# Widlar Circuit equation

Discussion in 'Homework Help' started by black_bear, Apr 14, 2010.

1. ### black_bear Thread Starter New Member

Apr 14, 2010
2
0
I have the following equation from a widlar circuit:

Vt[ln(.9927) -ln(Iout)] = 700(Iout-1)

Vt = 26mV
I(in) = 0.9927 mA
R in the circuit = 700 Ohms
I had a third transistor with V(be) ON = 700mV

Can someone help me find I(out)

Thanks

2. ### syed_husain Well-Known Member

Aug 24, 2009
61
5
as you know this type of equation is transcendental equation, u cant solve this like normal polynomial equation. if u want to do it manually, u have to use NEWTON's mehtod. it is an iterative process (quite tiresome). do some research u will able to figure it out.

here i calculated the value of Iout using Mathemtica.

NSolve[26*10^-3 (Log[0.9927*10^-3] - Log[Iout]) - 700 (Iout - 1) == 0,
Iout] // N

Iout-> 0.999743 A.

hope this will help.

3. ### black_bear Thread Starter New Member

Apr 14, 2010
2
0
Hello Syed
(AoA)

Thanks for the reply but unfortunately the answer to this problem is one of this 5 choices: I(out) equals

1. 0.2 mA
2. 0.5 mA
3. 1 mA
4. 1.7 mA
5. 3.2 mA

I think I have not set-up the equation properly. Please see the attached circuit diagram.

WaS
Faisal

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4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
As a rough guess I'd say it's going to be about 1mA.

I2*R2+VBE2=VBE1+VBE3

If VBE1≈VBE2≈VBE3=0.7V

I2≈0.7/700=1mA

A more careful analysis is difficult for me due to the presence of Q1.

If Q1 was omitted then normally one might try a closer solution using the relationship -

I2=(Vt/R2)*Ln(I3/I2)

- which is solved iteratively. I3 is the Q3 [&Q1 when included] emitter current which you work out as indicated in your analysis.

An better 'approximate' guess with Q1 included might be obtained using

I2=VBE1/700+(Vt/700)*Ln(I3/I2) - again solved iteratively

I3 would be about 1mA (depending on your assumed VBE's for Q1 & Q3 - which I took as 0.65V)

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
It's possible with the selected transistor that VBE at 1mA could be a lot less than 0.65V - perhaps as low as 0.57V

In this case the current in Q1 & Q3 [I3] would be approx 1.01mA.

The approx estimate for the resulting Q2 current would be determined from

I=0.57/700+(Vt/700)*Ln(1.01e-3/I) where Vt=26mV

When solved iteratively this would give

I=822uA.

Which isn't far removed from I=VBE/700=0.57/700=815uA.

So the iterative method in this case doesn't add much to the most basic & simpler solution.

Last edited: Apr 16, 2010