100% correct. An LED has little to no internal resistance, so if you connect it directly to a battery with a voltage higher than the LED forward voltage drop, you're theoretically allowing infinite current to pass through the LED. This will destroy it. For that reason, you use an external resistor to limit the current.I'll take a shot at this, and then let the experts chime in.
1. Whenever you see an LED and a resistor in series, the resistor is there to limit current flow so you don't torch the LED when it comes ON.
Overall this is correct. When the base voltage reaches a certain level, it will connect, allowing current to pass between the collector and emitter. It acts as a switch, allowing the LED to turn on.2. Whenever you see something connected in series to the collector of an NPN resistor, it is there to "turn on" when there is collector-emitter current (the transistor must be "turned on" before that current flows)
Close enough, though there's a bit more to it than that3. The arrow on the emitter points in the direction of current flow
In this case the resistor between the base and +V, along with the LDR, creates a voltage divider bias. This means that as the resistance of the LDR increases or decreases, so does the voltage on the base. Looking at the circuit, I would say that when the LDR has a low resistance, there will be little to no voltage on the base, and the LED will turn off.4. If you ever see a resistor connected to the base of a transistor, it is probably there to control the base current or voltage, and prepare it for "another signal". When that "other signal" comes, it will turn on the transistor and allow collector emitter current to flow
"LDR" stands for "Light Dependent Resistor". The more light it receives, the more resistive it becomes.5. That LDR1 looks like it might be "another signal". (maybe a sound or light detector?)
Though this may be true in a sense, I'd watch the terminology a bit. A battery supplies a voltage, which causes current to flow.6. A battery gives you "energy" - just like caffeine
Changing it to a Darlington Pair will increase the gain of the circuit, which means the value of the current limiting resistor on the LED would have to be recalculated. There really wouldn't be any advantage of doing this. I'd just stick with a single NPN transistor, such as the 2N3904.If i change the transistor into darlington pair, will the current increased at the output? The output nodes are at above R2 and below D2.
Thank you Brownout, I failed to mention that. I was going to expand on my statement about increasing gain, and I guess I forgotChanging to a darlington does nothing, as the transistor operates in cutoff/saturation. Darlingtons aren't meant to operate that way.
Sorry to be really pedantic, but surely here the LDR has less resistance as the light increases? Otherwise this circuit would be a "light" detector rather than a "dark" detector."LDR" stands for "Light Dependent Resistor". The more light it receives, the more resistive it becomes.
Matt
It's true that MOSFETs are usually a better choice than Darlingtons for switching. The big advantage of ULN2803A is density. You get 8 switches in a package. There a a few MOSFET arrays available, but I don't know of any with 8 to a package, except for an 8 bit shift register/latch/open drain driver combo.Those darlington switches operate in the linear region, a disadvantage in power usage.