Industrial Motors -- Start-Stop Push-Button Control

GOALS:

• Describe the operation of a start-stop relay control circuit.
• Describe the operation of the basic gates used in this section.
• Describe the operation of the solid-state control circuit.
• Discuss practical wiring techniques for connecting digital logic circuits.
• Connect a start-stop, push-button control using logic gates.

Start-stop, push-button circuit.

START button energizes "M" relay coil.

"M" contacts maintain the circuit.

STOP button breaks the circuit.

In this section, a digital circuit will be designed to perform the same function as a common relay circuit. The relay circuit is a basic stop-start, push-button circuit with overload protection.

Before beginning the design of an electronic circuit that will perform the same function as this relay circuit, the operation of the relay circuit should first be discussed. In the circuit, no current can flow to relay coil M because the normally open START button and the normally open contact are controlled by relay coil M.

When the START button is pushed, current flows through the relay coil and normally closed overload con tact to the power source. When current flows through relay coil M, the contacts connected parallel to the START button close. These contacts maintain the circuit to coil M when the START button re leases and returns to its open position.

The circuit will continue to operate until the STOP button is pushed and breaks the circuit to the coil. When the current flow to the coil stops, the relay de-energizes and contact M reopens. Since the START button is now open and contact M is open, there is no complete circuit to the relay coil when the STOP button is returned to its normally closed position. If the relay is to be restarted, the START button must be pushed again to provide a complete circuit to the relay coil.

The only other logic condition that can occur in this circuit is caused by the motor connected to the load contacts of relay M. Assume the motor is connected in series with the heater of an overload relay. When coil M energizes, it closes the load contact M. When the load contact closes, it connects the motor to the 120 volt AC power line.

Overload contacts break the circuit.

If the motor is overloaded, it will cause too much current to flow through the circuit. When a current greater than normal flows through the overload heater, the heater produces more heat than it does under nor mal conditions. If the current becomes high enough, it will cause the normally closed overload contact to open.

Notice that the overload contact is electrically isolated from the heater. The contact, therefore, can be connected to a different voltage source than the motor.

If the overload contact opens, the control circuit is broken and the relay de-energizes as if the STOP button had been pushed. After the overload contact has been reset to its normally closed position, the coil will remain de-energized until the START button is again pressed.

Now that the logic of the circuit is understood, a digital logic circuit that will operate in this manner can be designed. The first problem is to find a circuit that can be turned on with one push button and turned off with another. The circuit shown can perform this function. This circuit consists of an OR gate and an AND gate. Input A of the OR gate is connected to a normally open push button, which is connected to 5 volts DC. Input B of the OR gate is connected to the output of the AND gate. The output of the OR gate is connected to input A of the AND gate. Input B of the AND gate is connected through a normally closed push button to _5 volts DC. This normally closed push button is used as the STOP button. The output of the AND gate is the output of the circuit.

To understand the logic of this circuit, assume that the output of the AND gate is low. This produces a low at input B of the OR gate. Since the push button connected to input A is open, a low is produced at this input also. When all inputs of an OR gate are low, its output is also low. The low output of the OR gate is connected to input A of the AND gate. Input B of the AND gate is connected to a high through the normally closed push button switch. Since input A of the AND gate is low, the output of the AND gate is forced to remain in a low state.

When the START button is pushed, a high is connected to input A of the OR gate. This causes the output of the OR gate to change to high. This high output is connected to input A of the AND gate. The AND gate now has both of its inputs high, so its output changes from a low to a high state. When the output of the AND gate changes to a high state, input B of the OR gate becomes high also. Since the OR gate now has a high connected to its B input, its output will remain high when the push button is returned to its open condition and input A becomes low. Notice that this circuit operates the same as the relay circuit when the START button is pushed. The output changes from a low state to a high state and the circuit locks in this condition so the START button can be reopened.

When the normally closed STOP button is pushed, input B of the AND gate changes from high to low.

When input B changes to a low state, the output of the AND gate changes to a low state also. This causes a low to appear at input B of the OR gate. The OR gate now has both of its inputs low, so its output changes from a high state to a low state. Since input A of the AND gate is now low, the output is forced to remain low when the STOP button returns to its closed position and input B becomes high. The circuit designed here can be turned on with the START button and turned off with the STOP button.

The next design task is to connect the overload con tact to the circuit. The overload contact must be connected in such a manner that it will cause the output of the circuit to turn off if it opens. One's first impulse might be to connect the overload contact to the circuit as shown. In this circuit, the output of AND gate #1 has been connected to input A of AND gate #2.

