5\text{ A} = 100\text{ A}$) to a standard burden without exceeding a $10%$ ratio error.
ANSI C-CLASS RATING RELATIONSHIP:
$$
V_{\text{terminal,rating}} = 20 \times I_{nom} \times Z_{b,\text{std}} = 100\text{ A} \times Z_{b,\text{std}}
$$
- C100: Rated for 100 V across a 1.0 Ω burden (100 A * 1.0 Ω = 100 V)
- C200: Rated for 200 V across a 2.0 Ω burden (100 A * 2.0 Ω = 200 V)
- C400: Rated for 400 V across a 4.0 Ω burden (100 A * 4.0 Ω = 400 V)
- C800: Rated for 800 V across a 8.0 Ω burden (100 A * 8.0 Ω = 800 V)
Saturation Prevention Criteria
To guarantee that a CT will not saturate during a maximum symmetrical primary fault current $I_{f,\max}$:
4. The Open-Circuit CT Secondary Hazard & Safety Protocols
[!CAUTION] FATAL HAZARD: NEVER OPEN-CIRCUIT A LIVE CURRENT TRANSFORMER SECONDARY! Operating a CT with an open-circuited secondary ($I_s = 0$) while primary current flows produces lethal peak voltages, catastrophic dielectric breakdown, arc-flash explosions, and fatal electric shock hazard to personnel.
OPEN-CIRCUIT CT EXCITATION MECHANISM:
NORMAL OPERATING STATE: OPEN-CIRCUIT FAULT STATE:
Primary MMF (Np*Ip) ≈ Secondary MMF (Ns*Is) Secondary MMF Ns*Is = 0 (Open Circuit!)
Np*Ip ====> <==== Ns*Is Np*Ip ====================>
| |
v v
Net MMF = Np*Ip - Ns*Is Net MMF = Np*Ip (Massive Un-opposed!)
Core Flux Φ_m is MINUSCULE Core Flux Driven into DEEP SATURATION!
Induced Voltage E_s = 5 - 20 V e_s(t) = -Ns * (dΦ/dt) ==> MULTI-KILOVOLT SPIKES!
Physical Explanation of the Open-Circuit Spike
In normal operation, the secondary current produces a counter-MMF ($N_s I_s$) that directly opposes and cancels nearly $99%$ of the primary MMF ($N_p I_p$), leaving only a tiny net magnetizing MMF. When the secondary circuit is opened ($I_s = 0$), the entire primary load or fault current acts as an un-opposed magnetizing current. As the primary current passes through zero, the flux collapses and reverses with extreme speed ($d\phi/dt \to \infty$), inducing narrow, lethal peak voltage spikes of $2,000\text{ V}$ to $10,000+\text{ V}$ across the open secondary terminals.
Standard Operating Procedure: Before disconnecting any relay, meter, or ammeter from a CT circuit, operators must always close the integrated CT secondary shorting block to provide a solid low-impedance metallic path for secondary current.
5. Voltage Transformers (VTs/PTs) & Grounding Transformers
Voltage Transformers (Potential Transformers - VTs/PTs)
Voltage Transformers are high-accuracy parallel step-down transformers designed to step down high line-to-line or line-to-neutral voltages to standard secondary levels ($120\text{ V}$ line-to-line, or $120/\sqrt{3} = 69.3\text{ V}$ line-to-neutral).
- Burden: VTs operate near open-circuit conditions (connected to high-impedance relay voltage inputs, drawing $< 0.1\text{ A}$). Burdens are rated in Volt-Amperes at standard power factors (e.g., ANSI Accuracy Classes: $0.3W, 0.3X, 0.3Y, 0.3Z$).
- Safety: Secondary circuits are protected with secondary fuses and solidly grounded at one point to protect against primary-to-secondary insulation puncture.
Grounding Transformers (Zig-Zag & Wye-Delta Banks)
Industrial facilities and utility distribution substations often operate three-phase Delta systems that have no neutral connection. Connecting single-phase loads or detecting single line-to-ground (SLG) faults requires establishing an artificial system neutral using a Grounding Transformer.
