Every electronics guy knows that Light-emitting diode will not glow at a voltage below 2 V. Using this simple circuit you can make Light-emitting diode lights up from one 1.5 volt battery.
Materials:
- Ferrite ring.
- Winding wire 0.3 mm
- Transistor KT315.
- Resistor 100 Ohm.
If you use PNP transport instead of NPN, then you need to change the polarity of the power supply and LED. I took a ferrite ring from the ballast of an energy-saving lamp. The resistor can be used from 50-100 Ohms.
Scheme
Manufacturing
1. First, we wind the wire onto the ferrite ring. We wind 10 turns with two wires in parallel.
2. Then we find the middle winding point.
3. Solder the negative leg to the emitter of the transistor LED.
4. Solder the positive leg to the collector of the transistor LED.
5. We also solder one inductor leg to the transistor collector.
6. Solder the resistor to the base of the transistor.
7. Solder the second terminal of the inductor onto the resistor.
8. Solder plus power to the emitter of the transistor.
9. Solder the minus power supply to the middle point of the throttle.
Checking the work:
Advantages of the scheme:
With this scheme you can squeeze all the juice out of dead batteries. Using this scheme, you can make a flashlight that will be powered by one 1.5 volt battery.
