Single Stage BJT Common Emitter Amplifier

Wendy

Joined Mar 24, 2008
23,415
You still have a big problem, the total base resistance, that is the value of the 2 resistors R1 and R2 in parallel with each other, should be around 10 times your emitter resistance.

You need to either increase your emitter resistor (recommended) or decrease the base resistors to fit this criteria.

Remember where I said the transistor Base loading is negligable (β X Re)? With the emitter resistor so low it isn't negligable, it is equivalent to a 500Ω in parallel with R2, 1.2KΩ, assuming the transistors β is 50. See what I mean?

What are you trying to accomplish with your design? This sets the design criteria. You keep jumping all over the place with your values, instead of tweaking it. This creates a moving target, it is hard to state anything with certainty, conditions that were negligable with one layout become major on the next.



So lets go through the DC analysis on your latest design.



R1 = 12KΩ, R2 = 1.2KΩ, R3 = 3.9KΩ, R4 = 10Ω, Vcc = 12V, β = 50

Base to Ground Resistance = 353Ω
Base Voltage = 0.34V

At this point I'm stopping because you don't have enough voltage on the base to reliably turn the transistor on, for it to conduct. It isn't in a predictable state anymore. You've gone from saturated to barely turned on.

Gain = 390, Given the transistor has a max gain of 50 this doesn't work either.

The design formula for a basic transistor amp with no R2 and R4 are different. There the transistor Beta (β) rules.
 
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Audioguru

Joined Dec 20, 2007
11,248
Calculate the transistor's base voltage given by the voltage divider of R1 and R2 (1.36V).
Then subtract the 0.7V base-emitter voltage (0.66V).
Then calculate the emitter current (66mA).
Then calculate how many volts there might be across the collector resistor Rc (257V!). Impossible and your transistor is turned on hard and is saturated..

The value of your emitter resistor is much too low. If it is 390 ohms then the voltage across the collector resistor is 6.6v and the transistor would be biased near half the supply voltage (at its collector) and have a gain of about 7 (17dB).
 

Thread Starter

NiCeBoY

Joined Aug 20, 2008
59
Hello,
Finally i calculated these values for R1 & R2 & Rc & RE...

Now the problem i am getting is that my calculated values and measured values is not the same..
Here is my circuit :



Using Beta = 110 as on the datasheet :
and also using approximate analysis..
My calculated values are as follows:

Vb=1.597V
Vce= 5.13V
Ic~=IE=0.897mA
Vc=6.03V
Ib=8.08 micro.Amps
Input inpedance = 4791.96 Ohms
Output impedance = 5000 Ohms
Vo=4.44V

Can anyone check my circuit and let me know if its in good working condition?

Thanks in advance...
 

Wendy

Joined Mar 24, 2008
23,415
What were the measured values, or is it on the print you included?



R1 = 47KΩ, R2 = 5.6KΩ, R3 = 10KΩ, R4 = 1KΩ, Vcc = 12V, β = 50

Here is my DC analysis:

Base to Ground Resistance = 50KΩ
R2 is now equivalent to 5.04KΩ

Base Voltage = 1.16V
Emitter Voltage = 0.46V
Emitter Current = Collector Current = 0.46ma
R3 drop = 4.6
Collector Voltage = 7.4


Or using your beta,
R1 = 47KΩ, R2 = 5.6KΩ, R3 = 10KΩ, R4 = 1KΩ, Vcc = 12V, β = 110

Base to Ground Resistance = 110KΩ
R2 is now equivalent to 5.33KΩ

Base Voltage = 1.22V
Emitter Voltage = 0.52V
Emitter Current = Collector Current = 0.52ma
R3 drop = 5.2V
Collector Voltage = 6.8


Your calculations for Base Voltage seem off to me. Ignoring beta entirely...
R1 = 47KΩ, R2 = 5.6KΩ, R3 = 10KΩ, R4 = 1KΩ, Vcc = 12V

Base Voltage = 1.28V
Emitter Voltage = 0.58V
Emitter Current = Collector Current = 0.58ma
R3 drop = 5.8V
Collector Voltage = 6.2V

See what I mean?


So how does this compare to your measured values?
 