ZIG-ZAG (INTERCONNECTED STAR) GROUNDING TRANSFORMER TOPOLOGY:
System Line A o----------------+
|
( ( ( a1
( ( (
|
+---( ( ( c2 ---+
System Line B o----------------+ ( ( ( |
| |
( ( ( b1 |
( ( ( |
| +--- N (Neutral Grounded via Resistor/Solid)
+---( ( ( a2 ---+ |
System Line C o----------------+ ( ( ( |
| |
( ( ( c1 |
( ( ( |
| |
+---( ( ( b2 -------+
( ( (
- Zig-Zag (Interconnected Star) Transformer: Each phase consists of two winding halves wound on separate core legs in opposite magnetic polarity. Under balanced normal conditions, positive-sequence currents cancel magnetically, resulting in high magnetizing impedance and near-zero idling current. Under a single line-to-ground fault, zero-sequence fault currents flow in the same direction through all windings, encountering very low leakage impedance ($Z_0$) to safely return fault current to the system neutral.
- Wye-Delta Grounding Bank: A grounded Wye primary connected to the delta system with a closed, ungrounded Delta secondary that allows zero-sequence circulating currents to flow freely during ground faults.
6. Comprehensive Step-by-Step Worked Mathematical Example
Problem Scenario
(Part A: Autotransformer Rating & Fault Analysis) An electric utility interconnects a $13.8\text{ kV}$ distribution bus to a $12.0\text{ kV}$ sub-transmission feeder using a step-down autotransformer bank rated for $15.0\text{ MVA}$ at $13.8 / 12.0\text{ kV}$:
- Calculate the physical two-winding transformer capacity ($S_{two\text{-}winding}$) required to construct this $15\text{ MVA}$ autotransformer.
- Calculate the conducted power ($S_{cond}$) and transformed power ($S_{trans}$) when operating at full $15\text{ MVA}$ capacity.
- If the physical two-winding transformer has a per-unit impedance of $Z_{pu,2w} = 0.060\text{ pu}$, determine the per-unit impedance on the autotransformer base ($Z_{pu,auto}$) and the maximum symmetrical three-phase short-circuit current ($I_{sc}$) available at the $12.0\text{ kV}$ bus assuming an infinite $13.8\text{ kV}$ source.
(Part B: CT Burden & ANSI C-Class Saturation Verification) A $1200:5\text{ A}$ ($CTR = 240$), ANSI Class C400 bushing Current Transformer with internal secondary winding resistance $R_{ct} = 0.35,\Omega$ is connected to a microprocessor overcurrent relay with burden impedance $Z_R = 0.05,\Omega$ via $300\text{ feet}$ of #12 AWG copper lead wire ($1.93,\Omega / 1000\text{ ft}$):
- Calculate the total secondary loop burden resistance ($Z_{b,\text{total}}$).
- Determine the maximum symmetrical primary fault current ($I_{f,\max}$) the CT can sustain without exceeding its ANSI C400 saturation rating.
=========================================================================================
CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:
=========================================================================================
PART A: AUTOTRANSFORMER CALCULATIONS
Step 1: Compute Turns Ratio and Power Advantage Factor
V_H = 13.8 kV, V_L = 12.0 kV
Turns Ratio: a = V_H / V_L = 13.8 / 12.0 = 1.150
Ratio Factor: (1 - 1/a) = (1 - 1/1.150) = (1 - 0.869565) = 0.130435
Power Advantage Boost Factor: 1 / (1 - 1/a) = 1 / 0.130435 = 7.6667
Step 2: Calculate Two-Winding Equivalent Capacity
S_two_winding = S_auto * (1 - 1/a)
= 15,000 kVA * 0.130435
= 1,956.52 kVA approx 1.957 MVA
(A physical 1.957 MVA transformer core delivers 15.0 MVA as an autotransformer!)
