transistorsvoltagevoltage-drop
I've been reading different answers online and am trying to get my head around calculating transistors.
I have a circuit like this:
(Imagine the base goes to a microcontroller and is connected properly.)
After the red LEDs, the voltage will drop to 3.2V.
Will the transistor (2N2222) drop any more voltage?
If this circuit goes to ground (remove the transistor) I've calculated about 15mA goes to each LED. If there was more voltage drop, would I need to reduce the resistance?
Here are the voltages as they are described:
Now, if you want to calculate the resistor current, you can see that it is 9.4V / 1000-Ohm = 9.4 mA. Since the base is in series with the resistor, the resistor current = the base current.
Does this help?
Assuming the + and - of the batteries are correct and not the polarity shown by the battery symbols (the long line is the +, short line is -) and also assuming that the battery voltages are such that D1 isn't destroyed then:
It depends on the values of R1, R2 and R3 what mode of operation Q1 will be in.
If Q1 operates in active mode then it will have a large enough Vce to enable some current (coming from R3) to flow into the base of Q2. As the LEDs are indicted as not lighting up, we can assume that this is not the case.
If Q1 operates in saturation mode then Vce will be much smaller, typically less then 0.4 V which will prevent any current to flow into the base of Q2. For a current to flow into the base of Q2 there needs to be more than about 0.6 V across the base-emitter of Q2.
This isn't a well designed circuit. If Q2 would conduct then it also determined the current through the LEDs which (depending on Q1 and resistors) could be high enough for the LEDs to conduct too much current. There should have been resistors in series with the LEDs.
Best Answer
If the transistor base current is adequate the transistor will be driven into saturation and there will be about 0.2 V between the collector and emitter. This is small enough that it can be ignored in many calculations such as this.
That leaves about 3 V across the resistor so I = 3/75 = 40 mA shared between the three LEDs. Generally direct paralleling of the LEDs is avoided because the currents will vary quite a lot depending on their individual forward voltages, Vf.
See what I've written in Variations in Vf and binning for more on the topic.