Wendy

Joined Mar 24, 2008
23,415
One note, reduce R1 and R2 about half and keep them proportional to each other, then go through this experiment again, it will reduce beta effects even more.

Do you understand how I am arriving at these numbers?

I see the numbers 6 on each side of C2 on your print. Does this mean they are the same point in the simulator? Not so if so.
 

Thread Starter

NiCeBoY

Joined Aug 20, 2008
59
these are my calculated values:
Vb=1.597V
Vce= 5.13V
Ic~=IE=0.897mA
Vc=6.03V
Ib=8.08 micro.Amps
Input inpedance = 4791.96 Ohms
Output impedance = 5000 Ohms
Vo=4.44V


I am comparing them with the measure values on the print screen
 

Wendy

Joined Mar 24, 2008
23,415
Your Base voltage is WAY off. You have something wrong here. This is a basic voltage divider

Vbase = (12V X 5.6KΩ) / (5.6KΩ + 47KΩ) = 1.278V

Before you introduce a AC signal on the simulator, try the DC stuff first. Turn your AC off. I think I mentioned we'd go through AC analysis later.
 

Wendy

Joined Mar 24, 2008
23,415
Nope. There is a math error.

Base voltage is 1.28V.

Emitter voltage = Base voltage - 0.7V = 0.58V

Plenty of voltage on the emitter.
 

Wendy

Joined Mar 24, 2008
23,415
So it is. You keep changing parameters on me, I used this post as my reference. I take it you will stick with 15 Volts from now on? That's what I was talking about in my earlier post talking about tweaking values instead of starting from scratch every time. Only make changes to see what will happen.

We have a real chat going on here. :)

It is 1:42AM here, what time is it there in OZ?
 

Wendy

Joined Mar 24, 2008
23,415


R1 = 47KΩ, R2 = 5.6KΩ, R3 = 10KΩ, R4 = 1KΩ, Vcc = 15V, β = 110

Base to Ground resistance = 110KΩ
R2 is now equivalent to 5.33KΩ

Base Voltage = 1.53
Emitter Voltage = 0.83
Emitter Current = Collector Current = 0.83ma
R3 drop = 8.3V
Collector Voltage = 6.7V
 

Thread Starter

NiCeBoY

Joined Aug 20, 2008
59
sorry man.. may be i made a mistake..

These values should stay fixed :

RL= Load resistor 10K.
Transistor BC547B (Beta=110)
Vcc=15V
The gain should be between 10-20Db.

I should only change the ressitor values to get the gain between 10-20dB...

So you told me "Plenty of voltage on the emitter. "
Should i decrease the voltage ?

and what the values of about Rc (10K) and RE(1k)?
 

Wendy

Joined Mar 24, 2008
23,415
Nope, stick with what you have. How close to the measured values are the calculated numbers? Understand that these are somewhat appoximations, the BE voltage drop is between .6V - .7V, so most cases we use .7 as a base number.

Looking at it your collector voltage is a bit low, but I still have a question about the print I'm seeing from you.
 

Thread Starter

NiCeBoY

Joined Aug 20, 2008
59
ah ok.. here i am in australia its 5.02PM

here is a new screenshot..

for the number 6 i am also wondering why its like that ...

For the collector voltage how much it should be for the amplifier to be goood enough to get the required gain between 10 - 20 dB plz?

Here is another screenshot...

 

Wendy

Joined Mar 24, 2008
23,415
Drop R2 to 5.1KΩ, the collector voltage should rise. I'm too tired to go through the AC stuff, but it will push the collector towards the middle of its range, and reduce clipping of an AC signal.

Think about the transistor if it were fully on, and not dropping any voltage. In this case the collector and emitter would be the same voltage. Put this around 1.4 V. This is the minimum voltage the collector can reach.

Now figure the maximum voltage if the transistor were fully off, 15V. Mid range between these values is 8.2. That is what you are after. One either end, high voltage or low voltage on the collector, the amp will be non linear (this is bad), but you have a lot of leeway if you get it in the middle.

Looking at the diagram above there is no way the collector is below 1 volt. There is an error somewhere. Is that a SPICE program? I still suspect it is showing your C2 as shorted.
 
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