Buy this book The Art of Electronics by Horowitz and Hill (2nd edition).
It cost $US20 (which is a bargain). It's in New Delhi and they have a number of them. If you cannot afford the 1050 Rupee get several friends to buy it together, This is the best book on the subject that you will find.
- The Art of Electronics (Second Edition)
(ISBN: 0521689171 )
Paul Horowitz,Winfield Hill
Bookseller: BookVistas (New Delhi, DEL, India)
Bookseller Rating:
Quantity Available: > 20
WARNING" There are a lot of these also advertised in India. They cost typically the same or more as what I recommended and are not the same. Take due care. This the associated student manual by Horowitz and Hayes. If you can afford to buy one of these AS WELL do so but get the proper textbook first. Copy of workbook here for Rs484 including postage in India.
The easiest way of solving such problems is to use complex phasors. The total complex impedance is
$$Z_T=R_1+R_2||j\omega L=R_1+\frac{j\omega R_2L}{R_2+j\omega L}=877.18\cdot e^{30.67^{\circ}}\Omega$$
with \$\omega=2\pi\cdot 10,000 Hz\$, \$R_1=470\Omega\$, \$R_2=988\Omega\$, and \$L=10mH\$ (as shown in your circuit diagram).
This gives for the total current
$$I_s=\frac{V_0}{Z_T}=9.12\cdot e^{-30.67^{\circ}} mA$$
with \$V_0=8V\$.
The voltage across \$R_1\$ is
$$V_1=I_sR_1=4.28\cdot e^{-30.67^{\circ}} V$$
The voltage across \$R_2\$ and \$L\$ is
$$V_2=I_s(R_2||j\omega L)=I_s\frac{j\omega R_2L}{R_2+j\omega L}=4.83\cdot e^{26.88^{\circ}} V\quad(=V_0-V_1)$$
The current through \$R_2\$ is
$$I_2=\frac{V_2}{R_2}=4.89\cdot e^{26.88^{\circ}} mA$$
The current through the inductor is
$$I_L=\frac{V_2}{j\omega L}=7.70\cdot e^{-63.12^{\circ}} mA\quad(=I_s-I_2)$$
I realize that there are quite a few differences between these results and your calculations and measurements. I just used the values you gave in the circuit diagram, and I did my best to avoid any errors.
Best Answer
Here is a simplified approach that hopefully give you the concepts.
Assume that the base emitter voltage of the transistor is 0.7V and the gain is large enough that the base current is negligible.
First of all calculate the voltage at the junction of the 10k and 6.8k resistors as if the transistor was not present.
Vbase = 20*6.8/(10+6.8) = 8.1 volts.
AS Photon points out the transistor is in emitter follower configuration so the emitter will follow the base but be lower by the base to emitter voltage.
In that case the voltage at the emitter (Vout) will be 8.1 - 0.7 = 7.4V. This will be the output voltage Vout.
Now lets how much the base current will affect things if we have a more normal transistor.
The current through the 1K resistor is Vout/1K = 7.4/1 = 7.4mA. If we assume the Hfe of the transistor is 100 (This isa reasonable gain although many transistors these days are better than that). The base current will be 7.4/100 = 74uA.
How much will that affect the voltage at the base?
The effective impedance at the base is equals to the Thevenin equivalent resistance which is 6.8K in parallel with 10k.
This is (R1*R2)/(R1 + R2) (10*6.8)/(10+6.8) = 4.04k.
The voltage drop due to the base current will be Ib * 4.04k = 0.074*4.04 = 299mV.
So the output with this correction will be about 300mV less than our original assumption when using a very high gain transistor. i.e. 7.1V.
This approach is not 100% accurate but will be very close and is more straightforward than the full analytical method, especially when some of the parameters such as the gain of the transistor are not known and will vary significantly from unit to unit.