Electronics Made Easy - A Beginners Guide to Learning Electronics

Electronics Made Easy: The Beginners Electronics Guide

Design Example: Light Sensitive Switch

Using a light dependant resistor design a light sensitive switch to operate a security light, internal or external, during hours of darkness. The light switch should be capable of switching 500w at 240V ac, and utilise an existing 12v dc supply.

light sensitive switch

Component Data:

SPNO Relay Tr1 BC109 D1 IN4001

Contact Rating 6A/250v ac Ic = 100mA I = 1A

Coil Resistance 720W Vcbo = 30v Vb = 100v

Operating Volts 8.4v to 27.6v Vceo = 20v

hfe = 200 to 800

D2 LDR1 NORP12 NSL19

Red 10 Lux 9KW 20KW to 100KW

Vf = 1.9v 100 Lux ---- 5KW

If = 10mA typ 1000 Lux 400W ----

If = 30mA Max Dark 1MW 20MW

Calculating Ic (Tr1 collector current)

Vsupply = 12v, For D2 Vf = 1.9v, Therefore Voltage RL1 coil = 10.1v

Ic = Vc/Rc = 10.1/720 = 14mA (for diode D2 If 10mA to 30mA)

Tr1 base current Ib = Ic/hfe = 14/200 mA to 14/800 mA. = 70mA to 17.5mA

Option 1

Use voltage generated by resistor divider R1, VR1/LDR1 to switch transistor Tr1 (see fig 1).

For Tr1 to conduct: Vb ł 0.6v and Ib ł (17.5mA to 70mA)

Selecting LDR1 as NORP12

Nominal Switching point LDR1 = 9KW (Vb = 0.6v) so Ildr = 66mA

and I(R) = 66mA + 17.5mA = 83.5mA to 66mA + 70mA = 136mA

Where R is the combined resistance of R1 and VR1.

Therefore R = (Vsupply - Vb) / I at nominal switching point.

R = 11.4/0.0835 KW or 11.4/0.136 KW = 137KW to 84KW

Adjustment for switching point LDR1, say 4K5W to 100KW

At LDR1 = 4K5W, Ildr = 0.6/4K5 =133mA

At LDR1 = 100KW, Ildr = 0.6/100K = 6mA

At LDR1 = 4K5W, R = 11.4V/(133+17.5) mA = 76KW to 11.4V/(133+70) mA = 56KW

At LDR1 = 100KW, R = 11.4/(6+17.5) mA = 485KW to 11.4/(6+70) mA = 150KW

For full adjustment R must vary between 56KW and 485KW therefore select VR1 500KW and R1 47KW.

NOTE: Although it is possible to achieve adequate adjustment it would be difficult to set up for precise light levels as a high proportion of the adjustment is allowing for gain variations in the selection of the transistor. It is not practical to use the cheaper NSL19 LDR in this circuit.

Option 2

(see Fig 2) Use an Operational Amplifier as a comparator.

IC1 output (for on state) is 10v (supply voltage less 2V)

therefore R4 Ł 10/0.07 KW = 143KW

Let R4 = 20KW

If LDR1 Resistance Variation is 4K5W to 100KW make 100KW equiv to 6v into pin 3 of IC1, therefore R1 = 100KW at 4K5W and Vp3 = 12 * 4K5/104K5 = 517mV.

IC1 is configured as a differential amplifier with a gain equal to the open loop gain of the amplifier (a comparator). When the voltage on pin 2 is less than the voltage at pin 3 the output will rise to the saturation voltage of the device typically 2v below supply voltage.

To allow the specified adjustment, select VR1 = 10KW, R2 9K1W, R3 750W

As fig 2 is more flexible than fig 1 NSL19 can also be used.

LDR1 variation becomes 10KW to 1MW therefore Let R1 = 270KW.

Then maximum volts at IC1 pin3 = 9.5v.

Minimum voltage at pin 3, IC1 = 12 * 4K5/274K5 = 0.197v

Select VR1 10KW:

then voltage across VR1 = 9.5V-0.2V therefore

R3= 0.197 * 10/9.3 KW = 212W, preferred value 200W

R2 = 2.5*10.2/9.5 KW = 2K68W preferred value 2K7W

The power rating of the resistors needs to be determined. The easiest way is to calculate the value of resistance for standard power ratings at the supply voltage.

For Power 1W, P = V²/R therefore R min = 12*12/1= 144W

For Power 1/2W, P = V²/R therefore R min = 2*12*12/1 = 288W

For Power 1/4W, P = V²/R therefore R min = 4*12*12/1 = 576W

By inspection only R3 is below 576W and R3 is part of a voltage divider which is well within tolerance.