| Not too long ago I was retained as an engineering expert to testify on behalf of a plaintiff who owned a sports bar. The place was filled with flat screen televisions that were plugged into 120 volt alternating current (VAC) wall outlets. To make a long story short, the electric utility wires that fed power to the bar were hit by a passing vehicle, causing the voltage in the outlets to increase well beyond what the electronics in the televisions could handle. The delicate electronics were not suited to be connected with the high voltage that suddenly surged through them as a result of the hit, and they overloaded and failed.
Similarly, lower voltage microprocessor and digital logic chips are also not suited to directly connect with higher voltage devices like motors, electrical relays, and light bulbs. An interface between the two is needed to keep the delicate electronic circuits in the chips from overloading and failing like the ill fated televisions in my client’s sports bar. Let’s look now at how a field effect transistor (FET) acts as the interface between low and high voltages when put into operation within an industrial product. I was once asked to design an industrial product, a machine which developed medical x-ray films, utilizing a microprocessor chip to automate its operation. The design requirements stated that the product be powered by a 120 VAC, such as that available through the nearest wall outlet. In terms of functionality, upon startup the microprocessor chip was to be programmed to first perform a 40-minute warmup of the machine, then activate a 12 volt direct current (VDC) buzzer for two seconds, signaling that it was ready for use. This sequence was to be initiated by a human operator depressing an activation button. The problem presented by this scenario was that the microprocessor chip manufacturer designed it to operate on a mere 5 VDC. In additional, it was equipped with a digital output lead that was limited in functionality to either “on” or “off” and capable of only supplying either extreme of 0 VDC or 5 VDC, not the 12 VDC required by the buzzer. Figure 1 illustrates my solution to this voltage problem, although the diagram shown presents a highly simplified version of the end solution.
Figure 1The illustration shows the initial power supplied at the upper left to be 120 VAC. This then is converted down to 5 VDC and 12 VDC respectively by a power supply circuit. The 5 VDC powers the microprocessor chip and the 12 VDC powers the buzzer. The conversion from high 120 VAC voltage to low 5 and 12 VDC voltage is accomplished through the use of a transformer, a diode bridge, and special transistors that regulate voltage. Since this article is about FETs, we’ll discuss transistor power supplies in more depth in a future article. To make things a little easier to follow, the diagram in Figure 1 shows the microprocessor chip with only one input lead and one output lead. In actuality a microprocessor chip can have dozens of input and output leads, as was the case in my solution. The input leads collect information from sensors, switches, and other electrical components for processing and decision making by the computer program contained within the chip. Output leads then send out commands in the form of digital signals that are either 0 VDC or 5 VDC. In other words, off or on. The net result is that these signals are turned off or on by the program’s decision making process. Figure 1 shows the input lead is connected to a pushbutton activated by a human. The output lead is connected to the gate (G) of the FET. The FET is shown in symbolic form in green. The FET drain (D) lead is connected to the buzzer and its source (S) lead terminates in connection to electrical ground to complete the electrical circuit. Remember, electric current naturally likes to flow from the supply source to electrical ground within circuits, and our scenario is no exception. Next time we’ll see what happens when someone presses the button to put everything into action.
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Posts Tagged ‘electrical circuit’
Industrial Control Basics – Pushbuttons
Monday, December 26th, 2011| I always enjoy watching impatient people waiting for an elevator. They press the button, and if it doesn’t come within a few seconds they press it over and over again, as if this will hurry things up. In the end they must resign themselves to the fact that the elevator will operate in its own good time.
Pushbuttons, although simple in appearance like the big, red “Easy” button that’s featured in a certain business supply chain’s commercials, are actually complex behind the scenes. They perform important functions within the industrial control systems of a huge diversity of mechanized equipment. Last week we introduced ladder diagrams, used to design and document industrial control systems, and we’ll now see how they depict the action of pushbuttons within two commonly used industrial settings, the “normally open” and the “normally closed.”
Figure 1
Figure 1(a) shows a pushbutton hooked up to an electric motor. When no one is pressing it a spring in the pushbutton forces the button to rest in the up position, allowing an air gap to exist in the electrical circuit between hot and neutral and preventing current from flowing. This type of switch is characterized as a “normally open” switch in industrial control terminology. In Figure 1(b) someone depresses the button, compressing its spring and closing the air gap, which allows current to flow and the motor to operate Figure 1(c) shows the ladder diagram version of 1(a). Now let’s take a look at Figure 2 to see a different type of pushbutton, one that’s characterized as “normally closed.”
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