Question

Question 1 - Analogue Closed Loop Control Systems Many control systems involve processing analogue signals such as heat, light and movement Therefore analogue closed loop control system require walogue processing devices such as the operational amplifier (op-amp). One of the most common control applications involves using the op-amp as a comparator. In it's simplest form a comparator just compares two voltage signals, VI and V2. Ir VI is higher than V2 the output, Ve, is 'low', VI is lower than V2 the output is 'high'. The block diagram below simulates an automatic heating system. When the bulb heats up the temperature rise is detected by the bead thermistor, which sends a feedback signal back to the comparator A circuit diagram of the system is shown below. TERATURE COACK SINA The reference voltage is located at the non-inverting input of the op-amp. This sets the desired temperature of the bulb Design a circuit using discrete components to attain the following performance criteria: • When the bulb beats above the reference level the thermistor should see the temperature and send a signal to the comparator which will switch the bulbofl. • When the bulb cools below the reference level the bulb should switch on again. • The system should operate continuously. a) Describe how the circuit operates b) Explain clearly how you designed the system. c) What is your reference voltage at non inverting input of the Opamp?

Answer 1

A)B) Describing the circuit operation and its design: i vor high v,< V₂ Vo= low V, > - I Vec - RE bead thermistor Red Roloker R=loke e OP-AMP is used in open loop mode. Let the voltage at inverting terminal is V1 and voltage at non-inverting terminal is V2 . When V1 > V2 output voltage is low ; When V1 < V2 output voltage is high. Transistor is ON when OP-AMP output is high i.e. V1 < V2 Transistor is OFF when OP-AMP output is low i.e V1 > V2 The bulb is between collector of transistor and +Vcc . Bulb is ON when transistor is ON and it is OFF when transistor is OFF. When the bulb is ON it gets heated up. Because of this heat, thermistor resistance Rt gets reduced. Bead thermistor (thermistor are negative coefficient of resistance materials,whose resistance decreases when temperature increases) 115-202CDK-801 is 2 KS2 at room temperature (25° C) At 0° C the resistance is 5K22 and at 50° C the resistance is 91012

Therefore in any bead thermistor as temperature increases , resistance decreases The OP-AMP does not consume any current because of its infinite input impedance. Therefore, current into the inverting and non-inverting terminal is zero. As a result resistance R1 and R2 act as if it is in series. Let us assume the resistance Ri and R2 are of the value 10K22 each. So the voltage V2 is Vcc/2. V2 = pth 2 ( apply voltage divider rule to find voltage V2). when R3= R¢ = 10KS , V2="c. Therefore voltage V2 is fixed to "cc Let us assume the resistance R3 as 2K12. Bead thermistor resistance Rt value is 2K12 at room temperature. So at room temperature voltage Vi is also Vcc / 2. V1 = cchs (apply voltage divider rule to find voltage V1) R3+Rt when R3= R¢ , Ve="CC When temperature of the bulb increases, which is sensed by Bead thermistor its resistance Rt decreases. When Rt decreases, Vı increases. When V1> V2, opamp output becomes low ,which turn off transistor as a result bulb is OFF(resulting in cooling off the bulb temperature). This in turn increases the resistance Rt, which inturn reduces voltage V1. Now Vi< V2, opamp output becomes high, which turn on transistor as a result bulb in ON (resulting in increase the bulb temperature) which reduces Rt. The above process repeats, thereby keeping automatic heating system heat constant. C. Voltage V2 at Non Inverting terminal: Let us assume the resistance Ri and R2 are of the value 10K2 each. So the voltage V2 is Vcc/2. V2 = picha ( apply voltage divider rule to find voltage V2 ) VCC when R3= R+ = 10K22 , V2='cc. Therefore voltage V2 is fixed to AssumeV2 = 12V. , then V2 would be 6V