Setting The Voltage Controller Gain - Emerson SP1201 Installation Manual

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Introduction
Information information
There is one value of the scaling factor K for each drive voltage rating as shown in the table below.
Drive voltage rating
200V 415V
400V 830V
575V 990V
690V 1190V
The integral gain Ki (Pr 4.14) is less critical and should be set so that
Ki = Kp x 256 x T / τ
m
where
τ
is the time constant (L / R).
m
R is the resistance of the supply for one phase.
Therefore
Ki = (K x L x Kc) x 256 x 167µs x R / L = 0.0427 x K x R x Kc
The above equations give the gain values that should give the best response at all switching frequencies with minimal overshoot. If required the gains
can be adjusted to improve performance as follows:
1. The integral gain (Ki) can be used to improve the performance of the current controllers by reducing the effects of inverter non-linearity. These
effects become more significant with higher switching frequency. These effects will more significant for drives with higher current ratings and
higher voltage ratings. If Ki is increased by a factor of 4 it is possible to get up to 10% overshoot in response to a step change of current
reference. For high performance applications, it is recommended that Ki is increased by a factor of 4 from the auto-tuned values. As the inverter
non-linearity is worse with higher switching frequencies it is may be necessary to increase Ki by a factor of 8 for operation with 16kHz switching
frequency.
2. It is possible to increase the proportional gain (Kp) to reduce the response time of the current controllers. If Kp is increased by a factor of 1.5 then
the response to a step change of reference will give 12.5% overshoot. It is recommended that Ki is increased in preference to Kp.

Setting the voltage controller gain

Even when the gains are set correctly there will be a transient change of DC Bus voltage when there is a change in the load on any drive connected
to the Regen unit. This can be reduced substantially by using an analogue input for power feed forward compensation (see Pr 3.10). The following
discussion relates to a system without power feed-forward compensation.
If the power flow from the supply is increased (i.e. more power is taken from the supply or less power is fed back into the supply) the DC Bus voltage
will fall, but the minimum level will be limited to just below the peak rectified level of the supply provided the maximum rating of the unit is not
exceeded. If the power flow from the supply is reduced (i.e. less power is taken from the supply or more power is fed back into the supply) the DC Bus
voltage will rise. During a rapid transient the bus will rise and then fall as shown below.
ch3: dT= 194ms dV=2.24 V
Active regen
unit current
Regen unit
DC Bus
voltage
50mS/div
The example shown is for a very rapid load change where the torque reference of the motor drive has been changed instantly from one value to
another. The proportional gain of the voltage controller defines the voltage transient because the integral term is too slow to have an effect. (In
applications where the motor drive is operating under speed control, the speed controller may only require a limited rate of change of torque demand,
and so the transient voltage may be less than covered in the discussion below.) If the set point voltage (Pr 3.05) plus the transient rise exceed the
over-voltage trip level the Regen unit will trip.
When a 400V motor is operated above base speed from a drive in vector mode, fed from the Regen unit with the same rating supplying a DC voltage
of 700V, and an instantaneous change of torque is demanded (i.e. -100% to +100%) the peak of the voltage transient (∆V) is approximately 80V if the
current controllers are set up correctly and the voltage controller uses the default gain. (Operating with maximum voltage on the motor, i.e. above
base speed, gives the biggest transient of power and hence the biggest value of ∆V.)
98
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design installation
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