Hi All Readers of CircuitsToday,
As there is serious discussions going on about our previous Automatic LED Emergency Light Circuit using LM 317, Mr.Seetharaman has come up with a modified version of the same which answers many of the doubts raised in our Comments section.
Note: Mr.Seetharaman has developed a new version of Automatic LED emergency light. This one is more simple,more efficient and uses minimum components. Take a look: Simple Emergency Lamp Circuit
Here follows Seetharaman’s Description about the Modified Automatic LED Emergency Light using LM 317.
Dear John
Since there were lots of doubts from our readers on the LED emergency light, I have written a detailed letter to two of our readers. Of course few modifications are required, which I have indicated on the drawings and on calculations how I arrived at it. I thought of enclosing it to you so that it can be kept in some library of our site for any ones reference. For each circuit we can have a similar detailed theory, may not be with all calculation, just general operation theory of each part of the circuit. You may think of it. As a first feed back I am sending this. I have a feeling that most of the people would have made the mistake in base emitter of BD140, the lead out are against normal convention.
To understand the above circuit in a better way, it can be divided into two parts.
1. LED lamp circuit
2. The Battery charger circuit
LED Lamp circuit
1. All are white hi bright LEDs rated for 3Volt @ 25mA
2. The total current requirement is 12 X 25 = 300mA
3. This current has to flow through T2 – BD140 PNP transistor
4. The minimum current gain (hfe) of this transistor @ 500mA is 50
5. Hence the base current Ib requirement is Ic / hfe, 300 / 50 = 6mA
6. Base emitter drop of T2 at 500mA is 0.77 volt
7. With the fully charged battery at 6.9volt terminal voltage (for cycle operation use) the voltage available across the new bias resistance is (6.9 – 0.77)
8. Hence the bias resistance is = 6.13 / 6 = 1000ohms
9. As the battery drains the final terminal voltage will be 5.4volt
10. The bias resistance will be (5.4 – 0.77) / 6 = 770 ohms Hence a 680 ohms was preferred for bias resistance with drained battery also it will give enough brightness.
11. The very important information about BD140 is, as you view the pins, metal portion of the transistor facing down left is emitter centre collector and right is base. Most of the constructors make this mistake, relying on the convention that left base and right emitter. If you have made this mistake please correct it.
Once this portion is checked for reliable operation we will proceed to charger portion.
The Battery charger circuit
1. The battery requires a full terminal voltage of 6.9V at this point charger should cut off.
2. That is the voltage across the chain ZD1, R2 and T1 be should be 6.9 volt
3. T1 be voltage of 0.7 volt plus drop across R2 and zener voltage should be 6.9V
4. T1 be current = Ic / hfe
5. Ic is 1.25 / 180 = 7mA
6. Ibe = Ic / hfe of T1 i.e = 7 / 70 = 100uA
7. Drop across R2 =1.2 X .1 mA = 0.12volt
8. Hence Zener voltage = 6.9 – (0.7 + 0.12) = 6.08 the near by preferred zener voltage is 6.2 volt
9. Say the battery voltage at full charge will be 7 volt with 6.2 volt zener diode
10. To calculate R16 value for charging at 1 /10 th of the rated current of the battery 4.5AH / 10 = 450mAH
11. Transformer 9volt AC the voltage across C1 will be 9 X 1.414 = 12.6 volt
12. The drop across LM317 at 450mA current for good regulation is 3volt
13. The drop across protective diode D5 is 0.7 volt.
14. The voltage available at cathode of D5 is 12.6 – (3+0.7) = 8.9volt
15. The battery after fair discharge will be at 6 volt
16. Hence R16 = (8.9 – 7) / 0.45 = 6 ohms
17. The nearby standard value for operation is 5 ohms.
18. At the end point of battery 5.4 volt the maximum charging current can be of (8.9 – 5.4) / 5 = 0.7 amps well within the higher charging limit of the battery.
19. With this circuit over night the battery will get charged fully.
20. Over charging is taken care and protected by T1
Hope with the above guide line you can make your light work successfully.
