Overview: 208V vs 230V AC Drive Operation
When a 230V-rated AC drive or servo amplifier is installed on a 208V three-phase service, the question almost always becomes: is the drive derated, or does the motor derate, or does nothing derate at all? The short answer is that the drive electronics themselves are typically rated across the entire 200–250 VAC input range, but the motor does derate when the actual line voltage drops below its nameplate rating. Servo systems add a second constraint: high-speed torque headroom shrinks because the DC bus voltage is lower, which reduces the drive's ability to push current into the motor winding at high rotational speeds.
This reference covers the physics, the math, the field symptoms, and the commissioning checks needed to deploy a 230V drive on either a 208V or a 240V service without surprise faults. It draws on the official AutomationDirect FAQ on nominal system voltages and on standard induction-motor and servo-drive design relationships.
Why 208V and 230V Are Not the Same System
Per the official AutomationDirect FAQ on this subject, the terms 220V, 230V, and 240V all refer to the same nominal system voltage level. They are simply historical labels for a single utility service class. The 208V designation, however, refers to a different and distinctly lower system voltage level. See the AutomationDirect FAQ on 208/220/230/240 VAC distinction.
The underlying reason is the transformer secondary connection:
- 208V three-phase comes from a 120/208V wye secondary. Phase-to-neutral is 120V; phase-to-phase is √3 × 120 = 208V.
- 240V three-phase comes from a 120/240V delta secondary. The high leg is 240V phase-to-phase; one corner is center-tapped to produce 120V phase-to-neutral.
- 230V three-phase is a NEMA nominal rating used for motors and drives, falling inside the 220–240V utility band.
In practice, a 230V-rated drive will see:
- ~208V on a 120/208V wye service (about 10% low)
- ~230V on a 120/240V delta service at nominal tap
- ~240V on a 120/240V delta service at full tap
How VFDs Handle Input Voltage Range
A modern voltage-source PWM VFD first rectifies the incoming AC into a DC bus, then synthesizes a variable-frequency, variable-voltage output with an IGBT inverter stage. The DC bus voltage is approximately:
V_DC ≈ √2 × V_LL
For a nominal 230V input, the DC bus sits near 325V; for a 208V input, the DC bus sits near 294V. The inverter can only synthesize an output voltage whose peak line-to-neutral fundamental cannot exceed V_DC / 2 (and typically V_DC / √3 for a sinusoidal modulation scheme). The practical consequence:
| Input V_LL | Approx DC Bus | Max Output V_L-N | Max Output V_L-L |
|---|---|---|---|
| 208V | 294V | 147V | 255V |
| 230V | 325V | 163V | 282V |
| 240V | 339V | 170V | 294V |
A VFD running in volts/hertz (V/Hz) mode will scale its output voltage proportionally with frequency to keep the motor's air-gap flux constant. If the drive's input is 208V instead of 230V, the drive will simply operate at a slightly lower V/Hz ceiling, but a properly sized drive will still reach its rated output voltage on a 208V supply — within its specified input tolerance window.
Motor Torque Derating: Formulas and Physics
An induction motor's developed electromagnetic torque at a given slip depends on the square of the applied voltage, but for a drive-regulated current scenario (VFD in sensorless vector or servo-style current control), torque is proportional to current and limited by voltage drop across stator impedance plus back-EMF:
T = K_t × I
V_phase = I × R_s + K_e × ω
Where K_t is the motor torque constant (N·m/A), K_e is the back-EMF constant (V·s/rad), R_s is the per-phase stator resistance (Ω), and ω is the angular velocity (rad/s).
For a constant-current load, the mechanical output power delivered to the shaft is:
P_mech (W) = T × ω = √3 × V_LL × I × PF × η
Three-phase apparent power drawn from the line:
S (VA) = √3 × V_LL × I_line
Single-phase apparent power (for reference, e.g., a 120V control transformer):
S (VA) = V × I
If the drive continues to push the same rated current into a motor whose terminal voltage has dropped from 230V to 208V, the mechanical power delivered at the shaft falls in direct proportion:
P_208 / P_230 = 208 / 230 = 0.904
Torque at a fixed speed falls by the same 9.6% factor. This is exactly the derating relationship that AC induction motors experience on a 208V service — the drive electronics are not at fault; the motor simply cannot convert the same mechanical power from a lower voltage input.
