# RECHARGABLE LED FLASHLIGHT

by Le Magicien

This is the project of rechargeable flashlight using a high intensity red Jumbo Led. The main idea is to have a moderate source of light that will be ever at hand, without batteries to replace or light bulbs to change. It may be left plugged in the 110 vac wall outlet, ready for use. Click on the left to see the main circuit of the flashlight.

CIRCUIT DESCRIPTION

When the flashlight is not plugged in a AC outlet, diode D4 prevents any current flow from battery B to Q1. Therefore the circuit is reduced to the scheme shown in Fig 1. Thus according, the emitter of Q2 transistor is directly connected to the (-) terminal of the battery B and its base is positively polarized through R5. In this way the base-emitter junction of the transistor is directly polarized turning on the collector-emitter junction and allowing current to flow from battery B to ML (Main Led).

When the flashlight is plugged and charging we have the situation depicted in Fig 2. Charging led indicator CL is on while the main led ML is off. How Q2 is turned off? Note that Q1 is turned on as the base has a positive supply given by R2 and the charging led CL has a 1.9 junction drop (as any other red led) that remains constant, as it behaves as a zener diode. Thus the circuit around Q1 configures a constant current supply.
In order to turn off Q2 we need its base-emitter junction forward drop to be lower than 0.7 volt or even better: to have VbeQ2 negative.
While charging, the measured voltage (red values in Fig 2) at top of voltage divider given by R5 & R6 is 4.7 volt. Supply voltage is aprox 0.7 volts above (due to D4 forward drop), or 5.7 volt. At Q1's collector we have 5.4 - 3.9= 0.8 volts. And the voltage divider sets Q2 base at 0.2 volt.

With these values it's easy to calculate VbeQ2:

VbeQ2 = Vb - Ve (for Q2) and as VeQ2 = VcQ1 = 0.8 volt
VbeQ2 = 0.2 - 0.8 = - 0.5 volt inversely polarized and Q2 remains in cut-off condition.

The basic point here was to calculate the voltage divider values so as to inversely polarize the base-emitter junction of Q2.

CONSTRUCTION TIPS

This flashlight has the advantage that no bulb will ever burn, also the rechargeable NiCd battery may be of the camera type: 3.9 volt / 110 mAh with a size a little bit bigger than the regular AA batteries. With the current circuit configuration this flashlight will run for 3 to 5 hours and will be fully charged in approximately 6 hours or less.

But you may use a battery pack if you like, just follow the following example to determine new values:

• Let's say we use 3 x 1.25 volt/650 mAh NiCd batteries:
always consider a voltage drop of 2 volts for all jumbo leds, we calculate the value of resistor R4 for a given operating current Ip, for example 19 mA:

R4 = [(3x1.25) - 2)volt]/Ip = [3.75 - 2 ]/19mA = 1.75 v/ 19 mA = 0.92 kohm
R4 = 92 ohm

This will be nice since now you have 650 mAh and will give you more than 24 hrs of operation! Also you may increase ML current by lowering R4 value, but check first the maximum current for the type of led you have.

• You may modify the constant current supply section in order to increase charge current:

Let's say you want a charge current of Ic milliampers then:

R3 [kohms] = (CL diode voltage drop - 0.7 )/ Ic

for example a constant charge current max of 70 mA will need
a (1.9-0.7)/ 70 mA = 0.017 kohm R3 resistor. Thus R3 = 17 ohm.

You may use this examples to build your constant current supplies, replacing the charging led with a zener diode.

ACTUAL LED FLASHLIGHT       lemagicien@email.com Hosting by WebRing.