![]() ![]() When you put a current into the base, the transistor will allow a current of up to I base*gain to flow through the collector/emitter path. Now, the transistor is also a current-operated device. For 450mA through a 3V V f LED and 5V supply, you should have a resistor of about 5 ohms (and you should figure out how I got that.) You can find lots of tutorials on how to pick the series resistor for a particular LED and power supply none of the details change significantly just because you have a transistor acting as the switch instead of just wires. To limit the current, you need a resistor in series with the LED. Since LEDs are current driven, you actually want a particular current to flow through the LED when the transistor is on. ![]() If the transistor has a gain of 100 we are setting the collector to base current ratio atĢ0 - giving it fives times as much base current to ensure proper saturation. Thus you would use a 150ohm resistor (3V/20mA) betweenĬollector and LED, and the base current should be about 1mA, therefore 4k7 base There is 1.9V across LED, 0.1V across saturated transistor, leaving 3.0V across If your LED needs 20mA and has a forward voltage of 1.9V (typical red LED), then Needs to be many times higher than for linear circuits. Here we are switching, so saturation is needed. Gain, and much more determined by the feedback network. Transistor into the correct operating area - negative feedback trades gain forĪccuracy, in particular the resulting gain is much less sensitive to the transistor With analog circuitry feedback from collector to base is often used to bias the So we never rely on the gain being exact. You cannot do it with the base because the gain of a transistor is not at all stable or The current for the LED is set by having a current-determining For a switching transistor you want theīase current to be high enough to saturate the device, bringing the collector ![]() You cannot use the base current to control the collector voltage, ![]()
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