modulation (PWM) circuit, or they can ramped up and down as a simple linear circuit. The base of the transistors can be controlled by a controller section of an amplifier that is completely linear. Or the controller can be digital with a D/A converter to provide the analog control signal to the base of Q1 and Q2, or the controller can be completely digital and the base of transistors Q1 and Q2 can be pulsed directly by the digital controller. Typically IGBTs (insulated gate bipolar transistors) are used in modern servo amplifiers where PWM or other switching circuits are used. The IGBTs allow the transistors to be switched on and off at frequencies that limit the harmonic hum in a motor or amplifier. The high-pitched hum represents both audio and electronic noise that must be eliminated or controlled.
FIGURE 11-76 A typical linear amplifier for a servo system.
The amplifier circuits for DC servomotors are similar to the AC circuits used for pulse-width modulation or other switching systems. In fact the complete amplifier for AC servomotors will be similar to the variable-frequency drive amplifiers shown earlier in this chapter.
Early Amplifiers (Push-Pull Amps)
The design of amplifiers has changed rapidly over the last 15 years because transistors, triacs, and SCRs have become able to handle larger voltages and currents without damaging themselves. It is easy to see the advantages that the changes in these devices have brought to motor drive amplifiers, but you must keep in mind that the early amplifiers were built so well that you will run into them even today when you are asked to troubleshoot a drive system. For this reason it is prudent to leam their basic parts and functions so you will be able to troubleshoot and analyze them. It is also a good idea to understand their basic operation because this is what has been modified to make the newer drives more efficient and more powerful.
One of the earliest types of linear amplifiers is called a push-pull amplifier, which was designed so that two transistors switched on and off to share the current load for the motor. Figure 11 -77 shows an example of this type of amplifier, and you can see that Q2 and Q3 are the power transistors. They are connected to the primary winding of transformer T2. The servomotor winding is connected to the secondary winding of transformer T2.
The operation of the push-pull amplifier begins with a sine wave signal that enters the input of the push-pull amplifier through capacitor C1. Capacitor C1 makes sure that the input signal is a pure sine wave with no DC bias. The base circuit of transistor Q1 has a DC bias on it of approximately 13.5 volts. The base-emitter junction needs only 0.7 volt to turn it on, the rest of the DC bias voltage providing DC current through R2 and R3. This causes the sine-wave input
signal to practically turn transistor Q1 off at its minimum, causing Q2 to be driven almost into saturation at the sine wave's maximum. This causes current to flow through the primary winding of transformer T1. Notice the secondary of T1 is center tapped to ground. When the positive half of the sine wave appears on the secondary, it appears across the entire secondary. Because of the center tap, only the upper portion of the secondary sees a positive voltage, and this forward bias transistor Q2 allows it to conduct. Transistor Q3 is turned off because it sees a negative voltage at its base. With Q2 conducting, current flows up through the primary of T2, providing a positive pulse to the secondary of T2. When Q3 conducts, a negative pulse is provided to the primary of T2.
FIGURE 11-77 Push-pull amplifier for an AC ser-vomotor. This diagram shows the power stages of the amplifier.
Another type of early amplifier for a servomotor is called the chopper amplifier (see Fig. 11-78). In this type of amplifier the positive rectangular DC pulses arrive at the input of the amplifier circuit at capacitor C1. These pulses arrive at the base of Q1 as narrow spikes, which momentarily turn Q1 on. This in turn momentarily turns Q2 on, which allows current to flow through the primary of transformer T1. Now the primary of transformer T1 is really an L-C tank circuit. (Remember that the primary winding of the transformer is actually a big inductor.) When this tank circuit is hit by a pulse, it will produce a cycle or two of pure sine wave. When hit, in other words, the tank circuit will ring like a bell. The amplifier circuit is the clapper that rings the bell. Notice the secondary of T1 is center tapped to —60 Vp. The secondary of T1 sees a pure AC sine wave, and to this AC signal, the —60 Vp appears as a ground. This means that for the positive half-cycle of the sine wave, Q3 would see a positive pulse, and Q4 would see a negative pulse. Both power transistors are NPN transistors, so a positive bias is needed at the base to cause them to conduct. As both bases are grounded , Q4 would go into conduction because its emitter is lower than its base, giving it a forward base-emitter bias. The output of the tapped control winding would then be a sine wave. It should be noticed that the tapped control winding has +60 Vp on it, and the secondary of T1 has —60 Vp on it. This means that
the output of the tapped control winding is going to be a 120-Vp sine wave.
