This article shows the emergency lights system.

The current source in my circuit represents a photodiode. When lights are turned OFF the current source produces almost zero current and the circuit thus activates emergency lights by providing current to the light bulb. The light bulb remains ON for a few seconds in case the photodiode is damaged due to an emergency.

This circuit is suitable for a 1.5 V light bulb. If you are using bright LEDs then you need to increase the power supply to at least 3 volts. However, if you raise the voltage to 6 V or above then you should use MOSFET transistor instead of BJT transistors. MOSFETs also have a maximum gate-source voltage that should not be exceeded.

My circuit allows the driving of higher current LED lights or light bulbs.

Step 1: Design the Circuit

The circuit that I designed is able to drive high-current light bulbs. You can use fewer transistors if you are connecting a bright LED.

Q3 and Q4 should be power transistors. I used PSpice student edition simulation software that does not include power transistors.

Each photo diode has a dark current. Dark current is the current across diode in darkness (when no light enters the diode). We can now calculate the maximum dark current of the photodiode:

Idark = Ipd

= (Vs - Rb1 * Ib1 - Vbe1) / Rpd - Ib1

= (Vs - Rb1 * Ib1 - Vbe1) / Rpd - Ic1 / Beta

= (Vs  - Rb1 * Ib1 - Vbe1) / Rpd - (Vs - Vbe2 - VceSat1) / (Rc1 + Rb1) / Beta

Ib1 = (Vs - Vbe1 - VceSat1) / (Rc1 + Rb1) / Beta

= (1.5 V - 0.7 V - 0.2 V) / (1,000,000 ohms + 220,000 ohms) / 20 

= 133.5113nA

Idark = 637.1162nA

If the photodiode dark current is above this value, the circuit might not work.

This is the Matlab code:

clear all
Vs=1.5;
Beta=20
Vbe1=0.7;
VceSat1=0.2;
Vbe2=0.7;
Rc1=4700;
Rb1=220000;
Rpd=1000000;
Ib1=(Vs-Vbe1-VceSat1)/(Rc1+Rb1)/Beta;
disp(['Ib1 = ' num2str(Ib1*1000000000) ' nA'])
Idark=(Vs-Rb1*Ib1-Vbe1)/Rpd-(Vs-Vbe2-VceSat1)/(Rc1+Rb1)/Beta;
disp(['Idark = ' num2str(Idark*1000000000) ' nA'])

I used Octave software.

You can see in the circuit above that I assumed the photodiode dark current to be 100 nA.

Step 2: Simulations

When the Ct capacitor is charged the Vt voltage is almost zero (with reference to ground) and the lights are ON. When the Ct capacitor is discharged the Vt voltage is almost equal to the power supply voltage and the lights are OFF.