I don't think 10% resistors (E12 series) have been used since 50 years ago. Everybody uses 5% resistors.

Your simulation software has C2 shorted so the DC collector voltage is half what is should be.

 

Look on either side of C2, it is showing as test point 6. Until this is fixed we are spinning our wheels. Audioguru is right about the resistors, 5% is the normal value nowdays, I haven't seen 10% in a very long time, and I've been doing this for 30 years.

R1 = 47KΩ, R2 = 5.1KΩ, R3 = 10KΩ, R4 = 1KΩ, Vcc = 15V, β = 110

Base to Ground = 110KΩ
R2 equivalent = 4.87KΩ
Base Voltage = 1.41V
Emitter Voltage = 0.71V
Emitter Current = 0.71ma
Collector Voltage = 7.9V

BTW, I'm doing these numbers by hand on a calculator. The sequence is the order in which I crunch the numbers, no paper needed.

 

Yes you were right C2 was shorted...

I succeded in solving this error in the simulator..
Here is the new results.:

 

I don't think 10% resistors (E12 series) have been used since 50 years ago.

Nonsense! I was using them a mere 20 years ago! Never underestimate the antiquity of tertiary and quaternary backups used by the military!

Of course, no one has designed with them for half a century...

 

The AC analysis is pretty simple. You are loading the amp with 10KΩ, and it has a 10KΩ collector resistor, so the signal is cut almost exactly in half. If you remove the load you would see the signal double in voltage, which fits the calculated gain perfectly.

I'm not sure it is accurate to state the output impedance of the amp is 10KΩ, but I believe it is pretty close.

You're gain (with loading) is

db = 20 log (AC Volts out/AC Volts in) = 20 log (0.960V/0.200) = 13.6db

Because your output impedance is matched, you should be getting maximum power transfer between your amp output and the load.

Questions?

 

Hello My measured Ic is 0.8mA.
I come across in a book, it says that "Ic should be between 4mA to 8 mA as a practical value".
Is this statement true ?

What will it change if the collector current is higher or lower plz?

thx

 

The base voltage as set by R1 and R2, which in turn sets the voltage across Re. This programs the current.

By raising or lowering the voltage on the base you can raise or lower the voltage across Re.

Or, by changing Re (which will change the gain) you can change the collector current.

Remember, small changes when you're close to where you want to be.

There is no magical collector current however, it all depends on the application. Where did you quote that statement from? Off the top of my head I would rate it as false, but it is a convienent design goal.

 

hmmm.. just wanted to confirm that statement so as not to blow my transistor during the first test...

I will test the circuit with the 0.8mA IC and will see the result.. if its good.. i won't change anything in this circuit..

 

You calculate the current along with the voltage between the emitter and the collector, this gives you wattage. 2N2222s are pretty hardy devices, they can take 200ma as long as the wattage is low.


******************

Actually they can take a lot more, check this Wikipedia article out.

 

it will be a bc547B anyway gonna give it a test tomorrow and let u know.. i will work out all the calculations first then assemble the circuit tomorrow..

 

I don't know what is avaiable in your area, but something like the 2N2222A runs about 25 cents (4 for a dollar). Who is your source?

Let us know how it works out. You've never mentioned, are you going to use this for a larger project?

 

On all the circuit i posted the transistor is bc547B... dnt worry.. i will let u know... later about the result...

thx.

 

Hey man... everything worked fine...

I assembled the circuit and power it.. Here are my measured values:
Vc=6V
VCE=5.08V
VB=1.554V
VE=0.928V
VBE=0.596V

But i used the resistor R2 as 5.6K because i have only E12 series resistors available but this does not change thing lot..

On application of AC:
Input: 200mV, 10Khz (P-P)
Output: 960mV, 9.923Khz(P-P)

Everything was fine...

Do you know any goood math software goood for doing math? I want to do the calculation on that... they told me there is mathcad... if you know something else let me know...
thanks..

 

Not really, like I said, I'm using a scientific calculator and skull sweat. If I need something programable their's always GWBASIC.

 

Gain = R3/R4.

No.
The gain is R3/R4 only when there is no load. The circuit has a 10k collector resistor (R3) and a 10k load so the gain is half.

 

No.
The gain is R3/R4 only when there is no load. The circuit has a 10k collector resistor (R3) and a 10k load so the gain is half.

This I know, if you read the entirety of this thread I did mention loading would make a difference. Even showed it in my calculations. Picky picky, choosing context.

Try reading this post. There were others of course, but this one is pretty plain.

I've spent a lot of time and energy helping here. If you want to help you need to try harder.

Have any suggestions for the OP math package?

 

Bill_Marsden you are rite... i assembled the circuit.. and all gain and all was goood..


by the way what is the formula i should use to calculate the Current, IB at the base of the transistor please?

 

With an emitter resistor the input resistance of a trasistor is approximately beta X Re (β X Re). Since beta is only an approximation (meaining there is a wide variation even within the same family of transistor) there will be a large variation in input resistance of the transistor.

The way to handle this is to have enough current flowing through R1 and R2 that the transistor Base to ground isn't significant. Figure 1/10 of the parallel resistance equivalent of R1 and R2.

 

What X stand for?

 

2 X 2 = 4

Anyhow, if the Base to ground resistance is 110KΩ, then R1 and R2 in parallel should be around 11KΩ. Then figure out what the base voltage needs to be and work from there.

There is some leeway in this rule, as it isn't really a rule at all, just a guide to help.

Looking at your design, R1 & R2 parallel is 4.6KΩ, and the emitter resistor is 1KΩ, Base to Ground is 110KΩ, which is good enough.

Figure the R1//R2 resistance is under 1/20 the Base to Ground, yet it still interacts substantially.

The actual base to emitter current is Base Voltage / Base to Ground resistance. Again, this is an approximation, no 2 transistors are quite alike.