Bjt led driver

But what if we purposely put the transistor into saturation? Saturation is a fixed value. To begin with, the whole point of the driver is to offload the current demand from the prior circuit. For example, we may wish to light an LED from the output of a logic gate or microcontroller chip. The problem is that those circuits might only be able to deliver, say, 5 mA when we might need well over 10 mA to achieve the desired brightness.

The LED driver circuit is used to overcome this limitation. Note: The negative terminal of VCC is connected to ground not shown. With the driver, the logic circuit will only need to supply base current, not LED current. Here is how it works: If the logic input voltage is zero, there will be no base current. This means that there will be no collector current and therefore the LED will be off.

At this point the BJT is in cutoff. If properly designed, this current will be sufficient to put the BJT into saturation. A value of 10 or so would guarantee hard saturation. This will be the LED current. The ratio of these two currents is just over Quote from: T3sl4co1l on July 11, , am. R2 is LED drive resistor. Mechatrommer Super Contributor Posts: Country: reassessing directives Quote from: Zero on July 11, , am. It's extremely difficult to start life.. You may wonder Why? We simply have to accept it.

One could describe the situation by saying that It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. I will freely admit that I am mainly digital when it comes to electronics, so may help explain my confusion that or just general incompetence.

I am getting review feedback on a design, and one thing that keeps bugging me is that placing a BJT below the LED keeps being labelled as 'wrong'. Is this a major issue or just a convention or is the reviewer just being annoying? With a BJT it doesn't just "act like a switch" - that is the end result you see at the high level, not what it actually does. What actually happens is you apply a current to the base, and that then flows through the transistor and out of the emitter down to ground.

At the same time it allows proportionally more current to flow from the collector down through the emitter to ground. So from an easy to understand point of view the left-hand circuit is best. However, the right-hand circuit does have some advantages. That can save on parts, and if you have a lot of these circuits in you design that reduction in parts can soon add up. If your only goal is to get light out of the LED, then the circuit on the left is perfectly fine.

The reason is because the voltage needed to change the state of the transistor is very low - typically around 0. So what happens is that if you accidentally short the LED's cathode to ground, then it will immediately bypass the transistor and dump the full voltage across the device.

This is perfectly avoidable if your LED is encased in some plastic box or something, but a very bad idea if the user can freely manipulate the thing say, if your enclosure is a metal box. I've personally worked on applications where the LED is very expensive and this exact problem needs to be avoided. The alternative is your circuit on the right.

This is called the "low side" configuration. This avoids any potential damage to your device when it accidentally shorts out to other metal objects. Remember that the voltage from the base to the emitter needs to be 0. But if the emitter is at, say, 5 V when current is flowing, then your base needs to get up to 5.

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