More Modification!!!
Automatic LED Emergency Light with Under Voltage Cut Off Protection:
Dear Readers,
Mr. Seetharaman has further modified this LED Emergency Light with an under voltage cutoff protection to protect battery from deep discharge. Once the battery terminal voltage falls below 5.7 volts the LEDs will be switched off. Take a look at the modified circuit shown below.
To understand the above circuit in a better way, it can be divided into two parts.

 LED lamp circuit
 The Battery charger circuit
LED Lamp circuit

 All are white hi bright LEDs rated for 3Volt @ 25mA
 The total current requirement is 12 X 25 = 300mA
 This current has to flow through T2 – BD140 PNP transistor
 The minimum current gain (hfe) of this transistor @ 500mA is 50
 Hence the base current Ib requirement is Ic / hfe, 300 / 50 = 6mA
 Base emitter drop of T2 at 500mA is 0.77 volt
 With the fully charged battery at 6.9volt terminal voltage (for cycle operation use) the voltage available across the new bias resistance is (6.9 – 0.77)
 Hence the bias resistance is = 6.13 / 6 = 1000ohms
 As the battery drains the final terminal voltage will be 5.4volt
 The bias resistance will be (5.4 – 0.77) / 6 = 770 ohms Hence a 680 ohms was preferred for bias resistance with drained battery also it will give enough brightness.
 The very important information about BD140 is, as you view the pins, metal portion of the transistor facing down left is emitter centre collector and right is base. Most of the constructors make this mistake, relying on the convention that left base and right emitter. If you have made this mistake please correct it.
 Once this portion is checked for reliable operation we will proceed to charger portion.
 The battery requires a full terminal voltage of 6.9V at this point charger should cut off.
 That is the voltage across the chain ZD1, R2 and T1 be should be 6.9 volt
 T1 be voltage of 0.7 volt plus drop across R2 and zener voltage should be 6.9V
 T1 be current = Ic / hfe
 Ic is 1.25 / 180 = 7mA
 Ibe = Ic / hfe of T1 i.e = 7 / 70 = 100uA
 Drop across R2 =1.2 X .1 mA = 0.12volt
 Hence Zener voltage = 6.9 – (0.7 + 0.12) = 6.08 the near by preferred zener voltage is 6.2 volt
 Say the battery voltage at full charge will be 7 volt with 6.2 volt zener diode
 To calculate R16 value for charging at 1 /10 th of the rated current of the battery 4.5AH / 10 = 450mAH
 Transformer 9volt AC the voltage across C1 will be 9 X 1.414 = 12.6 volt
 The drop across LM317 at 450mA current for good regulation is 3volt
 The drop across protective diode D5 is 0.7 volt.
 The voltage available at cathode of D5 is 12.6 – (3+0.7) = 8.9volt
 The battery after fair discharge will be at 6 volt
 Hence R16 = (8.9 – 7) / 0.45 = 6 ohms
 The nearby standard value for operation is 5 ohms.
 At the end point of battery 5.4 volt the maximum charging current can be of (8.9 – 5.4) / 5 = 0.7 amps well within the higher charging limit of the battery.
 With this circuit over night the battery will get charged fully.
 Over charging is taken care and protected by T1
Hope with the above guide line you can make your light work successfully.
Dear Mr. Steathman,
Gently, I need your clarification regarding battery charger Cct as below:
• The output voltage at the secondary Tr output is it (RMS) value or (PEAK) if (RMS) also we should take the voltage drop of rectification diodes (D1D3) in consideration, in another word (( Vp= 1.414 × (Vrms – 1.4))
• The output voltage of the LM317 should be adjusted to be the voltage drop on the ZD1+R2+T1 chain become (6.9 or 7) Volts that means the cathode point of the ZD1 reference to ground and the battery terminal equal to the mentioned value, but in this case the zener diode will be permanently open and T1 is ON so how would we make the battery receive the charging current.