FIGURE 11-78 Output stages of a chopper amplifier for an AC servomotor.
FIGURE 11-79 Two-transistor amplifier for a DC servomotor.
伺服电机驱动的正确选择
This document explains how to determine the correct drive for your servo motor, and includes two specific examples.
Finding the correct drive for your servo motor
The servo nuDrive is a +/-48V pulse-width, modulated servo drive. However, you can apply this drive to a +/- 24V, 12V, or any other motor given that certain conditions are
satisfied. The main factor is the current specification. You can modify the nuDrive (by changing resistor packs) to limit the current output of the drive. As long as the current output of the drive does not exceed the specifications of the motor, you should be OK.
The next thing you need to verify is the response of the motor to ripple current. To do this, use the following derivation (which is a function of the inductance, L, of the motor) and solve for di:
V = L * di/dt V * dt = L * di di = (V/L) * dt
The nuDrive has a +/- 48 V pulse-width modulated signal. So, V = 48 (volts) The pulse-width modulation of the drive is 20KHz. Therefore, dt = 1/20,000 (sec.) = 0.00005 (sec.)
The inductance, L, is a function of the motor you are going to use.
Example 1
Let's say your motor specifications are: ? ? ? ?
Nominal current (I_n) = 7 Amps max Nominal voltage (V_n) = 14 Volts max Motor inductance (L) = 0.003 Henry
Motor winding resistance (R_w) = V_n / I_n = 2 ohms
So, the change in current is:
di = (48 / 0.003) * 0.00005 = 0.8 Amps
Now, the ripple current is dissipated as heat. So, power is:
P = I^2 * R_w = (0.8)^2 * 2 = 1.28 Watts
1.28 Watts is not very much power and the mass of your motor should be able to dissipate 1.28 Watts without a problem. Now, let's look at Example 2.
Example 2
Let's say your motor specifications are: ? ? ?
Nominal current (I_n) = 4 Amps max Nominal voltage (V_n) = 24 Volts max Motor inductance (L) = 0.001 Henry
? Motor winding resistance (R_w) = V_n / I_n = 8 ohms
Now, the new change in current is:
di = ( 48 / 0.001) * 0.00005 = 2.5 Amps
So, the power dissipation for this example is:
P = I^2 * R_w = (2.5)^2 * 8 = 50 Watts
50 Watts IS A LOT OF POWER! This will heat your motor up too much over time. In this case, the 48 Volt pulse-width modulated nuDrive is NOT a wise choice. Now, if the motor winding resistance was less, ~ 1-2 ohm, or the inductance was larger, ~ 0.003 Henrys, the nuDrive application would be fine.
You can add inductors to the motor leads to increase the effective inductance and reduce the amount of heat generated. (This procedure will add a negligible amount of resistance to the \winding resistance\
---| Motor |
L_1
_________-------------- ()()()()() ------------- + voltage wire lead -------------________ ()()()()() ________ - voltage wire lead
L_2
The inductors L_1 and L_2 should be balanced (that is, L_1 = L_2). By adding these inductors to the motor power leads, you increase the effective inductance of the motor and reduce the power generated due to ripple current. Selecting the appropriate inductors can make the nuDrive now applicable.
In conclusion, when selecting a drive for a servo motor, the current specification is the determining factor. You should not exceed the nominal current rating for the motor. Next, look at the effects of ripple current on the motor. Larger motor winding resistances is bad. Small motor inductance is likewise bad. Depending on the mass of your motor, you probably don't want your heat dissipation due to ripple current to be larger than 5-10 Watts.
直流无刷电机的工作原理
直流无刷电机的优越性
直流电机具有响应快速、较大的起动转矩、从零转速至额定转速具备可提供额定转矩的性能，但直流电机的优点也正是它的缺点，因为直流电机要产生