How to light an 11W lamp with a 6V battery? Please point your fingers high!

It is not very difficult to generate instantaneous high voltage by the inductor of the voltage change, so we know that with a certain ratio of turns, instantaneous high voltage can be generated, I will teach you a few small tests, you are using a 6V battery, so take the transformer apart, take out the coil with a silicon steel sheet outside, the thin end of the copper coin is connected to the high voltage, and the thick end is connected to the low voltage, which is the constant contact of your battery, and then study it yourself.

You can also use a force converter A device that converts low-voltage DC into high-voltage AC, and of course, you have to make sure that at high currents, your battery won't** and the light will be on a little bit when it's connected.

With inverter power supply, the principle is very simple, convert 6V DC into pulse current, and then step up through the transformer, you need a detailed principle and circuit diagram, leave me a message, and reply to ...... within two days

Isn't the 6V battery-powered emergency light an example? First, an oscillation signal is generated, and through power amplification, the step-up transformer is driven to generate high voltage. You can light the tube.

The 555 chip or unsteady circuit is the most commonly used signal source, and the power amplifier component uses 2N3055.

No, high voltage is required to ionize the phosphor in the lamp to make the lamp glow.

Unless the internal resistance of the lamp is small, r=u*p

It's okay to have an inductor, and the inductor can generate instantaneous high voltage so...

The current of this light-emitting diode is around 10mA and the voltage drop is around 2V. The battery pack in your diagram is. To light up these 20 LEDs, you need to take them apart, and each group of 2 is connected in a group of 100 ohm resistors, so that it is divided into 10 groups.

Then connect these 10 groups in parallel to the battery pack and it will be lit.

Note that if you use a commercially available dry battery, the resistance in series is 150 ohms.

It can't be lit up like this, twenty series connections need a voltage of about 60V, and your four dry batteries can't reach such a high voltage.

The four dry batteries can provide about 6V, and you need to reconnect the LEDs into 2 strings of 10 to turn on.

The voltage of ordinary light-emitting diodes is about 1V, but there are also small diodes of about 1V. See how you use it. You can't afford to use it, and even if you use it, it's very weak.

The DC-DC conversion module can be used to drive the LED after increasing the voltage

Ultra-low voltage boost converters.

DC DC is an important part of the fuel cell system, the fuel cell can give a large current, but the output voltage is relatively low (

They need to be connected in series to get the voltage required by the device. This not only reduces the reliability of the power supply, but also leads to the loss of power. We designed DC DC

Boost converter. It is possible to convert a very low input voltage to the voltage required by the device, effectively reducing the power.

of attrition. , and simplifies the fuel cell system, improves the reliability of the system, and reduces the cost.

What we produce.

DC DC converter series, the main parameters are as follows:

The range of input voltages is in:

Output voltage: 6V

12v24v、100v、300v

Maximum input power.

5W, 10W average efficiency.

It depends on the input voltage and the size of the load.

It can also be achieved by making a switching oscillator.

The minimum ignition voltage of the light-emitting diode is about volts, and the voltage of a new alkaline battery is only volt, so a battery cannot directly light the light-emitting diode, and a boost circuit must be added.

The link circuit can boost a battery to 9 volts, which is more than enough to light up an LED.

Connect the diode to the wires and connect them to the positive and negative poles of the battery respectively.

One battery voltage is too low, two are required.

Just connect the positive and negative poles

Energy-saving lamp voltage v1, energy-saving lamp power: p1 (the bulb says: for example, 6v15w, then the power is 15w).

The power of 6 lamps in parallel = 6*P1

Power required for 8 hours = 8 x 6 x P1 (wh).

Then the capacity of the battery is: 8*6*P1(Wh)*V1 (unitAH) is the margin.

For example: energy-saving lamp voltage 6V bulb power 15W

Then the battery capacity = 6 * 8h * 15w * need about 6v144ah battery to see the voltage and power of your energy-saving lamp, car voltage 12v, how much is the battery voltage of large trucks? I do not know.

Then if you want a formula, I'll ask you, how much do you pay for energy-saving lights, is how much v? Do you want an inverter?

How w's? And how many batteries are yours? Is it a battery pack or a 12v one?

If one is 20W, 20W x 6 x 8H = 960Wh

960w/48v=20ah

If you are using a 48V battery pack, you will need at least a 20Ah battery pack.

If it's a 12V battery, you need 80ah.

You can use the small spotlights on the electric car, 12V to 100V universal, it is recommended to connect in parallel, the rated power is 3 watts to save electricity, but it is very bright, and it is also LED inside, which is durable and water-proof and shockproof. About 15 yuan each. And the after-sales service is half a year.

The battery only needs to be more than 20ah. Very power efficient.

Its filament is a resistance wire, which is simply like a red-hot iron rod, do you say there will be no temperature.

That's a problem. You connect two LEDs in series, and then try 50 groups in parallel, the current limiting resistor is a must.

100 pcs? It should not be lit up or the light is too dim and the current is not enough.

As long as one can be lit, because as long as there is an electric current through the bulb, it will produce bright light, and as long as there is voltage, there will be current, as for the brightness is the voltage size.