tutorial design :a sensitive change in light, detector

This is a tutorial to build a transistorized light detector, it is assumed that a person choosing to build this circuit, has a basic understanding of electronics fundamentals, and are familiar with the basic theorems such as (ohms law, kirchoffs voltage and current laws, ect..) that way the steps shown in the design are not broken down to great detail of showing the theorems being used, but rather the resuts of using the theorems for calculations for the design.

This is to detect the change in light from a high to a low, therby turning on the LED when the light source drops in value.

LED: vf=2V. chose to operate LED at 5mA. for low battery drain.
Photo resistor (CDS) works around 8K ohms with light source around 32" away.
all resistor values are rounded from calculations to nearest nominal value.


Chose Vdrop across R1 to be 3v.

1) R1=(3v / 5mA)=620 ohms.

2) R2=(2v. / 5mA)=390 ohms.

Chose to turn off output (led) by dropping to 1V at the collector of Q1.
therefore approximately (6v will drop across R1) and a current will result of (6v / 620)=9.7mA.
Chose Vce to be 0.2v, and 0.8v will appear across R3.

3) R3= (0.8v / 9.7mA0=82 ohms.

Chose to use a min Beta value of 20 for calculations.
Base voltage divider current IDQ1 will be ((9.7mA / 20) x 10)=4.85mA
Assuming a value of 0.7v for Vbe for all transistors.

4) R4=(1.5v / 4.85mA)=300 ohms.

5) R5=(5.5v / 4.85mA)=1.1K ohms.

Now to turn off Q1 by turning on Q2 to bring the collector voltage of Q2 to 0.3v, which will cause Q1 to shut off. with 0.3v at the collector of Q2, puts 6.7v across R5, which makes a current of (6.7v / 1.1K)=6mA to flow through R5.
Choosing to bring VCE of Q2 to 0.2v puts a drop across R6 0.1v.

6) R6=(0.1v / 6mA)=16 ohms

base divider current for Q2 is (10 x (6mA / 20))=3mA

7) R7=(0.8v / 3mA)=270 ohms.

8) R8=(6.2v / 3mA)=2k ohms.

Q3 needs to turn off Q2 so VCQ3 will be 0.3v which puts the base of Q2 at a low voltage to shut off, so calculating current flow through R8 needed to drop the base voltage of Q2 to 0.3v
(6.7v / 2K)=3.3mA, with 0.2v chosen for VCEQ3 then 0.1v across R9.

9) R9=(0.1v / 3.3mA)=27 ohms.

With photo resistor working at 8K ohms, then current through this sensor will be calculated as ((VCC - VBQ3) / 8K ohms)=((7v - 0.8v) / 8K ohms)=775uA

10) R11=(0.8v / 775uA)=1k ohms.

When I gathered the data for the CDS cell, I took a resistance measurement with the CDS cell on top of my bench and measured the distance to the lamp overhead, around 32 inches.

The resistance measured at around 8K ohms, so the last stage in this design was done using the 8K ohm value for the calculations for Q3 to be activated.

When I tested this circuit, the LED remained off when the breadboard was sitting on top of my bench, when I lowered it off of my bench around 4 to 6 inches, the LED came on, indicating that the circuit being built in more than one stage, added some signal gain, thereby making it sensitive to light changes on the CDS cell.
If this was built using only one stage the sensitivity to light change would not be as sensitive, the more stages introduced the more signal gain is aquired thereby sharpening the sensitivity of the output to the CDS light sensitive resistor, input signal.


here is a video of it working.

thanks for watching...

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