NOTES 4

Recall that the DC chopper circuit for DC motor control is as shown below. If you try to operate this circuit such that VL,DC<Ea then the armature current will be negative, Ia<0 and the machine will act as a generator. If the current is reversed a torque will act on the machine to counteract the motor rotation and the machine will start to slow down or ``brake''. Unfortunately the circuit in Figure 1 cannot be used for braking because the switches used will not conduct current in the reverse direction which is required.


 
Figure 1:
\begin{figure}
\special{psfile=chop5.ps hoffset=30 voffset=-410 hscale=75 vscale=75}
\vspace{2.5in}\end{figure}

The circuit required for braking must interchange the transistor and diode as shown in Figure 2. Here the input voltage has become the output voltage since the machine is now supplying power and VOUT is absorbing it. Typically VOUT would be a battery voltage.


 
Figure 2:
\begin{figure}
\special{psfile=chop6.ps hoffset=30 voffset=-410 hscale=75 vscale=75}
\vspace{2.5in}\end{figure}

In this circuit the time average, or DC current across the switch is given by

\begin{displaymath}
V_{S2,DC}=\frac{1}{T}\int_0^T V_{S2} dt=\frac{1}{T}V_{OUT} (1-D)T=V_{OUT} (1-D)\end{displaymath}

\begin{displaymath}
D={\rm duty\;\; cycle}=\frac{t_{ON}}{T}\end{displaymath}

The DC value of the armature voltage, Ea is then given by

Ea=VS2,DC+IaRa=(1-D)VOUT+IaRa

Equivalently the armature current, Ia is given by

\begin{displaymath}
I_a=\frac{E_a-(1-D)V_{OUT}}{R_a}\end{displaymath}

In this circuit the input voltage is Ea and the output is VOUT and VOUT is greater than Ea. This circuit is therefore often referred to as a step-up chopper.

It is possible to achieve motor and braking operation with the same circuit if all four switches are used. The two quadrant chopper is shown below. The graph to the right of the figure illustrates why it is referred to as a two-quadrant chopper. If motor voltage is graphed versus motor current on a cartesian coordinate system, four quadrants of operation can be identified. If both Ea and Ia are positive the machine motors in the forward direction. If Ea is positive but Ia is negative the machine brakes in the forward direction. Likewise if both quantities are negative the machine motors in the reverse direction. If Ea is negative but Ia positive the machine brakes in the reverse direction.


 
Figure 3:
\begin{figure}
\special{psfile=chop7.ps hoffset=30 voffset=-410 hscale=75 vscale=75}
\vspace{3in}\end{figure}

Its possible to create a four-quadrant chopper which can control the motor in both the forward and reverse directions. This circuit is shown below. Please study this circuit to convince yourself of its operation. Make sure you know which switches need to be controlled to operate the motor in all four quadrants. You should also think about what the control signals should look like for thiscircuit and how they can be implemented.


 
Figure 4:
\begin{figure}
\special{psfile=chop8.ps hoffset=30 voffset=-460 hscale=75 vscale=75}
\vspace{2.5in}\end{figure}



 
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Wismer Margaret
10/1/1998