By Hurley Gill • Senior applications and systems engineer | Kollmorgen

*Please refer to part one and part two of this three-part series for background information on the functions and types of regenerative resistors for servo applications.*

Let’s now consider the worst-case deceleration of an axis needing to reach a zero velocity for a p-stop or e-stop. This simple exercise will identify the necessary regen resistor based on an Er(sf) value without delving into the more complicated calculations for a multi-axis system or even single axis required to make multiple decelerations.

**Design assumptions:** A 3Ø 480-Vac source powers a rotary electric servomotor driving a horizontal axis needing to decelerate to zero RPM. So, normal application operation for the motor and drive yield:

Trms = 20_Nm and Tpk = 50_Nm at Ipk = 28_Arms

J_load = 1_kg·m^{2} and Pc_req = 1041W

Ppk_req = 1,150 W.

**Selected motor and drive performance values:** Let’s assume we selected a frameless brushless motor with a Kt(rated temp) = 2.19_Nm/Arms; Kb(25°C) = 133_Vrms/kRPM; Jm = 0.00304_kg·m^{2}; Tpk(motor) = 76.1_Nm; Rm(L-L) = 1.41 Ω (for Rm cold or ambient temperature).

Assume we have selected a servo drive having an Ic = 12_Arms, Ipk(drive) = 30_Arms, DC-bus capacitance = 470µFd, Vmax(fault) = 840 VDC, an internal regen RBint = 100 W and 33 Ω. Regenerative output Regen(c)_output(480vac) = 6kW and Regen(pk)_output(480vac) = 21.4kW for 1sec; Regen(ext) = 33_ohm (Ω).

Now assume our stop-function (sf) linear deceleration is Tpk = 50_Nm at Ipk = 28_Arms for t_dec = 1.2_sec. For a stop-function analysis, we need to use the data from the application’s motor-drive selections to minimally identify five values:

**Irms(application)** = Trms/Kt(motor) = 20_Nm ÷ 2.19_Nm/Arms ≅ 9.13_Arms

**Nmax (maximum velocity)** = 570rpm, where the highest kinetic energy resides; typically, this will be the maximum velocity (just prior to any possible stop-function (sf)) with largest mass, with the least friction and any external loading aiding (-) or resisting (+) axis’ ability to stop.

**Total inertia (J_total) of the axis** — i.e. all reflected inertia & otherwise (J_load) at the motor plus motor’s rotor inertia (J_motor). J_total = J_load + J_motor = 1_kg·m^{2} + 0.00304_kg·m^{2} = 1.00304. Note with a J_load / Jm = 329:1 mismatch, this must be a directly driven load.

**Friction torque (Tf) at the motor** … noting that if Tf is relatively small, it’s often assumed that Tf = 0_Nm.

**Any external torque (T_ext) at the motor** during subject stop-function deceleration; for our example: T_ext = 20_Nm (positive (+) value meaning for this application: it adds to the E(k) making it harder to stop axis (as for a vertical axis trying to stop a downward move where gravity adds E(k) making stops more difficult).

The next step in our analysis is to identify the desired stop-function deceleration-time t_dec. Note that this t_dec value is typically that required to meet some customer stop-function requirement and/or that which is dictated by available drive-current limits — such as the drive’s I2t (* I^{2}t*) foldback algorithm. Identifying the most suitable stop-function deceleration time will likely take multiple calculation iterations. For deceleration times demanding Ipk(drive) for more than the available time as a function of the drive’s I

^{2}t algorithm, the Ic(drive) is often used. Otherwise, Ipk(drive) or Ipk(motor) becomes the limit. For our example here, Ipk(drive) is available for the t_dec = 1.2_sec.

Note that for linear applications, it’s necessary to replace rotary-inertia J units in kg·m^{2} with mass M units in kg … as well as velocity ω units in radians/sec with velocity V units in meters/sec.

**Er(sf)** = E(k) – E(el) ± E(ext-f) – E(f); that is: kinetic energy less losses ± any external forces, where each unknown is calculated & identified, by the axis’ stop-function (sf) conditions. For this example:

**E(k)** = Energy(kinetic) = ½(J_load + J_motor) ωm^{2} = ½(1.00304_kgm^{2}) (570rpm/9.55)^{ 2} = 1786.62_joules

**E(el)** = Energy(electrical losses) = 3(I^{2}_dec x Rm/2) t_dec. = 3(28^{2} x 1.41/2) x 1.2 = 1989.8_joules

**E(ext-f)** = Energy(external forces) = T_ext x Δωm/2) t_dec = (+20_Nm x 570rpm/2/9.55) x 1.2_sec = 716.3_joules

**E(f)** = Energy(friction) = (Tf x Δωm/2) t_dec … and because Tf is relatively small, we assume E(f) = 0_joules

Hence Er(sf) = ½(J_load + J_motor) ωm^{2} – 3(I^{2}_dec x Rm/2) t_dec ± (T_ext x Δωm/2) t_dec) – (Tf x Δωm/2) t_dec.

**So Er(sf) = 1786.62_joules – 1989.8_joules +716.3_joules – 0_joules = 513.12_joules.**

Because the calculated Er(sf) > 0, we need to determine whether the drive’s DC bus has the capability to absorb the 513.12 joules returned to the DC bus. The DC-bus capacitance E(caps) = Energy(additional capacity) = ½ C (VDC^{2}_max(fault) – VDC^{2}_bus). For our design, that’s equal to ½ × 470µFd × ((840-1)^{2} – (480vac√2)^{2}) = 57.13_joules.

