Voltage divider biasing

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I want to know is there any constant or suitable ratio for external two resistors which we use for the base of the BJT in 'voltage divider biasing'? I have heard that the ratio of those two resistors should be something like 3.. but im not sure...
 

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  • #2
berkeman
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I want to know is there any constant or suitable ratio for external two resistors which we use for the base of the BJT in 'voltage divider biasing'? I have heard that the ratio of those two resistors should be something like 3.. but im not sure...
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

.
 
  • #3
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The current through the bias network should probably be 5x or 10x the base current for a circuit design with repeatable performance.
 
  • #4
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I read the Wikipedia.. it says that the ratio should be selected as the way the operating point of the transistor is independent of β.. in our practical of biasing BJT in university we were given that the ratio as '3' or something like that..(constant value). my problem is , 'is that a constantly using value or just a value for the practical...?' (in our question paper one of a problem is regarding this potential divider biasing. to solve that(to derive sufficient equations) i think i have to use that constant value...or otherwise i'll have to make lot of assumptions to get sufficient equations...!)
 
  • #5
vk6kro
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If you have a supply voltage of 12 volts and a transistor with no emitter resistor, you can select the resistors in the ratio of 0.6 to (12 - 0.6) ie in the same ratio as the voltages below and above 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.

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.
 
  • #6
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I would think that your instructor gave you a ratio for the resistors as an appropriate starting point, to help you arrive more easily at a practicable design. There is nothing magic about 3:1, for instance circuits with high supply voltages might use bigger ratios.
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 VBE, but not too big, because the available collector voltage swing is reduced by the emitter voltage.
 
  • #7
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well i will tell my problem with detail.. it is a common emmiter amplifier with voltage divider biasing, included all the three capacitors(input, output and to short circuit the emmiter terminal(paralell with emmiter resistor)).. given datas are - supply=12v , beta-100, base emmiter resistance(internal) - 1.09k.. and it says the amplifier operates between a source with 10k and a load of 2k... it says again the small signal voltage gain of the amplifier should be -75(minus).. they ask to find dc biasing Rb1,Rb2,collector resistor,emmiter resistor,base current and collector current..Rb1 and Rb2 are the potential dividing resistors.. further they ask to use hybrid pi model when necessary... I found base current (0.7=Ix1090)... here 0.7 is an assumption( the voltage between B and E)... then found collector current(100xbase current).. then by using hybrid pi model I got an equation of coll.current,Rb1,Rb2.. and switching back again to the dc biasing circuit I again got some other few equations.. (assuming that the coll.emmit. viltage to be 6V.. (12/2).. because the Q point should be at the middle of the load line).. but the equations are not enough to find the all variables...! that is my problem.. may be I have to assume some other things too.. I thought that should be the ratio between Rb1 & Rb2...
 
  • #8
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Something to think about: designing a transistor stage may seem a daunting prospect at first sight, but it is clearly not impossible - it has been done many times before, and what others have done in the past, you can do now if you follow a suitable procedure.

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.
 
  • #9
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HI

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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This is homework.

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

transistor biasing.PNG

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.
 
  • #11
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Thank you friends... every advise & opinion you gave to me were really helpful.. I think i have to start solving the problem again with a different start... thank you a lot... but still if you people have something important regarding this matter it will be really really helpful to me & others to know... so if you can post further , i really appreciate it...
 
  • #12
15
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look ..
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
 
  • #13
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hey... urekaaaa.... I think i got it... there is a wonderful equation.. gm=Ic / Vt, here gm is the transconductance, Ic is the DC bias collector current, Vt is a constant voltage about 26 mA at 300 kelvin.. together with that equation we can use this one too, ro=(Vce + Va)/ Ic, ro is parallel resistor for the current source in hybrid pi model, Vce (you know what), Va is the early reverse voltage.. and again we have gm = beta / r, here r is the resistor between base & emitter of hybrid pi model..(r pi).. with those equations I think I can solve that problem ... ya ya ya ya ya.....
 

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