i don't have the basic of analog, but i tried to look for in the books. unfortunately, the books are using voltage divider biasing instead of feedback biasing. and i can't find it on the net.

i have a circuit like this:

would anyone tell me the flow direction of the current in the circuit, and how it makes gain possible?

and, i've tried to calculate the gain, but i think it is wrong.
i hope someone can help.

 

Originally posted by den@Mar 3 2006, 11:03 PM
i don't have the basic of analog, but i tried to look for in the books. unfortunately, the books are using voltage divider biasing instead of feedback biasing. and i can't find it on the net.

i have a circuit like this:

would anyone tell me the flow direction of the current in the circuit, and how it makes gain possible?

and, i've tried to calculate the gain, but i think it is wrong.
i hope someone can help.

[post=14616]Quoted post[/post]​

You said "On DC, C1 and C2 are short"
I think capacitors at DC are OPEN, which allowed you to ignore them for the DC analysis.

You have not calculated the gain of the circuit, but the DC gain, or beta of the 2N2222 transitor. The value that you calculated is not unreasonable. Always check the data sheet to be sure.

Another way to get the same result using your equations and your impeccable reasoning, but without all the algebra is:

Vc = 5 Volts and Ic = 5V/1K = 5 mA
Vce = 5V
Ib = (5 -0.7)/175K = 24.57 uV
Ic/Ib = 203.488 = beta of 2N2222

Now I have a question for you in return:

On a set of characteristic curves for the 2N2222, do you now know how to draw the DC load line? If you can do this then you will understand what is happening when your AC signal causes your quiescent point to move along the load line.

 

im not sure how to draw the DC load line.

im following a Pspice tutorial on displaying the characteristics curve for the transistor 2N2222.
it says that i can estimate the β for the transistor.

look at this graph

lowest line represents Ib = 0, followed by 10 μ A, 20μ A, and 30μ A.

it is exactly like the graph in the tutorial, but how to estimate the value of β . It is stated in the tutorial that β is approx. 100

 

Here is the graph you supplied with the load line for the case where the collector supply is 10V and the collector load resistor is 1K.

[attachmentid=1227]

Using the blue plot on this graph and assuming it corresponds to Ib = 20uamps then I get

                  3.5 ma
         Beta =   --------    = 175
                  0.02 ma

Not exactly the value obtained by you and papabravo but close.

hgmjr

 

tq. i have a question.

is 175 DC BETA or AC BETA?

 

Originally posted by den@Mar 4 2006, 07:49 AM
tq. i have a question.

is 175 DC BETA or AC BETA?

[post=14642]Quoted post[/post]​

The value for beta I obtained using the load line method would correspond to the DC beta.

hgmjr

 

tqvm hgmjr. but i would like to ask, how can i obtain the value of AC BETA?

is 325 correct?

 

Originally posted by den@Mar 4 2006, 08:45 AM
tqvm hgmjr. but i would like to ask, how can i obtain the value of AC BETA?

is 325 correct?

[post=14645]Quoted post[/post]​

I have a question.

Is it possible that what you are interested in calculating is the gain of the transistor stage you have posted?

             Vout
specifically  gain = ----------
             Vin

hgmjr

 

i'm actually working on a report of something that i never learn before. that's why i'm having so many questions. sorry for that.

it is actually a PSpice simulation project.
i need your help. from this graph (Gain curve), it means that the Gain, A is 165. am i correct?

a bit explaination here, V(in) is the current through the capacitor C1, identified as the input current to the amplifier.
V1 provide 1V a.c.

what else does the graph tell you.

and i found from the book saying that gain = Vout/Vin = Vc / Vb

my other question is, how to obtain Vb.

and Vc = Vce, am i right?
from my calculation, Vce = 5.35V
therefore, Vc = 5.35V.

correct me if i am wrong.

 

Originally posted by hgmjr@Mar 4 2006, 08:22 AM
Here is the graph you supplied with the load line for the case where the collector supply is 10V and the collector load resistor is 1K.

[attachmentid=1227]

Using the blue plot on this graph and assuming it corresponds to Ib = 20uamps then I get

                  3.5 ma
         Beta =   --------    = 175
                  0.02 ma

Not exactly the value obtained by you and papabravo but close.

hgmjr

[post=14641]Quoted post[/post]​

I think your load line is slightly misplaced. It should intersect the horizontal axis at (10V, 0A) for the transistor in cutoff. It should go through the Q-point at (5V, 5mA) and it should intersect the vertical axis where? Why at (0V, 10mA) when the transistor is in saturation. If you restimate beta with this change I think you should be closer.

 

When speaking of the gain of a transistor it is more appropriate to consider the change in base current due to the change in the input signal voltage.

In the case of your transistor stage, your input signal change in voltage is developing a current across the 1K resistor in series with the input signal. This change in current is being delivered to the base of the transistor. The transistor is a current gain device whose change in collector current caused by the change in base current is being converted to an output voltage by the collector load resistor.

A rough estimate of the magnitude of your transistor stage midband gain can be obtained by dividing the 175K feedback resistor by the 1K resistance in series with your input signal source to get 175. The actual estimated gain is -175 since the common emitter stage inverts the input signal.

hgmjr

 

Originally posted by Papabravo@Mar 4 2006, 11:40 AM
I think your load line is slightly misplaced.  It should intersect the horizontal axis at (10V, 0A) for the transistor in cutoff.  It should go through the Q-point at (5V, 5mA) and it should intersect the vertical axis where? Why at (0V, 10mA) when the transistor is in saturation.  If you restimate beta with this change I think you should be closer.

[post=14653]Quoted post[/post]​

I think I have the load line intersecting at 10V, 0A with a slope that is consistent with the 1K resistor in the collector. The y-axis of the graph actually runs from -2.0mA to +6.0mA.

Added: The load line I drew does pass through the point 5V, 5.0mA as you indicated.

Did I miss something?

hgmjr

 

Originally posted by hgmjr@Mar 4 2006, 12:59 PM
I think I have the load line intersecting at 10V, 0A with a slope that is consistent with the 1K resistor in the collector. The y-axis of the graph actually runs from -2.0mA to +6.0mA.

Added: The load line I drew does pass through the point 5V, 5.0mA as you indicated.

Did I miss something?

hgmjr

[post=14655]Quoted post[/post]​

No, because the 10V label was moved to the left slightly I did not count the ticks correctly and thought that the load line intersected at (10.25V, 0A). I'm sorry for the confusion.

 

so the gain is 165.

can you show me the calculation using the conventional way, to get an answer gain = 165? ( or close to that value )


edit : changed the gain value to 165, not 175.

 

Originally posted by Papabravo@Mar 4 2006, 12:46 PM
No, because the 10V label was moved to the left slightly I did not count the ticks correctly and thought that the load line intersected at (10.25V, 0A). I'm sorry for the confusion.

[post=14657]Quoted post[/post]​

No problem. The nature of some attachments can cause confusion. I have more than once misinterpreted a diagram, schematic, or graph.

Thanks for the check and balance. The last thing I want to do is mistakenly post a response that is inaccurate or misleading.

hgmjr