We need a base current of 2A / 750 = 3mA so R1 = (5V - 1.4V) / 3mA = 1.2K For this circuit I chose a power darlington type BDX33C (shown RIGHT) which has Icmax=10A, Vcemax=100V, Pt = 70W (on heat sink) and Hfe > 750. Suppliers often have selectors that let you choose the parameters you need. ![]() As before Diode D1 prevents a voltage spike that would damage the transistor. When this current ceases the mottor stops running. The transistor is turned on by a positive voltage applied to R1. Here the motor current is being switched by a transistor. The other factor to consider is that the motor may act as a generator if the shaft continues to spin. This is wired so that it is reverse biased when the motor is running, but conducts when the negative spike begins, and "shorts it out". adding a series capacitor C1and resistor R1 across the terminals (interference suppressor) OR.This can be provided as shown here (left) by: If you dont deal with this either the switch will arc (or if its a semiconductor it will fry) or the winding insulation will break down. The same applies to any inductive load such as a relay. When the supply to the inductor is interrupted the energy has nowhere to go, resulting in a negative voltage spike. The windings of the motor are inductors, and store energy. However it introduces two important factors firstly, what happens when you remove the drive current to a motor. The simplest form of control system can often be achieved with a simple switch or relay.
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