Voltage Divider CE amplifier

Status
Not open for further replies.
F

fireback

Newbie
Joined
Nov 18, 2020
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Visit site
Activity points
20
Hello i have this question and was wondering if someone could help me,
Lets say we are given a simple CE amplifier(with voltage divider on base and Rc, Re resistances)
If we are given Ic=3mA , Vcc=20V , β=100, Rth=10kΩ, Vce=10V, how can we calculate the values for R1 and R2?
Also, how can we calculate them if we are not givven the equivalent Rth value?
Thank you in advance.
(Rc instead of RL)
 

Attachments

  • Screenshot_2.png
    24.5 KB · Views: 384

For a place to start from, turn on the transistor sufficiently so that output voltage centers around a level about halfway between supply V and ground. Current changes through the 'totem pole ' creates voltage changes (the totem pole consisting of RL, RE and the transistor).

Notice bias V needs to be about 0.6 V greater than V at emitter leg.

With hardware it's practical to substitute a potentiometer for R1 R2. Say 100 kohm.
 

Usually, I use I2 to be between 5 to 10 times Ib.
R2 = (Ve+Vbe)/I2.
R1 = (Vcc-V2)/(Ib+I2)
 
Last edited by a moderator:

Akanimo, you did not explain what I2 is. The current in R1 and R2?
You also did not explain why is the current fairly high? Because the transistor hFE has a wide range?
 

Audioguru said:
Akanimo, you did not explain what I2 is. The current in R1 and R2?
You also did not explain why is the current fairly high? Because the transistor hFE has a wide range?
Click to expand...
Oh! I believed it would be obvious. I2 is the current flowing through R2. The current through R1 is (I2+Ib). I2 is selected to be much greater than Ib for bias stability purposes from instability which could result from a lot of factors including noise and so on.
 
Last edited:

Akanimo said:
The current through R1 is (I2+Ib). I2 is selected to be much greater than Ib for bias stability purposes from instability which could result from a lot of factors including noise, input signal and so on.
Click to expand...
No. The high current required in R2 is not affected by those AC things, it is DC and is because a transistor has a wide range of DC hFE. You do not want not enough base voltage and current for a transistor that has low hFE.
 

Audioguru said:
No. The high current required in R2 is not affected by those AC things, it is DC and is because a transistor has a wide range of DC hFE. You do not want not enough base voltage and current for a transistor that has low hFE.
Click to expand...

Okay. I've never seen the hFE connection to I2 before, and I've not read it fro somewhere. It just hasn't occurred to me.

I probably would need to ask a few questions just to clarify:

- considering the range of hFE specified in the datasheet for a particular transistor (we can use bc547b as an example), what value of better do you use, the lowest, midpoint or highest? Why?

- You mentioned base voltage in connection with hFE and I2. Please could you explain with some detail how they connect?
 

I agree with Akanimo. A base voltage divider current of five- to tenfold base current is a standard design rule taught in second year of electronics engineering study.
 


Your example of a BC547B transistor is unusual because it has its hFE selected. The BC547 has an hFE from 110 to 800 but the BC547B has its hFE selected to be from 200 to 450. The BC547A has low hFE and the BC547C has a high hFE.

A transistor with a low hFE loads down and reduces the voltage set by the base voltage divider resistors. Then if you decide to make current in the voltage divider 1/10th the mid hFE of a transistor then you should calculate that a transistor with minimum hFE will work properly.
 

You can say 5 times maximal Ib and save most of the verbose explanations. It's however still a rule of thumb. In a more accurate design procedure, you'd specify the acceptable bias point shift over hFE variation.
 


With the OP's specification of Rth=10k, I believe by Rth here, the OP means the amplifier's input impedance.

So, Rth = R1//R2//(hFE*Re) - recall that when resistances are connected in parallel, the resulting equivalent is lower than the minimum of the connected resistances.

Ie = Ic*(1 + 1/hFE) =
Deciding for Ve = 1V, Re = Ve/Re = 1V/[3mA*(1 + 1/100)] = 330 ohm
hFE*Re = 100*330 = 33k
Ib = Ic/hFE = 3mA/100 = 30uA
R2 = V2/(10*Ib) = (Ve + Vbe)/(10*Ib) = (1 + 0.7)/(10*30u) = 5667 ohm
Let's select R2 = 5.6k and back calculate for I2
I2 = V2/R2 = 1.7/5.6k = 304uA
R1 = (Vcc - V2)/(I2 + Ib) = (20 - 1.7)/(304u + 30u) = 54.8k (we can approximate this value to nearest standard value)

Rth = 54.8k//5.6k//33k
Merely looking at this, it is obvious that we cannot meet the requirement of Rth = 10k because of the 5.6k resistance which is already below 10k.
Two ways we can tackle this is to increase V2 by increasing Ve or we use I2 of around 5*Ib. Assuming that the emitter bypass capacitor is not in the diagram, the gain being determined by Vc/Ve could also be a limiting factor on how much we can increase Ve.

However let's go ahead and increase Ve to 2.8V.
R2 = V2/(10*Ib) = (Ve + Vbe)/(10*Ib) = (2.8 + 0.7)/(10*30u) = 11.67k (round off to 12k and back calculate for I2)
I2 = V2/R2 = 3.5/12k = 292uA
R1 = (Vcc - V2)/(Ib + I2) = (20 - 3.5)/(30u + 292u) = 51.2k (round to 51k)
Re = Ve/[Ic*(1 + 1/hFE)] = 924 ohm (Round to 910)

Rth = 51k//11k//91k is about 8.2k and still does not meet the requirement of 10k. So Be can be further increased or I2 can be reduced and the resistances recalculated.

I hope this would be helpful.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…