LED DRIVER


Posted on Feb 7, 2014

With an an input sinewave on the left amplifier base, here`s what happens: The collector output rises above or falls below 6. 6 volts. Of course, the amplifier increases signal level and inverts the signal. Because the output is roughly centered, the chances of clipping are small. (The emitter voltage shifts the collector voltage upward a bit, but not significantly. In return, emitter


LED DRIVER
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feedback tends to improve fidelity. ) Right Amplifier Analysis. Now, we apply an input sinewave to the right amplifier base. The collector output rises above or falls below 11 volts. Again, the amplifier increases signal level and inverts the signal. Unfortunately, the output is far off center. The chance of clipping is likely. (The large collector resistor upset the circuit`s base bias. ) A negative input signal easily cuts off the collector signal. Remember that the collector quiescent voltage is 11 volts. The collector only needs to rise by one volt. In the positive direction, this amplifier has a lot of gain. A positive input signal can cause the collector to drop 11 volts. Compare that to 5. 4 volts for the left amplifier. Unfortunately, the right amplifier`s output waveform is terribly assymmetrical. I can vouch for the clipping effect. I demonstrated it. I wired this amplifier to a signal tracer. Then I used the circuit as a microphone preamp. I could hear the output distortion. Now, back to the LED driver. By substituting one LED for my display, you introduce similar problems to those in the example above. I designed the LED driver to output 80 milliamperes average to 12 LEDs. You substituted one LED. This LED probably requires 20 mA. (That depends on what LED type you used. ) The result You overdrive the LED. Why Because you aren`t using the rated load. Driver gain doesn`t change, but load resistance increases. No...




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