We first need to verify whether the drive’s DC bus can absorb the returned joules through its internal regenerative resistor capabilities — with that for our selected component expressed as RBint = 100 W and 33 Ω for a t_dec = 1.2_sec … and E(int-reg) = 120_joules.

That means Er(sf) – E(caps) – E(int-reg) = 513.12_joules – 57.13_joules – 120_joules = 336_joules … so Er(sf) – E(caps) – E(int-reg) > 0. That in turn means that an external regenerative resistor is required with a minimal dissipation capability > 513.12_joules … or at the very least, we need to extend the time over which deceleration is allowed to occur. **One caveat here:** For a stop function event, *we cannot guarantee any DC-bus capacitance storage capability* — which is why E(caps) is not subtracted from Er(sf) for the final determination of minimum requirements.

**Calculated wattage for the stop function: **Ppk_req(sf) = Er(sf)/t_dec = 513.12_joules/1.2_sec = **428 W**.

Now let’s calculate the required minimum continuous power requirement. The minimal continuous power requirement is *Er(total) / t_total*; however, we need the maximum peak requirement for the selected motor and drive based on a specific event — ** the stop function** — and not for the continuous power requirement defined by the Application results (i.e. Pc_req = 1041W) based on normal axis operation (motion profile).

The continuous power requirement defined by the application results gives Pc_req = 1,041 W based on normal axis operation executing the actions plotted on its motion profile. However, here we are looking for the maximum peak requirement for the selected motor and drive *based on a specific event*— ** the stop function** — and not the continuous power requirement.

**Note:** Any Pc_req calculated from a stop-function motion profile will be bogus and of little value … unless the specific stop-function is modeled with Trms, Nrms, and Nmax (RPMs) based on worse-case normal operation *appropriate for the application and motor-drive selection*. The latter type of modeling isn’t typically done, because that information is determined during the original sizing. A slightly altered approach to modeling an axis’ stop function is to use an equivalent motion-profile segment for normal operation just prior to the defined stop function under evaluation — including RMS continuous conditions, maximum velocity, and any external loading. This will help determine if there are any I2t foldback algorithm issues.

For our example, peak regen power requirement with a Pc_req = 1,041 W, Ppk_req = 1,150 W, and Ppk_reg(sf) = 428 W. So for this application example, Pc_req is actually the dominant requirement for regen wattage selection. The standard regen resistor selection for the proposed drive is 33 Ω with wattage ratings of 250, 500, 1,500, and 3,000 W. A 33-Ω 1,500-W regen resistor (assuming Ppk_resistor = 10 x continuous) is our best selection from standard manufacturer offerings.

**So does the 33-ohm 1500-W regen resistor meet all the application’s conditions? **Well, let’s find out. The selected regen resistor’s maximum resistance that will continuously allow the DC bus to be maintained under its maximum value — VDC_max(fault) — is determined by:

However, for the selected next size up standard resistor having a capability: 1,500W, R_regen(Ω) must be less than 469 Ω.

Assuming we choose the 1,500-W 33-Ω regen resistor, the maximum possible regenerative shunt current **I_shunt(Rmax) **limited by the resistor’s peak capability is:

With this selected regenerative resistor, the maximum possible regen shunt current **I_shunt(Drive-max)** is limited by the drive’s capability — Given : Regen(pk)_output (480 Vac) = 21.4 kW for one second — and is:

The selected regen resistor’s minimal resistance shouldn’t exceed either the regen resistor’s shunt-current capability … or the drive’s maximum regen shunt-current capability. So I_shunt is equal to the lesser value between I_shunt(Rmax) and I_shunt(Drive-max).

However, for the chosen standard resistor having 1,500 W and 33 ohms, R_regen(Ω) should be larger than **3.3 Ω**.

Pursuant to data from previously calculated requirements, the regen peak-current requirements **I_pk(Reg-app)** for the application and selected 1,500-W 33-Ω regen resistor is:

Given this selected resistor, the resistor’s continuous current capability **Ic(Reg-Res)** is:

Pursuant to the given data (from previously calculated requirements), the regen continuous current requirements **Ic(Reg-app) **for the application and the selected 1,500-W 33-Ω regen resistor is:

The drive’s continuous regen shunt current capability **Ic(RegenCapacity of drive)** — based on a Regen(c)_output (480 Vac) of 6,000 W with the selected 1,500-W regen resistor is:

Finally, we run through the final confirmation of our regen-resistor selection.

- Pc_resistor (1,500 W) > Pc_req (1,041 W)
- Ppk_resistor (1,500w x 10) > Ppk_req (1,150 W)
- R_regen = 33 Ω — is a manufacturer’s standard offering for the selected drive.
- R_regen = 33 Ω < 610 Ω & < 469 Ω for the selected 1,500 W resistor
- R_regen = 33 Ω > 2.3 Ω & > 3.3 Ω — with the latter presenting the lesser peak shunt current.

So does the does the 33-W 1,500-W regen resistor meet all conditions? Indeed, the selected regen resistor meets all conditions and is a reasonable choice.

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