Step 3: Calculate Power Components
Transformed Power: S_trans = S_two_winding = 1,956.52 kVA (13.04% of total)
Conducted Power: S_cond = S_auto - S_trans = 15,000 - 1,956.52 = 13,043.48 kVA (86.96% of total)
Step 4: Compute Autotransformer Per-Unit Impedance and Short-Circuit Current
Z_pu,auto = Z_pu,2w * (1 - 1/a)
= 0.060 * 0.130435
= 0.007826 pu (0.783% on 15 MVA base)
Rated Low-Voltage Current (12.0 kV bus):
I_rated,LV = S_auto / (sqrt(3) * V_L)
= 15,000,000 VA / (1.73205 * 12,000 V)
= 15,000,000 / 20,784.6
= 721.69 A
Available Symmetrical Short-Circuit Current:
I_sc = I_rated,LV / Z_pu,auto
= 721.69 A / 0.007826
= 92,217 A approx 92.2 kA
(Note: The ultra-low Z_pu results in an enormous 92.2 kA fault current!)
-----------------------------------------------------------------------------------------
PART B: CURRENT TRANSFORMER C-CLASS SATURATION CALCULATIONS
Step 5: Calculate Total Secondary Loop Burden Resistance
Lead wire one-way distance = 300 ft ==> Loop distance = 2 * 300 = 600 ft
Lead Loop Resistance: R_lead = 600 ft * (1.93 ohms / 1000 ft) = 1.158 ohms
Internal CT Resistance: R_ct = 0.350 ohms
Relay Burden Resistance: Z_relay = 0.050 ohms
Total Secondary Circuit Resistance:
Z_b,total = R_ct + R_lead + Z_relay
= 0.350 + 1.158 + 0.050
= 1.558 ohms
Step 6: Compute Maximum Primary Fault Current for ANSI C400 Rating
By ANSI/IEEE C57.13 definition, a C400 CT develops 400 V across standard burden at 20x nominal (100 A):
Internal Saturation EMF Knee Voltage: E_s,max = 400 V + (100 A * R_ct) = 400 + (100 * 0.35) = 435.0 V
Maximum Permissible Secondary Fault Current:
I_s,max = E_s,max / Z_b,total
= 435.0 V / 1.558 ohms
= 279.20 A
CT Ratio: CTR = 1200 / 5 = 240
Maximum Symmetrical Primary Fault Current:
I_f,max = I_s,max * CTR
= 279.20 A * 240
= 67,008 A approx 67.0 kA
=========================================================================================
7. Common Exam Traps & Tactical Pitfalls
- The Lead Wire Loop Omission (Missing Factor of 2): Forgetting that CT wiring forms a complete loop from the CT terminal to the relay and back to the neutral bar ($R_{lead} = 2 \times L \times r$), underestimating lead resistance by $50%$.
- Direct Application of Two-Winding Impedance to Autotransformer Faults: Using $Z_{pu,2w}$ instead of reducing it to $Z_{pu,auto} = Z_{pu,2w}(1 - 1/a)$ when computing autotransformer short-circuit fault currents, drastically underestimating fault duty.
- Conflating C-Class Voltage with Primary Voltage: Assuming the "400" in C400 refers to primary system voltage or relay trip setting rather than the secondary terminal voltage capability at 20 times rated current ($100\text{ A}$).
- Overlooking Open-Circuit Peak Voltage Severity: Believing an open CT secondary only generates rated nominal voltage. Without secondary demagnetizing MMF, the open terminal voltage reaches lethal multi-kilovolt levels ($> 5000\text{ V}$). Always ensure CT shorting blocks are closed before servicing.
An autotransformer connects a 230 kV transmission line to a 115 kV transmission line to deliver 100 MVA of apparent power. What is the physical two-winding transformer power rating (transformed power) required for the magnetic core and windings?
A 2000:5 A Current Transformer (CTR = 400) with ANSI relaying class C800 has an internal secondary resistance of R_ct = 0.50 ohms. The CT is connected to an overcurrent relay with burden Z_R = 0.20 ohms located 500 feet away using #10 AWG copper wire (resistance = 1.00 ohm per 1000 ft). What is the total secondary loop burden impedance?
Why is operating a Current Transformer (CT) with an open-circuited secondary under live primary current hazardous?