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- Thread starter Asok
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berkeman

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Welcome to the PF. Your question is pretty general, so I'm going to refer you to the wikipedia article first. After you read through that article, please post any specific questions here that you still have.

http://en.wikipedia.org/wiki/Bipolar_transistor_biasing

.

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- #5

vk6kro

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Knowing the gain of the transistor, you can work out the base current.

Now work out the value of the resistor from base to ground assuming 4 times the base current flows in it.

Then work out the value of the upper bias resistor assuming that 5 times the base current flows in it. The difference in current is the actual base current.

If there is an emitter resistor, you can assume a current through it and hence estimate the voltage across it. Add 0.6 volts for the base-emitter junction and this will give you the voltage necessary at the base.

Knowing the gain of the transistor, you can work out the base current.

Now work out the value of the resistor from base to ground assuming 4 times the base current flows in it.

Then work out the value of the upper bias resistor assuming that 5 times the base current flows in it. The difference in current is the actual base current.

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Can you confirm whether you are referring to circuits which use emitter resistors? Can you post a diagram?

Where an emitter resistor is used, the aim is to get a reasonable voltage across that resistor. The voltage should be big enough so that the bias current is stabilised against changes in current gain and V

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There are interactions, it is true, particularly if the gain requirement includes full adjustment for the loading effect of the input resistance, but I do not think this is so for your example. For such cases (depending what your syllabus requires) you might be expected to iterate your design or use simultaneous equations.

That said, you do need to take another look at what you are doing. The value of 1090 ohms base-emitter resistance you have been given is most likely a small-signal input resistance for the hybrid-pi model. This can't be used to estimate base current directly in quite the way you have suggested, but possibly your instructor meant you to find them by using an equation which relates the input resistance, emitter current, and current gain?

Perhaps your best way forward at this point would be to look again at the Wikipedia article, and any relavant course material you may have. Try to work slowly through the material, making sure you understand each step before moving on to the next.

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this the first time i give my opinion ...

a bout ur question bother ... it is not always be with ratio of 3

if u are goiong to design voltag divider ... u can assume that

B RE >> 10 Rb2

and from the data sheet of the transistor do ur calculation

i think u can .. assum ratio between RE and Rc

coz i tried it on my project of amplifier

and it is work ...

hop i give somthing worth

my regards

- #10

vk6kro

Science Advisor

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In the following diagram, look at the ratio of voltages between the base and ground and base and the supply voltage.

Assuming the base current is negligible compared with the current in the resistors, then the resistors must also be in this ratio.

General rules like "this one must be 10 times that one" might still produce a working amplifier, but the proper functioning of the network will be lost.

If the supply was only 4 volts, then the "this one must be 10 times that one" rule would produce only produce 0.36 volts on the base and the transistor would not even turn on. So, it wouldn't work at all.

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- #12

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i am gona tell u the steps ..that u can design ..most of amplifier

first of all

choes the transistors let take for e.g 2n3904 ... it's soo good to save power ..

ok after we choes that .. go and get it's datasheet

the most important thing ...the digram that contain Vce vs Ic and Ib of cource

but ...ic is the y axis and Vce is the x axis

after u get that ..

let us design ...

choes the voltage source emmm can we take 12V ok

match from the x axis at 12V to the y axis at any point but chose that point that make ur desin since

let us take u match 12V to 8mA this is the saturation current not the Ic

this line is the line u can choes the Q point on it

just on this line ..got that ? so important point ...

after that u chose a point ...

at that point take the value of Ic , the value Vce and Ib

now after u take the Vce

then the remain volt on the resistor RE and RC is( Vcc-Vce)

then the 1st eqn is

Vcc-vce=Ic(Re+Rc)

now we can assum somthing and that is

B*RE>>10*R2

and u can assume the ration between Rc and Re

like 3:1 respictively

after that u can get the value of Rc and Re

by the last assumtion

that

B*RE>>10*R2

so

Vbe=Vb-Ve

now we got Vb

wat is Vb

Vb=(Vcc*R2)/(R2+R1)

now we got 2 unknown resistors value

remember that

B*RE>>10*R2

so

B*RE/10>>R2

B*RE/10 calculated

then choes a value for R2 les that B*RE/10

then from

Vb=(Vcc*R2)/(R2+R1)

find the value of R1

and .....

giss that enough for ur qestion

for the frequncy analysis ...

we'll talk about that later ...

bye

hop u got somthing from me

my regards ..

Em Jay

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- #14

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https://www.physicsforums.com/attachment.php?attachmentid=21960&d=1258770808

Do you understand why the ratio was about 2:1 with V

Bob S

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