#26 Having completed a few more modules of electronic..

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Question

#26 Having completed a few more modules of electronic engineering you decide to improve a system that ensures a plant gets the most sunlight by automatically adjusting its position. The amount of sunlight is obtained through two Light Dependent Resistors (LDRs), the resistance of the LDRs when occluded is 200 and when fully exposed to sunlight is 100 2. The original system for this shown in Figure 1 has THREE distinct behaviours or operation modes: 

1. When only ONE LDR is occluded it moves the plant in one direction.

2. When the other LDR is occluded it moves the plant in the opposite direction.

3. When both LDRs are occluded/fully exposed the plant retains its position.

Figure 1: Sunlight tracking system for a potted plant. This circuit makes use of the fact that a motor can be driven in either direction by changing the direction that current will flow through it. In the different operation modes of the circuit therefore, not more than ONE transistor will be ON at any particular time.

1. The first modification of the circuit in Figure 1 with this new knowledge results in the following circuit:

Figure 2: Modified sunlight tracking system for a potted plant. By going through each of the three operation modes given, describe how this modified circuit operates. (Hint:Look at where the GROUND reference has been placed in this circuit, and use the KVL to calculate the two inputs to the Op-Amp. Pay special attention to what happens when the light on the LDR's are equal, think here about what might be happening practically.)

2. The second modification of the circuit in Figure 1 with this new knowledge results in the following circuit: 

Figure 3: Modified sunlight tracking system for a potted plant. Firstly calculate appropriate values of R, and R; and then by going through each of the three operation modes given, describe how this modified circuit operates. (Hint: When choosing the gain value for this circuit, you want the op-amp output's maximum and minimum values to be just equal to the positive and negative supply rails that power the op-amp.)

Answer

1). In the given circuit, the motor rotates in one direction when npn transistor is ON and rotates in the other direction when pnp is ON.

The voltage drop across the 10K resistor at the inverting terminal is

Hence, the voltage at the inverting terminal of the op amp is 0V

When LDR1 is exposed and LDR2 is occluded, the voltage drop across LDR2 will be

Therefore voltage at the non inverting terminal will be 0.5V > inverting terminal voltage.

Therefore, opamp output will be +1.5V

It will make the npn transistor ON, making the motor rotate in one direction.

When LDR2 is exposed and LDR1 is occluded, the voltage drop across LDR2 will be

Therefore voltage at the non inverting terminal will be -0.5V < inverting terminal voltage.

Therefore, opamp output will be -1.5V

It will make the pnp transistor ON, making the motor rotate in the other direction.

When both LDR1 and LDR2 are exposed or occluded, the voltage drop across LDR2 will be

Therefore voltage at the non inverting terminal will be 0V = inverting terminal voltage.

Therefore, opamp output will be 0V

It will make both npn and pnp transistor OFF, making the motor rest in its current position.

2) Here, since output should be equal to the input, gain has to be 1. Therefore, the resistors Rf and Ri should be equal, say 1K.

The opamp circuit forms an inverting amplifier.

Let the voltage at the junction between the two LDRs be V.

When LDR1 is exposed and LDR2 is occluded, current flowing through LDR1 is distributed LDR2 and Ri. That is

On solving, V= 0.4687V

Therefore, output of the opamp will be -0.4687V (since inverting with unit gain)

which makes pnp ON, making the motor rotate in one direction.

When LDR2 is exposed and LDR1 is occluded, current flowing through LDR1 is distributed LDR2 and Ri. That is

On solving, V= -0.4687V

Therefore, output of the opamp will be +0.4687V (since inverting with unit gain)

which makes npn ON, making the motor rotate in the other direction.

When LDR1 and LDR2 are exposed, current flowing through LDR1 is distributed LDR2 and Ri. That is

On solving, V= 0V

Similarly when both LDRs are occluded, V=0V

Therefore, output of the opamp will be 0V (since inverting with unit gain)

which makes npn and pnp turn OFF, making the motor rest in the same position.