Input B of AND gate #2 has been connected to a high through the normally closed overload contact. If the overload contact remains closed, input B will remain high. The output of AND gate #2 is, therefore, controlled by input A. If the output of AND gate #1 changes to a high state, the output of AND gate #2 will also change to a high state. If the output of AND gate #1 becomes low, the output of AND gate #2 will become low also.

If the output of AND gate #2 is high and the over load contact opens, input B will become low and the output will change from a high to a low state. This circuit appears to operate with the same logic as the relay circuit until the logic is examined closely. Assume that the overload contacts are closed and the output of AND gate #1 is high. Since both inputs of AND gate #2 are high, the output is also high. Now assume that the over load contact opens and causes input B to change to a low condition. This forces the output of AND gate #2 to change to low state also. Input A of AND gate #2 is still high, however. If the overload contact is reset, the output will immediately change back to a high state. If the overload contact opens and is then reset in the relay circuit, the relay won’t restart itself. The START but ton must be pushed to restart the circuit. Although this is a small difference in circuit logic, it could become a safety hazard in some cases.

This fault can be corrected with a slight design change. In this circuit, the normally closed STOP button has been connected to input A of AND gate #2, and the normally closed overload switch has been connected to input B. As long as both of these inputs are high, the output of AND gate #2 will provide a high to input B of AND gate #1. If either the STOP button or the overload contact opens, the output of AND gate #2 will change to a low state. When input B of AND gate #2 changes to a low state, it will cause the output of AND gate #1 to change to a low state and unlock the circuit, just as pushing the STOP button did in the circuit. The logic of this digital circuit is now the same as the relay circuit.

Although the logic of this circuit is now correct, there are still some problems that must be corrected.

When gates are used, their inputs must be connected to a de finite high or low. When the START button is in its normal position, input A of the OR gate is not connected to anything. When an input is left in this condition, the gate may not be able to determine if the input should be high or low. The gate could, therefore, assume either condition. To prevent this, inputs must always be connected to a definite high or low.

Resistor used to lower the input of a gate. Resistor used to raise the input of a gate. Push button produces a high at the input.

When using TTL logic, inputs are always pulled high with a resistor as opposed to being pulled low. If a resistor is used to pull an input low, it will cause the gate to have a voltage drop at its output. This means that in the high state, the output of the gate may be only 3 or 4 volts instead of 5 volts. If this output is used as the input of another gate, and the other gate has been pulled low with a resistor, the output of the second gate may be only 2 or 3 volts. Notice that each time a gate is pulled through a resistor, its outcome voltage becomes low. It this were done through several steps, the output voltage would soon become so low it could not be used to drive the input of another gate.

--- shows a resistor used to pull the input of a gate high. In this circuit, the push button is used to connect the input of the gate to ground, or low.

The push button can be adapted to produce a high at the input instead of a low by adding an INVERTER as shown. In this circuit, a pull-up resistor is connected to the input of an INVERTER.

Since the input of the INVERTER is high, its output will produce a low at input A of the OR gate. When the normally open push button is pressed, a low will be produced at the input of the INVERTER. When the input of the INVERTER becomes low, its output becomes high.

Notice that the push button will now produce a high input A of the OR gate when it’s pushed.

Since both of the push buttons and the normally closed overload contact are used to provide high inputs, the circuit is changed.

Notice that the normally closed push button and the normally closed overload switch connected to the inputs of AND gate #2 are connected to ground instead of Vcc. When the switches are connected to ground, a low is provided to the input of the INVERTERS to which they are connected. The INVERTERS, there fore, produce a high at the input of the AND gate. If one of these normally closed switches opens, a high will be provided to the input of the INVERTER. This will cause the output of the INVERTER to become low. If the logic of the circuit is checked, it can be seen that it’s the same as the logic of the circuit shown.

The final design problem for this circuit concerns the output. So far, a light-emitting diode has been used as the load. The LED is used to indicate when the output is high and when it’s low. The original circuit, however, was used to control a 120 volt AC motor. This control can be accomplished by connecting a solid state relay to the output in place of the LED. In this circuit, the output of AND gate #1 is connected to the input of an opto-isolated, solid-state relay. When the output of the AND gate changes to a high condition, the solid-state relay turns on and connects the 120 volt AC load to the line.

Solid-state, start-stop, push-button control.

QUIZ:

1. In a relay circuit, what function is served by the holding contacts?

2. What is the function of the overload relay in a motor control circuit?

3. What conditions of input must exist if an OR gate is to produce a high output?

4. What conditions of input must exist if an AND gate is to produce a high output?

5. When connecting TTL logic, why are inputs pulled high instead of low?

6. Referring, how would this circuit operate if input B of the OR gate was reconnected to input A of AND gate #1 instead of its output?

7. Referring to what functions, does the INVERTER serve in this circuit?