Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

How to understand this Bias Circuit?

Status
Not open for further replies.
isaacnewton

isaacnewton

Full Member level 2
Full Member level 2
Joined
May 13, 2004
Messages
143
Helped
12
Reputation
24
Reaction score
8
Trophy points
1,298
Location
Earth Village
Activity points
1,134
There are 3 branches in the this circuit. In order to get Equation (14) and (15). The currents through those 3 branches should be equal. I don't understand why the currents through the 3 branches are equal.
 

Which book did you get this figure?
 

It's from an IEEE paper:

Evolution of high-speed operational amplifier architectures

Smith, D. Koen, M. Witulski, A.F.
Burr-Brown Corp., Tucson, AZ, USA ;

This paper appears in: Solid-State Circuits, IEEE Journal of
Publication Date: Oct. 1994
Volume: 29 , Issue: 10
On page(s): 1166 - 1179

**broken link removed**
drabos said:
Which book did you get this figure?
Click to expand...
 

There are 3 branches in the this circuit. In order to get Equation (14) and (15). The currents through those 3 branches should be equal. I don't understand why the currents through the 3 branches are equal.
Click to expand...

I think the current through 3 branches are not equal.
Assuming I1=first branch (current through Q2)
I2=2nd branch (current through Q3)
I3=3rd branch (current through Q7)

You can derive that:
I2*R + Vbe5 +I1R + Vbe2 = Vbe4 + Vbe3
Hence, Eq 14 = I1 + I2
Vbe4 + Vbe3 = Vbe7 +I3R6
Hence. Eq15 = I3

I1, I2 and I3 may not be equal, but the sum of Eq14 and Eq15 should give flat current over temperature, which is current flow through Q6.
 

    isaacnewton

    isaacnewton

    Points: 2
    Helpful Answer Positive Rating
Heres the explaination hope it can clear up some of your question
 

    isaacnewton

    isaacnewton

    Points: 2
    Helpful Answer Positive Rating
You are right. Thanks.

Do you know what's the point to put R2 there?

sengyee88 said:
I think the current through 3 branches are not equal.
Assuming I1=first branch (current through Q2)
I2=2nd branch (current through Q3)
I3=3rd branch (current through Q7)

You can derive that:
I2*R + Vbe5 +I1R + Vbe2 = Vbe4 + Vbe3
Hence, Eq 14 = I1 + I2
Vbe4 + Vbe3 = Vbe7 +I3R6
Hence. Eq15 = I3

I1, I2 and I3 may not be equal, but the sum of Eq14 and Eq15 should give flat current over temperature, which is current flow through Q6.
Click to expand...
 

R2 serves as the one to produce the reference current in the second bias...

As you can see, between the bias lines is double differential pairs...
 

Hi
I think you need to read about TransLinear Circuits (Barrie Gilbert)
Or the book : The currrent mode approach
(Toumazou & Lidgey)
 

I don't understand gundam's explanation. In the equation you write why do you assume that current "I" goes through R1, R2, R6, and R5? Other strings also carry current...
Suppose I1 flows in the 1st string, I2 in second and I3 in 3rd. Then we have the following equations:

Vbe4+Vbe3=Vbe2+Vbe5+I1R3+I2R4
i.e. Vt ln(I1/A4 Is) + Vt ln(I2/A3 Is) = Vt ln(I1/ A2 Is) + Vt ln(I2/A Is) + (I1+I2) R

this gives us:
I1 + I2 = Vt/R Ln(A2 A5/A3 A4)

Also
I3 ≈ Vbe/R6 (Approximate)
Then we have (I1+I2+I3) (R1+R5) + Vbe1 + I1 R2 + Vbe3 + Vbe4 + Vbe6 = (Vcc-Vee)

None of these equations set the condition that I1 and I2 should be the same.
 

    isaacnewton

    isaacnewton

    Points: 2
    Helpful Answer Positive Rating
Many thanks for your sharing!
 

aryajur is correct, very excellent answer!
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top