Worked Example: 10 HP Motor at 208V vs 230V
Consider a standard NEMA Design B induction motor with the following nameplate:
| Parameter | Value |
|---|---|
| Power | 10 HP (7.46 kW) |
| Speed | 1770 RPM |
| Voltage | 230V |
| Current | 13.6 A |
| Power factor | 0.85 |
| Efficiency | 0.90 |
| Rated torque | 30 lb·in (3.39 N·m) |
Verify the rated torque:
T_rated = (HP × 5252) / RPM = (10 × 5252) / 1770 = 29.7 lb·in ≈ 30 lb·in ✓
Verify the rated mechanical power:
P_mech = √3 × 230 × 13.6 × 0.85 × 0.90 = 4109 W ≈ 5.5 HP apparent
Now apply the same motor on a 208V three-phase service. With the drive continuing to deliver the same 13.6 A (limited by drive and motor thermal ratings):
P_208 = P_230 × (208 / 230) = 7.46 kW × 0.904 = 6.74 kW
T_208 = T_230 × (208 / 230) = 30 lb·in × 0.904 = 27.1 lb·in
The motor still produces 1770 RPM at no-load, but under full load it now delivers only 27.1 lb·in instead of 30 lb·in. If the application requires full rated torque at 1770 RPM, the drive will either fault on overcurrent (because the VFD commands more current to maintain torque), or the motor will overheat as I²R losses climb.
Servo Drive Behavior: Current-Regulated, Voltage-Limited
A servo drive operates very differently from a V/Hz VFD. The drive's current loop regulates stator current to follow a torque command, and the position/velocity loop cascades on top. The limiting factor is whether the DC bus can supply enough voltage to push the commanded current through the stator impedance and overcome the motor's back-EMF at the current speed.
From the per-phase voltage equation:
V_phase_available = (V_DC / √3) − I × R_s × (safety factor)
The motor requires:
V_phase_required = I × R_s + K_e × ω
For the system to operate, V_phase_available ≥ V_phase_required. When input line voltage drops from 230V to 208V, V_DC drops from ~325V to ~294V, and the maximum sustainable motor speed (with full rated torque) falls accordingly. The relationship at the maximum-speed corner:
ω_max ≈ (V_DC/√3 − I × R_s) / K_e
A practical rule of thumb: a 10% reduction in line voltage produces roughly a 10% reduction in maximum continuous speed at rated torque. If the application profile requires operation above the new ω_max, the drive will report a position-error or over-current fault because it cannot source the demanded current at that speed.
Even with a 230V nominal input, the actual voltage reaching the drive terminals falls as load increases, particularly with an undersized supply transformer or long branch circuit runs. The drive does not have unlimited "voltage headroom"; it has whatever margin the DC bus can provide at that instantaneous line condition.
AutomationDirect 1kW Servo System Input Range
The AutomationDirect 1 kW-class servo motors and amplifiers (the SureServo/SureStep family of brushless servo products) accept input voltage across the range of 200 to 255 VAC without internal derating of the drive electronics. Within this window, the drive regulates current to the motor windings and produces rated torque up to its speed-dependent voltage ceiling.
This means the same 1 kW amplifier will run on:
- A 208V three-phase wye service
- A 240V three-phase delta service
- A 230V nominal utility supply
without any parameter change or hardware reconfiguration. What changes between those services is the maximum achievable speed under load — the 208V installation will reach approximately 90.4% of the speed that the same drive/motor pair would reach on a 230V service when delivering rated torque.
