circuit analysisled
I'm a beginner to electronics. I want to find out how to calculate the current in the following circuit. (I know how to calculate total resister value in serial and parallel resisters in order to calculate the current, but is it possible to calculate without knowing the resister value of LED?)
simulate this circuit – Schematic created using CircuitLab
I thought \$V_S\$ will act like short circuit
This is why you're confused.
In this circuit, we don't have any control over \$V_S\$ - it is a constant, unchanging, DC voltage. However, we do have control over the switch in the circuit. When, we close the switch, the circuit splits into two separate pieces.
The left side of the circuit becomes:
simulate this circuit – Schematic created using CircuitLab
Since this circuit has nothing to do with the inductor, we can ignore it.
The right side of the circuit is what your teacher drew, so we can simply analyze that on its own. Note that \$V_S\$ hasn't changed - instead, we've totally taken it out of the circuit.
You are forgetting that you can overdrive a base so that it goes into saturation. This is desirable to get the transistor fully turned on.
Your calculation for I7 is correct:
$$ I_7 = \frac {V}{R_3} = \frac {5 - 0.6}{25k} = 0.176~mA $$
You then calculate the maximum collector current using \$ I_8 = \beta I_7 \$ Lets say \$ \beta \$ = 100, then \$ I_8 = 0.176 * 100 = 17.6~mA \$.
Now look at R1. The maximum current through it is given by
$$ I_1 = \frac {V}{R_1} = \frac {5 - 0.6}{1k} = 4.4~mA $$
Clearly the maximum possible value for \$ I_8 \$ is this 4.4 mA + 0.176 mA from \$ I_7 \$.
The calculations indicate that we could reduce the base current into Q1. In practice we want to ensure that the drive is adequate to cover variations in \$ \beta \$ due to production spread or even transistor substitution. Drive it hard and make sure it's fully 'on' is the normal approach.
Best Answer
You can use another technique that uses just circuit observation. For a simple circuit like this, I think it's easier than Thevenin.
The LED has a characteristic forward voltage, Vf. It varies with current, but the resistance in this circuit will keep the drop across the LED very close to the reported value in the datasheet. Let's use Vf = 2V for convenience.
Since the LED is parallel to R1, the voltage across R1 is also Vf. That must mean the drop across R2 is what's left over: $$V_{R2} = 5V-V_f = 5V-2V=3V$$
Using Ohm's Law on R2: $$I_{R2}=V_{R2}/100=30ma$$
Since R2 is in series with the source, that 30ma is the total current going into the circuit and is being split between R1 and the LED. We can figure out how much is going through R1 using Ohm's again.
$$I_{R1} = V_f/75= 2V/75 = 26.67ma$$
There's only 30ma going through the whole circuit and 26.67ma of it is going through the resistor. Therefore, only 3.3ma is left to go through the LED:
$$I_{LED} = 30ma - 26.67ma = 3.3ma$$