Practical Symptoms of Undersized Supply Voltage
When a drive or amplifier is operated at the low end of its input range — especially under heavy load — the following symptoms typically appear in the field:
| Symptom | Likely Root Cause |
|---|---|
| Position-error / following-error fault during high-speed moves | Drive cannot source commanded current at ω_max; back-EMF exceeds available bus voltage |
| Over-current trips only at high speed | Current loop saturates against voltage ceiling |
| Motor stalls under load at rated RPM | Insufficient V/Hz headroom on VFD; torque collapses |
| DC bus under-voltage fault at high load | Supply transformer sagging or branch circuit too long/undersized |
| Motor thermal overload trip after sustained operation | Drive pushing excess current to compensate for low voltage |
| Erratic velocity oscillation during acceleration | Current loop unstable due to reduced bus voltage and PWM headroom |
Supply Transformer Sizing and Voltage Drop
The voltage the drive actually sees is not the transformer secondary rating — it is the transformer secondary minus the voltage drop in the feeder conductors. For a three-phase feeder:
V_drop = √3 × I × L × (R cos φ + X sin φ) / 1000
Where I is line current in amperes, L is one-way feeder length in feet, R is AC resistance in Ω/1000 ft, X is reactance in Ω/1000 ft, and φ is the load power-factor angle.
For a 10 HP drive at 13.6 A on a 100 ft run of 10 AWG copper (R ≈ 0.99 Ω/1000 ft, X ≈ 0.05 Ω/1000 ft, PF 0.85):
V_drop = √3 × 13.6 × 100 × (0.99 × 0.85 + 0.05 × 0.527) / 1000 = √3 × 13.6 × 100 × 0.868 / 1000 = 2.05 V
That is a manageable ~0.9% drop on a 230V service. But on a 208V service the same drop represents ~1.0% of nominal, and if the transformer is also sagging under plant-wide load, the cumulative effect can push the drive terminals below its minimum input threshold. The NEMA MG-1 standard permits motors to operate successfully down to 90% of rated voltage, which is the de facto benchmark for drive input as well.
Commissioning Verification Checklist
- Measure line-to-line voltage at the drive input terminals with a true-RMS meter, both at no-load and at full process load.
- Confirm the voltage stays above the drive's minimum input spec under worst-case plant loading.
- If the supply is 208V, derate expected motor torque and maximum servo speed by the 208/230 = 0.904 factor (or run the full torque calculation for the application's exact speed/load profile).
- Capture a DC bus voltage reading from the drive's diagnostics page during a high-speed loaded move; verify it remains above the under-voltage threshold.
- Run the motor to its commanded maximum speed under load and verify position error stays within the controller's following-error window.
- If the load profile requires full rated torque at high speed, consider using a 240V delta service, a step-up transformer, or selecting a motor/drive pair rated for 208V nominal from the start.
- Verify supply transformer kVA rating is at least 1.25× the drive's continuous input kVA to limit voltage sag under motor starting transients.
Recommended transformer kVA ≥ 1.25 × (√3 × V_LL × I_rated / 1000)
Frequently Asked Questions
Does an AC drive derate when operated on 208V instead of 230V?
The drive electronics typically do not derate across their rated input range (commonly 200–255 VAC). The motor, however, loses mechanical output power proportional to the voltage ratio: a 208V supply delivers approximately 90.4% of the shaft power available at 230V when current is held constant.
Why does 208V exist if 230V is the motor standard?
208V is the line-to-line voltage of a 120/208V wye three-phase service, which is widely used in commercial buildings because it also provides 120V single-phase for receptacles and lighting. Industrial facilities with motors typically use 240V delta or 480V wye services to avoid the motor derating penalty. See the AutomationDirect FAQ on nominal system voltages for the underlying definitions.
How does lower line voltage affect a servo drive's maximum speed?
Maximum continuous speed at rated torque is limited by the DC bus voltage available to overcome motor back-EMF. A 10% reduction in line voltage (from 230V to 208V) reduces the DC bus from ~325V to ~294V and typically reduces maximum servo speed at rated torque by a similar ~10%.
Can the drive compensate by pushing more current at 208V?
The drive can command more current, but stator I²R losses scale with current squared. Sustained overcurrent to recover lost torque will trip the motor's thermal overload, especially in NEMA Design B motors whose nameplate current assumes 230V operation.
What is the minimum acceptable supply voltage at the drive terminals?
Most 230V-class drives specify a minimum continuous input of around 180–200V. Field practice, however, targets at least 90% of nominal (207V for a 230V rating) under full load, per the NEMA MG-1 allowance for motors. Anything below 90% nominal under load risks torque loss, DC bus under-voltage faults, and motor overheating.